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.


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.


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.


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.


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.


(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 “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (a)The SI unit of quantity of electricity is the coulomb(C).


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 “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (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.


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.


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 – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4.


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 – 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).


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 – 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.


(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 – 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).


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.


(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 “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode.


(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 “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(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 “P” are deposited by 965C 88g of of metal “P” 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.


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 “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(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 “P” are deposited by 965C 88g of of metal “P” 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 ½ hours.


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 “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g
44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(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 “P” are deposited by 965C 88g of of metal “P” 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 ½ hours. Calculate the amount of current that was passed.


Determine the charge on an ion of “P”(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 “P” are deposited by 965C 88g of of metal “P” 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 ½ 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.


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?


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).


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.


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. Electrode ionization take place where the cathode increase in mass form the erosion of the anode (ii)Calculate the quantity of electricity required to deposit one mole of copper.(Cu =63.5) Q =It => 2 x 15 x 60 = 1800 coulombs Method 1 0.60 g of copper ->1800 coulombs 63.5 g -> 63.5 x 1800 = 190500 Coulombs 0.60 Method 2 Moles of Copper = Mass => 0.60 = 9.4488 x10 -3 moles Molar mass 63.5 9.4488 x10 -3 moles -> 1800 coulombs 1 Mole -> 1 x 1800 coulombs = 190500.381 coulombs 9.4488 x10 -3 moles (iii)Determine the oxidation number of copper produced at the cathode and hence the formula of its nitrate (V)salt (1 Faraday = 96500 Coulombs) 96500 Coulombs -> 1 Faraday 190500.381 coulombs -> 190500.381 coulombs x 1 96500 Coulombs = 1.9741 Faradays => 2F(whole number) Charge of copper = 2+ = Oxidation number => Valency of copper = 2 hence chemical formula of nitrate (V)salt = Cu (NO3)2
1 22.0.0 METALS (20 LESSONS) a)Introduction to metals The rationale of studying metals cannot be emphasized.Since ages, the world over, metals like gold and silver have been used for commercial purposes. The periodicity of alkali and alkaline earth metals was discussed in year 2 of secondary school education. This topic generally deals with: (a)Natural occurrence of the chief ores of the most useful metals for industrial /commercial purposes. (b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals.


(b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals. The metals given detailed emphasis here are; Sodium, Aluminium, Iron, Zinc, Lead and Copper. 2 The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. 1.SODIUM a) Natural occurrence Sodium naturally occurs as: (i)Brine-a concentrated solution of sodium chloride(NaCl(aq)) in salty seas and oceans. (ii)Rock salt-solid sodium chloride(NaCl(s) (iii)Trona-sodium sesquicarbonate(NaHCO3.Na2CO3.2H2O) especially in lake Magadi in Kenya. (iv)Chile saltpeter-sodium nitrate(NaNO3) Position on the earth’s crust If near the surface ,open cast mining / quarrying is used If deep on the earth’s crust deep mining is used If the ore is low grade oil, water, and air is blown forming a froth(froth flotation) to concentrate The ore first roasted if it is a carbonate or sulphide of Zinc, Iron, Tin, Lead, and Copper to form the oxide Electrolysis of the ore is used for reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium The oxide is reduced using carbon/ carbon(II) oxide in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
3 b)(i) Extraction of Sodium from brine/Manufacture of Sodium hydroxide/The flowing mercury cathode cell/ TheCaster-Keller process I.Raw materials (i) Brine-concentrated solution of sodium chloride (NaCl (aq)) from salty seas and oceans. (ii)Mercury (iii)Water from river/lakes II. Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process.


Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. H2O(l) H+(aq) + OH-(aq) NaCl(aq) Na+(aq) + Cl-(aq) II. Name the ions present in brine that moves to the: (i)Mercury cathode; H+(aq) , Na+(aq) (ii)Titanium/graphite; OH-(aq), Cl-(aq) III. Write the equation for the reaction that take place during the electrolytic process at the; Cathode; 2Na+(aq) + 2e 2Na(s) Anode; 2Cl-(aq) Cl2(g) + 2e Note (i)Concentration of 2Cl-(aq) ions is higher than OH- ions causing overvoltage thus blocking OH- ions from being discharged at the anode. (ii)Concentration of Na+(aq) ions is higher than H+ ions causing overvoltage thus blocking H+ ions from being discharged at the cathode. 4 IV. Name the products of electrolysis in the flowing mercury-cathode cell. (i)Mercury cathode; Sodium metal as grey soft metal/solid (ii)Titanium/graphite; Chlorine gas as a pale green gas that turns moist blue/red litmus papers red then bleaches both. Chlorine gas is a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas.


(iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas. 2Na Hg (l) + 2H2O(l) 2NaOH (aq) + 2Hg(l) + H2(g) Hydrogen gas is a very useful by-product in; (i)making ammonia gas in the Haber process (ii)manufacture of hydrochloric acid (iii)in weather balloons to forecast weather (iv)as rocket fuel As the electrolysis of brine continues, the concentration of Cl-ions decreases and oxygen gas start being liberated. Continuous feeding of the electrolyte is therefore very necessary. III.Uses of sodium hydroxide The sodium hydroxide produced is very pure and is used mainly in: (i)Making soapy and soapless detergents. (ii)making cellulose acetate/rayon IV. Diagram showing the Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 5 3 V. Environmental effects of Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 1.Most of the Mercury used at the cathode is recycled ; (i)to reduce the cost because mercury is expensive (ii)to reduce pollution because mercury kills marine life. (iii)because it causes chromosomal/genetic mutation to human beings. 2.Chlorine produced at the anode; (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. b)(ii) Extraction of sodium from rock salt/The Downs cell/process
6 I. Raw materials (i)Rock salt/solid sodium chloride (ii)calcium(II)chloride II. Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC.


Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC. A little calcium (II) chloride is added to lower the melting point of the electrolyte to about 600oC. The molten electrolyte is the electrolyzed in a carbon graphite anode suspended at the centre and surrounded by steel cathode. Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. NaCl(l) Na+(l) + Cl-(l) Note: In absence of water, the ions are in liquid state. II. Name the ions present in molten rock salt that move to the; (i)Steel cathode -Na+(l) (ii)Carbon graphite anode- Cl-(l) III. Write the equation for the reaction that take place during the electrolytic process at the; (i)Steel cathode 2Na+(l) + 2e 2Na(l) (ii)Carbon graphite anode 2Cl-(l) Cl2(g) + 2e IV. Name the products of electrolysis in the Downs cell at; (i)Cathode: Grey solid Sodium metal is less dense than the molten electrolyte and therefore float on top of the cathode to be periodically tapped off. (ii)Anode: Pale green chlorine gas that turns moist/damp/wet blue/red litmus papers red then bleaches/decolorizes both. Chlorine gas is again a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. 7 A steel diaphragm/gauze is suspended between the electrodes to prevent recombination of sodium at the cathode and chlorine gas at the anode back to sodium chloride. III. Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting.


Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting. 2.Sodium is used in making very useful sodium compounds like; (i)Sodium hydroxide(NaOH) (ii)Sodium cyanide(NaCN) (iii)Sodium peroxide(Na2O2) (iv)Sodamide(NaNH2) 3.An alloy of Potassium and Sodium is used as coolant in nuclear reactors. V. Environmental effects of Downs cell. 1.Chlorine produced at the anode;
8 (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. 2.Sodium metal rapidly react with traces of water to form alkaline Sodium hydroxide(NaOH(aq))solution. This raises the pH of rivers/lakes killing aquatic lifein case of leakages. VI. Test for presence of Na. If a compound has Na+ ions in solid/molten/aqueous state then it changes a non-luminous clear/colourless flame to a yellow coloration but does not burn Experiment Scoop a portion of sodium chloride crystals/solution in a clean metallic spatula. Introduce it to a clear /colourless Bunsen flame. Observation Inference Yellow coloration Na+ Practice (i)Calculate the time taken in hours for 230kg of sodium to be produced in the Downs cell when a current of 120kA is used. (ii)Determine the volume of chlorine released to the atmosphere.


10 (a)Calculate the concentration of the resulting solution in moles per litre. (b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration.


(b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25
11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium.


25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25
11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium. Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C.


Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C. (i) Name the salt that is added (1mk) Calcium chloride (ii) State why it is necessary to lower the temperature(1mk) To reduce the cost of production
12 (c) Explain why aqueous sodium chloride is not suitable as an electrolyte for the manufacture of sodium in the Downs process( 2mk) The sodium produced react explosively/vigorously with water in the aqueous sodium chloride (d) Sodium metal reacts with air to form two oxide. Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities.


Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities. The mixture is first dissolved in hot concentrated sodium/potassium hydroxide solution. The alkalis dissolve both bauxite and silicon(IV)oxide. This is because bauxite is amphotellic while silicon(IV)oxide is acidic. Iron(III)oxide (Fe2O3) is filtered of /removed as a residue. Carbon(IV)oxide is bubbled into the filtrate to precipitate aluminium (III) hydroxide (Al(OH)3) as residue. The aluminium (III) hydroxide (Al(OH)3) residue is filtered off. Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)
13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide.


Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)
13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide. It is therefore dissolved first in molten cryolite /sodium hexafluoroaluminate (III)/Na3AlF6 to lower the melting point to about 800oC. The molten electrolyte is put in the Hall cell made up of a steel tank lined with carbon graphite and an anode suspended into the electrolyte. During the electrolysis: (i)At the cathode; 4Al3+(l) + 12e 4Al(l) (ii) At the anode; 6O2-(l) 3O2(g) + 12e Aluminium is denser than the electrolyte therefore sink to the bottom of the Hall cell. At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere. C(s) + O2(g) CO2(g) The anode thus should be continuously replaced from time to time. Flow chart summary of extraction of aluminium from Bauxite Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2 Powdered mixture Crush (increase surface area) Hot concentrated sodium hydroxide
14 c) Diagram showing the Hall cell / process for extraction of Bauxite Iron(III)oxide- Fe2O3 as residue Sodium aluminate (NaAl(OH)4) and sodium silicate (Na2SiO3) as filtrate Carbon(IV)oxide Aluminium hydroxide (Al(OH)3) as residue Sodium silicate (Na2SiO3) Aluminium (III) Oxide Roast at 1000oC Cryolite Electrolysis Oxygen gas at anode Pure aluminium sinks in Hall cell
15 d)Uses of aluminium (i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light.


At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere. C(s) + O2(g) CO2(g) The anode thus should be continuously replaced from time to time. Flow chart summary of extraction of aluminium from Bauxite Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2 Powdered mixture Crush (increase surface area) Hot concentrated sodium hydroxide
14 c) Diagram showing the Hall cell / process for extraction of Bauxite Iron(III)oxide- Fe2O3 as residue Sodium aluminate (NaAl(OH)4) and sodium silicate (Na2SiO3) as filtrate Carbon(IV)oxide Aluminium hydroxide (Al(OH)3) as residue Sodium silicate (Na2SiO3) Aluminium (III) Oxide Roast at 1000oC Cryolite Electrolysis Oxygen gas at anode Pure aluminium sinks in Hall cell
15 d)Uses of aluminium (i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light. (ii)Making duralumin-an alloy which is harder and has a higher tensile strength (iii)Making utensils,sauce pans,spoons because it is light and good conductor of electricity. (iv)Making overhead electric cables because it is light,ductile and good conductor of electricity. (iv)Used in the thermite process for production of Manganese, Chromium amd Titanium. e) Environmental effects of extracting aluminium from Bauxite. Carbon(IV)oxide gas that escape to the atmosphere is a green house gas that causes global warming. Bauxite is extracted by open caste mining that causes soil/environmental degradation. 16 f) Test for presence of Al3+ If an ore is suspected to contain Al3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia.


Bauxite is extracted by open caste mining that causes soil/environmental degradation. 16 f) Test for presence of Al3+ If an ore is suspected to contain Al3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia. Observation Inference White precipitate in excess 2M NaOH(aq) Pb2+ , Al3+, Zn2+ White precipitate in excess 2M NH3(aq) Pb2+ , Al3+ No black precipitate on adding Na2S(aq) Al3+ No white precipitate on adding either NaCl(aq),HCl(aq),H2SO4(aq),Na2SO4(aq) Al3+ Practice 1.An unknown rock X was discovered in Ukraine. Test with dilute sulphuric (VI)acid shows rapid effervescence with production of a colourless gas A that forms a white precipitate with lime water and colourless solution B. On adding 3cm3 of 2M sodium hydroxide, a white precipitate C is formed that dissolves to form a colourless solution D on adding more sodium hydroxide. On adding 2M aqueous ammonia, a white precipitate E is formed which persist in excess aqueous ammonia.On which on adding 5cm3 of 1M Lead(II)nitrate(V) to F a white precipitate G is formed which remains on heating. Identify: A Hydrogen/H2 B Aluminium sulphate(VI)/Al2(SO4) 3 C Aluminium hydroxide/ Al(OH4) 3 D Tetrahydroxoaluminate(III)/ [Al(OH4) 3] - E Aluminium hydroxide/ Al(OH) 3 F Aluminium chloride/ AlCl3
17 2.Aluminium is obtained from the ore with the formula Al2O3. 2H2O. The ore is first heated and refined to obtain pure aluminium oxide (Al2O3). The oxide is then electrolysed to get Aluminium and oxygen gas using carbon anodes and carbon as cathode.


2H2O. The ore is first heated and refined to obtain pure aluminium oxide (Al2O3). The oxide is then electrolysed to get Aluminium and oxygen gas using carbon anodes and carbon as cathode. Give the common name of the ore from where aluminium is extracted from ½ mark What would be the importance of heating the ore first before refining it?1 mark To remove the water of crystallization The refined ore has to be dissolved in cryolite first before electrolysis. Why is this necessary? 1½ mark To lower the melting point of aluminium oxide from about 2015oC to 900oC so as to lower /reduce cost of production Why are the carbon anodes replaced every now and then in the cell for electrolysing aluminium oxide? 1 mark Oxygen produced at anode react with carbon to form carbon(IV)oxide gas that escape State two uses of aluminium In making aeroplane parts, buses, tankers, utensils, sauce pans,spoons Making overhead electric cables Making duralumin
18 3. IRON a)Natural occurrence Iron is the second most common naturally occurring metal. It makes 4% of the earths crust as: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) b)The blast furnace for extraction of iron from Haematite and Magnetite a)Raw materials: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) (iv)Coke/charcoal/ carbon (v)Limestone b)Chemical processes: Iron is usually extracted from Haematite (Fe2O3), Magnetite(Fe3O4) Siderite (FeCO3).These ores contain silicon(IV)oxide(SiO2) and aluminium(III)oxide (Al2O3) as impurities. When extracted from siderite, the ore must first be roasted in air to decompose the iron(II)Carbonate to Iron(II)oxide with production of carbon(IV)oxide gas: FeCO3(s) FeO(s) + CO2(g) Iron(II)oxide is then rapidly oxidized by air to iron(III)oxide(Haematite).


IRON a)Natural occurrence Iron is the second most common naturally occurring metal. It makes 4% of the earths crust as: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) b)The blast furnace for extraction of iron from Haematite and Magnetite a)Raw materials: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) (iv)Coke/charcoal/ carbon (v)Limestone b)Chemical processes: Iron is usually extracted from Haematite (Fe2O3), Magnetite(Fe3O4) Siderite (FeCO3).These ores contain silicon(IV)oxide(SiO2) and aluminium(III)oxide (Al2O3) as impurities. When extracted from siderite, the ore must first be roasted in air to decompose the iron(II)Carbonate to Iron(II)oxide with production of carbon(IV)oxide gas: FeCO3(s) FeO(s) + CO2(g) Iron(II)oxide is then rapidly oxidized by air to iron(III)oxide(Haematite). 4FeO(s) + O2(g) 2Fe2O3(s) Haematite (Fe2O3), Magnetite(Fe3O4), coke and limestone are all then fed from top into a tall (about 30metres in height) tapered steel chamber lined with refractory bricks called a blast furnace. The furnace is covered with inverted double cap to prevent/reduce amount of any gases escaping . Near the base/bottom, blast of hot air at about 1000K (827oC) is driven/forced into the furnace through small holes called Tuyeres. As the air enters ,it reacts with coke/charcoal/carbon to form carbon(IV)oxide gas. This reaction is highly exothermic. C(s)+ O2(g) CO2 (g) ∆H = -394kJ
19 This raises the temperature at the bottom of the furnace to about 2000K(1650oC).As Carbon(IV)oxide gas rises up the furnace it reacts with more coke to form carbon(II)oxide gas.This reaction is endothermic.


As the air enters ,it reacts with coke/charcoal/carbon to form carbon(IV)oxide gas. This reaction is highly exothermic. C(s)+ O2(g) CO2 (g) ∆H = -394kJ
19 This raises the temperature at the bottom of the furnace to about 2000K(1650oC).As Carbon(IV)oxide gas rises up the furnace it reacts with more coke to form carbon(II)oxide gas.This reaction is endothermic. CO2 (g) + C(s) 2CO (g) ∆H = +173kJ Carbon(II)oxide gas is a strong reducing agent that reduces the ores at the upper parts of the furnace where temperatures are about 750K(500oC) i.e. For Haematite; Fe2O3 (s) + 3CO(g) 2Fe(s) + CO2(g) For Magnetite; Fe3O4 (s) + 4CO(g) 3Fe(s) + 4CO2(g) Iron is denser than iron ore. As it falls to the hotter base of the furnace it melts and can easily be tapped off. Limestone fed into the furnace decomposes to quicklime/calcium oxide and produce more carbon(IV)oxide gas. CaCO3(s) CaO(s) + CO2(g) Quicklime/calcium oxide reacts with the impurities silicon(IV)oxide(SiO2) and aluminium(III)oxide(Al2O3)in the ore to form calcium silicate and calcium aluminate. CaO(s) + SiO2(s) CaSiO3 (l) CaO(s) + Al2O3(s) Ca Al2O4 (l) Calcium silicate and calcium aluminate mixture is called slag.Slag is denser than iron ore but less dense than iron therefore float on the pure iron. It is tapped at different levels to be tapped off for use in: (i)tarmacing roads (ii) cement manufacture (iii)as building construction material (c)Uses of Iron Iron obtained from the blast furnace is hard and brittle. It is called Pig iron. It is remelted, added scrap steel then cooled. This iron is called cast iron. Iron is mainly used to make: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes,lamp posts because it is cheap.


It is remelted, added scrap steel then cooled. This iron is called cast iron. Iron is mainly used to make: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes,lamp posts because it is cheap. 20 (ii)nails, cutlery, scissors, sinks, vats, spanners,steel rods, and railway points from steel. Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium. It does not rust/corrode like iron. e) Environmental effects of extracting Iron from Blast furnace (i)Carbon(IV)oxide(CO2) gas is a green house gas that causes/increases global warming if allowed to escape/leak from the furnace. (ii)Carbon(II)oxide(CO)gas is a highly poisonous/toxic odourless gas that can kill on leakage. It is preferentially absorbed by the haemoglobin in mammals instead of Oxygen to form a stable compound that reduce free hemoglobin in the blood. (iii) Haematite (Fe2O3), Magnetite(Fe3O4) and Siderite (FeCO3) are extracted through quarrying /open cast mining that cause soil / environmental degradation . 21 f) Test for the presence of Iron Iron naturally exist in its compound as Fe2+ /Fe3+ If an ore is suspected to contain Fe2+ /Fe3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. I) an insoluble brown precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present.


21 f) Test for the presence of Iron Iron naturally exist in its compound as Fe2+ /Fe3+ If an ore is suspected to contain Fe2+ /Fe3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. I) an insoluble brown precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. Observation Inference green precipitate in excess 2M NaOH(aq) Fe2+ green precipitate in excess 2M NH3(aq) Fe2+ brown precipitate in excess 2M NaOH(aq) Fe3+ brown precipitate in excess 2M NH3(aq) Fe3+ Practice questions
22 4.COPPER a)Natural occurrence Copper is found as uncombined element/metal on the earths crust in Zambia, Tanzania, USA and Canada .The chief ores of copper are: (i)Copper pyrites(CuFeS2) (ii)Malachite(CuCO3.Cu(OH)2) (iii)Cuprite(Cu2O) b)Extraction of copper from copper pyrites. Copper pyrites are first crushed into fine powder. The powdered ore is the added water and oil. The purpose of water is to dissolve hydrophilic substances/particle. The purpose of oil is to make cover copper ore particle so as to make it hydrophobic Air is blown through the mixture. Air creates bubbles that stick around hydrophobic copper ore. The air bubbles raise through buoyancy small hydrophobic copper ore particles to the surface. A concentrated ore floats at the top as froth. This is called froth flotation. The concentrated ore is then skimmed off.The ore is then roasted in air to form copper(I)sulphide ,sulphur(IV)oxide and iron (II) oxide.


A concentrated ore floats at the top as froth. This is called froth flotation. The concentrated ore is then skimmed off.The ore is then roasted in air to form copper(I)sulphide ,sulphur(IV)oxide and iron (II) oxide. 2CuFeS2(s) + 4O2(g) Cu2S(s) + 3SO2(g) + 2FeO(s) Limestone (CaCO3) and silicon(IV)oxide (SiO2) are added and the mixture heated in absence of air.Silicon(IV)oxide (SiO2) reacts with iron (II) oxide to form Iron silicate which constitutes the slag and is removed. FeO(s) + SiO2(s) FeSiO3(s) The slag separates off from the copper(I)sulphide. Copper(I)sulphide is then heated in a regulated supply of air where some of it is converted to copper (I) oxide. 2Cu2S (s) + 3O2(g) 2Cu2S(s) + 2SO2(g) The mixture then undergo self reduction in which copper(I)oxide is reduced by copper(I)sulphide to copper metal. Cu2S (s) + 2Cu2O (s) 6Cu (s) + SO2(g) The copper obtained has Iron, sulphur and traces of silver and gold as impurities.It is therefore about 97.5% pure. It is refined by electrolysis/electrolytic method. During the electrolysis of refining copper, the impure copper is made the anode and a small pure strip is made the cathode. Electrode ionization takes place where: At the anode;
23 Cu(s) Cu2+ (aq) + 2e Note: Impure copper anode dissolves/erodes into solution and decreases in size. At the Cathode; Cu2+ (aq) + 2e Cu(s) Note: The copper ions in the electrolyte(CuSO4) are reduced and deposited as copper metal at the cathode. The copper obtained is 99.98% pure. Valuable traces of silver and gold collect at the bottom of the electrolytic cell as sludge. It is used to finance the extraction of copper pyrites.


The copper obtained is 99.98% pure. Valuable traces of silver and gold collect at the bottom of the electrolytic cell as sludge. It is used to finance the extraction of copper pyrites. (c)Flow chart summary of extraction of copper from Copper pyrites Copper pyrites(CuFeS2) ore with impurities Fe2O3 and SiO2 Froth Crush (increase surface area) Oil Concentration chamber 1st roasting chamber Silicon(IV) oxide Smelting furnace 2nd roasting furnace Calcium aluminate (CaAl2O4)slag Limestone Sulphur(IV)Oxide Iron Silicate (FeSiO3)Slag Water Excess air Limited air Sulphur(IV)Oxide Self reduction Impure copper Rocky impurities Cu2S Cu2S, Cu2O Electrolysis using Copper electrodes
24 Electrolytic purification of impure copper d) Uses of copper Copper is mainly used in: (i)making low voltage electric cables,contact switches, cockets and plugs because it is a good conductor of electricity. (ii)Making solder because it is a good thermal conductor. (iii)Making useful alloys e.g. -Brass is an alloy of copper and Zinc(Cu/Zn) -Bronze is an alloy of copper and Tin(Cu/Sn) -German silver is an alloy of copper ,Zinc and Nickel(Cu/Zn/Ni) (iv)Making coins and ornaments. e) Environmental effects of extracting copper from Copper pyrites (i)Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes head ache to human in high concentration. (ii)Sulphur(IV)oxide gas if allowed to escape dissolves in water /rivers/rain to form weak sulphuric(IV)acid lowering the pH of the water leading to marine pollution, accelerated corrosion/rusting of metals/roofs and breathing problems to human beings. Anode; Impure Copper eroded. Cathode; Pure Copper deposited. 25 (iii)Copper is extracted by open caste mining leading to land /environmental /soil degradation.


Anode; Impure Copper eroded. Cathode; Pure Copper deposited. 25 (iii)Copper is extracted by open caste mining leading to land /environmental /soil degradation. f) Test for the presence of copper in an ore Copper naturally exist in its compound as Cu2+ /Cu+ Copper (I) / Cu+ is readily oxidized to copper(II)/ Cu2+ If an ore is suspected to contain Cu2+ /Cu+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent; 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble blue precipitate in excess of 2M sodium hydroxide if Cu2+ ions are present. I) an insoluble blue precipitate in 2M aqueous ammonia that dissolve to royal/deep blue solution in excess if Cu2+ ions are present. Observation Inference blue precipitate in excess 2M NaOH(aq) Cu2+ blue precipitate,dissolve to royal/deep blue solution in excess 2M NH3(aq) Cu2+ g)Sample questions Copper is extracted from copper pyrites as in the flow chart outlined below. Study it and answer the questions that follow
26 5.ZINC and LEAD a)Natural occurrence Zinc occurs mainly as: (i)Calamine-Zinc carbonate(ZnCO3) (ii)Zinc blende-Zinc sulphide(ZnS) Lead occurs mainly as Galena-Lead(II)Sulphide mixed with Zinc blende: b)Extraction of Zinc/Lead from Calamine ,Zinc blende and Galena. During extraction of Zinc , the ore is first roasted in air: For Calamine Zinc carbonate decompose to Zinc oxide and carbon(IV) oxide gas. ZnCO3(s) ZnO(s) + CO2(g) Zinc blende does not decompose but reacts with air to form Zinc oxide and sulphur(IV) oxide gas. Galena as a useful impurity also reacts with air to form Lead(II) oxide and sulphur(IV) oxide gas.


During extraction of Zinc , the ore is first roasted in air: For Calamine Zinc carbonate decompose to Zinc oxide and carbon(IV) oxide gas. ZnCO3(s) ZnO(s) + CO2(g) Zinc blende does not decompose but reacts with air to form Zinc oxide and sulphur(IV) oxide gas. Galena as a useful impurity also reacts with air to form Lead(II) oxide and sulphur(IV) oxide gas. 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) (Zinc blende) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) (Galena) The oxides are mixed with coke and limestone/Iron(II)oxide/ Aluminium (III) oxide and heated in a blast furnace. At the furnace temperatures limestone decomposes to quicklime/CaO and produce Carbon(IV)oxide gas. CaCO3(s) CaO(s) + CO2 (g) Carbon(IV)oxide gas reacts with more coke to form the Carbon(II)oxide gas. C(s) + CO2 (g) 2CO (g) Both Carbon(II)oxide and carbon/coke/carbon are reducing agents. 27 The oxides are reduced to the metals by either coke or carbon (II)oxide. ZnO(s) + C(s) Zn(g) + CO (g) PbO(s) + C(s) Pb(l) + CO (g) PbO(s) + CO(s) Pb(l) + CO2 (g) PbO(s) + CO(s) Pb(g) + CO2 (g) At the furnace temperature: (i)Zinc is a gas/vapour and is collected at the top of the furnace. It is condensed in a spray of molten lead to prevent reoxidation to Zinc oxide. On further cooling , Zinc collects on the surface from where it can be tapped off (ii)Lead is a liquid and is ale to trickle to the bottom of the furnace from where it is tapped off. Quicklime/CaO, Iron(II)Oxide, Aluminium(III)oxide are used to remove silica/silicon(IV)oxide as silicates which float above Lead preventing its reoxidation back to Lead(II)Oxide.


It is condensed in a spray of molten lead to prevent reoxidation to Zinc oxide. On further cooling , Zinc collects on the surface from where it can be tapped off (ii)Lead is a liquid and is ale to trickle to the bottom of the furnace from where it is tapped off. Quicklime/CaO, Iron(II)Oxide, Aluminium(III)oxide are used to remove silica/silicon(IV)oxide as silicates which float above Lead preventing its reoxidation back to Lead(II)Oxide. CaO(s) + SiO2(s) CaSiO3(s/l) (Slag-Calcium silicate) FeO(s) + SiO2(s) FeSiO3(s/l) (Slag-Iron silicate) Al2O3(s) + SiO2(s) Al2SiO4(s/l) (Slag-Aluminium silicate) c)Flow chart on extraction of Zinc from Calamine ,Zinc blende. Zinc ore (calamine /Zinc blende Powdered ore Froth flotation Roasting chamber Oil Water CO2 from calamine SO2 from Zinc blende Reduction chamber Iron/aluminium/ Limestone Coke Slag (Iron silicate/ aluminium silicate/calcium Condenser Filtration Gl ikWater
28 d) Flow chart on extraction of Lead from Galena e) Uses of Lead Lead is used in: (i)making gun-burettes. (ii)making protective clothes against nuclear (alpha rays/particle) radiation in a nuclear reactor. (iii)Mixed with tin(Sn) to make solder alloy f) Uses of Zinc Zinc is used in: Lead ore/Galena Powdered ore Froth flotation Roasting chamber oil Water Reduction chamber Iron/Limestone coke SO2(g) Slag(Iron silicate) Condenser Filtration LEAD VAPOUR Zinc residue
29 (i)Galvanization-when iron sheet is dipped in molten Zinc ,a thin layer of Zinc is formed on the surface.Since Zinc is more reactive than iron ,it reacts with elements of air(CO2/ O2 / H2O) to form basic Zinc carbonate(ZnCO3.Zn(OH)2).This sacrificial method protects iron from corrosion/rusting. (ii)As negative terminal and casing in dry/Laclanche cells. (iii)Making brass alloy with copper(Cu/Zn) g) Environmental effects of extracting Zinc and Lead.


(iii)Mixed with tin(Sn) to make solder alloy f) Uses of Zinc Zinc is used in: Lead ore/Galena Powdered ore Froth flotation Roasting chamber oil Water Reduction chamber Iron/Limestone coke SO2(g) Slag(Iron silicate) Condenser Filtration LEAD VAPOUR Zinc residue
29 (i)Galvanization-when iron sheet is dipped in molten Zinc ,a thin layer of Zinc is formed on the surface.Since Zinc is more reactive than iron ,it reacts with elements of air(CO2/ O2 / H2O) to form basic Zinc carbonate(ZnCO3.Zn(OH)2).This sacrificial method protects iron from corrosion/rusting. (ii)As negative terminal and casing in dry/Laclanche cells. (iii)Making brass alloy with copper(Cu/Zn) g) Environmental effects of extracting Zinc and Lead. (i) Lead and Lead salts are carcinogenic/causes cancer (ii)Carbon(IV)oxide is a green house gas that causes/accelerate global warming. (iii)Carbon(II)oxide is a colourless odourless poisonous /toxic gas that combines with haemoglobin in the blood to form stable carboxyhaemoglobin reducing free haemoglobin leading to death. (iv) Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes headache to human if in high concentration. (v)Any leakages in Sulphur(IV)oxide gas escapes to the water bodies to form weak sulphuric(VI)acid lowering the pH of the water. This causes marine pollution /death of aquatic life, accelerated rusting/corrosion of metals/roofs and breathing problems to human beings. h) Test for presence of Zinc/ Lead. If an ore is suspected to contain Zinc/Lead it is: I.added hot concentrated Nitric(V)acid to free the ions present. Note: Concentrated Sulphuric(VI)acid forms insoluble PbSO4 thus cannot be used to free the ions in Lead salts. II.the free ions are then added a precipitating reagent mostly 2M sodium hydroxide or 2M aqueous ammonia with the formation of; (i)a soluble precipitate in excess of 2M sodium hydroxide if Zn2+, Pb2+, Al3+ions are present.


If an ore is suspected to contain Zinc/Lead it is: I.added hot concentrated Nitric(V)acid to free the ions present. Note: Concentrated Sulphuric(VI)acid forms insoluble PbSO4 thus cannot be used to free the ions in Lead salts. II.the free ions are then added a precipitating reagent mostly 2M sodium hydroxide or 2M aqueous ammonia with the formation of; (i)a soluble precipitate in excess of 2M sodium hydroxide if Zn2+, Pb2+, Al3+ions are present. (ii)a white precipitate that dissolves to form a colorless solution in excess 2M aqueous ammonia if Zn2+ions are present. (iii)an insoluble white precipitate in excess 2M aqueous ammonia if Pb2+, Al3+ions are present. (iv) Pb2+ ions form a white precipitate when any soluble SO42-, SO32-, CO32-, Cl-, is added while Al3+ ions do not form a white precipitate
30 (v) Pb2+ ions form a yellow precipitate when any soluble I-(e.g. Potassium/sodium Iodide) is added while Al3+ ions do not form a yellow precipitate. (vi) Pb2+ ions form a black precipitate when any soluble S-(e.g.


(iv) Pb2+ ions form a white precipitate when any soluble SO42-, SO32-, CO32-, Cl-, is added while Al3+ ions do not form a white precipitate
30 (v) Pb2+ ions form a yellow precipitate when any soluble I-(e.g. Potassium/sodium Iodide) is added while Al3+ ions do not form a yellow precipitate. (vi) Pb2+ ions form a black precipitate when any soluble S-(e.g. Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions
31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
32 b) Summary of extraction of common metal.


(vi) Pb2+ ions form a black precipitate when any soluble S-(e.g. Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions
31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3.


Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions
31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S.


Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air.


Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air. Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)
33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal.


(Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air. Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)
33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal. Alloy name Constituents of the alloy Uses of the alloy Brass Copper and Zinc Making scews and bulb caps Bronze Copper and Tin Making clock springs,electrical contacts and copper coins Soldier Lead and Tin Soldering, joining electrical contacts because of its low melting points and high thermal conductivity Duralumin Aluminium, Copper and Magnesium Making aircraft , utensils ,windows frames because of its light weight and corrosion resistant. Steel Iron, Carbon ,Manganese and other metals Railway lines , car bodies girders and utensils.


Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)
33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal. Alloy name Constituents of the alloy Uses of the alloy Brass Copper and Zinc Making scews and bulb caps Bronze Copper and Tin Making clock springs,electrical contacts and copper coins Soldier Lead and Tin Soldering, joining electrical contacts because of its low melting points and high thermal conductivity Duralumin Aluminium, Copper and Magnesium Making aircraft , utensils ,windows frames because of its light weight and corrosion resistant. Steel Iron, Carbon ,Manganese and other metals Railway lines , car bodies girders and utensils. Nichrome Nichrome and Chromium Provide resistance in electric heaters and ovens German silver Copper,Zinc and Nickel Making coins d) Physical properties of metal. Metals form giant metallic structure joined by metallic bond from electrostatic attraction between the metallic cation and free delocalized electrons. This makes metals to have the following physical properties: (i)High melting and boiling points The giant metallic structure has a very close packed metallic lattice joined by strong electrostatic attraction between the metallic cation and free delocalized electrons.The more delocalized electrons the higher the melting/boiling points e.g. Aluminium has a melting point of about 2015oC while that of sodium is about 98oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one.


Metals form giant metallic structure joined by metallic bond from electrostatic attraction between the metallic cation and free delocalized electrons. This makes metals to have the following physical properties: (i)High melting and boiling points The giant metallic structure has a very close packed metallic lattice joined by strong electrostatic attraction between the metallic cation and free delocalized electrons.The more delocalized electrons the higher the melting/boiling points e.g. Aluminium has a melting point of about 2015oC while that of sodium is about 98oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one. Aluminium has a boiling point of about 2470oC while that of sodium is about 890oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one. 34 (ii)High thermal and electrical conductivity All metals are good thermal and electrical conductors as liquid or solids. The more delocalized electrons the higher the thermal and electrical conductivity. e.g. Aluminium has an electrical conductivity of about 3.82 x 19-9 ohms per metre. Sodium has an electrical conductivity of about 2.18 x 19-9 ohms per metre. (iii)Shiny/Lustrous The free delocalized electrons on the surface of the metal absorb, vibrate and then scatter/re-emit/lose light energy. All metals are therefore usually shades of grey in colour except copper which is shiny brown.e.g. Zinc is bluish grey while iron is silvery grey. (iv)High tensile strength The free delocalized electrons on the surface of the metal atoms binds the surface immediately when the metal is coiled/folded preventing it from breaking /being brittle. (v)Malleable. Metals can be made into thin sheet. The metallic crystal lattice on being beaten/pressed/hammered on two sides extend its length and width/bredth and is then immediately bound by the delocalized electrons preventing it from breaking/being brittle. (vi)Ductile. Metals can be made into thin wires. The metallic crystal lattice on being beaten/pressed/hammered on all sides extend its length is then immediately bound by the delocalized electrons preventing it from breaking/being brittle.