FeO _ 4 ^ 2 - + 2 . 2V ⟶ Fe ^ 3 + + 0 . 70V ⟶ Fe ^ 2 + - 0 . 45V ⟶...

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

$∆ {G}^{o} = - {nFE}_{cell}^{o}$

${FeO}_{4}^{2 -}\overset{+ 2 . 2V}{⟶}{Fe}^{3 +}{\Delta G}_{1}^{o} = - 6 . 6F$(3 electrons are involved)

${Fe}^{3 +}\overset{+ 0 . 70V}{⟶}{Fe}^{2 +}{\Delta G}_{2}^{o} = - 0 . 7F$(one electron is involved)

Hence, for

${FeO}_{4}^{2 -} ⟶ {Fe}^{2 +}{\Delta G}_{3}^{o} = - 7 . 3F$

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

${E}_{{FeO}_{4}^{2 -} / {Fe}^{+ 2}}^{0} = \dfrac{- 7 . 3F}{- 4F} = 1 . 825, n = 4$

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

$n =$ Electron exchange of that half cell reaction.