What is the correct Nernst equation for the following cell? Zn(s)|Zn^…

The correct answer is Option (2) → \( E_{cell}=E^{\circ}_{cell}-\frac{2.303\,RT}{nF}\log\frac{[Zn^{2+}]}{[Ag^+]^2} \)

For a cell reaction, the Nernst equation is written from the balanced equation, with the reaction quotient Q = [products]/[reactants] (solids are taken as 1):

    \( \displaystyle E_{cell}=E^{\circ}_{cell}-\frac{2.303\,RT}{nF}\log Q \)

Each concentration is raised to the power of its coefficient.

In this cell zinc is the anode and silver the cathode:

    \( \displaystyle \text{Anode: }\ce{Zn -> Zn^2+ + 2e^-} \)

    \( \displaystyle \text{Cathode: }\ce{2Ag^+ + 2e^- -> 2Ag} \)

    \( \displaystyle \text{Cell: }\ce{Zn + 2Ag^+ -> Zn^2+ + 2Ag},\quad n=2 \)

So:

    \( \displaystyle E_{cell}=E^{\circ}_{cell}-\frac{2.303\,RT}{2F}\log\frac{[\ce{Zn^2+}]}{[\ce{Ag^+}]^2} \)

This is Option (2), with n = 2.

Why the other options are incorrect

• Option (1) — Incorrect. The factor 2.303 (needed with log) and n are missing, the ratio is reactants over products, and \( \ce{[Ag^{+}]} \) is not squared.

• Option (3) — Incorrect. n = 2 is missing from the denominator; with a minus sign the ratio must be products over reactants, but here it is inverted, and the square on \( \ce{[Ag^{+}]} \) is also missing.

• Option (4) — Incorrect. The ratio is the right way up, but \( \ce{[Ag^{+}]} \) is not squared. Two \( \ce{Ag^{+}} \) ions take part in the balanced reaction.

NCERT (2026-27 edition):
Nernst Equation – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: