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