Consider the following reaction:

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

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

The above question can be solved as:

Step 1: The reaction given is: ${{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

Step 2: In this reaction, $1mole$ ${{\mathrm{MnO}}_4}^{-}$ reduces to form ${\mathrm{Mn}}^{2+}$and for this reaction $5F{e}^{-}$ are required.

$$\begin{gathered} Mn{O_4}^{-}\to M{n}^{2+} \\ +7+2 \end{gathered}$$

Step 3: Now,

$$\begin{gathered} 1moleMn{O_4}^{-}=5F \\ 5moleMn{O_4}^{-}=\left(5\times 5\right)F \\ =25F \end{gathered}$$

Therefore, the quantity of electricity required to reduce five moles of ${{\mathrm{MnO}}_4}^{-}$ is $\mathbf{25}\mathbf{}\mathit{F}$.