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.