# In the electrolysis of potassium iodide k e k l 1mole

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In the electrolysis of potassium iodide K + + e K (l) 1mole 1 mole 1 mole FARADAY’S SECOND LAW : States that the number of faradays required to discharge one mole of an ion at an electrode equals the number of charges on the ion. Example 1 1.K + + e K Single charge on potassium requires one faraday 2.Cu 2+ + 2e Cu 1 mole of copper requires 2 faradays 3.Al 3+ + 3e Al 1 mole of all ions requires 3 faraday 4.2CL CL 2 + 2e 2 moles of chlorine ion requires 2 faraday Example 2 What mass of copper is produced at the cathode by a current of 2 amps flowing for 10min (Cu = 64) Q = I X t I = 2A t = 10 X 60 = 600s Q = I X t = 2 x 600= 1200C Cu 2+ + 2e Cu i.e. 2C 1 mole of Cu (64g) 2 x 96 000 64g
Prepared By: ChilesheMwenyaMporokoso Page 85 1200C х X = ଵଶ଴଴ ௫ ଺ସ ଶ ௫ ଽ଺ ଴଴଴ = 0.4g of Cu 2.What quantity of electricity: (i)In faraday (ii) In Coulombs is required to produce 3.2g of copper from Cu 2+ ions [ Cu = 64] (i) Cu 2+ + 2e Cu 1 mole 64g X 3.2g X = ଷ.ଶ ଺ସ = 0.05moles 3.2g = 0.05moles 2F 1 mol x 0.05mol x = 2 x 0.05 = 0.1F (a) From 3.2g we need 0.1F (b)1F 96 000C 0.1F x X = 0.1 x 96 000 = 9600C 3.2g requires 9600C Exercise What thickness of Zinc Valence = 2, density = 7150 kg/m 3 will be deposited on an iron sheet with a total surface area = 25.0cm 2 in 2.00Hrs if the sheet is the cathode in a Zinc sulphate electrolyte where the average current is 5.00A? [Faraday constant F = 96 500C/mol, relative atomic mass of Zinc = 65.38kg/mol]. Zn 2+ + 2e Zn Q = I x t
Prepared By: ChilesheMwenyaMporokoso Page 86 = 5 x (2 x 3600) = 36 000C 2F (2 x 96 500) 1 mol (65.38kg) 36 000 x X = ଷ଺ ଴଴଴ ௫ ଺ହ.ଷ଼ ଶ ௫ ଽ଺ ହ଴଴ = 0.0122kg Density = ௠௔௦௦ ௩௢௟௨௠௘ Volume = ௠௔௦௦ ௗ௘௡௦௜௧௬ = ଴.଴ଵଶଶ ଻ଵହ଴ = 1.7063 x 10 6 m 3 Volume = surface area x thickness V = s x L L = = ଵ.଻଴଺ଷ ௫ ଵ଴ షల ଴.଴଴ଶହ = 6.825 x 10 4 m

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