Consider the following reaction 

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}^{\mathrm{o}}=1.51\mathrm{V}.$

The quantity of electricity required in Faraday to reduce five moles of ${\mathrm{MnO}}_4^{-}$ is __________

For the reaction ${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5e^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O},E^{\circ }=1.51\mathrm{V}$

No. of faradays required = Change in oxidation state 

$\begin{array}{l}{\mathrm{MnO}}_{(+7)}^{-}\to \underset{(+2)}{{\mathrm{Mn}}^{2+}}\\ {\mathrm{MnO}}_4^{-}+5{\mathrm{e}}^{-}\mathrm{s}\to {\mathrm{Mn}}^{2+}\\ 1\text{ mole }=5{\mathrm{e}}^{-}\mathrm{s}\\ 5\text{ mole }=25{\mathrm{e}}^{-}\mathrm{s}\\ \text{ No. of faradays required }=25\end{array}$

The quantity of electricity required in Faraday to reduce5 moles of ${\mathrm{MnO}}_4^{-}$ is 25 F.