Chemistry - Electrochemistry Question with Solution | TestHub

The relation between $∆{\mathrm{G}}^{\mathrm{o}}$ and ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ is 

$∆{\mathrm{G}}^{\mathrm{o}}=-{\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}$

${\mathrm{FeO}}_4^{2-}\stackrel{+2.2 \mathrm{V}}{⟶}{\mathrm{Fe}}^{3+} {\mathrm{\Delta G}}_1^{\mathrm{o}}=-6.6 \mathrm{F}$(3 electrons are involved)

${\mathrm{Fe}}^{3+}\stackrel{+0.70 \mathrm{V}}{⟶}{\mathrm{Fe}}^{2+}{\mathrm{\Delta G}}_2^{\mathrm{o}}=-0.7 \mathrm{F}$(one electron is involved)

Hence, for

${\mathrm{FeO}}_4^{2-}⟶{\mathrm{Fe}}^{2+}{\mathrm{\Delta G}}_3^{\mathrm{o}}=-7.3 \mathrm{F}$

$=-{\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}$(Four electrons are involved)

${\mathrm{E}}_{{\mathrm{FeO}}_4^{2-}/{\mathrm{Fe}}^{+2}}^0=\frac{-7.3 \mathrm{F}}{-4 \mathrm{F}}=1.825, \mathrm{n}=4$

$=1825\times {10}^{-3} \mathrm{V}$

$\mathrm{n}=$ Electron exchange of that half cell reaction.