Calculate the charge in Faradays required to reduce $1\text{ mole}$ of ${\text{MnO}}_4^{-}$ to ${\text{Mn}}^{2+}$.

Solution

1. The oxidation state of Mn in ${\text{MnO}}_4^{-}$ is $+7$.
2. The oxidation state of Mn in ${\text{Mn}}^{2+}$ is $+2$.
3. The reduction half-reaction is ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$.
4. The number of electrons transferred is $5$.
5. Therefore, $5\text{ F}$ of charge is required.
6. Hence the answer is (C).