JEE Main 2023 — Chemistry Physical Chemistry

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In an electrochemical reaction of lead, at standard temperature, if ${E}_{({\mathrm{Pb}}^{2+}/\mathrm{Pb})}^{o}=m\text{ Volt }$ and ${E}_{({\mathrm{Pb}}^{4+}/\mathrm{Pb})}^{o}=n\mathrm{Volt},$ then the value of $E^{\circ}({\mathrm{Pb}}^{2+}/{\mathrm{Pb}}^{4+})$ is given by $m–\mathrm{xn}$ . The value of x is _______. (Nearest integer)

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

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Solution

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

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

The electrode reactions can be written as follows,

${\mathrm{Pb}}^{2+}+2{e}^{-}\overset{}{\rightarrow }\mathrm{Pb}\Delta {G}_{1}=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^{o}$

${\mathrm{Pb}}^{4+}+4{e}^{-}\overset{}{\rightarrow }\mathrm{Pb}\Delta {G}_{2}=-4{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/\mathrm{Pb}}^{o}$

${\mathrm{Pb}}^{4+}+2{e}^{-}\overset{}{\rightarrow }{\mathrm{Pb}}^{2+}\Delta {G}_{3}=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/{\mathrm{Pb}}^{2+}}^{o}$

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

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

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

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

Hence, the value of $x=2$

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Solution

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

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

The electrode reactions can be written as follows,

${\mathrm{Pb}}^{2+}+2{e}^{-}\overset{}{\rightarrow }\mathrm{Pb}\Delta {G}_{1}=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^{o}$

${\mathrm{Pb}}^{4+}+4{e}^{-}\overset{}{\rightarrow }\mathrm{Pb}\Delta {G}_{2}=-4{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/\mathrm{Pb}}^{o}$

${\mathrm{Pb}}^{4+}+2{e}^{-}\overset{}{\rightarrow }{\mathrm{Pb}}^{2+}\Delta {G}_{3}=-2{\mathrm{FE}}_{{\mathrm{Pb}}^{4+}/{\mathrm{Pb}}^{2+}}^{o}$

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

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

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

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

Hence, the value of $x=2$

JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry