For the cell represented by the following reaction
\( Mg(s) + 2Ag^+ (0.0001M) → Mg^{2+} (0.2M) + 2Ag(s) \)
If \( E^°_{(cell)} = 3.17 V \), its \( E_{(cell)} \) at \( 298\, K \) will be:
(Given: \( \log_{10} 2 = 0.301 \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.96 V
The emf of a cell at concentrations other than 1 M is given by the Nernst equation. In this cell magnesium is oxidised at the anode and silver ions are reduced at the cathode:
\( \displaystyle \text{Anode: } \ce{Mg(s) -> Mg^2+(aq) + 2e^-} \)
\( \displaystyle \text{Cathode: } \ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \)
\( \displaystyle \text{Cell: } \ce{Mg(s) + 2Ag^+(aq) -> Mg^2+(aq) + 2Ag(s)},\quad n = 2 \)
At 298 K:
\( \displaystyle E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\frac{[\ce{Mg^2+}]}{[\ce{Ag+}]^2} \)
Substituting the concentrations:
\( \displaystyle \frac{[\ce{Mg^2+}]}{[\ce{Ag+}]^2} = \frac{0.2}{(10^{-4})^2} = \frac{0.2}{10^{-8}} = 2 \times 10^{7} \)
\( \displaystyle \log(2 \times 10^{7}) = 0.301 + 7 = 7.301 \)
\( \displaystyle E_{\text{cell}} = 3.17 - \frac{0.059}{2} \times 7.301 = 3.17 - 0.215 = 2.955\ \mathrm{V} \)
So \( E_{\text{cell}} \approx \) 2.96 V.
Shortcut: The reaction quotient is far greater than 1, so \( E_{\text{cell}} \) must be a little less than 3.17 V; the correction is about 0.03 × 7 ≈ 0.2 V, which points to 2.96 V.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Nernst Equation – …
