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d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area. An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions. Reaction of chalk/calcium carbonate on dilute hydrochloric acid Procedure Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 . Using a watch glass weigh three separate 2.5g a piece of white chalk. Place the conical flask C1 on an electronic balance. Reset the balance scale to 0.0. Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0. Simultaneously start of the stop watch. Determine and record the scale reading after every 30 seconds to complete Table I . Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieces/chips for C2 and crushing the chalk to powder for C3 Sample results:Table 1. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 2.0 1.8 1.4 1.2 1.0 0.8 0.5 0.5 Loss in mass 0.0 0.5 0.7 1.1 1.3 1.5 1.7 2.0 2.0 Sample results:Table 1I.
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Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5 Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0 Sample questions: 1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III 2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III. 3.Explain why there is a loss in mass in all experiments. Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere. 4.Write an ionic equation for the reaction that take place CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g) 5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion.
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Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion. 6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0) Method I Mole ratio CaCO3(s) : CO2(g) = 1:1 Moles CaCO3(s) used = Mass CaCO3(s) = 0.025 moles Molar mass CaCO3(s) Moles CO2(g) = 0.025 moles Volume of CO2(g) = moles x molar gas volume =>0.025 moles x 24 dm3 = 0.600 dm3/600cm3
Method II Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g) Mass of CaCO3(s) =2.5 g produce 2.5 x 24 = 0.600dm3 100 7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve. From tangent at 180 seconds on curve I Rate = M2-M1 => 2.08 β 1.375 = 0.625 = 0.006944g sec-1 T2- T1 222-132 90 8.What is the effect of particle size on the rate of reaction? A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency. Theoretical examples 1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass?
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Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass? Carbon (IV) oxide gas was produced that escape to the surrounding (b)Calculate the average rate of loss in mass between: (i) 0 to 2 minutes Average rate =M2-M1 => 1.80 β 0.0 = 1.8 = 9.00g min-1 T2- T1 2.0 β 0.0 2 (i) 6 to 8 minutes Average rate =M2-M1 => 3.20 β 2.95 = 0.25 = 0.125g min-1 T2- T1 8.0 β 6.0 2 (iii) Explain the difference between the average rates of reaction in (i) and(ii) above. Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products. Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products. (c)Write the equation for the reaction that takes place. CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.
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CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster. (iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles. (e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen. It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent. (f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed . (i)Name the white precipitate. Calcium(II)sulphate(VI) (ii)Write an ionic equation for the formation of the white precipitate Ca2+(aq) + SO42-(aq)->CaSO4(s) (iii)State one use of the white precipitate -Making plaster for building -Manufacture of plaster of Paris
-Making sulphuric(VI)acid (g)(i) Plot a graph of total loss in mass(y-axes) against time (ii)From the graph, determine the rate of reaction at time 2 minutes. From a tangent/slope at 2 minutes; Rate of reaction = Average rate =M2-M1 => 2.25 β 1.30 = 0.95 = 0.3958g min-1 T2- T1 3.20 β 0.8 2.4 (iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II e) Influence of catalyst on rate of reaction Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder.
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Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(βHa)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(βHa)/activation energy (Ea) by: (i) forming short lived intermediate compounds called activated complex that break up to form the final product/s (ii) being absorbed by the reactants thus providing the surface area on which reaction occurs. A catalyst has no effect on the enthalpy of reaction βHr but only lowers the Enthalpy of activation(βHa)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Ea Catalysed A A B B Energy kJ Ea uncatalysed A-A B-B A-B A-B
Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities.
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Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities. (b)Ostwalds process A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B βHr
Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) ----Pt/Rh--> 4NO (g) + 6H2O(l) (c)Haber process Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas N2(g) + 3H2(g) ---Pt or Fe---> 2NH3(g) (d)Hydrogenation/Hardening of oil to fat Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing - C=C- or βC=C- to saturated compounds without double or triple bond This process is used is used in hardening oil to fat. (e)Decomposition of hydrogen peroxide Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas. This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen.
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This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen. H2 SO4 (aq) + Zn(s) ----CuSO4--> ZnSO4 (aq) + H2(g) (g) Substitution reactions When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow. e.g. CH4(g) + Cl2(g) ---u.v. light--> CH3Cl(g) + HCl(g) (h)Photosynthesis Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet light. 6CO2(g) + 6H2O(l) ---u.v. light--> C6H12O6(g) + O2(g) (i)Photography
Photographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight. 2AgBr(s) ---u.v/sun light--> 2Ag(s) + Br2(l) When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred. Practical determination of effect of catalyst on decomposition of hydrogen peroxide Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water. Divide it into two equal portions. (i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.
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Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask. (ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask. Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below. Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1. (iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide. Mass of MnO2 before reaction(g) Mass of MnO2 after reaction(g) 2.0 2.0 Plot a graph of volume of gas produced against time(x-axes) b) On the same axes, plot a graph of the uncatalysed reaction. (c) Explain the changes in mass of manganese(IV)oxide before and after the reaction. The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Volume of gas (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 96.0 Catalysed reaction Uncatalysed reaction
B.EQUILIBRIA (CHEMICAL CYBERNETICS) Equilibrium is a state of balance. Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e.
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Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e. A + B -> C + D Reactions in which the reactants form products and the products can reform the reactants are said to be reversible. A + B C + D Reversible reactions may be: (a)Reversible physical changes (b)Reversible chemical changes (c)Dynamic equilibrium (a)Reversible physical changes Reversible physical change is one which involves: (i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original. (ii) colour changes. Some substances/compounds change their colours without change in chemical substance. Examples of reversible physical changes (i) colour change on heating and cooling: I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back. ZnO(s) ZnO(s) (white when cold) (yellow when hot) II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back. PbO(s) PbO(s)
(brown when hot) (yellow when cold) (ii)Sublimation I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid. I2(s) I2(g) (grey crystalline solid (purple vapour undergo sublimation) undergo deposition) II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced CO2(s) CO2(g) (white powdery solid (colourless/odourless gas undergo sublimation) undergo deposition) (iii)Melting/ freezing and boiling/condensation Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.
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Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid. Melting boiling Freezing condensing (iv)Dissolving/ crystallization/distillation Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g. NaCl(s) + aq NaCl(aq) (b)Reversible chemical changes These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products. Examples of Reversible chemical changes (i)Heating Hydrated salts/adding water to anhydrous salts. H2O(s) H2O(l) H2O(s)
When hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated. When anhydrous salts are added (back) some water they form hydrated compound/salts. Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate (i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes. (ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white.
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Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white. (ii)colourless liquid forms on the cooler parts of boiling / test tube (i)colour changes from white to blue (ii)boiling tube becomes warm /hot. Cobalt(II)chloride hexahydrate Pink crystalline solid/solution (i)colour changes from pink to blue. (ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used) (i)colour changes from blue to pink (ii)boiling tube becomes warm/hot. When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube. This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is
formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively. CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq) (white/anhydrous) (blue/hydrated) CoCl2(s) + 6H2 O(l) CoCl2.6H2 O(s/aq) (blue/anhydrous) (pink/hydrated) (ii)Chemical sublimation Some compounds sublime from solid to gas by dissociating into new different compounds. e.g. Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.
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Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made. When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes. This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl- ions) and ammonia gas (and hence NH4+ ions) (ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made. When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change. NH3(g) + HCl(g) NH4Cl(s) (Turns moist (Turns moist (forms white fumes) litmus paper blue) litmus paper red)
(c)Dynamic equilibria For reversible reactions in a closed system: (i) at the beginning; -the reactants are decreasing in concentration with time -the products are increasing in concentration with time (ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium. Sketch showing the changes in concentration of reactants and products in a closed system For a system in equilibrium: (i) a reaction from left to right (reactants to products) is called forward reaction. (ii) a reaction from right to left (products to reactants) is called backward reaction. (iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium. A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants.
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A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants. Concentration Mole dm-3 Reaction progress/path/coordinate
The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that: βif a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/changeβ Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium: (a)Concentration (b)Pressure (c)Temperature (d)Catalyst (a)Influence of concentration on dynamic equilibrium An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration. Examples of influence of concentration on dynamic equilibrium (i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation: 2H+ (aq) + 2CrO42- (aq) Cr2O72- (aq) + H2O(l) (Yellow) (Orange) I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More Cr2O72- ions formed in the solution mixture make it to be more orange in colour. II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water.
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II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH- (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour. 2OH- (aq) + 2Cr2O72- (aq) CrO42- (aq) + H2O(l) (Orange) (Yellow) I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH- (aq). H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the 2OH- (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour. II. If a base /OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH- (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH- (aq) ions added. More of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour. (i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube. Note that the solution mixture is orange. Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully. Note that the solution mixture is remains orange. Add about six drops of 2M sodium hydroxide solution. Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium.
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Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:
OH- H+ When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist. When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist. (ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture Measure 2cm3 of bromine water into a boiling tube. Note its colour. Bromine water is yellow Add three drops of 2M sulphuric(VI)acid. Note any colour change Colour becomes more yellow Add seven drops of 2M sodium hydroxide solution. Note any colour change. Solution mixture becomes colourless/Bromine water is decolourized. Explanation When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br-), hydrobromite ion(OBr-) and hydrogen ion(H+) as in the equation: H2O(l) + Br2(aq) OBr- (aq) + H+ (aq) + Br- (aq) If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow. If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.
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If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes. To each test tube add two drops of water. Record your observations in Table 1 below. To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below. To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below. To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1 Indicator Colour of indicator in Water Acid(2M sulphuric (VI) acid) Base(2M sodium hydroxide) Phenolphthalein Colourless Colourless Pink Methyl orange Yellow Red Orange Litmus solution Colourless Red Blue Explanation An indicator is a substance which shows whether another substance is an acid , base or neutral. Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water.
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i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In-) molecule shows/appears. The following examples illustrate the above. (i)Phenolphthalein indicator exist as: HPh H+ (aq) + Ph-(aq) (colourless molecule) (Pink anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph-) molecule shows/appears. (ii)Methyl Orange indicator exists as: HMe H+ (aq) + Me-(aq) (Red molecule) (Yellow/Orange anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules.
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The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed. If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium The following examples show the influence of pressure on dynamic equilibrium: (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture
Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) Gay Lussacs law 2Volume 1Volume Avogadros law 2molecule 1molecule 2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown. (ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture. Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas. Chemical equation : I2(g) + H2(g) ===== 2HI (g) Gay Lussacs law 1Volume 1Volume 2Volume Avogadros law 1molecule 1molecule 2molecule (1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas. Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.
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Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process. The yield of ammonia is thus favoured by high pressures Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g) Gay Lussacs law 1Volume 3Volume 2Volume Avogadros law 1molecule 3molecule 2molecule (1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of ammonia increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of ammonia decrease. (iv)Contact process. Increase in pressure of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures. Chemical equation : 2SO2(g) + O2 (g) -> 2SO3 (g) Gay Lussacs law 2Volume 1Volume 2Volume Avogadros law 2molecule 1molecule 2molecule (2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease. (v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.
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(v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures. Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (g) Gay Lussacs law 4Volume 5Volume 4Volume 6Volume Avogadros law 4molecule 5molecule 4molecule 6Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour. Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease. Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase. Note If the water vapour is condensed on cooling, then: Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (l) Gay Lussacs law 4Volume 5Volume 4Volume 0Volume Avogadros law 4molecule 5molecule 4molecule 0Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour. Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase. Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΞH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΞH).
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The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΞH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΞH). Endothermic reaction are thus favoured by high temperature/heating Exothermic reaction are favoured by low temperature/cooling. If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position. (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow. This show that (i)the forward reaction to the right is exothermic(-ΞH). On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat. (ii)the backward reaction to the right is endothermic(+ΞH). On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat. (c)Influence of Catalyst on dynamic equilibrium A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g. Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase. CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.
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CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount. The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions Optimum condition thus require understanding the effect of various factors on: (i)rate of reaction(Chemical kinetics) (ii)dynamic equilibrium(Chemical cybernetics)
1.Optimum condition in Haber process Chemical equation N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΞH = -92kJ Equilibrium/Reaction rate considerations (i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΞH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.
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(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation. 2.Optimum condition in Contact process Chemical equation 2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΞH = -197kJ Equilibrium/Reaction rate considerations
(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΞH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.
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Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation. 3.Optimum condition in Ostwalds process Chemical equation 4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΞH = -950kJ Equilibrium/Reaction rate considerations (i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained. (ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules. An optimum pressure of about 9 atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΞH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 900oC is normally used. (iv)Platinum can be used as catalyst. Platinum is very expensive.It is: -promoted with Rhodium to increase the surface area/area of contact. -added/coated on the surface of asbestos to form platinized βasbestos to reduce the amount/quantity used. The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation. C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2.
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C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2. Explain the effect of increase in pressure on the following: (i) N2(g) + O2(g) ===== 2NO(g) Gay Lussacs law 1Volume 1Volume 2 Volume Avogadros law 1 molecule 1 molecule 2 molecule 2 volume on reactant side produce 2 volume on product side. Increase in pressure thus have no effect on position of equilibrium. (ii) 2H2(g) + CO(g) ===== CH3OH (g)
Gay Lussacs law 2Volume 1Volume 1 Volume Avogadros law 2 molecule 1 molecule 1 molecule 3 volume on reactant side produce 1 volume on product side. Increase in pressure shift the equilibrium forward to the left. More yield of CH3OH is formed. 4. Explain the effect of increasing temperature on the following: 2SO2(g) + O2 (g) ===== 2SO3 (g) ΞH = -189kJ Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat. 5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below Time(Seconds) Mass of flask(grams) Loss in mass(grams) 0 600 20 599.50 40 599.12 60 598.84 80 598.66 100 598.54 120 598.50 140 598.50 160 598.50 (a)Complete the table by calculating the loss in mass (b)What does the β600β gram reading on the balance represent The initial mass of brass and the acid before any reaction take place. (c)Plot a graph of Time (x-axes) against loss in mass. (d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant.
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(d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant. (d)At what time was the loss in mass equal to: (i)1.20g Reading from a correctly plotted graph = (ii)1.30g Reading from a correctly plotted graph = (iii)1.40g Reading from a correctly plotted graph = (e)What was the loss in mass at: (i)50oC Reading from a correctly plotted graph = (ii) 70oC Reading from a correctly plotted graph = (iii) 90oC g Reading from a correctly plotted graph =
1 21.0.0 ELECTROCHEMISTRY (25 LESSONS) Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with: i) Reduction and oxidation ii) Electrochemical (voltaic) cell iii) Electrolysis (electrolytic) cell (i)REDUCTION AND OXIDATION (REDOX) 1. In teams of oxygen transfer: i) Reduction is removal of oxygen. ii) Oxidation is addition of oxygen. iii) Redox is simultaneous addition and removal of oxygen. iv) Reducing agent is the species that undergoes oxidation, therefore gains oxygen. v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates oxygen. e.g. When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below: CuO (s) + H2 (g) -> Cu (s) + H2O (l)
2 (Oxidising agent) (Reducing agent) 2. In terms of hydrogen transfer: i) Oxidation is the removal of hydrogen. ii) Reduction is the addition of hydrogen. iii) Redox is simultaneous addition and removal of hydrogen. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates hydrogen. v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid).
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v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas. Cl2 (g) + H2S (g) -> S(S) + 2HCl (g) (Oxidizing agent) (Reducing agent) 3. In terms of electron transfer: i) Oxidation is donation/ loss/ removal of electrons. ii) Reduction is gain/ accept/ addition of electrons. iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates electrons. v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts electrons. Example a) Displacement of metals from their solutions: Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below. Metal added to Iron (II) sulphate (VI) solution Colour changes Copper Solution remains green Zinc Green colour fades Magnesium Green colour fades Explanation -When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution. 3 -When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution. -Copper is less reactive than iron therefore cannot displace iron its solution. -Zinc is more reactive than iron therefore can displace iron from its solution. -Magnesium is more reactive than iron therefore can displace iron from its solution. In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1.
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In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Zn(s) -> Zn2+(aq) + Fe(s) (redox/both donation and gain of electrons) 2. Mg(s) -> Mg2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Mg(s) -> Mg2+(aq) + Fe(s) (redox/both donation and gain of electrons) 3. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6.
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Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6. 2Mg(s) -> 2Mg2+(aq) + 4e (oxidation/donation of electrons) O2(g) + 4e -> 2O2-(aq) (reduction/gain of electrons) O2(g) + 2Mg(s) -> 2Mg2+(aq) + 2O2-(aq) (redox/both donation and gain of electrons)
4 Note (i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired. (i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions from their solution. b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution.
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b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions from their solution to form Fe3+(aq). (iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement. (iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent. 4. (a)In terms of oxidation number: i) Oxidation is increase in oxidation numbers. ii) Reduction is decrease in oxidation numbers. iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance. iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.
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iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g.
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v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 β (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 β (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II.
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(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 β (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 β (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 β (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III.
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Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 β (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 β (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 β (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 β -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 β (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 β +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 β +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.
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Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 β (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 β (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 β (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 β -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 β (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 β +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 β +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion.
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Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 β (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 β -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 β (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 β +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 β +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 β (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II.
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Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 β -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 β (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 β (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 β +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 β +2 +-8 = +12 / 2= +6
7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 β (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III.
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Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 β (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 β +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1.
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Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 β (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 β +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e.
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Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 β +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate.
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Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 β (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 β +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)
8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents
9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL
10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged.
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When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms M+(aq) + e -> M(s) ( monovalent metal) M2+(aq) + 2e -> M(s) (divalent metal) M3+(aq) + 3e -> M(s) (Trivalent metal) 3. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure: To set up an electrochemical /voltaic cell To compare the relative tendency of metals to ionize Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks.
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Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades.
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Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons.
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Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod.
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Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions.
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Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+
12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells.
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When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells. For the negative half cell the NO3- /Cl- from salt bridge decrease/neutralise the increased positive(Zn2+) ion. For the positive half cell the Na+ / K+ from salt bridge increase the decreased positive(Cu2+) ion. The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed. A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell. By convention a voltaic/electrochemical cell is represented; M(s) / M2+(aq) // N2+ (aq) / N(s) (metal rod of M)(solution ofM)(solution ofN)(metal rod ofN) Note; a)(i)Metal M must be the one higher in the reactivity series. (ii)It forms the negative terminal of the cell. (iii)It must diagrammatically be drawn first on the left hand side when illustrating the voltaic/electrochemical cell. b)(i)Metal N must be the one lower in the reactivity series. (ii)It forms the positive terminal of the cell. (iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell. Illustration of the voltaic/electrochemical cell. (i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation.
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(i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation. Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +1.10 V 5.cell diagram (ii)Mg/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Mg(s) -> Mg2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Mg(s) -> Mg2+(aq) + Cu(s) 4.cell representation. Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +2.04 V 5.cell diagram. Voltmeter
14 (iii)Fe/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Fe(s) -> Fe2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Fe(s) -> Fe2+(aq) + Cu(s) 4.cell representation. Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2.
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Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2. Silver ions in solution gain the donated electrons to form silver atoms/metal 2Ag+(aq) + 2e -> 2Ag(s) 3.Overall redox equation 2Ag+(aq) + Cu(s) -> Cu2+(aq) + 2Ag(s) 4.cell representation. Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s) E0 = +0.46 V 5.cell diagram. Standard electrode potential (EαΆΏ) The standard electrode potential (EαΆΏ) is obtained if the hydrogen half cell is used as reference. The standard electrode potential (EαΆΏ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.
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Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell. Β½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,Β½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (EαΆΏ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (EαΆΏ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (EαΆΏ) values.
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Β½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,Β½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (EαΆΏ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (EαΆΏ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (EαΆΏ) values. Table showing the standard electrode potential (EαΆΏ) of some reactions Reaction (EαΆΏ) values in volts F2 (g)+ 2e -> 2F- (aq) +2.87 H2 O2 (aq)+ H+ (aq) +2e -> H2 O (l) +1.77 Mn O4- (aq)+ 4H+ (aq) +3e -> MnO2 (s) +H2 O (l) +1.70 2HClO (aq)+ 2H+ (aq) +2e -> Cl2 (aq) +2H2 O (l) +1.59 Mn O4- (aq)+ 4H+ (aq) +5e -> Mn2+ (aq) +H2 O (l) +1.51 Cl2 (g)+ 2e -> 2Cl- (aq) +1.36 Mn O2 (s)+ 4H+ (aq) +2e -> Mn2+ (aq) +2H2 O (l) +1.23 Br2 (aq)+ 2e -> 2Br- (aq) +1.09 NO3- (aq)+ 2H+ (aq) + e -> NO2 (g) + H2 O (l) +0.80 Ag+ (aq) + e -> Ag(s) +0.80 Fe3+ (aq) + e -> Fe2+ (aq) +0.77 2H+ (aq)+ O2 (g) -> H2 O2 (aq) +0.68 I2 (aq)+ 2e -> 2I- (aq) +0.54 Cu2+ (aq) + 2e -> Cu(s) +0.34 2H+ (aq) + 2e -> H2(g) +0.00 Pb2+ (aq) + 2e -> Pb(s) -0.13 Fe2+ (aq) + 2e -> Fe(s) -0.44 Zn2+ (aq) + 2e -> Zn(s) -0.77 Al3+ (aq) + 3e -> Al(s) -1.66 Mg2+ (aq) + 2e -> Mg(s) -2.37 Na+ (aq) + e -> Na(s) -2.71 K+ (aq) + e -> K(s) -2.92 Note: (i)EαΆΏ values generally show the possibility/feasibility of a reduction process/oxidizing strength.
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(ii)The element/species in the half cell with the highest negative EαΆΏ value easily gain / acquire electrons. It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive EαΆΏ value easily donate / lose electrons. 17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) EαΆΏ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) EαΆΏ Calculation examples on EαΆΏ Calculate the EαΆΏ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) EαΆΏ = -0.77V(lower EαΆΏ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e EαΆΏ = +0.77(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏoxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) EαΆΏ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall EαΆΏ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution.
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17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) EαΆΏ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) EαΆΏ Calculation examples on EαΆΏ Calculate the EαΆΏ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) EαΆΏ = -0.77V(lower EαΆΏ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e EαΆΏ = +0.77(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏoxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) EαΆΏ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall EαΆΏ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) EαΆΏ = -2.37V(lower EαΆΏ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e EαΆΏ = +2.37(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) EαΆΏ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) EαΆΏ = +0.80V(higher EαΆΏ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) EαΆΏ = -0.13V(lower EαΆΏ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e EαΆΏ = +0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.80 β (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) EαΆΏ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) EαΆΏ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) EαΆΏ = +1.36V(higher EαΆΏ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) EαΆΏ = +0.13V(lower EαΆΏ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e EαΆΏ = -0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = - 0.13 β (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) EαΆΏ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) EαΆΏ = +1.23V Chlorine displaces bromine from bromine water.
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(iii)The overall redox reaction is possible/feasible is it has a positive (+) EαΆΏ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) EαΆΏ Calculation examples on EαΆΏ Calculate the EαΆΏ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) EαΆΏ = -0.77V(lower EαΆΏ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e EαΆΏ = +0.77(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏoxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) EαΆΏ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall EαΆΏ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) EαΆΏ = -2.37V(lower EαΆΏ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e EαΆΏ = +2.37(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) EαΆΏ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) EαΆΏ = +0.80V(higher EαΆΏ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) EαΆΏ = -0.13V(lower EαΆΏ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e EαΆΏ = +0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.80 β (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) EαΆΏ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) EαΆΏ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) EαΆΏ = +1.36V(higher EαΆΏ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) EαΆΏ = +0.13V(lower EαΆΏ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e EαΆΏ = -0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = - 0.13 β (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) EαΆΏ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) EαΆΏ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed.
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If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) EαΆΏ Calculation examples on EαΆΏ Calculate the EαΆΏ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) EαΆΏ = -0.77V(lower EαΆΏ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e EαΆΏ = +0.77(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏoxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) EαΆΏ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+
18 Zinc and copper reaction has a positive(+) overall EαΆΏ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) EαΆΏ = -2.37V(lower EαΆΏ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e EαΆΏ = +2.37(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) EαΆΏ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) EαΆΏ = +0.80V(higher EαΆΏ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) EαΆΏ = -0.13V(lower EαΆΏ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e EαΆΏ = +0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.80 β (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) EαΆΏ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) EαΆΏ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) EαΆΏ = +1.36V(higher EαΆΏ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) EαΆΏ = +0.13V(lower EαΆΏ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e EαΆΏ = -0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = - 0.13 β (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) EαΆΏ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) EαΆΏ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive EαΆΏ value.
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b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) EαΆΏ = +0.34V(higher EαΆΏ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) EαΆΏ = -2.37V(lower EαΆΏ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e EαΆΏ = +2.37(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.34 β (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) EαΆΏ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) EαΆΏ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) EαΆΏ = +0.80V(higher EαΆΏ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) EαΆΏ = -0.13V(lower EαΆΏ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e EαΆΏ = +0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +0.80 β (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) EαΆΏ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) EαΆΏ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) EαΆΏ = +1.36V(higher EαΆΏ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) EαΆΏ = +0.13V(lower EαΆΏ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e EαΆΏ = -0.13(reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = - 0.13 β (- 1.36) = +1.23V Overall redox equation:
19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) EαΆΏ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) EαΆΏ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive EαΆΏ value. e)Strongest oxidizing agent and the strongest reducing agent.
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When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive EαΆΏ value. e)Strongest oxidizing agent and the strongest reducing agent. From the table above: 2e + F2(g) ->2F- (aq) EαΆΏ = +2.87V(highest EαΆΏ /strongest oxidizing agent) 2e + 2K+ (aq) ->2K (aq) EαΆΏ = -2.92V(lowest EαΆΏ/ strongest reducing agent) 2K (aq) -> 2K+ (aq) + 2e EαΆΏ = +2.92V (reverse lower EαΆΏ to derive cell reaction / representation) Overall EαΆΏ = EαΆΏ higher- EαΆΏ lower / EαΆΏ RHS - EαΆΏ LHS/ EαΆΏ oxidized- EαΆΏ reduced Substituting: Overall EαΆΏ = +2.87 β (-2.92) = +5.79V Overall redox equation: F2(g) + 2K(s) -> 2F- (aq) + 2K+ (aq) EαΆΏ = +5.79V Overall conventional cell representation: 2K(s) / 2K+ (aq),1M, // 1M, 2F- (aq) / F2(g) EαΆΏ = +5.79V The redox reactions in an electrochemical/voltaic is commercially applied to make the: (a)Dry /primary/Laclanche cell. (b)Wet /secondary /accumulators. (a)Dry/primary/Laclanche cell Examine a used dry cell. Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon.
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Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon. Zinc acts/serve as the negative terminal where it ionizes/dissociates:
20 Zn(s) -> Zn2+(aq) + 2e Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas. 2NH4+(aq) + 2e -> 2NH3(g) + H2(g) Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste. Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell. Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1.
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Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of: (i)Lead plate that forms the negative terminal (ii)Lead(IV) oxide that forms the positive terminal 2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3 3.At the negative terminal,lead ionizes /dissolves; Pb(s) -> Pb2+ + 2e 4.At the positive terminal,
21 (i) Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions; PbO2(s) + 4H+(aq) + 2e -> Pb2+ (aq) + H2O(l) (ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/ SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI). Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) EαΆΏ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7.
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Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) EαΆΏ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7. During recharging, the electrode reaction is reversed as below: 2PbSO4(s) + 2H2O(l) ->PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq) 8. A car battery has six Lead-acid cells making a total of 12 volts. (iii)ELECTROLYSIS (ELECTROLYTIC CELL) 1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity. A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong. 2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include: (i)all mineral acids (ii)all strong alkalis/sodium hydroxide/potassium hydroxide. (iii)all soluble salts
22 3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions. Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g.
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A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (ii)all hydrocarbons(alkanes /alkenes/alkynes) (iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar) 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state. 6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. An electrode that does not influence/alter the products of electrolysis is called an inert electrode. Common inert electrodes include: (i)Platinum (ii)Carbon graphite Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells). 7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte 8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e. M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.
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M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode
23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules. X+ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water) 2X+ (l) + 2e -> X (l) (for cations from molten electrolytes) The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode. 11. The below set up shows an electrolytic cell. BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas.
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BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)
25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas.
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II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas.
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II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas. 26 d)To determine the products of electrolysis of molten Aluminium (III)oxide (i)Decomposition of electrolyte into free ions; Al2O3 (l) -> 2Al 3+(l) + 3O2-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 4Al 3+ (l) + 12e -> 4Al (l) (Cation / Al 3+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 6O2-(l) -> 3O2 (g) + 12e (Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid aluminium metal. II.At the anode colourless gas that relights/rekindles glowing splint. 13. For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell.
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For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell. Table I showing the relative ease of discharge of cations in an electrolytic cell K+(aq) + e -> K(s) (least readily/easily discharged) Na+(aq) + e -> Na(s) Ca2+(aq) + 2e -> Ca(s) Mg2+(aq) + 2e -> Mg(s) Al3+(aq) + 3e -> Al(s) Zn2+(aq) + 2e -> Zn(s) Fe2+(aq) + 2e -> Fe(s) Pb2+(aq) + 2e -> Pb(s) 2H+(aq) + 2e -> H2(g) (hydrogen is usually βmetallicβ) Cu2+(aq) + 2e -> Cu(s) Hg2+(aq) + 2e -> Hg(s) Ag+(aq) + e -> Ag(s) (most readily/easily discharged) 2.The OH- ion is the most readily/easily discharged anion . All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged.
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All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged. 3.(a)When two or more cations are attracted to the cathode, the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above. This is called selective/preferential discharge at cathode. (b)When two or more anions are attracted to the anode, the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above. This is called selective/preferential discharge at anode. 4.The following experiments shows the influence /effect of selective/preferential discharge on the products of electrolysis: (i)Electrolysis of acidified water/dilute sulphuric(VI) acid Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II.
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Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)
28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II. Name the ions in acidified water that are attracted/move to: Cathode- H+(aq) from either sulphuric(VI) acid (H2 SO4) or water (H2O) Anode- SO42-(aq) from sulphuric (VI) acid (H2 SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of acidified water. Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called βelectrolysis of (acidified) waterβ?
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The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called βelectrolysis of (acidified) waterβ? The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. This implies/means that water in the electrolyte is being decomposed into hydrogen and Oxygen gases. The electrolysis of dilute sulphuric acid is therefore called βelectrolysis of acidified water.β VI. Explain the changes in concentration of the electrolyte during electrolysis of acidified waterβ The concentration of dilute sulphuric (VI) acid increases. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape. The concentration /mole of acid present in a given volume of solution thus continue increasing/rising. (ii)Electrolysis of Magnesium sulphate(VI) solution Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Answer the following questions:
29 I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III.
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Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Mg2+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of Magnesium sulphate(VI) solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water.
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VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /mole of acid present in a given volume of Magnesium sulphate(VI) solution thus continue increasing/rising. 30 The set β up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes. Name a suitable pair of electrodes for this experiment Identify the ions and cations in the solution On the diagram label the cathode Write ionic equations for the reactions that took place at the anode. Explain the change that occurred to the concentration of magnesium sulphate solution during the experience. During the electrolysis a current of 2 amperes was passed through the solution for 4 hours. Calculate the volume of the gas produced at the anode.(1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3) One of the uses of electrolysis is electroplating What is meant by electroplating? Give tow reasons why electroplating is necessary. b) Concentration of the electrolytes 1.High concentrations of cations and/or anions at the electrodes block the ion/s that is likely to be discharged at the electrode. This is called over voltage. A concentrated solution therefore produces different products of electrolysis from a dilute one. 2. The following experiments show the influence/effect of concentration of electrolyte on the products of electrolysis. (i)Electrolysis of dilute and concentrated(brine)sodium chloride solution
31 I. Dissolve about 0.5 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Transfer the set up into a fume chamber/open and continue to make observations for a further 10 minute. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II.
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Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodiumchloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute sodium chloride solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution The concentration of dilute sodium chloride solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II.
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Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations continuously at each electrode for 30 minutes in a fume chamber/open. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodium chloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2H+(aq) + 2e -> H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2(g) + 4e (Cl- ions with a higher concentration block the discharge of OH- ions at the anode) IV. Name the products of electrolysis of concentrated sodium chloride solution/brine Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Chlorine gas(pale green gas that bleaches damp/moist/wet litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI.
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Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solution/brine The concentration of concentrated sodium chloride solution/brine increases. The ratio of Cl2 (g): H2 (g) is 1:1 as they are combined in water. Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode. 33 The concentration /moles of OH- (aq) and Na+ ion (as NaOH) present in a given volume of electrolyte continue increasing/rising. This makes the electrolyte strongly alkaline with high pH. As the electrolysis of brine continues the concentration of Cl- ions decrease and oxygen gas start being liberated at anode. The electrolyte pH is thus lowered and the concentration of brine starts again increasing. (ii)Electrolysis of dilute and concentrated Hydrochloric acid solution I. Prepare about 50cm3 of 0.05 M of dilute Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III.
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Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> H2O(l) +O2+ 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. 34 VI. Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products.
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Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of HCl present in a given volume of dilute Hydrochloric acid continue increasing/rising. II. Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 30 minutes CautionThis experiment should be done in the open/fume chamber. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in 2M Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode.
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Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ βpopβ sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two(2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode. 35 VI. Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid The concentration of Hydrochloric acid decreases. The ratio of H2 (g): Cl2 (g) is 1:1 as they are combined in Hydrochloric acid. Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode. There is a net accumulation of excess OH- (aq) ions in solution. This makes the electrolyte strongly alkaline with high pH. c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I.
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c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I. Using carbon-graphite electrodes Weigh Carbon -graphite electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 22.4 g Brown solid deposit at the cathode after electrolysis - Bubbles of colourless gas that relight splint - Blue colour of electrolyte fades/become less blue - Blue colour of electrolyte fades /become less blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)
36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III.
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Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)
36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2Cu2+ (aq) + 4e -> 2Cu(g) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode 4OH- (aq) -> H2O(l) + O2+ 4e (OH- ions ions are higher than SO42- ions in the electrochemical series therefore selectively discharged at the cathode.)) IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution Cathode-2 moles of copper metal as brown solid coat Anode-Oxygen gas (Colourless gas that relights /rekindles glowing splint) V. Explain the changes that take place at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of Oxygen (O2)gas at the anode are gained/acquired/accepted by two Cu2+(aq) ions to form 2 moles of brown copper solid that deposit itself at the cathode. The moles of oxygen gas at the anode is equal to the moles of copper produced at the cathode VI. Explain the changes in electrolyte during electrolysis of 1M copper (II) sulphate(VI) solution. (i)The pH of copper(II) sulphate(VI) solution lowers/decreases. The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution.
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The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution. This makes the electrolyte strongly acidic with low pH. (ii) Cu2+ (aq) ions are responsible for the blue colour of the electrolyte/ copper(II) sulphate (VI) solution. As electrolysis continues, blue Cu2+ (aq) ions gain electrons to form brown Copper. The blue colour of electrolyte therefore fades/become less blue. (iii)Copper is deposited at the cathode. This increases the mass of the cathode.OH- ions that produce Oxygen gas at anode come from water. Oxygen escapes out/away without increasing the mass of anode. 37 II. Using copper electrodes Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II.
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Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate because less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. 38 IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.
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Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode. Cu2+ ions have greater tendency to accept/gain/acquire electrons to form brown copper atoms/solid that deposit itself and increase the mass/size of the cathode.The copper deposited at the cathode is pure -H+ ions accumulate around the cathode. Electrolyte thus becomes strongly acidic around the cathode. -Cu2+ ions in solution are responsible for the blue colour of electrolyte. Blue colour of electrolyte fade around the cathode. (ii)Anode Copper atom at the anode easily ionizes to release electrons. The anode therefore keeps decreasing in mass/eroding. The amount of copper that dissolve/erode is equal to the mass of copper deposited. This is called electrode ionization. Electrode ionization is where the anode erodes/decrease and the cathode deposits/increase during electrolysis. The overall concentration of the electrolyte remains constant 14.In industries electrolysis has the following uses/applications: (a)Extraction of reactive metals from their ores. Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods. (b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode.
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(b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there: (i)At the cathode; Cu2+ (aq) + 2e -> Cu(s) (Pure copper deposits on the strip (ii)At the anode; Cu(s) ->Cu2+ (aq) + 2e (impure copper erodes/dissolves) (c)Electroplating The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that βthe mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.β (a)The SI unit of quantity of electricity is the coulomb(C).
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During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that βthe mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.β (a)The SI unit of quantity of electricity is the coulomb(C). The coulomb may be defined as the quantity of electricity passed/used when a current of one ampere flow for one second.i.e; 1Coulomb = 1 Ampere x 1Second The Ampere is the SI unit of current(I) The Second is the SI unit of time(t) therefore;
40 Quantity of electricity(in Coulombs) = Current(I) x time(t) Practice examples 1. A current of 2 amperes was passed through an electrolytic cell for 20 minutes. Calculate the quantity of electric charge produced. Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x (20 x 60) = 2400 C 2. A current of 2 amperes was passed through an electrolytic.96500 coulombs of charge were produced. Calculate the time taken. Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used.
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Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used. Working: Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 96500 10 x 60 = 160.8333 Amperes (b)The quantity of electricity required for one mole of electrons at the anode/cathode is called the Faraday constant(F). It is about 96500 Coulombs.i.e The number of Faradays used /required is equal to the number of electrons used at cathode/anode during the electrolytic process. e.g. Cu2+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode. Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas 2O2- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas. 4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell.
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Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas 2O2- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas. 4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass β mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4.
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4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass β mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI).
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41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass β mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5.
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(Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e
42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass β mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI).
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Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used.
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(a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element βPβ has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of βPβ for 32 minutes and 10seconds ,0.44 g
44 of βPβ was deposited at the cathode.
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(Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element βPβ has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of βPβ for 32 minutes and 10seconds ,0.44 g
44 of βPβ was deposited at the cathode. Determine the charge on an ion of βPβ(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal βPβ are deposited by 965C 88g of of metal βPβ are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8.
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Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element βPβ has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of βPβ for 32 minutes and 10seconds ,0.44 g
44 of βPβ was deposited at the cathode. Determine the charge on an ion of βPβ(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal βPβ are deposited by 965C 88g of of metal βPβ are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 Β½ hours.
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Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element βPβ has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of βPβ for 32 minutes and 10seconds ,0.44 g
44 of βPβ was deposited at the cathode. Determine the charge on an ion of βPβ(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal βPβ are deposited by 965C 88g of of metal βPβ are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 Β½ hours. Calculate the amount of current that was passed.
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Determine the charge on an ion of βPβ(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal βPβ are deposited by 965C 88g of of metal βPβ are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 Β½ hours. Calculate the amount of current that was passed. (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu2+ (aq) + 2e-> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 63.5 g -> 2 x 96500C 1.48 g -> 1.48 x 2 x 96500 = 4255.1181 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 4255.1181C (( 2 x 60) + 30) x60 = 0.4728 Amperes 17. Practically Faraday 1st law of electrolysis can be verified as below. Verifying Faraday 1st law of electrolysis Procedure. Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry.
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Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 7.00 g Mass of anode before electrolysis 7.75 g Mass of cathode after 8.25 g Mass of anode after 6.50 g
45 electrolysis electrolysis Change in mass at cathode after electrolysis 1.25 g Change in mass at anode after electrolysis 1.25 g Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate as less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode?
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Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode= 2 moles (ii)How many Faradays are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode : 2 moles = 2 Faradays (iii)Calculate the quantity of electric charge used Working:
46 Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x 20 x 60 = 2400C VI. (i) Calculate the quantity of electricity required to deposit/erode one mole of copper at the cathode/anode(Cu=63.5) Since 1.25 g of copper -> 2400C Then 63.5 g (1mole of copper) -> 63.5 x 2400 = 121920 C 1.25 (ii)Determine the Faraday constant from the results in V(i) above From the equation at; Cathode Cu2+ (aq) + 2e -> Cu(s) Anode Cu (s) -> Cu2+(aq) + 2e 2 moles = 2 Faradays -> 121920 C 1 moles = 1 Faradays -> 121920 = 60960 C 2 (iii) The faraday constant obtained above is far lower than theoretical.Explain -high resistance of the wires used. -temperatures at 25oC were not kept constant -plates/electrodes used were not made of pure copper -plates/electrodes used were not thoroughly clean copper Further practice 1.An element P has a relative atomic mass of 88. When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs).
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When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs). 2.During electrolysis of aqueous copper (II) sulphate, 144750 coulombs of electricity were used. Calculate the mass of copper metal that was obtained (Cu = 64 ;1 Faraday = 96500 coulombs) ( 3 mks) 3.A nitrate of a metal M was electrolysed .1.18 g of metal was deposited when a current of 4 ampheres flow for 16 minutes.Determine the formula of the sulphate(VI)salt of the metal. (Faraday constant = 96500 , RAM of X = 59.0) Working Q = It =>( 4 x 16 x 60) = 3840 C 1.18 g of X => 3840 C 59.0 g => 59.0 x 3840 = 192000 C 1.18 96500 C = 1Faraday 192000 C= 192000 C x1 = 2F thus charge of M = M2+ 96500 C Valency of M is 2 thus formula of sulphate(VI)salt MSO4
47 4. Below is the results obtained when a current of 2.0ampheres is passed through copper(II)sulphate(VI)solution for 15 minutes during electrolysis using copper electrode. Initial mass of cathode = 1.0 g Final mass of cathode = 1.6 g Change in mass of cathode = 0.60 g (i)Determine the change in mass at the anode. Explain your answer. Mass decrease = 0.6g.
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