When 0.1 mole of ${\mathrm{MnO}}_4^{2-}$ is oxidised, the quantity of electricity required to completely oxidise ${\mathrm{MnO}}_4^{2-}$ to ${\mathrm{MnO}}_4^{-}$

$\underset{+6}{{\mathrm{MnO}}_4^{2-}}\rightleftharpoons \underset{(+7)}{{\mathrm{MnO}}_4^{-}}$
As per the equation, for 1 mole of ${\mathrm{MnO}}_4^{2-},1$ F of electricity is required. Thus, for 0.1 mole of ${\mathrm{MnO}}_4^{2-} \mathrm{to} {\mathrm{MnO}}_4^{-}$F of electricity is required.
Since; 1 F = 96500 C
$\therefore 0.1 \mathrm{F}=0.1 \times 96500 \mathrm{C}=9650 \mathrm{C}$
Hence, 9650 C of electricity is required to completely oxidise ${\mathrm{MnO}}_4^{2-} \mathrm{to} {\mathrm{MnO}}_4^{-}$.