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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 29 January · Morning Shift · Chemistry · Electrochemistry

For a (Mg|Mg^ 2+ (aq)||Ag^ + (aq)| Ag) the correct Nernst Equation...

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

Options

  1. 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]})
  2. 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]})
  3. 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]})
  4. 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]})

Correct answer

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]})

Step-by-step solution

 Cathode (Ag+(aq)+e−→Ag)×2 Anode Mg→Mg2+(aq)+2e−\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} For a (Mg|Mg^ 2+ (aq)||Ag^ + (aq)| Ag) the correct Nernst Equation is :
By Nernst equation Ecell =Ecell∘−RTnFln⁡QEcell =Ecell∘−RTnFln⁡[Mg2+][Ag+]2=Ecell ∘+RT2 Fln⁡[Ag+]2[Mg2+]\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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