Question
How much charge is required for the following reduction of
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of MnO4– to Mn2+
Solution
(i) Al3+ + 3e– → Al
1 mol Al3+ for reduction requires 3 mol e– or 3F electricity
1F = 96500C
∴ 3F = 3 x 96500 = 289500 C
(ii) Reduction of 1 mol Cu2+ to Cu requires 2 mol electrons
2 mol electrons = 2F = 2 x 96500 = 193000 C
(iii) In the reduction of 1 mol MnO–4 to Mn2+, there is net gain of 5e–
MnO4 + 5e– + 8H+ → Mn2+ + 4H2O
5F = 5 x 96500 = 482500 C
