JEE Main 2023 — Chemistry Physical Chemistry

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The logarithm of equilibrium constant for the reaction ${\mathrm{Pd}}^{2+}+4{\mathrm{Cl}}^{-}\rightleftharpoons {\mathrm{PdCl}}_{4}^{2-}$ is

(Nearest integer)

Given: $\frac{2.303\mathrm{RT}}{F}=0.06V$

${\mathrm{Pd}}_{(\mathrm{aq})}^{2+}+2{e}^{-}\rightleftharpoons \mathrm{Pd}(s){E}^{o}=0.83V$

${\mathrm{PdCl}}_{4}^{2-}(\mathrm{aq})+2{e}^{-}\rightleftharpoons \mathrm{Pd}(s)+4{\mathrm{Cl}}^{-}(\mathrm{aq})$

${E}^{o}=0.65V$

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given, $\frac{2.303\mathrm{RT}}{F}=0.06V$

Using gibbs free energy equation,

$\Delta G^{\circ}=-\mathrm{RT}\mathrm{lnK}$

$-{\mathrm{nFE}}_{\text{cell }}^{o}=-\mathrm{RT}\times 2.303({\mathrm{log}}_{10}K)$

$\frac{{E}_{\text{Cell}}^{0}}{0.06}\times n=\mathrm{log}K---------(i)$

${\mathrm{Pd}}^{2+}(\mathrm{aq})+2{e}^{-}\rightleftharpoons \mathrm{Pd}(s),{E}_{\text{cathode, reduction }}^{0}=0.83V$

$\mathrm{Pd}(s)+4{\mathrm{Cl}}^{-}(\mathrm{aq}.)\rightleftharpoons {\mathrm{PdCl}}_{4}^{2-}(\mathrm{aq})+2{e}^{-},{E}_{\text{anode, oxidation}}^{o}=0.65V$

Net Reaction $\rightarrow {\mathrm{Pd}}^{2+}(\mathrm{aq}.)+4{\mathrm{Cl}}^{-}(\mathrm{aq}.)\rightleftharpoons {{\mathrm{PdCl}}_{4}}^{2-}(\mathrm{aq}.)$

${E}_{\text{cell }}^{o}={E}_{\text{cathode,reduction}}^{o}-{E}_{\text{anode, oxidation }}^{o}$

$\Rightarrow {E}_{\mathrm{cell}}^{o}=0.83-0.65 \Rightarrow {E}_{\mathrm{cell}}^{o}=0.18--------(\mathrm{ii})$

Also $n=2--------(\mathrm{iii})$

Using equation $(i),(\mathrm{ii})$ & $(\mathrm{iii})$ , we get

$\frac{{E}_{\text{Cell}}^{0}}{0.06}\times n=\mathrm{log}K$

$\Rightarrow \frac{0.18}{0.06}\times 2=\mathrm{log}K$

$\Rightarrow \mathrm{logK}=6$

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Solution

Given, $\frac{2.303\mathrm{RT}}{F}=0.06V$

Using gibbs free energy equation,

$\Delta G^{\circ}=-\mathrm{RT}\mathrm{lnK}$

$-{\mathrm{nFE}}_{\text{cell }}^{o}=-\mathrm{RT}\times 2.303({\mathrm{log}}_{10}K)$

$\frac{{E}_{\text{Cell}}^{0}}{0.06}\times n=\mathrm{log}K---------(i)$

${\mathrm{Pd}}^{2+}(\mathrm{aq})+2{e}^{-}\rightleftharpoons \mathrm{Pd}(s),{E}_{\text{cathode, reduction }}^{0}=0.83V$

$\mathrm{Pd}(s)+4{\mathrm{Cl}}^{-}(\mathrm{aq}.)\rightleftharpoons {\mathrm{PdCl}}_{4}^{2-}(\mathrm{aq})+2{e}^{-},{E}_{\text{anode, oxidation}}^{o}=0.65V$

Net Reaction $\rightarrow {\mathrm{Pd}}^{2+}(\mathrm{aq}.)+4{\mathrm{Cl}}^{-}(\mathrm{aq}.)\rightleftharpoons {{\mathrm{PdCl}}_{4}}^{2-}(\mathrm{aq}.)$

${E}_{\text{cell }}^{o}={E}_{\text{cathode,reduction}}^{o}-{E}_{\text{anode, oxidation }}^{o}$

$\Rightarrow {E}_{\mathrm{cell}}^{o}=0.83-0.65 \Rightarrow {E}_{\mathrm{cell}}^{o}=0.18--------(\mathrm{ii})$

Also $n=2--------(\mathrm{iii})$

Using equation $(i),(\mathrm{ii})$ & $(\mathrm{iii})$ , we get

$\frac{{E}_{\text{Cell}}^{0}}{0.06}\times n=\mathrm{log}K$

$\Rightarrow \frac{0.18}{0.06}\times 2=\mathrm{log}K$

$\Rightarrow \mathrm{logK}=6$

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

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