In an electrochemical reaction of lead, at standard temperature, if...

The relation between Gibbs Free energy and emf of the cell can be related as follows,

$∆ G = - {nFE}_{cell}$

The electrode reactions can be written as follows,

${Pb}^{2 +} + 2{e}^{-}\overset{}{\to}Pb ∆ {G}_{1} = - 2{FE}_{{Pb}^{2 +} / Pb}^{o}$

${Pb}^{4 +} + 4{e}^{-}\overset{}{\to}Pb ∆ {G}_{2} = - 4{FE}_{{Pb}^{4 +} / Pb}^{o}$

${Pb}^{4 +} + 2{e}^{-}\overset{}{\to}{Pb}^{2 +} ∆ {G}_{3} = - 2{FE}_{{Pb}^{4 +} / {Pb}^{2 +}}^{o}$

Now, $∆ {G}_{3} = ∆ {G}_{2} - ∆ {G}_{1}$

$- 2{FE}_{{Pb}^{4 +} / {Pb}^{2 +}}^{o} = - 4{FE}_{{Pb}^{4 +} / Pb}^{o} + 2{FE}_{{Pb}^{2 +} / Pb}^{o}$

$⇒ 4{E}_{{Pb}^{+ 4} / Pb}^{o} = 2{E}_{{Pb}^{+ 2} / Pb}^{o} + 2{E}_{{Pb}^{+ 4} / {Pb}^{+ 2}}^{o}$

$⇒ 4n = 2m + 2{E}_{{Pb}^{+ 4} / {Pb}^{+ 2}}^{o} ⇒ {E}_{{Pb}^{+ 4} / {Pb}^{+ 2}}^{o} = 2n - m ⇒ {E}_{{Pb}^{+ 2} / {Pb}^{+ 4}}^{o} = m - 2n$

Hence, the value of $x = 2$