JEE Main 2019 — Electrochemistry Question with Solution

$\stackrel{\mathrm{A}}{\mathrm{F}{\mathrm{e}}^{+2}}+\stackrel{\mathrm{C}}{\mathrm{A}{\mathrm{g}}^{+}}\to \mathrm{F}{\mathrm{e}}^{3+}+\mathrm{A}\mathrm{g}$

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\mathrm{o}}=\mathrm{S}\mathrm{R}\mathrm{P}$ of cathode $-\mathrm{ }\mathrm{S}\mathrm{R}\mathrm{P}$ of anode
$={\mathrm{E}}_{\mathrm{A}{\mathrm{g}}^{+}|\mathrm{A}\mathrm{g}}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{F}{\mathrm{e}}^{+3}|\mathrm{F}{\mathrm{e}}^{+2}}^{\mathrm{o}}$
$=\mathrm{x}-{\mathrm{E}}_{\mathrm{F}{\mathrm{e}}^{3+}|\mathrm{F}{\mathrm{e}}^{+2}}^{\mathrm{o}}$

Since $\mathrm{E}°$ cell is an intensive property, the $\mathrm{E}°$ cell values are not additive by nature. But the standard Gibbs free energy change is an extensive property and hence can be added.

(i) $\mathrm{F}{\mathrm{e}}^{+3}\to \mathrm{F}\mathrm{e}+3{\mathrm{e}}^{-}\to \mathrm{Z}={\mathrm{E}}_1^{\mathrm{o}};\mathrm{\Delta }{\mathrm{G}}_1=-3\times \mathrm{F}\times z$

(ii) $\mathrm{F}{\mathrm{e}}^{+2}\to \mathrm{F}\mathrm{e}+2{\mathrm{e}}^{-}$ $\mathrm{Y}={\mathrm{E}}_2^{\mathrm{o}};$ $\mathrm{ }\mathrm{\Delta }{\mathrm{G}}_2=-2\times \mathrm{F}\times y$

Subtracting (ii) from (i)
$\mathrm{F}{\mathrm{e}}^{+3}\to \mathrm{F}{\mathrm{e}}^{+2}$ $\mathrm{\Delta }{\mathrm{G}}_3=\mathrm{\Delta }{\mathrm{G}}_1-\mathrm{\Delta }{\mathrm{G}}_2$

$\left({\mathrm{E}}^{\mathrm{o}}=-2\mathrm{y}+3\mathrm{z}\right)$

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\mathrm{o}}=\mathrm{x}-\left(-2\mathrm{y}+3\mathrm{z}\right)$

$=\mathrm{x}+2\mathrm{y}-3\mathrm{z}$