Q18 / 75 · JEE Main 2025 (29 Jan Shift 1) · Chemistry
For a \(\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}\) the correct Nernst Equation is :
- A\(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}\)
- Correct answerB\(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}\)
- 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\(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}\)
Solution
$\begin{array}{lll}\text { Cathode } & \left(\mathrm{Ag}+(\mathrm{aq})+\mathrm{e}^{-}\right. & \rightarrow \mathrm{Ag}) \times 2 \\ \text { Anode } & \mathrm{Mg} & \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\end{array}$

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

By Nernst equation $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{Q} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & =\mathrm{E}_{\text {cell }}^{\circ}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]} \end{aligned}$
Hence, the correct answer is option: 'B'.