Daniell Cell
Electrochemistry is the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations. The cell below represents Daniell cell which converts the chemical energy liberated during the redox reaction to electrical energy.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
The reaction is a combination of two half-cell reactions, one called the oxidation half-cell and the other called the reduction half-cell. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected through a metallic wire and switch externally. The electrolyte of the two half cells is connected internally through a salt bridge. At each electrode-electrolyte interface there is a tendency for metal ions from the solution to deposit on the metal electrode, trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode, trying to make it negatively charged. A potential difference develops between the electrode and the electrolyte, which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity, then the electrode potential is known as the standard electrode potential.
For a general half-cell; Mn+(aq) + ne− → M(s), the Nernst equation is written as
\( E_{(M^{n+}/M)}=E^{o}_{(M^{n+}/M)}-\frac{RT}{nF}\ln\frac{1}{[M^{n+}]} \)
where \( E^{o}_{(M^{n+}/M)} \) is the standard reduction potential of the half-cell, R is gas constant (8.314 J K−1 mol−1), F is Faraday constant (96487 C mol−1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+.
Which of the following is the correct Nernst equation for Daniell cell?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( E_{cell}=E^{o}_{cell}-\frac{RT}{2F}\ln\frac{[Zn^{2+}]}{[Cu^{2+}]} \)
The Nernst equation for a cell reaction is written with the number of electrons transferred (n) and the reaction quotient, in which products appear in the numerator and reactants in the denominator. Pure solids have unit activity and are left out:
\( \displaystyle E_{cell}=E^{o}_{cell}-\frac{RT}{nF}\ln\frac{[\text{products}]}{[\text{reactants}]} \)
For the Daniell cell, Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), n = 2, with \( \ce{Zn^{2+}} \) the product ion and \( \ce{Cu^{2+}} \) the reactant ion:
\( \displaystyle E_{cell}=E^{o}_{cell}-\frac{RT}{2F}\ln\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
Shortcut: Two electrons are transferred, so the term must be RT/2F, which leaves Options (3) and (4). The product ion \( \ce{Zn^{2+}} \) goes in the numerator, which gives Option (3).
Why the other options are incorrect
Option (1) — Incorrect. It uses n = 1 (RT/F) instead of n = 2; the ratio [Zn2+]/[Cu2+] is right.
Option (2) — Incorrect. It uses n = 1 (RT/F) instead of n = 2, and the ratio is inverted, with the reactant ion \( \ce{Cu^{2+}} \) in the numerator.
Option (4) — Incorrect. It uses n = 2, but the ratio is inverted, with the reactant ion \( \ce{Cu^{2+}} \) in the numerator.
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:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 30-05 S1 26-05 S1 | – | – | – | – |