How much charge is required for the following reductions:
$(i)$ $1 \, mol$ of $Al^{3+}$ to $Al$.
$(ii)$ $1 \, mol$ of $Cu^{2+}$ to $Cu$.
$(iii)$ $1 \, mol$ of $MnO_{4}^{-}$ to $Mn^{2+}$.

$(i)$ $Al^{3+} + 3e^{-} \longrightarrow Al$
$\therefore$ Charge required $= 3 \, F = 3 \times 96487 \, C = 289461 \, C$
$(ii)$ $Cu^{2+} + 2e^{-} \longrightarrow Cu$
$\therefore$ Charge required $= 2 \, F = 2 \times 96487 \, C = 192974 \, C$
$(iii)$ $MnO_{4}^{-} \longrightarrow Mn^{2+}$
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $+7$. The reduction to $Mn^{2+}$ involves a change of $5$ electrons.
$Mn^{7+} + 5e^{-} \longrightarrow Mn^{2+}$
$\therefore$ Charge required $= 5 \, F = 5 \times 96487 \, C = 482435 \, C$