Relation between equilibrium constant and standard potential of the c…

The correct answer is Option (2) → \( E°_{cell}=\frac{2.303\, RT}{nF}\log K_c \)

The relation comes from two expressions for the standard Gibbs energy of the cell reaction. Electrical work gives \( \Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}} \), and thermodynamics gives \( \Delta_r G^{\circ} = -RT\ln K_C \). Equating the two:

    \( \displaystyle -nFE^{\circ}_{\text{cell}} = -RT\ln K_C \)

    \( \displaystyle E^{\circ}_{\text{cell}} = \frac{RT}{nF}\ln K_C \)

    \( \displaystyle E^{\circ}_{\text{cell}} = \frac{2.303\,RT}{nF}\log K_C \)

The same result follows from the Nernst equation: at equilibrium \( E_{\text{cell}} = 0 \) and the reaction quotient equals \( K_C \). At 298 K, \( 2.303RT/F = 0.059 \) V, so \( E^{\circ}_{\text{cell}} = \frac{0.059}{n}\log K_C \).

Shortcut: \( RT/F \) has the unit of volts (J mol⁻¹ divided by C mol⁻¹), so the factor in front of \( \log K_c \) must be \( 2.303RT/nF \) for the right side to be a potential. Only Option (2) has the whole factor.

Why the other options are incorrect

• Option (1) \( E°_{cell}=\frac{2.303\, RT}{F}\log K_c \) — Incorrect. n is missing; since \( \Delta_r G^{\circ} = -nFE^{\circ} \), the right side is divided by nF, not by F alone.

• Option (3) \( E°_{cell}=\frac{2.303}{nF}\log K_c \) — Incorrect. RT is missing; it comes from \( \Delta_r G^{\circ} = -RT\ln K_C \), and without it the right side is not a potential.

• Option (4) \( E°_{cell}=\frac{2.303\, R}{nF}\log K_c \) — Incorrect. The temperature T is missing; the factor comes from \( RT\ln K_C \), so it must be \( 2.303RT/nF \).

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: