Consider the following reaction MnO_4^-+8 H^++5 e^- → Mn^+2+4 H_2 O, E^°=1.51 V. The quantity of electricity required in Faraday to reduce five moles…

Consider the following reaction $$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{~V} . $$ The quantity of electricity required in Faraday to reduce five moles of $\displaystyle \mathrm{MnO}_4^{-}$ is \_\_\_\_. (Integer answer)