The value of log Kc for the given cell reaction at 298 K is
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
E°(cell) = 0.46 V
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15.6
The standard cell potential is related to the equilibrium constant through the standard Gibbs energy change:
\( \displaystyle \Delta_rG^\circ=-nFE^\circ_{cell} \quad\text{and}\quad \Delta_rG^\circ=-2.303\,RT\log K_c \)
\( \displaystyle E^\circ_{cell}=\frac{2.303\,RT}{nF}\log K_c \)
At 298 K, 2.303RT/F = 0.059 V, so:
\( \displaystyle \log K_c=\frac{n\,E^\circ_{cell}}{0.059} \)
In the reaction \( \ce{Cu(s) + 2Ag^+(aq) -> Cu^2+(aq) + 2Ag(s)} \), each copper atom loses two electrons, which are taken by two \( \ce{Ag^{+}} \) ions, so n = 2.
\( \displaystyle \log K_c=\frac{2\times0.46}{0.059} \)
\( \displaystyle \log K_c=\frac{0.92}{0.059}=15.6 \)
So the value of log Kc is 15.6.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Equilibrium Constant from Nernst Equation – …
