Question

How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO-4 to Mn2+ .

Answer 1

(i) Al3+ 3e- $\rightarrow$ Al
∴ Required charge = 3 F
= 3 × 96487 C
= 289461 C

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(ii) Cu2+ +2e- $\rightarrow$ Cu
∴ Required charge = 2 F
= 2 × 96487 C
= 192974 C

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(iii) MnO-4 $\rightarrow$ Mn2+
i.e., Mn7+ +5e- $\rightarrow$ Mn2+
∴ Required charge = 5 F
= 5 × 96487 C
= 482435 C