JEE Main 2025 — Chemistry Physical Chemistry

2025mcqeasy

For a $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}$ the correct Nernst Equation is :

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

A
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}$
B
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$
C
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$
D
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}$

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Solution

$\begin{array}{lll}\text { Cathode } & \left(\mathrm{Ag}+(\mathrm{aq})+\mathrm{e}^{-}\right. & \rightarrow \mathrm{Ag}) \times 2 \\ \text { Anode } & \mathrm{Mg} & \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\end{array}$ JEE Main 2025 Chemistry, Electrochemistry — solution figure

By Nernst equation $\begin{aligned}
& \mathrm{E}{\text {cell }}=\mathrm{E}{\mathrm{cell}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{Q} \
& \mathrm{E}{\text {cell }}=\mathrm{E}{\mathrm{cell}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \
& =\mathrm{E}_{\text {cell }}^{\circ}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}
\end{aligned}$

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Options

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}$

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}$

Solution

JEE Main 2025 — Chemistry Physical Chemistry

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JEE Main 2025 — Chemistry Physical Chemistry

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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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