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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ .
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in the last question , the length i put in was marked as wrong , so i went to the `` help '' button and followed my calculations , and the answer was the same as the answer i wrote , so why was it marked wrong ?
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor .
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how to find gcd of a polynomial ?
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor .
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why do i divide the gcf from the length of the two rectangles ?
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ .
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so the 14x^4 and the 6x^2 represent square meters and we have to find the lengths of the two rectangles to find the length of the bigger rectangle ?
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ .
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then we go on to find the width of the rectangle using area=length times width but i 'm confused with the part were we used gcf to get 2x^2 why would we do that and how do we get length ?
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what you should be familiar with before this lesson the gcf ( greatest common factor ) of two or more monomials is the product of all their common prime factors . for example , the gcf of $ 6x $ and $ 4x^2 $ is $ 2x $ . if this is new to you , you 'll want to check out our greatest common factors of monomials article . what you will learn in this lesson in this lesson , you will learn how to factor out common factors from polynomials . the distributive property : $ a ( b+c ) =ab+ac $ to understand how to factor out common factors , we must understand the distributive property . for example , we can use the distributive property to find the product of $ 3x^2 $ and $ 4x+3 $ as shown below : notice how each term in the binomial was multiplied by a common factor of $ \teald { 3x^2 } $ . however , because the distributive property is an equality , the reverse of this process is also true ! if we start with $ 3x^2 ( 4x ) +3x^2 ( 3 ) $ , we can use the distributive property to factor out $ ~\teald { 3x^2 } $ and obtain $ 3x^2 ( 4x+3 ) $ . the resulting expression is in factored form because it is written as a product of two polynomials , whereas the original expression is a two-termed sum . check your understanding factoring out the greatest common factor ( gcf ) to factor the gcf out of a polynomial , we do the following : find the gcf of all the terms in the polynomial . express each term as a product of the gcf and another factor . use the distributive property to factor out the gcf . let 's factor the gcf out of $ 2x^3-6x^2 $ . step 1 : find the gcf $ 2x^3=\maroond2\cdot \goldd { x } \cdot \goldd { x } \cdot x $ $ 6x^2=\maroond2\cdot 3\cdot \goldd { x } \cdot \goldd { x } $ so the gcf of $ 2x^3-6x^2 $ is $ \maroond2 \cdot \goldd x \cdot \goldd x=\teald { 2x^2 } $ . step 2 : express each term as a product of $ \teald { 2x^2 } $ and another factor . $ 2x^3= ( \teald { 2x^2 } ) ( { x } ) $ $ 6x^2= ( \teald { 2x^2 } ) ( { 3 } ) $ so the polynomial can be written as $ 2x^3-6x^2= ( \teald { 2x^2 } ) ( x ) - ( \teald { 2x^2 } ) ( 3 ) $ . step 3 : factor out the gcf now we can apply the distributive property to factor out $ \teald { 2x^2 } $ . verifying our result we can check our factorization by multiplying $ 2x^2 $ back into the polynomial . since this is the same as the original polynomial , our factorization is correct ! check your understanding can we be more efficient ? if you feel comfortable with the process of factoring out the gcf , you can use a faster method : once we know the gcf , the factored form is simply the product of that gcf and the sum of the terms in the original polynomial divided by the gcf . see , for example , how we use this fast method to factor $ 5x^2+10x $ , whose gcf is $ \teald { 5x } $ : $ 5x^2+10x=\teald { 5x } \left ( \dfrac { 5x^2 } { \teald { 5x } } +\dfrac { 10x } { \teald { 5x } } \right ) =\teald { 5x } ( x+2 ) $ factoring out binomial factors the common factor in a polynomial does not have to be a monomial . for example , consider the polynomial $ x ( 2x-1 ) -4 ( 2x-1 ) $ . notice that the binomial $ \teald { 2x-1 } $ is common to both terms . we can factor this out using the distributive property : check your understanding different kinds of factorizations it may seem that we have used the term `` factor '' to describe several different processes : we factored monomials by writing them as a product of other monomials . for example , $ 12x^2= ( 4x ) ( 3x ) $ . we factored the gcf from polynomials using the distributive property . for example , $ 2x^2+12x=2x ( x+6 ) $ . we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors . so in all three examples , we indeed factored the polynomial . challenge problems
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we factored out common binomial factors which resulted in an expression equal to the product of two binomials . for example $ x ( x+1 ) +2 ( x+1 ) = ( x+1 ) ( x+2 ) $ . while we may have used different techniques , in each case we are writing the polynomial as a product of two or more factors .
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i am doing a math assignment where i have to factor x^4+4x^2 and the answer choices have imaginary numbers ?
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background multivariable functions the idea of transformations in all of our methods for visualizing multivariable functions , the goal is to somehow see the connection between the input and the output of a function . with graphs , this means plotting points whose coordinates include both input and output information . with contour maps this means marking which input values will go to certain output values . with parametric functions , you mark where the input lands in the output space . with vector fields you plot the output as a vector whose tail sits at the input . the thought behind transformations is to simply watch ( or imagine ) each input point moving to its corresponding output point . it can be a bit of a mind-warp to view functions as transformations if you never have before , so if it feels confusing at first , that 's okay . to whet your appetite for what this might look like , here 's a video from the parametric surface article which shows how a certain function transforms a square into a torus ( doughnut shape ) : concept over precision thinking about functions as transformations can be very powerful for a few reasons : we are not constrained as much by dimension . both the input and the output can have either one , two or three dimensions , and there will be a way to concretely think about what the function is doing . even when the dimensions are too big to look at , thinking in terms of a transformation at least allows for a vague idea of what 's happening in principle . for example , we can know that a function from $ 100 $ -dimensional space to $ 20 $ -dimensional space is `` flattening '' down $ 80 $ dimensions , perhaps analogous to squishing three-dimensional space onto the line . this idea generalizes more easily to functions with different types of inputs and outputs , such as functions of the complex numbers , or functions that map points of the sphere onto the $ xy $ -plane . understanding functions in this capacity will make it easier to see the connections between multivariable calculus and linear algebra . however , with all that said , it should be stressed that transformations are most powerful as an understanding of what functions do , not as a precise description . it would be rare to learn the properties of a given function by observing what it looks like as a transformation . example 1 : from line to line let 's start simple , with a single-variable function . $ f ( x ) = x^2-3 $ consider all the input-output pairs . $ x $ ( input ) | $ x^2 - 3 $ ( output ) -|- $ -2 $ | $ \ ; \ ; \ ; 1 $ $ -1 $ | $ -2 $ $ \ ; \ ; \ ; 0 $ | $ -3 $ $ \ ; \ ; \ ; 1 $ | $ -2 $ $ \ ; \ ; \ ; 2 $ | $ \ ; \ ; \ ; 1 $ $ \quad\vdots $ | $ \quad\vdots $ what would it look like for all the inputs on the number line to slide over onto their corresponding output ? if we pictured the input space as one number line , and the output space as another number line , we might get a motion like this : alternatively , since in this case the input space and output space are really the same thing , a number line , we could think of the line transforming onto itself , dragging each point $ x $ to where the point $ x^2 - 3 $ started off , like this : example 2 : from line to plane now let 's take a function with a one-dimensional input and a two-dimensional output , like $ \begin { align } \quad f ( x ) = \left ( \cos ( x ) , \dfrac { x } { 2 } \sin ( x ) \right ) \end { align } $ again we consider all input-output pairs . inputs $ x $ | outputs $ \left ( \cos ( x ) , \frac { x } { 2 } \sin ( x ) \right ) $ -|- $ 0 $ | $ ( 1 , 0 ) $ $ \dfrac { \pi } { 2 } $ | $ \left ( 0 , \dfrac { \pi } { 4 } \right ) $ $ \pi $ | $ ( -1 , 0 ) $ $ \vdots $ | $ \quad\ ; \vdots $ imagine all possible inputs on the number line sliding onto their corresponding outputs . this time , since the outputs have two coordinates , they live in the $ xy $ -plane . notice , the final image of the warped and twirled number line inside the $ xy $ -plane is what we would have drawn if we interpreted $ f $ as a parametric function , but this time , we can actually see which input points end up where on the final curve . let 's take a moment to watch it again and follow some specific inputs as they move to their outputs . $ \begin { align } \quad \bluee { 0 } & amp ; \to f ( 0 ) = ( \cos ( 0 ) , 0\sin ( 0 ) ) = \bluee { ( 1 , 0 ) } \ \ \greene { \frac { \pi } { 2 } } & amp ; \to f\left ( \frac { \pi } { 2 } \right ) = \left ( \cos\left ( \frac { \pi } { 2 } \right ) , \frac { \pi } { 4 } \sin\left ( \frac { \pi } { 2 } \right ) \right ) = \greene { ( 0 , \pi/4 ) } \ \ \rede { \pi } & amp ; \to f ( \pi ) = ( \cos ( \pi ) , \frac { \pi } { 2 } \sin ( \pi ) ) = \rede { ( -1 , 0 ) } \ \ \end { align } $ example 3 : simple plane to plane transformation consider a $ 90^\circ $ rotation of the plane ( arrows are pictured just to help follow the transformation ) : this could be considered a way to visualize a certain function with a two-dimensional input and a two-dimensional output . why ? this transformation moves points in two-dimensional space to other points in two-dimensional space . for example , the point that starts at $ ( 1 , 0 ) $ ends at $ ( 0 , 1 ) $ . the point that starts at $ ( 1 , 2 ) $ ends at $ ( -2 , 1 ) $ , etc . the function describing this transformation is $ f ( x , y ) = ( -y , x ) $ for any given point , like $ ( 3 , 4 ) $ , this function $ f $ tells you where that point lands after you rotate the plane $ 90^\circ $ counterclockwise , ( in this case $ ( -4 , 3 ) $ ) . example 4 : more complicated plane to plane transformation now let 's look at a more complicated function with a two-dimensional input and a two-dimensional output : $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ . each input is a point on the plane , such as $ ( 1 , 2 ) $ , and it moves to another point on the plane , such as $ ( 1^2 + 2^2 , 1^2 - 2^2 ) = ( 5 , -3 ) $ . when we watch every point on the plane slide over to its corresponding output point , it looks as if a copy of the plane is morphing : notice , all the points end up on the right side of the plane . this is because the first coordinate of the output is $ x^2 + y^2 $ , which must always be positive . challenge question : in the transformation above , representing the function $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ , notice that all points end up in the sideways- $ v $ -shaped region between the lines $ x = y $ and $ x = -y $ . which of the following numerical facts explains this ? example 5 : from plane to line next think of a function with a two-dimensional input and a one-dimensional output . $ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way . for instance , this gives a new way to interpret the level sets in a contour map : they are all the points of the plane which scrunch together into a common point on the line . example 6 : from plane to space functions with a two-dimensional input and three-dimensional output map the plane into three-dimensional space . for instance , such a transformation might look like this ( the red and blue lines are just to help keep track of what happens to the $ x $ and $ y $ directions ) : analogous to the one-to-two dimensions example above , our final image reflects the surface we would get by interpreting the function as a parametric function . example 7 : from space to space functions from three dimensions to three dimensions can be seen as mapping all three-dimensional space onto itself . with this many variables , actually looking at the transformation can be a combination of horrifying , beautiful , and confusing . for instance , consider this function : $ f ( x , y , z ) = ( yz , xz , xy ) $ here 's what it looks like as a transformation . it might be pretty , but it 's a serious spaghettified mess to actually try to follow . final thoughts transformations can provide wonderful ways to interpret properties of a function once you learn them . for instance , constant functions squish their input space to a point , and discontinuous functions must tear apart the input space during the movement . these physical interpretations can become particularly helpful as we venture into the topics of multivariable calculus , in which one runs the risk of learning concepts and operations symbolically without an underlying understanding of what 's happening .
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background multivariable functions the idea of transformations in all of our methods for visualizing multivariable functions , the goal is to somehow see the connection between the input and the output of a function . with graphs , this means plotting points whose coordinates include both input and output information .
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what test book do you recommend ?
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background multivariable functions the idea of transformations in all of our methods for visualizing multivariable functions , the goal is to somehow see the connection between the input and the output of a function . with graphs , this means plotting points whose coordinates include both input and output information . with contour maps this means marking which input values will go to certain output values . with parametric functions , you mark where the input lands in the output space . with vector fields you plot the output as a vector whose tail sits at the input . the thought behind transformations is to simply watch ( or imagine ) each input point moving to its corresponding output point . it can be a bit of a mind-warp to view functions as transformations if you never have before , so if it feels confusing at first , that 's okay . to whet your appetite for what this might look like , here 's a video from the parametric surface article which shows how a certain function transforms a square into a torus ( doughnut shape ) : concept over precision thinking about functions as transformations can be very powerful for a few reasons : we are not constrained as much by dimension . both the input and the output can have either one , two or three dimensions , and there will be a way to concretely think about what the function is doing . even when the dimensions are too big to look at , thinking in terms of a transformation at least allows for a vague idea of what 's happening in principle . for example , we can know that a function from $ 100 $ -dimensional space to $ 20 $ -dimensional space is `` flattening '' down $ 80 $ dimensions , perhaps analogous to squishing three-dimensional space onto the line . this idea generalizes more easily to functions with different types of inputs and outputs , such as functions of the complex numbers , or functions that map points of the sphere onto the $ xy $ -plane . understanding functions in this capacity will make it easier to see the connections between multivariable calculus and linear algebra . however , with all that said , it should be stressed that transformations are most powerful as an understanding of what functions do , not as a precise description . it would be rare to learn the properties of a given function by observing what it looks like as a transformation . example 1 : from line to line let 's start simple , with a single-variable function . $ f ( x ) = x^2-3 $ consider all the input-output pairs . $ x $ ( input ) | $ x^2 - 3 $ ( output ) -|- $ -2 $ | $ \ ; \ ; \ ; 1 $ $ -1 $ | $ -2 $ $ \ ; \ ; \ ; 0 $ | $ -3 $ $ \ ; \ ; \ ; 1 $ | $ -2 $ $ \ ; \ ; \ ; 2 $ | $ \ ; \ ; \ ; 1 $ $ \quad\vdots $ | $ \quad\vdots $ what would it look like for all the inputs on the number line to slide over onto their corresponding output ? if we pictured the input space as one number line , and the output space as another number line , we might get a motion like this : alternatively , since in this case the input space and output space are really the same thing , a number line , we could think of the line transforming onto itself , dragging each point $ x $ to where the point $ x^2 - 3 $ started off , like this : example 2 : from line to plane now let 's take a function with a one-dimensional input and a two-dimensional output , like $ \begin { align } \quad f ( x ) = \left ( \cos ( x ) , \dfrac { x } { 2 } \sin ( x ) \right ) \end { align } $ again we consider all input-output pairs . inputs $ x $ | outputs $ \left ( \cos ( x ) , \frac { x } { 2 } \sin ( x ) \right ) $ -|- $ 0 $ | $ ( 1 , 0 ) $ $ \dfrac { \pi } { 2 } $ | $ \left ( 0 , \dfrac { \pi } { 4 } \right ) $ $ \pi $ | $ ( -1 , 0 ) $ $ \vdots $ | $ \quad\ ; \vdots $ imagine all possible inputs on the number line sliding onto their corresponding outputs . this time , since the outputs have two coordinates , they live in the $ xy $ -plane . notice , the final image of the warped and twirled number line inside the $ xy $ -plane is what we would have drawn if we interpreted $ f $ as a parametric function , but this time , we can actually see which input points end up where on the final curve . let 's take a moment to watch it again and follow some specific inputs as they move to their outputs . $ \begin { align } \quad \bluee { 0 } & amp ; \to f ( 0 ) = ( \cos ( 0 ) , 0\sin ( 0 ) ) = \bluee { ( 1 , 0 ) } \ \ \greene { \frac { \pi } { 2 } } & amp ; \to f\left ( \frac { \pi } { 2 } \right ) = \left ( \cos\left ( \frac { \pi } { 2 } \right ) , \frac { \pi } { 4 } \sin\left ( \frac { \pi } { 2 } \right ) \right ) = \greene { ( 0 , \pi/4 ) } \ \ \rede { \pi } & amp ; \to f ( \pi ) = ( \cos ( \pi ) , \frac { \pi } { 2 } \sin ( \pi ) ) = \rede { ( -1 , 0 ) } \ \ \end { align } $ example 3 : simple plane to plane transformation consider a $ 90^\circ $ rotation of the plane ( arrows are pictured just to help follow the transformation ) : this could be considered a way to visualize a certain function with a two-dimensional input and a two-dimensional output . why ? this transformation moves points in two-dimensional space to other points in two-dimensional space . for example , the point that starts at $ ( 1 , 0 ) $ ends at $ ( 0 , 1 ) $ . the point that starts at $ ( 1 , 2 ) $ ends at $ ( -2 , 1 ) $ , etc . the function describing this transformation is $ f ( x , y ) = ( -y , x ) $ for any given point , like $ ( 3 , 4 ) $ , this function $ f $ tells you where that point lands after you rotate the plane $ 90^\circ $ counterclockwise , ( in this case $ ( -4 , 3 ) $ ) . example 4 : more complicated plane to plane transformation now let 's look at a more complicated function with a two-dimensional input and a two-dimensional output : $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ . each input is a point on the plane , such as $ ( 1 , 2 ) $ , and it moves to another point on the plane , such as $ ( 1^2 + 2^2 , 1^2 - 2^2 ) = ( 5 , -3 ) $ . when we watch every point on the plane slide over to its corresponding output point , it looks as if a copy of the plane is morphing : notice , all the points end up on the right side of the plane . this is because the first coordinate of the output is $ x^2 + y^2 $ , which must always be positive . challenge question : in the transformation above , representing the function $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ , notice that all points end up in the sideways- $ v $ -shaped region between the lines $ x = y $ and $ x = -y $ . which of the following numerical facts explains this ? example 5 : from plane to line next think of a function with a two-dimensional input and a one-dimensional output . $ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way . for instance , this gives a new way to interpret the level sets in a contour map : they are all the points of the plane which scrunch together into a common point on the line . example 6 : from plane to space functions with a two-dimensional input and three-dimensional output map the plane into three-dimensional space . for instance , such a transformation might look like this ( the red and blue lines are just to help keep track of what happens to the $ x $ and $ y $ directions ) : analogous to the one-to-two dimensions example above , our final image reflects the surface we would get by interpreting the function as a parametric function . example 7 : from space to space functions from three dimensions to three dimensions can be seen as mapping all three-dimensional space onto itself . with this many variables , actually looking at the transformation can be a combination of horrifying , beautiful , and confusing . for instance , consider this function : $ f ( x , y , z ) = ( yz , xz , xy ) $ here 's what it looks like as a transformation . it might be pretty , but it 's a serious spaghettified mess to actually try to follow . final thoughts transformations can provide wonderful ways to interpret properties of a function once you learn them . for instance , constant functions squish their input space to a point , and discontinuous functions must tear apart the input space during the movement . these physical interpretations can become particularly helpful as we venture into the topics of multivariable calculus , in which one runs the risk of learning concepts and operations symbolically without an underlying understanding of what 's happening .
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$ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way .
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what software are you using to make all these neat visualization ?
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background multivariable functions the idea of transformations in all of our methods for visualizing multivariable functions , the goal is to somehow see the connection between the input and the output of a function . with graphs , this means plotting points whose coordinates include both input and output information . with contour maps this means marking which input values will go to certain output values . with parametric functions , you mark where the input lands in the output space . with vector fields you plot the output as a vector whose tail sits at the input . the thought behind transformations is to simply watch ( or imagine ) each input point moving to its corresponding output point . it can be a bit of a mind-warp to view functions as transformations if you never have before , so if it feels confusing at first , that 's okay . to whet your appetite for what this might look like , here 's a video from the parametric surface article which shows how a certain function transforms a square into a torus ( doughnut shape ) : concept over precision thinking about functions as transformations can be very powerful for a few reasons : we are not constrained as much by dimension . both the input and the output can have either one , two or three dimensions , and there will be a way to concretely think about what the function is doing . even when the dimensions are too big to look at , thinking in terms of a transformation at least allows for a vague idea of what 's happening in principle . for example , we can know that a function from $ 100 $ -dimensional space to $ 20 $ -dimensional space is `` flattening '' down $ 80 $ dimensions , perhaps analogous to squishing three-dimensional space onto the line . this idea generalizes more easily to functions with different types of inputs and outputs , such as functions of the complex numbers , or functions that map points of the sphere onto the $ xy $ -plane . understanding functions in this capacity will make it easier to see the connections between multivariable calculus and linear algebra . however , with all that said , it should be stressed that transformations are most powerful as an understanding of what functions do , not as a precise description . it would be rare to learn the properties of a given function by observing what it looks like as a transformation . example 1 : from line to line let 's start simple , with a single-variable function . $ f ( x ) = x^2-3 $ consider all the input-output pairs . $ x $ ( input ) | $ x^2 - 3 $ ( output ) -|- $ -2 $ | $ \ ; \ ; \ ; 1 $ $ -1 $ | $ -2 $ $ \ ; \ ; \ ; 0 $ | $ -3 $ $ \ ; \ ; \ ; 1 $ | $ -2 $ $ \ ; \ ; \ ; 2 $ | $ \ ; \ ; \ ; 1 $ $ \quad\vdots $ | $ \quad\vdots $ what would it look like for all the inputs on the number line to slide over onto their corresponding output ? if we pictured the input space as one number line , and the output space as another number line , we might get a motion like this : alternatively , since in this case the input space and output space are really the same thing , a number line , we could think of the line transforming onto itself , dragging each point $ x $ to where the point $ x^2 - 3 $ started off , like this : example 2 : from line to plane now let 's take a function with a one-dimensional input and a two-dimensional output , like $ \begin { align } \quad f ( x ) = \left ( \cos ( x ) , \dfrac { x } { 2 } \sin ( x ) \right ) \end { align } $ again we consider all input-output pairs . inputs $ x $ | outputs $ \left ( \cos ( x ) , \frac { x } { 2 } \sin ( x ) \right ) $ -|- $ 0 $ | $ ( 1 , 0 ) $ $ \dfrac { \pi } { 2 } $ | $ \left ( 0 , \dfrac { \pi } { 4 } \right ) $ $ \pi $ | $ ( -1 , 0 ) $ $ \vdots $ | $ \quad\ ; \vdots $ imagine all possible inputs on the number line sliding onto their corresponding outputs . this time , since the outputs have two coordinates , they live in the $ xy $ -plane . notice , the final image of the warped and twirled number line inside the $ xy $ -plane is what we would have drawn if we interpreted $ f $ as a parametric function , but this time , we can actually see which input points end up where on the final curve . let 's take a moment to watch it again and follow some specific inputs as they move to their outputs . $ \begin { align } \quad \bluee { 0 } & amp ; \to f ( 0 ) = ( \cos ( 0 ) , 0\sin ( 0 ) ) = \bluee { ( 1 , 0 ) } \ \ \greene { \frac { \pi } { 2 } } & amp ; \to f\left ( \frac { \pi } { 2 } \right ) = \left ( \cos\left ( \frac { \pi } { 2 } \right ) , \frac { \pi } { 4 } \sin\left ( \frac { \pi } { 2 } \right ) \right ) = \greene { ( 0 , \pi/4 ) } \ \ \rede { \pi } & amp ; \to f ( \pi ) = ( \cos ( \pi ) , \frac { \pi } { 2 } \sin ( \pi ) ) = \rede { ( -1 , 0 ) } \ \ \end { align } $ example 3 : simple plane to plane transformation consider a $ 90^\circ $ rotation of the plane ( arrows are pictured just to help follow the transformation ) : this could be considered a way to visualize a certain function with a two-dimensional input and a two-dimensional output . why ? this transformation moves points in two-dimensional space to other points in two-dimensional space . for example , the point that starts at $ ( 1 , 0 ) $ ends at $ ( 0 , 1 ) $ . the point that starts at $ ( 1 , 2 ) $ ends at $ ( -2 , 1 ) $ , etc . the function describing this transformation is $ f ( x , y ) = ( -y , x ) $ for any given point , like $ ( 3 , 4 ) $ , this function $ f $ tells you where that point lands after you rotate the plane $ 90^\circ $ counterclockwise , ( in this case $ ( -4 , 3 ) $ ) . example 4 : more complicated plane to plane transformation now let 's look at a more complicated function with a two-dimensional input and a two-dimensional output : $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ . each input is a point on the plane , such as $ ( 1 , 2 ) $ , and it moves to another point on the plane , such as $ ( 1^2 + 2^2 , 1^2 - 2^2 ) = ( 5 , -3 ) $ . when we watch every point on the plane slide over to its corresponding output point , it looks as if a copy of the plane is morphing : notice , all the points end up on the right side of the plane . this is because the first coordinate of the output is $ x^2 + y^2 $ , which must always be positive . challenge question : in the transformation above , representing the function $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ , notice that all points end up in the sideways- $ v $ -shaped region between the lines $ x = y $ and $ x = -y $ . which of the following numerical facts explains this ? example 5 : from plane to line next think of a function with a two-dimensional input and a one-dimensional output . $ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way . for instance , this gives a new way to interpret the level sets in a contour map : they are all the points of the plane which scrunch together into a common point on the line . example 6 : from plane to space functions with a two-dimensional input and three-dimensional output map the plane into three-dimensional space . for instance , such a transformation might look like this ( the red and blue lines are just to help keep track of what happens to the $ x $ and $ y $ directions ) : analogous to the one-to-two dimensions example above , our final image reflects the surface we would get by interpreting the function as a parametric function . example 7 : from space to space functions from three dimensions to three dimensions can be seen as mapping all three-dimensional space onto itself . with this many variables , actually looking at the transformation can be a combination of horrifying , beautiful , and confusing . for instance , consider this function : $ f ( x , y , z ) = ( yz , xz , xy ) $ here 's what it looks like as a transformation . it might be pretty , but it 's a serious spaghettified mess to actually try to follow . final thoughts transformations can provide wonderful ways to interpret properties of a function once you learn them . for instance , constant functions squish their input space to a point , and discontinuous functions must tear apart the input space during the movement . these physical interpretations can become particularly helpful as we venture into the topics of multivariable calculus , in which one runs the risk of learning concepts and operations symbolically without an underlying understanding of what 's happening .
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$ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way .
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can someone explain to me better what exactly the reflection question is ?
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background multivariable functions the idea of transformations in all of our methods for visualizing multivariable functions , the goal is to somehow see the connection between the input and the output of a function . with graphs , this means plotting points whose coordinates include both input and output information . with contour maps this means marking which input values will go to certain output values . with parametric functions , you mark where the input lands in the output space . with vector fields you plot the output as a vector whose tail sits at the input . the thought behind transformations is to simply watch ( or imagine ) each input point moving to its corresponding output point . it can be a bit of a mind-warp to view functions as transformations if you never have before , so if it feels confusing at first , that 's okay . to whet your appetite for what this might look like , here 's a video from the parametric surface article which shows how a certain function transforms a square into a torus ( doughnut shape ) : concept over precision thinking about functions as transformations can be very powerful for a few reasons : we are not constrained as much by dimension . both the input and the output can have either one , two or three dimensions , and there will be a way to concretely think about what the function is doing . even when the dimensions are too big to look at , thinking in terms of a transformation at least allows for a vague idea of what 's happening in principle . for example , we can know that a function from $ 100 $ -dimensional space to $ 20 $ -dimensional space is `` flattening '' down $ 80 $ dimensions , perhaps analogous to squishing three-dimensional space onto the line . this idea generalizes more easily to functions with different types of inputs and outputs , such as functions of the complex numbers , or functions that map points of the sphere onto the $ xy $ -plane . understanding functions in this capacity will make it easier to see the connections between multivariable calculus and linear algebra . however , with all that said , it should be stressed that transformations are most powerful as an understanding of what functions do , not as a precise description . it would be rare to learn the properties of a given function by observing what it looks like as a transformation . example 1 : from line to line let 's start simple , with a single-variable function . $ f ( x ) = x^2-3 $ consider all the input-output pairs . $ x $ ( input ) | $ x^2 - 3 $ ( output ) -|- $ -2 $ | $ \ ; \ ; \ ; 1 $ $ -1 $ | $ -2 $ $ \ ; \ ; \ ; 0 $ | $ -3 $ $ \ ; \ ; \ ; 1 $ | $ -2 $ $ \ ; \ ; \ ; 2 $ | $ \ ; \ ; \ ; 1 $ $ \quad\vdots $ | $ \quad\vdots $ what would it look like for all the inputs on the number line to slide over onto their corresponding output ? if we pictured the input space as one number line , and the output space as another number line , we might get a motion like this : alternatively , since in this case the input space and output space are really the same thing , a number line , we could think of the line transforming onto itself , dragging each point $ x $ to where the point $ x^2 - 3 $ started off , like this : example 2 : from line to plane now let 's take a function with a one-dimensional input and a two-dimensional output , like $ \begin { align } \quad f ( x ) = \left ( \cos ( x ) , \dfrac { x } { 2 } \sin ( x ) \right ) \end { align } $ again we consider all input-output pairs . inputs $ x $ | outputs $ \left ( \cos ( x ) , \frac { x } { 2 } \sin ( x ) \right ) $ -|- $ 0 $ | $ ( 1 , 0 ) $ $ \dfrac { \pi } { 2 } $ | $ \left ( 0 , \dfrac { \pi } { 4 } \right ) $ $ \pi $ | $ ( -1 , 0 ) $ $ \vdots $ | $ \quad\ ; \vdots $ imagine all possible inputs on the number line sliding onto their corresponding outputs . this time , since the outputs have two coordinates , they live in the $ xy $ -plane . notice , the final image of the warped and twirled number line inside the $ xy $ -plane is what we would have drawn if we interpreted $ f $ as a parametric function , but this time , we can actually see which input points end up where on the final curve . let 's take a moment to watch it again and follow some specific inputs as they move to their outputs . $ \begin { align } \quad \bluee { 0 } & amp ; \to f ( 0 ) = ( \cos ( 0 ) , 0\sin ( 0 ) ) = \bluee { ( 1 , 0 ) } \ \ \greene { \frac { \pi } { 2 } } & amp ; \to f\left ( \frac { \pi } { 2 } \right ) = \left ( \cos\left ( \frac { \pi } { 2 } \right ) , \frac { \pi } { 4 } \sin\left ( \frac { \pi } { 2 } \right ) \right ) = \greene { ( 0 , \pi/4 ) } \ \ \rede { \pi } & amp ; \to f ( \pi ) = ( \cos ( \pi ) , \frac { \pi } { 2 } \sin ( \pi ) ) = \rede { ( -1 , 0 ) } \ \ \end { align } $ example 3 : simple plane to plane transformation consider a $ 90^\circ $ rotation of the plane ( arrows are pictured just to help follow the transformation ) : this could be considered a way to visualize a certain function with a two-dimensional input and a two-dimensional output . why ? this transformation moves points in two-dimensional space to other points in two-dimensional space . for example , the point that starts at $ ( 1 , 0 ) $ ends at $ ( 0 , 1 ) $ . the point that starts at $ ( 1 , 2 ) $ ends at $ ( -2 , 1 ) $ , etc . the function describing this transformation is $ f ( x , y ) = ( -y , x ) $ for any given point , like $ ( 3 , 4 ) $ , this function $ f $ tells you where that point lands after you rotate the plane $ 90^\circ $ counterclockwise , ( in this case $ ( -4 , 3 ) $ ) . example 4 : more complicated plane to plane transformation now let 's look at a more complicated function with a two-dimensional input and a two-dimensional output : $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ . each input is a point on the plane , such as $ ( 1 , 2 ) $ , and it moves to another point on the plane , such as $ ( 1^2 + 2^2 , 1^2 - 2^2 ) = ( 5 , -3 ) $ . when we watch every point on the plane slide over to its corresponding output point , it looks as if a copy of the plane is morphing : notice , all the points end up on the right side of the plane . this is because the first coordinate of the output is $ x^2 + y^2 $ , which must always be positive . challenge question : in the transformation above , representing the function $ f ( x , y ) = ( x^2 + y^2 , x^2 - y^2 ) $ , notice that all points end up in the sideways- $ v $ -shaped region between the lines $ x = y $ and $ x = -y $ . which of the following numerical facts explains this ? example 5 : from plane to line next think of a function with a two-dimensional input and a one-dimensional output . $ f ( x , y ) = x^2 + y^2 $ , the corresponding transformation will squish the $ xy $ -plane onto the number line . such squishification can make it hard to follow everything that 's going on , so for the sake of a precise and clear description , you would be better off using a graph or a contour map . nevertheless , it can be a helpful concept to keep in the back of your mind that what function from two dimensions to one dimension does is squish the plane onto the line in a certain way . for instance , this gives a new way to interpret the level sets in a contour map : they are all the points of the plane which scrunch together into a common point on the line . example 6 : from plane to space functions with a two-dimensional input and three-dimensional output map the plane into three-dimensional space . for instance , such a transformation might look like this ( the red and blue lines are just to help keep track of what happens to the $ x $ and $ y $ directions ) : analogous to the one-to-two dimensions example above , our final image reflects the surface we would get by interpreting the function as a parametric function . example 7 : from space to space functions from three dimensions to three dimensions can be seen as mapping all three-dimensional space onto itself . with this many variables , actually looking at the transformation can be a combination of horrifying , beautiful , and confusing . for instance , consider this function : $ f ( x , y , z ) = ( yz , xz , xy ) $ here 's what it looks like as a transformation . it might be pretty , but it 's a serious spaghettified mess to actually try to follow . final thoughts transformations can provide wonderful ways to interpret properties of a function once you learn them . for instance , constant functions squish their input space to a point , and discontinuous functions must tear apart the input space during the movement . these physical interpretations can become particularly helpful as we venture into the topics of multivariable calculus , in which one runs the risk of learning concepts and operations symbolically without an underlying understanding of what 's happening .
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why ? this transformation moves points in two-dimensional space to other points in two-dimensional space . for example , the point that starts at $ ( 1 , 0 ) $ ends at $ ( 0 , 1 ) $ .
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is there a way of following the space to space transformation ?
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ? how can we evaluate these ? finding $ i^3 $ and $ i^4 $ the properties of exponents can help us here ! in fact , when calculating powers of $ i $ , we can apply the properties of exponents that we know to be true in the real number system , so long as the exponents are integers . with this in mind , let 's find $ i^3 $ and $ i^4 $ . we know that $ i^3=i^2\cdot i $ . but since $ { i^2=-1 } $ , we see that : $ \begin { align } i^3 & amp ; = { { i^2 } } \cdot i\ \ & amp ; = { ( -1 ) } \cdot i\ \ & amp ; = \purpled { -i } \end { align } $ similarly $ i^4=i^2\cdot i^2 $ . again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table . $ i^1 $ | $ i^2 $ | $ i^3 $ | $ i^4 $ | $ i^5 $ | $ i^6 $ | $ i^7 $ | $ i^8 $ : - : |- : |- : |- : |- : |- : |- : |- : $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ | $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ an emerging pattern from the table , it appears that the powers of $ i $ cycle through the sequence of $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ and $ \goldd1 $ . using this pattern , can we find $ i^ { 20 } $ ? let 's try it ! the following list shows the first $ 20 $ numbers in the repeating sequence . $ \quad $ $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ according to this logic , $ i^ { 20 } $ should be equal to $ \goldd 1 $ . let 's see if we can support this by using exponents . remember , we can use the properties of exponents here just like we do with real numbers ! $ \begin { align } i^ { 20 } & amp ; = ( i^4 ) ^5 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; = ( 1 ) ^5 & amp ; & amp ; \small { \gray { i^4=1 } } \\ & amp ; = \goldd 1 & amp ; & amp ; \small { \gray { \text { simplify } } } \end { align } $ either way , we see that $ i^ { 20 } =1 $ . larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ . we can use this fact along with the properties of exponents to help us simplify $ i^ { 138 } $ . example simplify $ i^ { 138 } $ . solution while $ 138 $ is not a multiple of $ 4 $ , the number $ 136 $ is ! let 's use this to help us simplify $ i^ { 138 } $ . $ \begin { align } i^ { 138 } & amp ; =i^ { 136 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( i^ { 4\cdot 34 } ) \cdot i^2 & amp ; & amp ; \small { \gray { 136=4\cdot 34 } } \\ & amp ; = ( i^ { 4 } ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( 1 ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { $ i^4=1 $ } } } \\ & amp ; =1\cdot -1 & amp ; & amp ; \small { \gray { \text { $ i^2=-1 $ } } } \\ & amp ; =-1 \end { align } $ so $ i^ { 138 } =-1 $ . now you might ask why we chose to write $ i^ { 138 } $ as $ i^ { 136 } \cdot i^2 $ . well , if the original exponent is not a multiple of $ 4 $ , then finding the closest multiple of $ 4 $ less than it allows us to simplify the power down to $ i $ , $ i^2 $ , or $ i^3 $ just by using the fact that $ i^4=1 $ . this number is easy to find if you divide the original exponent by $ 4 $ . it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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how to solve the problem i^x when x is a very huge prime number ?
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ? how can we evaluate these ? finding $ i^3 $ and $ i^4 $ the properties of exponents can help us here ! in fact , when calculating powers of $ i $ , we can apply the properties of exponents that we know to be true in the real number system , so long as the exponents are integers . with this in mind , let 's find $ i^3 $ and $ i^4 $ . we know that $ i^3=i^2\cdot i $ . but since $ { i^2=-1 } $ , we see that : $ \begin { align } i^3 & amp ; = { { i^2 } } \cdot i\ \ & amp ; = { ( -1 ) } \cdot i\ \ & amp ; = \purpled { -i } \end { align } $ similarly $ i^4=i^2\cdot i^2 $ . again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table . $ i^1 $ | $ i^2 $ | $ i^3 $ | $ i^4 $ | $ i^5 $ | $ i^6 $ | $ i^7 $ | $ i^8 $ : - : |- : |- : |- : |- : |- : |- : |- : $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ | $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ an emerging pattern from the table , it appears that the powers of $ i $ cycle through the sequence of $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ and $ \goldd1 $ . using this pattern , can we find $ i^ { 20 } $ ? let 's try it ! the following list shows the first $ 20 $ numbers in the repeating sequence . $ \quad $ $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ according to this logic , $ i^ { 20 } $ should be equal to $ \goldd 1 $ . let 's see if we can support this by using exponents . remember , we can use the properties of exponents here just like we do with real numbers ! $ \begin { align } i^ { 20 } & amp ; = ( i^4 ) ^5 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; = ( 1 ) ^5 & amp ; & amp ; \small { \gray { i^4=1 } } \\ & amp ; = \goldd 1 & amp ; & amp ; \small { \gray { \text { simplify } } } \end { align } $ either way , we see that $ i^ { 20 } =1 $ . larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ . we can use this fact along with the properties of exponents to help us simplify $ i^ { 138 } $ . example simplify $ i^ { 138 } $ . solution while $ 138 $ is not a multiple of $ 4 $ , the number $ 136 $ is ! let 's use this to help us simplify $ i^ { 138 } $ . $ \begin { align } i^ { 138 } & amp ; =i^ { 136 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( i^ { 4\cdot 34 } ) \cdot i^2 & amp ; & amp ; \small { \gray { 136=4\cdot 34 } } \\ & amp ; = ( i^ { 4 } ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( 1 ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { $ i^4=1 $ } } } \\ & amp ; =1\cdot -1 & amp ; & amp ; \small { \gray { \text { $ i^2=-1 $ } } } \\ & amp ; =-1 \end { align } $ so $ i^ { 138 } =-1 $ . now you might ask why we chose to write $ i^ { 138 } $ as $ i^ { 136 } \cdot i^2 $ . well , if the original exponent is not a multiple of $ 4 $ , then finding the closest multiple of $ 4 $ less than it allows us to simplify the power down to $ i $ , $ i^2 $ , or $ i^3 $ just by using the fact that $ i^4=1 $ . this number is easy to find if you divide the original exponent by $ 4 $ . it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table .
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do you have to use powers of 4 to find the solution ?
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ? how can we evaluate these ? finding $ i^3 $ and $ i^4 $ the properties of exponents can help us here ! in fact , when calculating powers of $ i $ , we can apply the properties of exponents that we know to be true in the real number system , so long as the exponents are integers . with this in mind , let 's find $ i^3 $ and $ i^4 $ . we know that $ i^3=i^2\cdot i $ . but since $ { i^2=-1 } $ , we see that : $ \begin { align } i^3 & amp ; = { { i^2 } } \cdot i\ \ & amp ; = { ( -1 ) } \cdot i\ \ & amp ; = \purpled { -i } \end { align } $ similarly $ i^4=i^2\cdot i^2 $ . again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table . $ i^1 $ | $ i^2 $ | $ i^3 $ | $ i^4 $ | $ i^5 $ | $ i^6 $ | $ i^7 $ | $ i^8 $ : - : |- : |- : |- : |- : |- : |- : |- : $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ | $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ an emerging pattern from the table , it appears that the powers of $ i $ cycle through the sequence of $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ and $ \goldd1 $ . using this pattern , can we find $ i^ { 20 } $ ? let 's try it ! the following list shows the first $ 20 $ numbers in the repeating sequence . $ \quad $ $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ according to this logic , $ i^ { 20 } $ should be equal to $ \goldd 1 $ . let 's see if we can support this by using exponents . remember , we can use the properties of exponents here just like we do with real numbers ! $ \begin { align } i^ { 20 } & amp ; = ( i^4 ) ^5 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; = ( 1 ) ^5 & amp ; & amp ; \small { \gray { i^4=1 } } \\ & amp ; = \goldd 1 & amp ; & amp ; \small { \gray { \text { simplify } } } \end { align } $ either way , we see that $ i^ { 20 } =1 $ . larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ . we can use this fact along with the properties of exponents to help us simplify $ i^ { 138 } $ . example simplify $ i^ { 138 } $ . solution while $ 138 $ is not a multiple of $ 4 $ , the number $ 136 $ is ! let 's use this to help us simplify $ i^ { 138 } $ . $ \begin { align } i^ { 138 } & amp ; =i^ { 136 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( i^ { 4\cdot 34 } ) \cdot i^2 & amp ; & amp ; \small { \gray { 136=4\cdot 34 } } \\ & amp ; = ( i^ { 4 } ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( 1 ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { $ i^4=1 $ } } } \\ & amp ; =1\cdot -1 & amp ; & amp ; \small { \gray { \text { $ i^2=-1 $ } } } \\ & amp ; =-1 \end { align } $ so $ i^ { 138 } =-1 $ . now you might ask why we chose to write $ i^ { 138 } $ as $ i^ { 136 } \cdot i^2 $ . well , if the original exponent is not a multiple of $ 4 $ , then finding the closest multiple of $ 4 $ less than it allows us to simplify the power down to $ i $ , $ i^2 $ , or $ i^3 $ just by using the fact that $ i^4=1 $ . this number is easy to find if you divide the original exponent by $ 4 $ . it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ .
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how would i be used ?
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ? how can we evaluate these ? finding $ i^3 $ and $ i^4 $ the properties of exponents can help us here ! in fact , when calculating powers of $ i $ , we can apply the properties of exponents that we know to be true in the real number system , so long as the exponents are integers . with this in mind , let 's find $ i^3 $ and $ i^4 $ . we know that $ i^3=i^2\cdot i $ . but since $ { i^2=-1 } $ , we see that : $ \begin { align } i^3 & amp ; = { { i^2 } } \cdot i\ \ & amp ; = { ( -1 ) } \cdot i\ \ & amp ; = \purpled { -i } \end { align } $ similarly $ i^4=i^2\cdot i^2 $ . again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table . $ i^1 $ | $ i^2 $ | $ i^3 $ | $ i^4 $ | $ i^5 $ | $ i^6 $ | $ i^7 $ | $ i^8 $ : - : |- : |- : |- : |- : |- : |- : |- : $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ | $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ an emerging pattern from the table , it appears that the powers of $ i $ cycle through the sequence of $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ and $ \goldd1 $ . using this pattern , can we find $ i^ { 20 } $ ? let 's try it ! the following list shows the first $ 20 $ numbers in the repeating sequence . $ \quad $ $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ according to this logic , $ i^ { 20 } $ should be equal to $ \goldd 1 $ . let 's see if we can support this by using exponents . remember , we can use the properties of exponents here just like we do with real numbers ! $ \begin { align } i^ { 20 } & amp ; = ( i^4 ) ^5 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; = ( 1 ) ^5 & amp ; & amp ; \small { \gray { i^4=1 } } \\ & amp ; = \goldd 1 & amp ; & amp ; \small { \gray { \text { simplify } } } \end { align } $ either way , we see that $ i^ { 20 } =1 $ . larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ . we can use this fact along with the properties of exponents to help us simplify $ i^ { 138 } $ . example simplify $ i^ { 138 } $ . solution while $ 138 $ is not a multiple of $ 4 $ , the number $ 136 $ is ! let 's use this to help us simplify $ i^ { 138 } $ . $ \begin { align } i^ { 138 } & amp ; =i^ { 136 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( i^ { 4\cdot 34 } ) \cdot i^2 & amp ; & amp ; \small { \gray { 136=4\cdot 34 } } \\ & amp ; = ( i^ { 4 } ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( 1 ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { $ i^4=1 $ } } } \\ & amp ; =1\cdot -1 & amp ; & amp ; \small { \gray { \text { $ i^2=-1 $ } } } \\ & amp ; =-1 \end { align } $ so $ i^ { 138 } =-1 $ . now you might ask why we chose to write $ i^ { 138 } $ as $ i^ { 136 } \cdot i^2 $ . well , if the original exponent is not a multiple of $ 4 $ , then finding the closest multiple of $ 4 $ less than it allows us to simplify the power down to $ i $ , $ i^2 $ , or $ i^3 $ just by using the fact that $ i^4=1 $ . this number is easy to find if you divide the original exponent by $ 4 $ . it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ?
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with larger exponents , do i always have to find a multiple of 4 in the exponent or can i use numbers less than 4 ?
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ? $ i^4 $ ? other integer powers of $ i $ ? how can we evaluate these ? finding $ i^3 $ and $ i^4 $ the properties of exponents can help us here ! in fact , when calculating powers of $ i $ , we can apply the properties of exponents that we know to be true in the real number system , so long as the exponents are integers . with this in mind , let 's find $ i^3 $ and $ i^4 $ . we know that $ i^3=i^2\cdot i $ . but since $ { i^2=-1 } $ , we see that : $ \begin { align } i^3 & amp ; = { { i^2 } } \cdot i\ \ & amp ; = { ( -1 ) } \cdot i\ \ & amp ; = \purpled { -i } \end { align } $ similarly $ i^4=i^2\cdot i^2 $ . again , using the fact that $ { i^2=-1 } $ , we have the following : $ \begin { align } i^4 & amp ; = { { i^2\cdot i^2 } } \ \ & amp ; = ( { -1 } ) \cdot ( { -1 } ) \ \ & amp ; = \goldd { 1 } \end { align } $ more powers of $ i $ let 's keep this going ! let 's find the next $ 4 $ powers of $ i $ using a similar method . $ \begin { align } \large i^5 & amp ; = { i^4\cdot i } ~~~~~ & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot i & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; = \blued i \end { align } $ $ \begin { align } \large i^6 & amp ; = { i^4\cdot i^2 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -1 ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^2=-1 $ } } } \ \ & amp ; =\greend { -1 } \end { align } $ $ \begin { align } \large i^7 & amp ; = { i^4\cdot i^3 } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot ( -i ) & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ and $ i^3=-i $ } } } \ \ & amp ; =\purpled { -i } \end { align } $ $ \begin { align } \large i^8 & amp ; = { i^4\cdot i^4~~~~ } & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; =1\cdot 1 & amp ; & amp ; \small { \gray { \text { since $ i^4=1 $ } } } \ \ & amp ; =\goldd 1 \end { align } $ the results are summarized in the table . $ i^1 $ | $ i^2 $ | $ i^3 $ | $ i^4 $ | $ i^5 $ | $ i^6 $ | $ i^7 $ | $ i^8 $ : - : |- : |- : |- : |- : |- : |- : |- : $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ | $ \blued i $ | $ \greend { -1 } $ | $ \purpled { -i } $ | $ \goldd 1 $ an emerging pattern from the table , it appears that the powers of $ i $ cycle through the sequence of $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ and $ \goldd1 $ . using this pattern , can we find $ i^ { 20 } $ ? let 's try it ! the following list shows the first $ 20 $ numbers in the repeating sequence . $ \quad $ $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ according to this logic , $ i^ { 20 } $ should be equal to $ \goldd 1 $ . let 's see if we can support this by using exponents . remember , we can use the properties of exponents here just like we do with real numbers ! $ \begin { align } i^ { 20 } & amp ; = ( i^4 ) ^5 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \ \ & amp ; = ( 1 ) ^5 & amp ; & amp ; \small { \gray { i^4=1 } } \\ & amp ; = \goldd 1 & amp ; & amp ; \small { \gray { \text { simplify } } } \end { align } $ either way , we see that $ i^ { 20 } =1 $ . larger powers of $ i $ suppose we now wanted to find $ i^ { 138 } $ . we could list the sequence $ \blued i $ , $ \greend { -1 } $ , $ \purpled { -i } $ , $ \goldd 1 $ , ... out to the $ 138^\text { th } $ term , but this would take too much time ! notice , however , that $ i^4=1 $ , $ i^8=1 $ , $ i^ { 12 } =1 $ , etc. , or , in other words , that $ i $ raised to a multiple of $ 4 $ is $ 1 $ . we can use this fact along with the properties of exponents to help us simplify $ i^ { 138 } $ . example simplify $ i^ { 138 } $ . solution while $ 138 $ is not a multiple of $ 4 $ , the number $ 136 $ is ! let 's use this to help us simplify $ i^ { 138 } $ . $ \begin { align } i^ { 138 } & amp ; =i^ { 136 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( i^ { 4\cdot 34 } ) \cdot i^2 & amp ; & amp ; \small { \gray { 136=4\cdot 34 } } \\ & amp ; = ( i^ { 4 } ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { properties of exponents } } } \\ & amp ; = ( 1 ) ^ { 34 } \cdot i^2 & amp ; & amp ; \small { \gray { \text { $ i^4=1 $ } } } \\ & amp ; =1\cdot -1 & amp ; & amp ; \small { \gray { \text { $ i^2=-1 $ } } } \\ & amp ; =-1 \end { align } $ so $ i^ { 138 } =-1 $ . now you might ask why we chose to write $ i^ { 138 } $ as $ i^ { 136 } \cdot i^2 $ . well , if the original exponent is not a multiple of $ 4 $ , then finding the closest multiple of $ 4 $ less than it allows us to simplify the power down to $ i $ , $ i^2 $ , or $ i^3 $ just by using the fact that $ i^4=1 $ . this number is easy to find if you divide the original exponent by $ 4 $ . it 's just the quotient ( without the remainder ) times $ 4 $ . let 's practice some problems problem 1 problem 2 problem 3 challenge problem
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we know that $ i=\sqrt { -1 } $ and that $ i^2=-1 $ . but what about $ i^3 $ ?
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if we know that imaginary unit is square root of -1 , then why is it called imaginary unit ?
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röttgen pietà , c. 1300-25 , painted wood , 34 1/2 inches high ( rheinisches landesmuseum , bonn ) ( photo : ralf heinz , with permission ) an emotional response it is hard to look at the röttgen pietà and not feel something—perhaps revulsion , horror , or distaste . it is terrifying and the more you look at it , the more intriguing it becomes . this is part of the beauty and drama of gothic art , which aimed to create an emotional response in medieval viewers . earlier medieval representations of christ focused on his divinity ( left ) . in these works of art , christ is on the cross , but never suffers . these types of crucifixion images are a type called christus triumphans or the triumphant christ . his divinity overcomes all human elements and so christ stands proud and alert on the cross , immune to human suffering . triumphant christ / patient christ in the later middle ages , a number of preachers and writers discussed a different type of christ who suffered in the way that humans suffered . this was different from catholic writers of earlier ages , who emphasized christ ’ s divinity and distance from humanity . late medieval devotional writing ( from the 13th-15th centuries ) leaned toward mysticism and many of these writers had visions of christ ’ s suffering . francis of assisi stressed christ ’ s humanity and poverty . several writers , such as st. bonaventure , st. bridget of sweden , and st. bernardino of siena , imagined mary ’ s thoughts as she held her dead son . it wasn ’ t long before artists began to visualize these new devotional trends . crucifixion images influenced by this body of devotional literature are called christus patiens , the patient christ . the effects of this new devotional style , which emphasized the humanity of christ , quickly spread throughout western europe through the rise of new religious orders ( the franciscans , for example ) and the popularity of their preaching . it isn ’ t hard to see the appeal of the idea that god understands the pain and difficulty of being human . in the röttgen pietà , christ clearly died from the horrific ordeal of crucifixion , but his skin is taut around his ribs , showing that he also led a life of hunger and suffering . pietà statues appeared in germany in the late 1200s and were made in this region throughout the middle ages . many examples of pietàs survive today . many of those that survive today are made of marble or stone but the röttgen pietà is made of wood and retains some of its original paint . the röttgen pietà is the most gruesome of these extant examples . many of the other pietàs also show a reclining dead christ with three dimensional wounds and a skeletal abdomen . one of the unique elements of the röttgen pietà is mary ’ s response to her dead son . she is youthful and draped in heavy robes like many of the other marys , but her facial expression is different . in catholic tradition , mary had a special foreknowledge of the resurrection of christ and so to her , christ ’ s death is not only tragic . images that reflect mary ’ s divine knowledge show her at peace while holding her dead son . mary in the röttgen pietà appears to be angry and confused . she doesn ’ t seem to know that her son will live again . she shows strong negative emotions that emphasize her humanity , just as the representation of christ emphasizes his . all of these pietàs were devotional images and were intended as a focal point for contemplation and prayer . even though the statues are horrific , the intent was to show that god and mary , divine figures , were sympathetic to human suffering , and to the pain , and loss experienced by medieval viewers . by looking at the röttgen pietà , medieval viewers may have felt a closer personal connection to god by viewing this representation of death and pain . essay by dr. nancy ross additional resources : the pietà at harvard art museums images of the röttgen pietà
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röttgen pietà , c. 1300-25 , painted wood , 34 1/2 inches high ( rheinisches landesmuseum , bonn ) ( photo : ralf heinz , with permission ) an emotional response it is hard to look at the röttgen pietà and not feel something—perhaps revulsion , horror , or distaste . it is terrifying and the more you look at it , the more intriguing it becomes .
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could the artist have depicted jesus in this manner for other reasons ?
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what you should be familiar with before this lesson a monomial is an expression that is the product of constants and nonnegative integer powers of $ x $ , like $ 3x^2 $ . a polynomial is a sum of monomials . you can write the complete factorization of a monomial by writing the prime factorization of the coefficient and expanding the variable part . check out our factoring monomials article if this is new to you . what you will learn in this lesson in this lesson , you will learn about the greatest common factor ( gcf ) and how to find this for monomials . review : greatest common factors in integers the greatest common factor of two numbers is the greatest integer that is a factor of both numbers . for example , the gcf of $ 12 $ and $ 18 $ is $ 6 $ . we can find the gcf for any two numbers by examining their prime factorizations : $ 12=\blued { 2 } \cdot 2\cdot \goldd { 3 } $ $ 18=\blued { 2 } \cdot \goldd3\cdot 3 $ notice that $ 12 $ and $ 18 $ have a factor of $ \blued { 2 } $ and a factor of $ \goldd { 3 } $ in common , and so the greatest common factor of $ 12 $ and $ 18 $ is $ \blued { 2 } \cdot \goldd { 3 } =6 $ . greatest common factors in monomials the process is similar when you are asked to find the greatest common factor of two or more monomials . simply write the complete factorization of each monomial and find the common factors . the product of all the common factors will be the gcf . for example , let 's find the greatest common factor of $ 10x^3 $ and $ 4x $ : $ 10x^3=\blued { 2 } \cdot 5\cdot \goldd { x } \cdot x\cdot x $ $ 4x=\blued { 2 } \cdot 2\cdot \goldd { x } $ notice that $ 10x^3 $ and $ 4x $ have one factor of $ \blued { 2 } $ and one factor of $ \goldd { x } $ in common . therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ . check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf . you could then find the gcf of $ \blued6 $ and $ \blued4 $ , which is $ \blued2 $ , and multiply this by $ \goldd { x^2 } $ to obtain $ \blued2\goldd { x^2 } $ , the gcf of the monomials ! this is especially helpful to understand when finding the gcf of monomials with very large powers of $ x $ . for example , it would be very tedious to completely factor monomials like $ 32x^ { 100 } $ and $ 16x^ { 88 } $ ! challenge problems what 's next ? to see how we can use these skills to factor polynomials , check out our next article on factoring out the greatest common factor !
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therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ . check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf .
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can a2 ( variable 'a ' square ) - 400 be factored ?
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what you should be familiar with before this lesson a monomial is an expression that is the product of constants and nonnegative integer powers of $ x $ , like $ 3x^2 $ . a polynomial is a sum of monomials . you can write the complete factorization of a monomial by writing the prime factorization of the coefficient and expanding the variable part . check out our factoring monomials article if this is new to you . what you will learn in this lesson in this lesson , you will learn about the greatest common factor ( gcf ) and how to find this for monomials . review : greatest common factors in integers the greatest common factor of two numbers is the greatest integer that is a factor of both numbers . for example , the gcf of $ 12 $ and $ 18 $ is $ 6 $ . we can find the gcf for any two numbers by examining their prime factorizations : $ 12=\blued { 2 } \cdot 2\cdot \goldd { 3 } $ $ 18=\blued { 2 } \cdot \goldd3\cdot 3 $ notice that $ 12 $ and $ 18 $ have a factor of $ \blued { 2 } $ and a factor of $ \goldd { 3 } $ in common , and so the greatest common factor of $ 12 $ and $ 18 $ is $ \blued { 2 } \cdot \goldd { 3 } =6 $ . greatest common factors in monomials the process is similar when you are asked to find the greatest common factor of two or more monomials . simply write the complete factorization of each monomial and find the common factors . the product of all the common factors will be the gcf . for example , let 's find the greatest common factor of $ 10x^3 $ and $ 4x $ : $ 10x^3=\blued { 2 } \cdot 5\cdot \goldd { x } \cdot x\cdot x $ $ 4x=\blued { 2 } \cdot 2\cdot \goldd { x } $ notice that $ 10x^3 $ and $ 4x $ have one factor of $ \blued { 2 } $ and one factor of $ \goldd { x } $ in common . therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ . check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf . you could then find the gcf of $ \blued6 $ and $ \blued4 $ , which is $ \blued2 $ , and multiply this by $ \goldd { x^2 } $ to obtain $ \blued2\goldd { x^2 } $ , the gcf of the monomials ! this is especially helpful to understand when finding the gcf of monomials with very large powers of $ x $ . for example , it would be very tedious to completely factor monomials like $ 32x^ { 100 } $ and $ 16x^ { 88 } $ ! challenge problems what 's next ? to see how we can use these skills to factor polynomials , check out our next article on factoring out the greatest common factor !
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the product of all the common factors will be the gcf . for example , let 's find the greatest common factor of $ 10x^3 $ and $ 4x $ : $ 10x^3=\blued { 2 } \cdot 5\cdot \goldd { x } \cdot x\cdot x $ $ 4x=\blued { 2 } \cdot 2\cdot \goldd { x } $ notice that $ 10x^3 $ and $ 4x $ have one factor of $ \blued { 2 } $ and one factor of $ \goldd { x } $ in common . therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ .
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in the last two problems , why could n't the solutions be 2x^76 and 4x^2y^2 ?
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what you should be familiar with before this lesson a monomial is an expression that is the product of constants and nonnegative integer powers of $ x $ , like $ 3x^2 $ . a polynomial is a sum of monomials . you can write the complete factorization of a monomial by writing the prime factorization of the coefficient and expanding the variable part . check out our factoring monomials article if this is new to you . what you will learn in this lesson in this lesson , you will learn about the greatest common factor ( gcf ) and how to find this for monomials . review : greatest common factors in integers the greatest common factor of two numbers is the greatest integer that is a factor of both numbers . for example , the gcf of $ 12 $ and $ 18 $ is $ 6 $ . we can find the gcf for any two numbers by examining their prime factorizations : $ 12=\blued { 2 } \cdot 2\cdot \goldd { 3 } $ $ 18=\blued { 2 } \cdot \goldd3\cdot 3 $ notice that $ 12 $ and $ 18 $ have a factor of $ \blued { 2 } $ and a factor of $ \goldd { 3 } $ in common , and so the greatest common factor of $ 12 $ and $ 18 $ is $ \blued { 2 } \cdot \goldd { 3 } =6 $ . greatest common factors in monomials the process is similar when you are asked to find the greatest common factor of two or more monomials . simply write the complete factorization of each monomial and find the common factors . the product of all the common factors will be the gcf . for example , let 's find the greatest common factor of $ 10x^3 $ and $ 4x $ : $ 10x^3=\blued { 2 } \cdot 5\cdot \goldd { x } \cdot x\cdot x $ $ 4x=\blued { 2 } \cdot 2\cdot \goldd { x } $ notice that $ 10x^3 $ and $ 4x $ have one factor of $ \blued { 2 } $ and one factor of $ \goldd { x } $ in common . therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ . check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf . you could then find the gcf of $ \blued6 $ and $ \blued4 $ , which is $ \blued2 $ , and multiply this by $ \goldd { x^2 } $ to obtain $ \blued2\goldd { x^2 } $ , the gcf of the monomials ! this is especially helpful to understand when finding the gcf of monomials with very large powers of $ x $ . for example , it would be very tedious to completely factor monomials like $ 32x^ { 100 } $ and $ 16x^ { 88 } $ ! challenge problems what 's next ? to see how we can use these skills to factor polynomials , check out our next article on factoring out the greatest common factor !
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check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf . you could then find the gcf of $ \blued6 $ and $ \blued4 $ , which is $ \blued2 $ , and multiply this by $ \goldd { x^2 } $ to obtain $ \blued2\goldd { x^2 } $ , the gcf of the monomials !
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at number 5 ) it says that 4x ( to the power of 2 ) y ( to the power of 2 ) is not correct , why is it not since 4 fits into both 40 and 32 and x ( to the power of 2 ) and y ( to the power of 2 ) are both at there highest ?
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what you should be familiar with before this lesson a monomial is an expression that is the product of constants and nonnegative integer powers of $ x $ , like $ 3x^2 $ . a polynomial is a sum of monomials . you can write the complete factorization of a monomial by writing the prime factorization of the coefficient and expanding the variable part . check out our factoring monomials article if this is new to you . what you will learn in this lesson in this lesson , you will learn about the greatest common factor ( gcf ) and how to find this for monomials . review : greatest common factors in integers the greatest common factor of two numbers is the greatest integer that is a factor of both numbers . for example , the gcf of $ 12 $ and $ 18 $ is $ 6 $ . we can find the gcf for any two numbers by examining their prime factorizations : $ 12=\blued { 2 } \cdot 2\cdot \goldd { 3 } $ $ 18=\blued { 2 } \cdot \goldd3\cdot 3 $ notice that $ 12 $ and $ 18 $ have a factor of $ \blued { 2 } $ and a factor of $ \goldd { 3 } $ in common , and so the greatest common factor of $ 12 $ and $ 18 $ is $ \blued { 2 } \cdot \goldd { 3 } =6 $ . greatest common factors in monomials the process is similar when you are asked to find the greatest common factor of two or more monomials . simply write the complete factorization of each monomial and find the common factors . the product of all the common factors will be the gcf . for example , let 's find the greatest common factor of $ 10x^3 $ and $ 4x $ : $ 10x^3=\blued { 2 } \cdot 5\cdot \goldd { x } \cdot x\cdot x $ $ 4x=\blued { 2 } \cdot 2\cdot \goldd { x } $ notice that $ 10x^3 $ and $ 4x $ have one factor of $ \blued { 2 } $ and one factor of $ \goldd { x } $ in common . therefore , their greatest common factor is $ \blued2\cdot \goldd { x } $ or $ 2x $ . check your understanding a note on the variable part of the gcf in general , the variable part of the gcf for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $ x $ . for example , consider the monomials $ \blued { 6 } \goldd { x^5 } $ and $ \blued { 4 } \goldd { x^2 } $ : since the lowest power of $ x $ is $ \goldd { x^2 } $ , that will be the variable part of the gcf . you could then find the gcf of $ \blued6 $ and $ \blued4 $ , which is $ \blued2 $ , and multiply this by $ \goldd { x^2 } $ to obtain $ \blued2\goldd { x^2 } $ , the gcf of the monomials ! this is especially helpful to understand when finding the gcf of monomials with very large powers of $ x $ . for example , it would be very tedious to completely factor monomials like $ 32x^ { 100 } $ and $ 16x^ { 88 } $ ! challenge problems what 's next ? to see how we can use these skills to factor polynomials , check out our next article on factoring out the greatest common factor !
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for example , the gcf of $ 12 $ and $ 18 $ is $ 6 $ . we can find the gcf for any two numbers by examining their prime factorizations : $ 12=\blued { 2 } \cdot 2\cdot \goldd { 3 } $ $ 18=\blued { 2 } \cdot \goldd3\cdot 3 $ notice that $ 12 $ and $ 18 $ have a factor of $ \blued { 2 } $ and a factor of $ \goldd { 3 } $ in common , and so the greatest common factor of $ 12 $ and $ 18 $ is $ \blued { 2 } \cdot \goldd { 3 } =6 $ . greatest common factors in monomials the process is similar when you are asked to find the greatest common factor of two or more monomials .
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what is the greatest common factor of 20a^ { 2 } 20a 2 20 , a , start superscript , 2 , end superscript , 12a^ { 2 } 12a 2 12 , a , start superscript , 2 , end superscript , and 8a8a8 , a ?
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ancestor portraits we all know portraits can be made of ancestors , but can a portrait be an ancestor ? in te ao māori , the māori world , they can . paintings like this one—and even photographs—do two important things . they record likenesses and bring ancestral presence into the world of the living . in other words , this portrait is not merely a representation of tamati waka nene , it can be an embodiment of him . portraits and other taonga tuku iho ( treasures passed down from the ancestors ) are treated with great care and reverence . after a person has died their portrait may be hung on the walls of family homes and in the wharenui ( the central building of a community center ) , to be spoken to , wept over , and cherished by people with genealogical connections to them . even when portraits like this one , kept in the collection of the auckland art gallery , are absent from their families , the stories woven around them keep them alive and present . auckland art gallery acknowledges these living links through its relationships with descendants of those whose portraits it cares for . the gallery seeks their advice when asked for permission to reproduce such portraits . this portrait has been published in the google art project , which is why we can look at it here . tamati waka nene māori are the indigenous people of new zealand . the subject of this portrait , tamati waka nene , was a rangatira or chief of the ngāti hao people in hokianga , of the ngāpuhi iwi or tribe , and an important war leader . he was probably born in the 1780s , and died in 1871 . he lived through a time of rapid change in new zealand , when the first british missionaries and settlers were arriving and changing the māori world forever . an astute leader and businessman , nene exemplified the types of changes that were occurring when he converted to the wesleyan faith and was baptised in 1839 , choosing to be named tamati waka after thomas walker , who was an english merchant patron of the church missionary society . he was revered throughout his life as a man with great mana or personal efficacy . in this portrait , nene wears a kahu kiwi , a fine cloak covered in kiwi feathers , and an earring of greenstone or pounamu . both of these are prestigious taonga or treasures . he is holding a hand weapon known as a tewhatewha , which has feathers adorning its blade and a finely carved hand grip with an abalone or paua eye . all of these mark him as man of mana or personal efficacy and status . but perhaps the most striking feature for an international audience is his intricate facial tattoo , called moko . gottfried lindauer and his patron lindauer was a czech artist who arrived in new zealand in 1873 after a decade of painting professionally in europe . he had studied at the academy of fine art in vienna from 1855 to 1861 , and learned painting techniques rooted in renaissance naturalism . when he left the academy he began working as a portrait painter , and established his own portrait studio in 1864 . just ten years later he arrived in new zealand and quickly became acquainted with a man called henry partridge who became his patron . partridge commissioned lindauer to paint portraits of well-known māori , and three years later , in 1877 , lindauer held an exhibition in wellington . the exhibition was important because it demonstrated lindauer ’ s abilities and he was soon being commissioned by māori chiefs to paint their portraits . lindauer took different approaches to his commissions depending on who was paying . he tended to paint well-known māori in māori clothing for european purchasers , but painted unknown māori in everyday european clothing when commissioned by their families to do so . his paintings are realistic , convincingly three-dimensional , and play beautifully with the contrast between light and shadow , causing his subjects to glow against their dark backgrounds . as his patron , partridge amassed a large collection of portraits as well as large scale depictions that re-enacted māori ways of life that were thought to be disappearing . in 1915 , partridge gave his collection of 62 portraits to the auckland art gallery—the largest collection of lindauer paintings in the world . painting tamati waka nene if you ’ ve been paying attention to dates you will have noticed that nene died in 1871 but lindauer didn ’ t arrive in new zealand until 1873 , and didn ’ t paint his portrait until 1890 . it is likely that lindauer based this portrait on a photograph taken by john crombie , who had been commissioned to produce 12 photographic portraits of māori chiefs for the london illustrated news ( image above ) . there are several other photographs of nene , and in 1934 charles f goldie—another famous portrayer of māori—painted yet another portrait of him from a photograph . so nene didn ’ t sit for either of his famous painted portraits , but clearly sat for photographic portraits in the later years of his life . these were becoming more common by 1870 , due to developments in photographic methods that made the whole process easier and cheaper . many māori had their portraits taken photographically and produced as a carte de visite , roughly the size of a playing card , and some had larger , postcard sized images made , called cabinet portraits . lindauer is thought to have used a device called an epidiascope to enlarge and project small photographs such as these so he could paint them . lindauer didn ’ t make many sketches . he worked straight onto stretched canvas , outlining his subjects in pencil over a white background before applying translucent paints and glazes . through the thinly painted surface of some of his works you can still see traces of pencil lines that may be evidence of his practice of outlining projected images . but lindauer wasn ’ t simply copying photographs . in the 1870s , color photography had yet to be invented—lindauer was working from black and white images and reimagining them in color . moreover , sometimes he dressed his sitters—and those he painted from photos—in borrowed garments and adorned them with taonga that were not necessarily theirs . thus some of his works contain artistic interventions rather than being entirely documentary . essay by dr. billie lythberg additional resources this painting at the auckland art gallery lindauer online behind the brush—tv documentary series on lindauer 's māori portraits moko , encyclopedia of new zealand
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this portrait has been published in the google art project , which is why we can look at it here . tamati waka nene māori are the indigenous people of new zealand . the subject of this portrait , tamati waka nene , was a rangatira or chief of the ngāti hao people in hokianga , of the ngāpuhi iwi or tribe , and an important war leader .
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how was new zealand made ?
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ancestor portraits we all know portraits can be made of ancestors , but can a portrait be an ancestor ? in te ao māori , the māori world , they can . paintings like this one—and even photographs—do two important things . they record likenesses and bring ancestral presence into the world of the living . in other words , this portrait is not merely a representation of tamati waka nene , it can be an embodiment of him . portraits and other taonga tuku iho ( treasures passed down from the ancestors ) are treated with great care and reverence . after a person has died their portrait may be hung on the walls of family homes and in the wharenui ( the central building of a community center ) , to be spoken to , wept over , and cherished by people with genealogical connections to them . even when portraits like this one , kept in the collection of the auckland art gallery , are absent from their families , the stories woven around them keep them alive and present . auckland art gallery acknowledges these living links through its relationships with descendants of those whose portraits it cares for . the gallery seeks their advice when asked for permission to reproduce such portraits . this portrait has been published in the google art project , which is why we can look at it here . tamati waka nene māori are the indigenous people of new zealand . the subject of this portrait , tamati waka nene , was a rangatira or chief of the ngāti hao people in hokianga , of the ngāpuhi iwi or tribe , and an important war leader . he was probably born in the 1780s , and died in 1871 . he lived through a time of rapid change in new zealand , when the first british missionaries and settlers were arriving and changing the māori world forever . an astute leader and businessman , nene exemplified the types of changes that were occurring when he converted to the wesleyan faith and was baptised in 1839 , choosing to be named tamati waka after thomas walker , who was an english merchant patron of the church missionary society . he was revered throughout his life as a man with great mana or personal efficacy . in this portrait , nene wears a kahu kiwi , a fine cloak covered in kiwi feathers , and an earring of greenstone or pounamu . both of these are prestigious taonga or treasures . he is holding a hand weapon known as a tewhatewha , which has feathers adorning its blade and a finely carved hand grip with an abalone or paua eye . all of these mark him as man of mana or personal efficacy and status . but perhaps the most striking feature for an international audience is his intricate facial tattoo , called moko . gottfried lindauer and his patron lindauer was a czech artist who arrived in new zealand in 1873 after a decade of painting professionally in europe . he had studied at the academy of fine art in vienna from 1855 to 1861 , and learned painting techniques rooted in renaissance naturalism . when he left the academy he began working as a portrait painter , and established his own portrait studio in 1864 . just ten years later he arrived in new zealand and quickly became acquainted with a man called henry partridge who became his patron . partridge commissioned lindauer to paint portraits of well-known māori , and three years later , in 1877 , lindauer held an exhibition in wellington . the exhibition was important because it demonstrated lindauer ’ s abilities and he was soon being commissioned by māori chiefs to paint their portraits . lindauer took different approaches to his commissions depending on who was paying . he tended to paint well-known māori in māori clothing for european purchasers , but painted unknown māori in everyday european clothing when commissioned by their families to do so . his paintings are realistic , convincingly three-dimensional , and play beautifully with the contrast between light and shadow , causing his subjects to glow against their dark backgrounds . as his patron , partridge amassed a large collection of portraits as well as large scale depictions that re-enacted māori ways of life that were thought to be disappearing . in 1915 , partridge gave his collection of 62 portraits to the auckland art gallery—the largest collection of lindauer paintings in the world . painting tamati waka nene if you ’ ve been paying attention to dates you will have noticed that nene died in 1871 but lindauer didn ’ t arrive in new zealand until 1873 , and didn ’ t paint his portrait until 1890 . it is likely that lindauer based this portrait on a photograph taken by john crombie , who had been commissioned to produce 12 photographic portraits of māori chiefs for the london illustrated news ( image above ) . there are several other photographs of nene , and in 1934 charles f goldie—another famous portrayer of māori—painted yet another portrait of him from a photograph . so nene didn ’ t sit for either of his famous painted portraits , but clearly sat for photographic portraits in the later years of his life . these were becoming more common by 1870 , due to developments in photographic methods that made the whole process easier and cheaper . many māori had their portraits taken photographically and produced as a carte de visite , roughly the size of a playing card , and some had larger , postcard sized images made , called cabinet portraits . lindauer is thought to have used a device called an epidiascope to enlarge and project small photographs such as these so he could paint them . lindauer didn ’ t make many sketches . he worked straight onto stretched canvas , outlining his subjects in pencil over a white background before applying translucent paints and glazes . through the thinly painted surface of some of his works you can still see traces of pencil lines that may be evidence of his practice of outlining projected images . but lindauer wasn ’ t simply copying photographs . in the 1870s , color photography had yet to be invented—lindauer was working from black and white images and reimagining them in color . moreover , sometimes he dressed his sitters—and those he painted from photos—in borrowed garments and adorned them with taonga that were not necessarily theirs . thus some of his works contain artistic interventions rather than being entirely documentary . essay by dr. billie lythberg additional resources this painting at the auckland art gallery lindauer online behind the brush—tv documentary series on lindauer 's māori portraits moko , encyclopedia of new zealand
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this portrait has been published in the google art project , which is why we can look at it here . tamati waka nene māori are the indigenous people of new zealand . the subject of this portrait , tamati waka nene , was a rangatira or chief of the ngāti hao people in hokianga , of the ngāpuhi iwi or tribe , and an important war leader .
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did the influence of the chief 's with their people begin to wane during this period ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar .
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what are the mentioned a and b levels is that a different time period in the neolithic , or is it a reference to the level of the excavation ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar .
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what is the most widely excepted theroy surronding these skulls ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning .
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could the plastered skulls possibly be akin to sumerian votive figures like the ones found at tell asmar ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century .
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how respectable is it for a modern archaeologist to go looking for proof of something in the bible ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall .
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why was obsidian a `` particularly useful kind of stone '' ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey .
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do archaeologists think that the plaster covering the skulls was simply for adornment purposes , or were people trying to preserve or mummify them ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e .
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what kinds of non-human predators were common during this time ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement .
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are there extinct kinds of animals that preyed on humans that the inhabitants of jericho would need to construct a wall to keep out ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement .
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despite having no proof of joshuah 's existence , is there archaeological evidence that proves the wall had a major collapse whilst under a major siege ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho .
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or is the supposed siege a metaphor like the majority of the christian bible ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered .
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if it is believed that jericho existed during joshua 's time in the bible then why is there not any evidence to support such claims ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity .
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was jericho part of an ancient civilization before joshua 's conquest ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e .
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i ca n't imagine the people were looking hundreds of years in advance to avoid the formation of tells , so does this theory have to do with dust storms in the area piling sediment ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs .
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how many levels of excavation are there in jericho ?
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a natural oasis the site of jericho , just north of the dead sea and due west of the jordan river , is one of the oldest continuously lived-in cities in the world . the reason for this may be found in its arabic name , ārīḥā , which means fragrant ; jericho is a natural oasis in the desert where countless fresh water springs can be found . this resource , which drew its first visitors between 10,000 and 9000 b.c.e. , still has descendants that live there today . biblical reference the site of jericho is best known for its identity in the bible and this has drawn pilgrims and explorers to it as early as the 4th century c.e . ; serious archaeological exploration didn ’ t begin until the latter half of the 19th century . what continues to draw archaeologists to jericho today is the hope of finding some evidence of the warrior joshua , who lead the israelites to an unlikely victory against the canaanites ( `` the walls of the city fell when joshua and his men marched around them blowing horns '' joshua 6:1-27 ) . although unequivocal evidence of joshua himself has yet to be found , what has been uncovered are some 12,000 years of human activity . the most spectacular finds at jericho , however , do not date to the time of joshua , roughly the bronze age ( 3300-1200 b.c.e . ) , but rather to the earliest part of the neolithic era , before even the technology to make pottery had been discovered . old walls the site of jericho rises above the wide plain of the jordan valley , its height the result of layer upon layer of human habitation , a formation called a tell . the earliest visitors to the site who left remains ( stone tools ) came in the mesolithic period ( around 9000 b.c.e . ) but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs . as the name of this era implies , these early people at jericho had not yet figured out how to make pottery , but they made vessels out of stone , wove cloth and for tools were trading for a particularly useful kind of stone , obsidian , from as far away as çiftlik , in eastern turkey . the settlement grew quickly and , for reasons unknown , the inhabitants soon constructed a substantial stone wall and exterior ditch around their town , complete with a stone tower almost eight meters high , set against the inner side of the wall . theories as to the function of this wall range from military defense to keeping out animal predators to even combating the natural rising of the level of the ground surrounding the settlement . however , regardless of its original use , here we have the first version of the walls joshua so ably conquered some six thousand years later . plastered human skulls the pre-pottery neolithic a period is followed by the pre-pottery neolithic b ( 7000-5200 bce ) , which was different from its predecessor in important ways . houses in this era were uniformly rectangular and constructed with a new kind of rectangular mud bricks which were decorated with herringbone thumb impressions , and always laid lengthwise in thick mud mortar . this mortar , like a plaster , was also used to create a smooth surface on the interior walls , extending down across the floors as well . in this period there is some strong evidence for cult or religious belief at jericho . archaeologists discovered one uniquely large building dating to the period with unique series of plastered interior pits and basins as well as domed adjoining structures and it is thought this was for ceremonial use . other possible evidence of cult practice was discovered in several homes of the pre-pottery neolithic town , in the form of plastered human skulls which were molded over to resemble living heads . shells were used for eyes and traces of paint revealed that skin and hair were also included in the representations . the largest group found together were nine examples , buried in the fill below the plastered floor of one house . jericho isn ’ t the only site at which plastered skulls have been found in pre-pottery neolithic b levels ; they have also been found at tell ramad , beisamoun , kfar hahoresh , ‘ ain ghazal and nahal hemar . among the some sixty-two skulls discovered among these sites , we know that older and younger men as well as women and children are represented , which poses interesting questions as to their meaning . were they focal points in ancestor worship , as was originally thought , or did they function as images by which deceased family members could be remembered ? as we are without any written record of the belief system practiced in the neolithic period in the area , we will never know . essay by dr. senta german
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but the first settlement at the site , around the ein as-sultan spring , dates to the early neolithic era , and these people , who built homes , grew plants , and kept animals , were among the earliest to do such anywhere in the world . specifically , in the pre-pottery neolithic a levels at jericho ( 8500-7000 b.c.e . ) archaeologists found remains of a very large settlement of circular homes made with mud brick and topped with domed roofs .
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what kind of jobs could you get in ancient jericho ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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why did the people in the great depression blame hoover for everything ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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how would hoover 's presidency been different had he been able to help mitigate the effects of the great depression ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy .
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who did hoover run against in the 1928 election ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ?
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if tariffs caused a trade war , then it caused further damage to the economy , is that is why jobs were lost and the depression grew deeper ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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or is that the ball ( depression ) was sliding down the hillside and just kept growing until millions were out of work ?
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overview herbert hoover was the 31st president of the united states . he served one term , from 1929 to 1933 . before becoming president , hoover directed relief efforts to supply war-torn europe and russia during and after the first world war . after the 1929 stock market crash , the hoover administration attempted to mitigate the negative effects of the great depression but was unable to significantly improve the economy . the early life of herbert hoover herbert clark hoover was born in 1874 in iowa , and was the first us president to have been born west of the mississippi river . he worked as a mining engineer and an independent mining consultant , traveling the world and building a sizable personal fortune. $ ^1 $ when world war i broke out , hoover became active in humanitarian work , and chaired the commission for relief in belgium , which provided relief to that country as it faced a food crisis brought on by the german invasion in 1914. $ ^2 $ he also served as the director of the american relief administration , which was formed in 1919 and supplied relief to war-torn europe and russia. $ ^3 $ during the first world war , president woodrow wilson appointed hoover head of the us food administration , which sought to reduce consumption and avoid wartime food rationing . hoover went on to serve as secretary of commerce in the administration of warren g. harding . the presidency of herbert hoover hoover won the republican nomination in the 1928 presidential election after republican president calvin coolidge announced that he would not be running for reelection . hoover campaigned on coolidge ’ s legacy of economic prosperity , pledging to support business , improve the quality of life of the nation ’ s farmers , and conduct a relatively isolationist foreign policy . hoover was ambivalent about prohibition , referring to it as a “ great social and economic experiment , ” but failing to back it wholeheartedly. $ ^4 $ hoover won the election in a landslide against his democratic opponent . once in office , hoover sought to reform the nation ’ s regulatory system . he was not an advocate of a laissez-faire economy , but instead encouraged the voluntary cooperation of the federal government and big business . at hoover ’ s direction , the internal revenue service and the justice department prosecuted gangsters , including al capone , for tax evasion . hoover considered himself a progressive , and this was reflected in some of his administration ’ s policies , including the reorganization of the bureau of indian affairs , the organization of the federal bureau of prisons , the closing of tax loopholes for the wealthiest americans , the expansion of national park lands , and the strengthening of protections for labor . hoover ’ s most notable foreign policy achievements were in latin america . his administration laid the groundwork for what became the good neighbor policy under hoover 's successor franklin d. roosevelt . the good neighbor policy explicitly disavowed military interventionism in hemispheric relations . hoover himself embarked upon a goodwill tour in latin america , traveling to ten countries and delivering pledges to reduce us political and military interference in the domestic affairs of latin american countries . hoover also successfully mediated a dispute between chile and peru over land . hoover and the great depression in 1929 , the stock market crash catalyzed the onset of the great depression. $ ^5 $ though hoover has gained a reputation for dithering in the face of economic peril , his administration actually pursued measures that helped lay the basis for roosevelt ’ s new deal . hoover launched a massive public works program , part of which included funding for construction of the hoover dam on the colorado river . his administration implemented stronger protections for labor and substantially increased federal subsidies for agriculture. $ ^6 $ hoover also played a key role in passing the glass-steagall act of 1932 , which limited the activities of commercial banks in an attempt to stabilize the banking sector . however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt . what do you think ? did hoover ’ s early career effectively prepare him for the presidency ? why or why not ? what were hoover ’ s greatest accomplishments as president ? what were his most consequential shortcomings ? how would you characterize the hoover administration ’ s response to the great depression ?
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however , many of these policies were not immediately effective , and some of the administration ’ s actions actually worsened the effects of the depression . the smoot-hawley tariff act , for instance , which hoover signed reluctantly , raised tariffs on thousands of imported goods and initiated a trade war between the united states and europe , thereby exacerbating the global economic downturn. $ ^7 $ although hoover ran for reelection in 1932 , his inability to mitigate the negative economic consequences of the great depression had made him widely unpopular . he lost the election to democrat franklin d. roosevelt .
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is it true then that by placing tariffs we bought less of their goods , put their workers out of a job and the same happened here ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones .
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would n't lipid ligands generally be hydrophobic ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones .
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why do n't hydrophobic ligands get stuck in the membrane ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases .
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what are considered kinases : an rtk receptor or an rtk 's specialized ligand ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand .
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why do they [ hormones ] need to bind to carrier proteins ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes .
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is it also via genes - > rna - > codes for proteins to `` make '' the receptor in a ribosome ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules .
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is it just that a chain of proteins is brought to the membrane via vesicles ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize .
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where does the energy come from to create signaling cascades from an extracellular signal ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand .
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are hydrophobicity and receptors affinity towards the ligand indirectly proportional ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) .
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do g proteins bind to gtp which will be turned naturally into gdp after a period of time by the g protein and then have it 's gdp returned to a gtp when the gpcr is activated ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus .
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what types of signal molecule that are likely to use intracellular receptors ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane .
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are cell surface receptors also called 'extra cellular receptors '' ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ .
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`` the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane '' how would that get past the phosphorate heads of the phospholipds ?
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introduction just as a journey of a thousand miles begins with a single step , so a complex signaling pathway inside of a cell begins with a single key event – the binding of a signaling molecule , or ligand , to its receiving molecule , or receptor . receptors and ligands come in many forms , but they all have one thing in common : they come in closely matched pairs , with a receptor recognizing just one ( or a few ) specific ligands , and a ligand binding to just one ( or a few ) target receptors . binding of a ligand to a receptor changes its shape or activity , allowing it to transmit a signal or directly produce a change inside of the cell in this section , we ’ ll look at different types of receptors and ligands , seeing how they interact to turn information from outside the cell into a change inside the cell . types of receptors receptors come in many types , but they can be divided into two categories : intracellular receptors , which are found inside of the cell ( in the cytoplasm or nucleus ) , and cell surface receptors , which are found in the plasma membrane . intracellular receptors intracellular receptors are receptor proteins found on the inside of the cell , typically in the cytoplasm or nucleus . in most cases , the ligands of intracellular receptors are small , hydrophobic ( water-hating ) molecules , since they must be able to cross the plasma membrane in order to reach their receptors . for example , the primary receptors for hydrophobic steroid hormones , such as the sex hormones estradiol ( an estrogen ) and testosterone , are intracellular $ ^ { 1,2 } $ . when a hormone enters a cell and binds to its receptor , it causes the receptor to change shape , allowing the receptor-hormone complex to enter the nucleus ( if it wasn ’ t there already ) and regulate gene activity . hormone binding exposes regions of the receptor that have dna-binding activity , meaning they can attach to specific sequences of dna . these sequences are found next to certain genes in the dna of the cell , and when the receptor binds next to these genes , it alters their level of transcription . many signaling pathways , involving both intracellular and cell surface receptors , cause changes in the transcription of genes . however , intracellular receptors are unique because they cause these changes very directly , binding to the dna and altering transcription themselves . cell-surface receptors cell-surface receptors are membrane-anchored proteins that bind to ligands on the outside surface of the cell . in this type of signaling , the ligand does not need to cross the plasma membrane . so , many different kinds of molecules ( including large , hydrophilic or `` water-loving '' ones ) may act as ligands . a typical cell-surface receptor has three different domains , or protein regions : a extracellular ( `` outside of cell '' ) ligand-binding domain , a hydrophobic domain extending through the membrane , and an intracellular ( `` inside of cell '' ) domain , which often transmits a signal . the size and structure of these regions can vary a lot depending on the type of receptor , and the hydrophobic region may consist of multiple stretches of amino acids that criss-cross the membrane . there are many kinds of cell-surface receptors , but here we ’ ll look at three common types : ligand-gated ion channels , g protein-coupled receptors , and receptor tyrosine kinases . ligand-gated ion channels ligand-gated ion channels are ion channels that can open in response to the binding of a ligand . to form a channel , this type of cell-surface receptor has a membrane-spanning region with a hydrophilic ( water-loving ) channel through the middle of it . the channel lets ions to cross the membrane without having to touch the hydrophobic core of the phospholipid bilayer . when a ligand binds to the extracellular region of the channel , the protein ’ s structure changes in such a way that ions of a particular type , such as $ \text { ca } ^ { 2+ } $ or $ \text { cl } ^- $ , can pass through . in some cases , the reverse is actually true : the channel is usually open , and ligand binding causes it to close . changes in ion levels inside the cell can change the activity of other molecules , such as ion-binding enzymes and voltage-sensitive channels , to produce a response . neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) . gpcrs are diverse and bind many different types of ligands . one particularly interesting class of gpcrs is the odorant ( scent ) receptors . there are about $ 800 $ of them in humans , and each binds its own “ scent molecule ” – such as a particular chemical in perfume , or a certain compound released by rotting fish – and causes a signal to be sent to the brain , making us smell a smell ! $ ^3 $ when its ligand is not present , a g protein-coupled receptor waits at the plasma membrane in an inactive state . for at least some types of gpcrs , the inactive receptor is already docked to its signaling target , a g protein $ ^4 $ . g proteins come in different types , but they all bind the nucleotide guanosine triphosphate ( gtp ) , which they can break down ( hydrolyze ) to form gdp . a g protein attached to gtp is active , or “ on , ” while a g protein that ’ s bound to gdp is inactive , or “ off. ” the g proteins that associate with gpcrs are a type made up of three subunits , known as heterotrimeric g proteins . when they ’ re attached to an inactive receptor , they ’ re in the “ off ” form ( bound to gdp ) . ligand binding , however , changes the picture : the gpcr is activated and causes the g protein to gdp for gtp . the now-active g protein separates into two pieces ( one called the α subunit , the other consisting of the β and γ subunits ) , which are freed from the gpcr . the subunits can interact with other proteins , triggering a signaling pathway that leads to a response . eventually , the α subunit will hydrolyze gtp back to gdp , at which point the g protein becomes inactive . the inactive g protein reassembles as a three-piece unit associated with a gpcr . cell signaling using g protein-coupled receptors is a cycle , one that can repeat over and over in response to ligand binding . g protein-coupled receptors play many different roles in the human body , and disruption of gpcr signaling can cause disease . receptor tyrosine kinases enzyme-linked receptors are cell-surface receptors with intracellular domains that are associated with an enzyme . in some cases , the intracellular domain of the receptor actually is an enzyme that can catalyze a reaction . other enzyme-linked receptors have an intracellular domain that interacts with an enzyme $ ^5 $ . receptor tyrosine kinases ( rtks ) are a class of enzyme-linked receptors found in humans and many other species . a kinase is just a name for an enzyme that transfers phosphate groups to a protein or other target , and an receptor tyrosine kinase transfers phosphate groups to specifically to the amino acid tyrosine . how does rtk signaling work ? in a typical example , signaling molecules first bind to the extracellular domains of two nearby receptor tyrosine kinases . the two neighboring receptors then come together , or dimerize . the receptors then attach phosphates to tyrosines in each others ' intracellular domains . the phosphorylated tyrosine can transmit the signal to other molecules in the cell . in many cases , the phosphorylated receptors serve as a docking platform for other proteins that contain special types of binding domains . a variety of proteins contain these domains , and when one of these proteins binds , it can initiate a downstream signaling cascade that leads to a cellular response $ ^ { 6,7 } $ . receptor tyrosine kinases are crucial to many signaling processes in humans . for instance , they bind to growth factors , signaling molecules that promote cell division and survival . growth factors include platelet-derived growth factor ( pdgf ) , which participates in wound healing , and nerve growth factor ( ngf ) , which must be continually supplied to certain types of neurons to keep them alive $ ^8 $ . because of their role in growth factor signaling , receptor tyrosine kinases are essential in the body , but their activity must be kept in balance : overactive growth factor receptors are associated with some types of cancers . types of ligands ligands , which are produced by signaling cells and interact with receptors in or on target cells , come in many different varieties . some are proteins , others are hydrophobic molecules like steroids , and others yet are gases like nitric oxide . here , we ’ ll look at some examples of different types of ligands . ligands that can enter the cell small , hydrophobic ligands can pass through the plasma membrane and bind to intracellular receptors in the nucleus or cytoplasm . in the human body , some of the most important ligands of this type are the steroid hormones . familiar steroid hormones include the female sex hormone estradiol , which is a type of estrogen , and the male sex hormone testosterone . vitamin d , a molecule synthesized in the skin using energy from light , is another example of a steroid hormone . because they are hydrophobic , these hormones don ’ t have trouble crossing the plasma membrane , but they must bind to carrier proteins in order to travel through the ( watery ) bloodstream . nitric oxide ( no ) is a gas that acts as a ligand . like steroid hormones , it can diffuse directly across the plasma membrane thanks to is small size . one of its key roles is to activate a signaling pathway in the smooth muscle surrounding blood vessels , one that makes the muscle relax and allows the blood vessels to expand ( dilate ) . in fact , the drug nitroglycerin treats heart disease by triggering the release of no , dilating vessels to restore blood flow to the heart . no has become better-known recently because the pathway that it affects is targeted by prescription medications for erectile dysfunction , such as viagra . ligands that bind on the outside of the cell water-soluble ligands are polar or charged and can not readily cross the plasma membrane . so , most water-soluble ligands bind to the extracellular domains of cell-surface receptors , staying on the outer surface of the cell . peptide ( protein ) ligands make up the largest and most diverse class of water-soluble ligands . for instance , growth factors , hormones such as insulin , and certain neurotransmitters fall into this category . peptide ligands can range from just a few amino acids long , as in the pain-suppressing enkephalins , to a hundred or more amino acids in length $ ^9 $ . as mentioned above , some neurotransmitters are proteins . many other neurotransmitters , however , are small , hydrophilic ( water-loving ) organic molecules . some neurotransmitters are standard amino acids , such as glutamate and glycine , and others are modified or non-standard amino acids .
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neurons , or nerve cells , have ligand-gated channels that are bound by neurotransmitters . g protein-coupled receptors g protein-coupled receptors ( gpcrs ) are a large family of cell surface receptors that share a common structure and method of signaling . the members of the gpcr family all have seven different protein segments that cross the membrane , and they transmit signals inside the cell through a type of protein called a g protein ( more details below ) .
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how does the structure of the chemical relate to the effect ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ?
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in example 3 , is there an error ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ .
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if h is an infinitsimal why does the magnitude of v matter ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ .
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even if it would matter would n't it be better to aproach the vector 's magnitude to zero too ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space .
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whats the difference between directional derivative and slope ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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is n't the directional derivative just computing the rate at which f will change while the input moves along v , which is a lengthier way of describing the slope ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient .
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what does it mean to `` normalize '' a vector ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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$ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative .
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are there any mathematical prove of the directional derivatives ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ .
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in example 3 , should n't the unit vector u in the direction of v be 2/ ( 13 ) ^ ( 1/2 ) i + 3/ ( 13 ) ^ ( 1/2 ) j ?
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background partial derivatives gradient what we 're building to if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first . generalizing partial derivatives consider some multivariable function : $ f ( x , y ) = x^2 - xy $ we know that the partial derivatives with respect to $ x $ and $ y $ tell us the rate of change of $ f $ as we nudge the input either in the $ x $ or $ y $ direction . the question now is what happens when we nudge the input of $ f $ in a direction which is not parallel to the $ x $ or $ y $ axes . for example , the image below shows the graph of $ f $ along with a small step along a vector $ \vec { \textbf { v } } $ in the input space , meaning the $ xy $ -plane in this case . is there an operation which tells us how the height of the graph above the tip of $ \vec { \textbf { v } } $ compares to the height of the graph above its tail ? as you have probably guessed , there is a new type of derivative , called the directional derivative , which answers this question . just as the partial derivative is taken with respect to some input variable—e.g. , $ x $ or $ y $ —the directional derivative is taken along some vector $ \vec { \textbf { v } } $ in the input space . one very helpful way to think about this is to picture a point in the input space moving with velocity $ \vec { \textbf { v } } $ . the directional derivative of $ f $ along $ \vec { \textbf { v } } $ is the resulting rate of change in the output of the function . so , for example , multiplying the vector $ \vec { \textbf { v } } $ by two would double the value of the directional derivative since all changes would be happening twice as fast . notation there are quite a few different notations for this one concept : $ \nabla_\vec { \textbf { v } } f $ $ \dfrac { \partial f } { \partial \vec { \textbf { v } } } $ $ f'_\vec { \textbf { v } } $ $ d_\vec { \textbf { v } } f $ $ \partial_\vec { \textbf { v } } f $ all of these represent the same thing : the rate of change of $ f $ as you nudge the input along the direction of $ \vec { \textbf { v } } $ . we 'll use the $ \nabla_\vec { \textbf { v } } f $ notation , just because it subtly hints at how you compute the directional derivative using the gradient , which you 'll see in a moment . example 1 : $ \vec { \textbf { v } } = \hat { \textbf { j } } $ before jumping into the general rule for computing $ \nabla_ { \vec { \textbf { v } } } f $ , let 's look at how we can rewrite the more familiar notion of a partial derivative as a directional derivative . for example , the partial derivative $ \dfrac { \partial f } { \partial y } $ tells us the rate at which $ f $ changes as we nudge the input in the $ y $ direction . in other words , as we nudge it along the vector $ \hat { \textbf { j } } $ . therefore , we could equivalently write the partial derivative with respect to $ y $ as $ \dfrac { \partial f } { \partial y } = \nabla_ { \hat { \textbf { j } } } f $ . this is all just fiddling with different notation . what 's more important is to have a clear mental image of what all this notation represents . reflection question : suppose $ \vec { \textbf { v } } = \hat { \textbf { i } } + \hat { \textbf { j } } $ , what is your best guess for $ \nabla_ { \vec { \textbf { v } } } f\ , $ ? how to compute the directional derivative let 's say you have a multivariable $ f ( x , y , z ) $ which takes in three variables— $ x $ , $ y $ and $ z $ —and you want to compute its directional derivative along the following vector : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { 2 } \ \rede { 3 } \ \greene { -1 } \end { array } \right ] $ the answer , as it turns out , is $ \nabla_\vec { \textbf { v } } f = \bluee { 2 } \dfrac { \partial f } { \partial x } + \rede { 3 } \dfrac { \partial f } { \partial y } + \greene { ( -1 ) } \dfrac { \partial f } { \partial z } $ this should make sense because a tiny nudge along $ \vec { \textbf { v } } $ can be broken down into $ \bluee { two } $ tiny nudges in the $ x $ -direction , $ \rede { three } $ tiny nudges in the $ y $ -direction , and a tiny nudge backwards , by $ \greene { -1 } $ , in the $ z $ -direction . we 'll go through the rigorous reasoning behind this much more thoroughly in the next article . more generally , we can write the vector $ \vec { \textbf { v } } $ abstractly as follows : $ \vec { \textbf { v } } = \left [ \begin { array } { c } \bluee { v_1 } \ \rede { v_2 } \ \greene { v_3 } \end { array } \right ] $ the directional derivative looks like this : $ \nabla_\vec { \textbf { v } } f = \bluee { v_1 } \dfrac { \partial f } { \partial x } + \rede { v_2 } \dfrac { \partial f } { \partial y } + \greene { v_3 } \dfrac { \partial f } { \partial z } $ that is , a tiny nudge in the $ \vec { \textbf { v } } $ direction consists of $ \bluee { v_1 } $ times a tiny nudge in the $ x $ -direction , $ \rede { v_2 } $ times a tiny nudge in the $ y $ -direction , and $ \greene { v_3 } $ times a tiny nudge in the $ z $ -direction . this can be written in a super-pleasing compact way using the dot product and the gradient : this is why the notation $ \nabla_\vec { \textbf { v } } $ is so suggestive of the way we compute the directional derivative : $ \begin { align } \nabla_\maroond { \vec { \textbf { v } } } f = \nabla f \cdot \maroond { \vec { \textbf { v } } } \ \end { align } $ take a moment to delight in the fact that one single operation , the gradient , packs enough information to compute the rate of change of a function in every possible direction ! that 's so many directions ! left , right , up , down , north-north-east , 34.8 $ ^\circ $ clockwise from the $ x $ -axis ... madness ! example 2 : problem : take a look at the following function . $ f ( x , y ) = x^2 - xy $ , what is the directional derivative of $ f $ at the point $ ( 2 , -3 ) $ along the vector $ \begin { align } \vec { \textbf { v } } = \bluee { 0.6 } \hat { \textbf { i } } + \rede { 0.8 } \hat { \textbf { j } } \end { align } $ ? solution : you can think of the direction derivative either as a weighted sum of partial derivatives , as below : $ \begin { align } \nabla_\vec { \textbf { v } } f = \bluee { 0.6 } \dfrac { \partial f } { \partial x } + \rede { 0.8 } \dfrac { \partial f } { \partial y } \end { align } $ or , you can think of it as a dot product with the gradient , as you see here : $ \begin { align } \nabla_\vec { \textbf { v } } f = \nabla f \cdot \vec { \textbf { v } } \end { align } $ the first is faster , but just for practice , let 's see how the gradient interpretation unfolds . we start by computing the gradient itself : next , plug in the point $ ( x , y ) = ( 2 , -3 ) $ since this is the point the question asks us about . $ \begin { align } \nabla f ( 2 , -3 ) = \left [ \begin { array } { c } 2 ( 2 ) - ( -3 ) \ - ( 2 ) \end { array } \right ] = \left [ \begin { array } { c } 7 \ -2 \end { array } \right ] \end { align } $ to get the desired directional derivative , we take the dot product between this gradient and $ \textbf { v } $ : finding slope how do you find the slope of a graph intersected with a plane that is not parallel to the $ x $ or $ y $ axes ? you can use the directional derivative , but there is one important thing to remember : if the directional derivative is used to compute slope , either $ \vec { \textbf { v } } $ must be a unit vector or you must remember to divide by $ ||\vec { \textbf { v } } || $ at the end . in the definition and computation above , doubling the length of $ \vec { \textbf { v } } $ would double the value of the directional derivative . in terms of the computation , this is because $ \nabla f \cdot ( 2\vec { \textbf { v } } ) = 2 ( \nabla f \cdot v ) $ . however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why . how can we imagine this slope ? slice the graph of $ f $ with a vertical plane that cuts the $ xy $ -plane in the direction of $ \vec { \textbf { v } } $ . the slope in question is that of a line tangent to the resulting curve . as with any slope , we look for the rise over run . in this case , the run will be the distance of a small nudge in the direction of $ \vec { \textbf { v } } $ . we can express such a nudge as an addition of $ h\vec { \textbf { v } } $ to an input point $ \textbf { x } _0 $ , where $ h $ is thought of as some small number . the magnitude of this nudge is $ h ||\vec { \textbf { v } } || $ . the resulting change in the output of $ f $ can be approximated by multiplying this little value $ h $ by the directional derivative : $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ in fact , the rise of the tangent line—as opposed to the graph of the function— is precisely $ h \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) $ due to this run of size $ h ||\vec { \textbf { v } } || $ . for full details on why this is true , see the formal definition of the directional derivative in the next article . therefore , the rise-over-run slope of our graph is $ \begin { align } \quad \dfrac { h\nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { h||v|| } = \boxed { \dfrac { \nabla_\vec { \textbf { v } } f ( x_0 , y_0 ) } { ||v|| } } \end { align } $ notice , if $ \vec { \textbf { v } } $ is a unit vector , meaning $ ||\vec { \textbf { v } } || = 1 $ , then the directional derivative does give the slope of a graph along that direction . otherwise , it is important to remember to divide out by the magnitude of $ \vec { \textbf { v } } $ . some authors even go so far as to include normalization in the definition of $ \nabla_\vec { \textbf { v } } f $ . alternate definition of directional derivative : $ \begin { align } \quad \nabla_ { \vec { \textbf { v } } } f ( \textbf { x } ) = \lim_ { h \to 0 } \dfrac { f ( \textbf { x } + h\vec { \textbf { v } } ) - f ( \textbf { x } ) } { h\bluee { ||\vec { \textbf { v } } || } } \end { align } $ personally , i think this definition puts too much emphasis on the particular use case of finding slope , so i prefer to use the original definition and normalize $ \vec { \textbf { v } } $ when necessary . example 3 : slope problem : on the stage for this problem we have three players . player 1 , the function : $ f ( x , y ) = \sin ( xy ) $ player 2 , the point : $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ player 3 , the vector : $ \vec { \textbf { v } } = 2 \hat { \textbf { i } } + 3 \hat { \textbf { j } } $ what is the slope of the graph of $ f $ at the point $ ( x_0 , y_0 ) $ along the vector $ \vec { \textbf { v } } $ ? answer : since we are finding slope , we must first normalize the vector in question . the magnitude $ ||\textbf { v } || $ is $ \sqrt { 2^2 + 3^2 } = \sqrt { 13 } $ , so we divide each term by $ \sqrt { 13 } $ to get the resulting unit vector $ \hat { \textbf { u } } $ in the direction of $ \textbf { v } $ : next , find the gradient of $ f $ : plug in the point $ ( x_0 , y_0 ) = \left ( \dfrac { \pi } { 3 } , \dfrac { 1 } { 2 } \right ) $ to this gradient . finally , take the dot product between $ \hat { \textbf { u } } $ and $ \nabla f ( \pi/3 , 1/2 ) $ : summary if you have some multivariable function , $ f ( x , y ) $ and some vector in the function 's input space , $ \vec { \textbf { v } } $ , the directional derivative of $ f $ along $ \vec { \textbf { v } } $ tells you the rate at which $ f $ will change while the input moves with velocity vector $ \vec { \textbf { v } } $ . the notation here is $ \nabla_ { \vec { \textbf { v } } } f $ , and it is computed by taking the dot product between the gradient of $ f $ and the vector $ \vec { \textbf { v } } $ , that is , $ \nabla f \cdot \vec { \textbf { v } } $ . when the directional derivative is used to compute slope , be sure to normalize the vector $ \vec { \textbf { v } } $ first .
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however , this might not always be what you want . the slope of a graph in the direction of $ \vec { \textbf { v } } $ , for example , depends only on the direction of $ \vec { \textbf { v } } $ , not the magnitude $ |\vec { \textbf { v } } | $ . let 's see why .
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can you cite an example where i can find the rate of change of a certain temperature in the direction towards a point if the direction of a vector is not given ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side .
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what is the `` aq '' in the parentheses on problem 1 ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water .
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where does the water from the combustion example go ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction .
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can a oxidation number be +1.6667 or decimal of some sort as with 3clo- o= -2 therefore 3xo = -6 therefore 3xcl must = +5 so 1cl equals 1.6666667 or +5 divided by the 3cl or have i lost the plot ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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$ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon .
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what do you mean by saying `` we can check to see if any reactants and products appear on both side '' ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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$ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ?
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how 2fe2o3 is oxidising agent it gaining electrons it should reducing agent ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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problem 1 - on the oxidation half-reaction , how did the electrons become 2 ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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( ex : 2h+ + 2e- - > h2 ) - i understand why h+ became 2 to balance the h2 in products but are n't the electrons only to balance the oxidation state ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale .
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can someone please explain redox reactions in photosynthesis ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction .
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how can electrons be shared within an atom ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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how come the oxidation agent is not simply fe aswell ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions .
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is there a way of finding the oxidation number of a constituent without having to remember the oxidation number of oxygen , florine or other elements ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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how do you find the oxidation numbers on the c8h18 in the combustion reaction ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction .
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should n't 1 electron be added to the 2h^+ because the oxidation number is 1 ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides .
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electrons can not be transferred in fractions right ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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so would the oxidation half reaction be co- > co2+ +2e ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ .
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is oxidation number of o always 2- ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced .
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for the combustion of octane , what is the oxidation number of carbon in octane ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions .
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can an oxidation number be a fraction ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions .
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so you do n't consider the coefficient in a molecule when finding an oxidation number and only think about the ratio of subscripts ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ?
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how to calculate ph of a solution without using log ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done !
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it 's been mentioned a few times now , and i have a vague idea of what it is , but what does it actually mean by species ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale .
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what is the significance of balancing redox reactions ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions .
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when do you multiply by the subscript to find the oxidation number ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species .
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could someone explain why the combustion of octane is a redox reaction ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side .
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is loss of electrons from one atom in a compound only possible if the other atom is more electronegative ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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$ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon .
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how to certain reactants and certain products have these charges and others do n't ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction .
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can we calculate these charges ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound .
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how does oxygen always have an oxidation number of -2 ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction .
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is it not possible that in a compound it gains only one electron or gives away its electrons ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ .
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does it always have to gain 2 electrons ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced .
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in the combustion reaction of octane does the carbon atom in the octane molecule have an oxidation number of -2.25 to neutralize the hyrdrogen ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ?
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in the above example h has a charge of 1+ why are 2 electrons added to the half reaction of h to get a charge of 0 ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species .
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how can it be used to predict if a reaction is redox or not ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ .
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can oxidation number be decimal , in the above combustion reaction example oxidation number of carbon is -2.25 in c8h18 if so then what does it means since oxidation number shows transfer of electrons ?
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what is a redox reaction ? a redox ( or oxidation-reduction ) reaction is a type of chemical reaction that involves a transfer of electrons between two species . we can tell there has been a transfer of electrons if there is any change in the oxidation number between the reactants and the products . redox reactions are everywhere ! your body uses redox reactions to convert food and oxygen to energy plus water and $ \text { co } _2 $ , which we then exhale . the batteries in your electronics also rely on redox reactions , which you will hear more about when we learn electrochemistry . can you find other examples of redox reactions happening around you ? an example and important terms redox reactions have some associated terms you should be comfortable using . we will go over these terms using the following example reaction : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ here are some questions we want to be able to answer : $ 1.~ $ is this reaction a redox reaction , and how do we know ? $ 2.~ $ if this is a redox reaction , what is being reduced and oxidized ? $ 3.~ $ what is the reducing agent in this reaction ? $ 4.~ $ what is the oxidizing agent in this reaction ? question $ 1 $ : based on the title of this article , we can make an educated guess about the first part of this question . yes , this is probably a redox reaction , but how do we know that for sure ? we need to show there is an electron transfer occurring , and we can do that by checking if any oxidation numbers change from the reactants to the products . if we find the oxidation numbers for each atom in the reactants and products , we get the following : $ 2 \text { fe } _2 \text o_3 ( s ) + 3 \text c ( s ) \rightarrow 4 \text { fe } ( s ) + 3 \text { co } _2 ( g ) $ $ \purplec { ~~~\downarrow~~~~\downarrow ~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~\downarrow~~~\downarrow } ~~~~~~~~~ $ $ \purplec { +3 , -2~~~~~~~~~~~~~0~~~~~~~~~~~~~~~0~~~~~~~~+4 , -2 } ~~~~~~~~~\text { ( oxidation numbers ) } $ we can use the oxidation numbers to answer the second part of question $ 1 $ , because we can show that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons . question $ 2 $ : carbon being oxidized because it is losing electrons as the oxidation number increases from $ 0 $ to $ +4 $ . iron is being reduced because it is gaining electrons when the oxidation number decreases from $ +3 $ to $ 0 $ . question $ 3 $ : the reducing agent is the reactant that is being oxidized ( and thus causing something else to be reduced ) , so $ \text c ( s ) $ is the reducing agent . question $ 4 $ : the oxidizing agent is reactant that is being reduced ( and thus causing something else to be oxidized ) , so $ \text { fe } _2 \text o_3 ( s ) $ is the oxidizing agent . common types of redox reactions since redox reactions are an important class of reactions , we want to be able to recognize them . there are a few special types of redox reactions you might want to be familiar with . for each of these examples , take a minute to figure out what is getting reduced and oxidized ! $ 1.~ $ combustion reactions a combustion reaction is a redox reaction between a compound and molecular oxygen ( $ \text o_2 $ ) to form oxygen-containing products . when one of the reactants is a hydrocarbon , the products include carbon dioxide and water . the following reaction is the combustion of octane , a hydrocarbon . octane is a component of gasoline , and this combustion reaction occurs in the engine of most cars : $ 2\text c_8 \text h_ { 18 } + 25 \text o_2 \rightarrow 16 \text { co } _2 ( g ) + 18\text h_2 \text o $ $ 2.~ $ disproportionation reactions a disproportionation reaction ( or auto-oxidation reaction ) is a reaction in which a single reactant is both oxidized and reduced . the following reaction is for the disproportionation of hypochlorite , $ \text { clo } ^- $ : $ 3\text { clo } ^- ( aq ) \rightarrow \text { clo } _3^- ( aq ) + 2\text { cl } ^- ( aq ) $ if we analyze the oxidation numbers for chlorine , we see that the reactant $ \text { clo } ^- $ is being oxidized to $ \text { clo } _3^- $ ( where the oxidation number increases from $ +1 $ to $ +5 $ ) . at the same time , the chlorine in some other molecules of $ \text { clo } ^- $ are being reduced to $ \text { cl } ^- $ ( where the oxidation number decreases from $ +1 $ to $ -1 $ ) . oxygen has an oxidation number of $ -2 $ in both $ \text { clo } ^- $ and $ \text { clo } _3^- $ , so it does not get oxidized or reduced in the reaction . $ 3.~ $ single replacement reactions a single replacement reaction ( or single displacement reaction ) involves two elements trading places within a compound . for example , many metals react with dilute acids to form salts and hydrogen gas . the following reaction shows zinc replacing hydrogen in the single replacement reaction between zinc metal and aqueous hydrochloric acid : $ \text { zn } ( s ) +2\text { hcl } ( aq ) \rightarrow \text { zncl } _2 ( aq ) +\text h_2 ( g ) $ balancing a simple redox reaction using the half-reaction method redox reactions can be split into reduction and oxidation half-reactions . chemists use half-reactions to make it easier to see the electron transfer , and it also helps when balancing redox reactions . let 's write the half-reactions for another example reaction : $ \text { al } ( s ) + \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { al } ^ { 3+ } ( aq ) + \text { cu } ( s ) $ is the above reaction balanced ? our atoms appear to be balanced : we have $ 1\ , \text { al } $ atom and $ 1\ , \text { cu } $ atom on each side of the arrow . however , when we add up the charges on the reactant side we get a $ 2+ $ charge , which is not the same as the $ 3+ $ charge on the product side . we need to make sure both the atoms and the charges are balanced ! we will use the half-reaction method to balance the reaction . reduction half-reaction : the reduction half-reaction shows the reactants and products participating in the reduction step . since $ \text { cu } ^ { 2+ } $ is being reduced to $ \text { cu } ( s ) $ , we might start by writing out that step : $ \text { cu } ^ { 2+ } ( aq ) \rightarrow \text { cu } ( s ) $ however , this is not the correct half-reaction , because it is not charge-balanced . there is a net charge of $ 2+ $ on the reactant side and $ 0 $ on the product side . we can balance the charges by including the electrons being transferred , and then we will get our reduction half-reaction : $ \text { cu } ^ { 2+ } ( aq ) +2e^-\rightarrow \text { cu } ( s ) ~~~~~~~~~~~~~~~~~~\blued { \text { reduction half-reaction } } $ the balanced half-reaction tells us that $ \text { cu } ^ { 2+ } $ is gaining $ 2 e^- $ per copper atom to form $ \text { cu } ^0 $ . so where are those electrons coming from ? we can follow the trail of electrons to the oxidation half-reaction . oxidation half-reaction : the oxidation half-reaction shows the reactants and products participating in the oxidation step . this reaction will include the oxidation of $ \text { al } ( s ) $ to $ \text { al } ^ { 3+ } $ , and we will also want to make sure the half-reaction is charge-balanced : $ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) +3e^-~~~~~~~~~~~~~~~~~~~\blued { \text { oxidation half-reaction } } $ the oxidation half reaction tells us that each atom of $ \text { al } ( s ) $ is losing $ 3e^- $ to form $ \text { al } ^ { 3+ } $ . we will combine the balanced half-reactions to get the balanced overall equation , but there is one more thing to check . the electrons must cancel out in the overall equation . another way to think of this that we want to make sure that any electrons that are released in the oxidation half-reaction get used up in the reduction half-reaction . otherwise we would have stray electrons floating around ! that means we need the number of electrons being transferred in each half-reaction to be equal . we can multiply the reduction half-reaction by $ 3 $ and multiply the oxidation half-reaction by $ 2 $ so both reactions involve the transfer of $ 6 $ electrons : $ \redd { 3 } \times [ \text { cu } ^ { 2+ } ( aq ) + { 2e^- } \rightarrow \text { cu } ( s ) ] ~~~~~~~~~~~~~~~~~~~3 \times \text { reduction half-reaction } \ \greend { 2 } \times [ \text { al } ( s ) \rightarrow \text { al } ^ { 3+ } ( aq ) + { 3e^- } ] ~~~~~~~~~~~~~~~~~~~~~2 \times \text { oxidation half-reaction } $ now that we have the same number of electrons in each half-reaction , we can add them together to get our overall balanced equation : $ \begin { align } \cancel { 6e^- } +\redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow ~~~~~~~~~~~~~~~~~~~~~~\redd { 3 } \text { cu } ( s ) ~~~~~~3 \times \text { reduction half-reaction } \ \\greend { 2 } \text { al } ( s ) ~~~~~~~~~~~~~~~~~~~~~~~ & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) +\cancel { 6e^- } ~~~~~~~~~2 \times \text { oxidation half-reaction } \ \\hline \greend { 2 } \text { al } ( s ) + \redd { 3 } \text { cu } ^ { 2+ } ( aq ) & amp ; \rightarrow \greend { 2 } \text { al } ^ { 3+ } ( aq ) + \redd { 3 } \text { cu } ( s ) ~~~~~~~~\text { overall balanced reaction } \end { align } $ lastly , we can check to see if any reactants and products appear on both sides . since that is not the case here , we are done ! our reaction is balanced for both mass and charge . summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced . three common types of redox reactions are combustion , disproportionation , and single replacement reactions . try it !
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summary we can identify redox reactions by checking for changes in oxidation number . redox reactions can be split into oxidation and reduction half-reactions . we can use the half-reaction method to balance redox reactions , which requires that both mass and charge are balanced .
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what are the examples of oxidation and reduction in daily life ?
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