Electrochemical cells
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge.
For the cell
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag
The cell reaction is Ni(s) + 2Ag+ (aq) → Ni2+(aq) + 2Ag(s).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
The relationship between the equilibrium constant of the reaction and the standard electrode potential of the cell in which that reaction takes place is given by
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( E^{\circ}_{cell}=\displaystyle\frac{2.303RT}{nF}\log K_c \)
The standard emf of a cell and the equilibrium constant of its reaction are both linked to the standard Gibbs energy change:
\( \displaystyle \Delta_rG^{\circ}=-nFE^{\circ}_{cell} \)
\( \displaystyle \Delta_rG^{\circ}=-RT\ln K_c=-2.303\,RT\log K_c \)
Equating the two expressions:
\( \displaystyle nFE^{\circ}_{cell}=2.303\,RT\log K_c \)
\( \displaystyle E^{\circ}_{cell}=\frac{2.303\,RT}{nF}\log K_c \)
At 298 K, 2.303RT/F = 0.059 V, so the relation becomes:
\( \displaystyle E^{\circ}_{cell}=\frac{0.059}{n}\log K_c \)
For the Ni–Ag cell of the passage, n = 2.
Why the other options are incorrect
Option (1) \( E^{\circ}_{cell}=\displaystyle\frac{RT}{2.303\times nF}\log K_c \) — Incorrect. The factor 2.303 converts ln into log, so it must multiply RT, not divide it.
Option (3) \( E^{\circ}_{cell}=\displaystyle\frac{2.303RT}{nF}\ln K_c \) — Incorrect. The factor 2.303 is used only with log (base 10); with ln the relation is E°cell = (RT/nF) ln Kc.
Option (4) \( E^{\circ}_{cell}=2.303RT\ln K_c \) — Incorrect. Without nF the right side is an energy (J mol−1), not a potential (V).
NCERT (2026-27 edition):
Equilibrium Constant from Nernst Equation – “If the circuit in Daniell cell (Fig. 2.1) is closed then we note that the reaction … takes place and as time passes, the concentration of Zn2+ keeps on increasing while the concentration of Cu2+ keeps on decreasing. … After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. … In general,
\( \displaystyle E^{\circ}_{cell}=\frac{2.303RT}{nF}\log K_C \qquad (2.14) \)
Thus, Eq. (2.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place.”