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) $1 \mathrm{~mol}$ of $\mathrm{MnO}_{4}^{-}$to $\mathrm{Mn}^{2+}$.
Solution:
(i) $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
$\therefore$ Required charge $=3 \mathrm{~F}$
= 3 × 96487 C
= 289461 C
(ii) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$
$\therefore$ Required charge $=2 \mathrm{~F}$
= 2 × 96487 C
= 192974 C
(iii) $\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}$
i.e., $\mathrm{Mn}^{7+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}$
$\therefore$ Required charge $=5 \mathrm{~F}$
= 5 × 96487 C
= 482435 C