Chemistry - Electrochemistry Question with Solution | TestHub

The oxidation state of $\mathrm{Mn}$ in ${\mathrm{KMnO}}_4$ is $+7$.

The oxidation state of $\mathrm{Mn}$ in ${\mathrm{MnSO}}_4$ is $+2$.

The reduction half-reaction is:

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

This shows that $1\mathrm{mol}$ of ${\mathrm{KMnO}}_4$ requires $5\mathrm{mol}$ of electrons for reduction.

Therefore, $2\mathrm{mol}$ of ${\mathrm{KMnO}}_4$ will require $2\times 5=10\mathrm{mol}$ of electrons.

Since $1\mathrm{mol}$ of electrons corresponds to $1\mathrm{F}$ of charge, the total charge required is $10\mathrm{F}$.

Answer: 10 F