Question

Consider the following reaction
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O,$
$E^o=1.51V.$

The quantity of electricity required in Faraday to reduced five moles of $Mn{O}_4^{-}$ is
(Integer answer)

( <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> JEE MAIN 2021)

Answer 1

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O;E^o=1.51V$

$\because$ 1 mole of $Mn{O}_4^{-}$ requires $5$ moles of electrons or $5F$ electricity.

$\therefore$ $5$ moles of $Mn{O}_4^{-}$ requires $25F$ electricity.