Identify the correct equations for the emf of a cell at equilibrium.
(A) \(E^\theta_{cell} = \frac{2.303RT}{nF}\log K_C\)
(B) \(E^\theta_{cell} = \frac{\Delta G}{nF}\)
(C) \(E_{cell} = E^\theta_{cell} - \frac{RT}{nF}\ln\frac{[Products]}{[Reactants]}\)
(D) \(E^\theta_{cell} = E^\theta_{left} - E^\theta_{right}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
The Nernst equation gives the emf of a cell at any composition, where Q is the reaction quotient:
\( \displaystyle E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]} \)
At equilibrium the cell does no work, Ecell = 0, and Q becomes KC:
\( \displaystyle 0 = E^\circ_{cell} - \frac{2.303RT}{nF}\log K_C \;\Rightarrow\; E^\circ_{cell} = \frac{2.303RT}{nF}\log K_C \)
(A) \( E^\circ_{cell} = \frac{2.303RT}{nF}\log K_C \) — Correct.
(B) \( E^\circ_{cell} = \frac{\Delta G}{nF} \) — Incorrect. The relation is ΔrG° = −nFE°cell, so E°cell = −ΔrG°/nF; the minus sign is missing.
(C) \( E_{cell} = E^\theta_{cell} - \frac{RT}{nF}\ln\frac{[Products]}{[Reactants]} \) — Correct. This is the Nernst equation. It holds at any composition, including equilibrium, where Ecell = 0 and the ratio becomes KC; this gives (A).
(D) \( E^\circ_{cell} = E^\circ_{left} - E^\circ_{right} \) — Incorrect. It is E°right − E°left (cathode minus anode).