Chemistry - Electrochemistry Question with Solution | TestHub

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

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

The electrode reactions can be written as follows,

${\mathrm{Pb}}^{2+}+2{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Pb} ∆{\mathrm{G}}_1=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^{\mathrm{o}}$

${\mathrm{Pb}}^{4+}+4{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Pb} ∆{\mathrm{G}}_2=-4{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/\mathrm{Pb}}^{\mathrm{o}}$

${\mathrm{Pb}}^{4+}+2{\mathrm{e}}^{-}\stackrel{}{\to }{\mathrm{Pb}}^{2+} ∆{\mathrm{G}}_3=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/{\mathrm{Pb}}^{2+}}^{\mathrm{o}}$

Now, $∆{\mathrm{G}}_3=∆{\mathrm{G}}_2-∆{\mathrm{G}}_1$

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

$\Rightarrow 4{\mathrm{E}}_{{\mathrm{Pb}}^{+4}/\mathrm{Pb}}^{\mathrm{o}}=2{\mathrm{E}}_{{\mathrm{Pb}}^{+2}/\mathrm{Pb}}^{\mathrm{o}}+2{\mathrm{E}}_{{\mathrm{Pb}}^{+4}/{\mathrm{Pb}}^{+2}}^{\mathrm{o}}$

$$\begin{gathered} \Rightarrow 4\mathrm{n}=2\mathrm{m}+2{\mathrm{E}}_{{\mathrm{Pb}}^{+4}/{\mathrm{Pb}}^{+2}}^{\mathrm{o}} \\ \Rightarrow {\mathrm{E}}_{{\mathrm{Pb}}^{+4}/{\mathrm{Pb}}^{+2}}^{\mathrm{o}}=2\mathrm{n}-\mathrm{m} \\ \Rightarrow {\mathrm{E}}_{{\mathrm{Pb}}^{+2}/{\mathrm{Pb}}^{+4}}^{\mathrm{o}}=\mathrm{m}-2\mathrm{n} \end{gathered}$$

Hence, the value of $\mathrm{x}=2$