For the following reaction, at equilibrium the Nernst equation may be written as:
\( Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(E_{cell}^0=\frac{2.303RT}{2F}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
The Nernst equation gives the cell potential at any concentration. For the Daniell cell two electrons are transferred (n = 2):
\( \displaystyle \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \qquad \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
\( \displaystyle E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303RT}{2F}\log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
As the cell runs, \( [\ce{Zn^2+}] \) rises and \( [\ce{Cu^2+}] \) falls until the reaction reaches equilibrium; then \( E_{\text{cell}} = 0 \) and the equation becomes:
\( \displaystyle E^\circ_{\text{cell}} = \frac{2.303RT}{2F}\log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
At equilibrium the ratio \( [\ce{Zn^2+}]/[\ce{Cu^2+}] \) is the equilibrium constant \( K_c \), so this is the relation \( E^\circ_{\text{cell}} = \frac{2.303RT}{nF}\log K_c \).
Shortcut: Two electrons are transferred, so the factor must be 2F (Options (2) and (4)). The log term is products over reactants, \( [\ce{Zn^2+}]/[\ce{Cu^2+}] \), which leaves Option (2).
NCERT (2026-27 edition):
Equilibrium Constant from Nernst Equation
