Consider the following hypothetical redox reaction
\( \text{A}(s)+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B}(s):E_{\text{cell}}^0=0.295\ \text{V} \)
The equilibrium constant of the reaction at 298 K will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{10} \)
The standard cell potential and the equilibrium constant are linked through the standard Gibbs energy change of the reaction:
\( \displaystyle \Delta_r G^{\circ} = -nFE^{\circ}_{cell} \quad \text{and} \quad \Delta_r G^{\circ} = -RT\ln K \)
\( \displaystyle E^{\circ}_{cell} = \frac{2.303RT}{nF}\log K \)
At 298 K, \( 2.303RT/F = 0.059\ \mathrm{V} \), so
\( \displaystyle E^{\circ}_{cell} = \frac{0.059\ \mathrm{V}}{n}\log K \)
Here n is the number of electrons transferred in the reaction as written. A(s) loses two electrons to become \( \ce{A^2+} \), and the two \( \ce{B+} \) ions gain one electron each:
\( \displaystyle \ce{A(s) -> A^2+(aq) + 2e-} \qquad \ce{2B+(aq) + 2e- -> 2B(s)} \)
So n = 2.
\( \displaystyle \log K = \frac{n\,E^{\circ}_{cell}}{0.059\ \mathrm{V}} = \frac{2 \times 0.295\ \mathrm{V}}{0.059\ \mathrm{V}} = 10 \)
\( \displaystyle K = 10^{10} \)
So the equilibrium constant is \( 10^{10} \).
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):
Equilibrium Constant from Nernst Equation – …
