The Nernst equation corresponding to the cell reaction given below, is
\( \text{Ni}(\text{s})+2\text{Ag}^{+}(\text{aq})\rightarrow\text{Ni}^{2+}(\text{aq})+2\text{Ag}(\text{s}) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{2F}\ln\frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2}\)
The Nernst equation gives the cell potential at any concentration:
\( \displaystyle E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF}\ln Q \)
where n is the number of electrons transferred in the reaction as written and Q is the reaction quotient. In Q, each concentration is raised to the power of its stoichiometric coefficient, and pure solids are left out because their activity is 1.
For this cell, nickel loses two electrons and the two silver ions gain one each, so n = 2:
\( \displaystyle \ce{Ni(s) -> Ni^2+(aq) + 2e-} \qquad \ce{2Ag+(aq) + 2e- -> 2Ag(s)} \)
\( \displaystyle Q = \frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}} \)
Putting n = 2 and this Q into the general equation gives Option (1). At 298 K it becomes \( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2}\log\frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}} \).
Why the other options are incorrect
Option (2) \( E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{2F}\log\frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2} \) — Incorrect. It uses log with \( \frac{RT}{2F} \). Changing ln to log brings in the factor 2.303, so the term should be \( \frac{2.303RT}{2F}\log \).
Option (3) \( E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{2F}\ln\frac{[\text{Ni}^{2+}]}{2[\text{Ag}^{+}]^2} \) — Incorrect. It puts a factor 2 in front of \( [\ce{Ag+}]^{2} \). The coefficient 2 of \( \ce{Ag+} \) appears only once, as the power of its concentration.
Option (4) \( E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{F}\ln\frac{[\text{Ni}^{2+}][\text{Ag}]^2}{[\text{Ag}^{+}]^2[\text{Ni}]} \) — Incorrect. It uses \( \frac{RT}{F} \), that is, n = 1, though two electrons are transferred; writing [Ag] and [Ni] for the solids adds nothing, since each is taken as 1.
NCERT (2026-27 edition):
Nernst Equation – “… By converting the natural logarithm in Eq. (2.11) to the base 10 and substituting the values of R, F and T = 298 K, it reduces to
\( \displaystyle E_{(cell)} = E^{\circ}_{(cell)} - \frac{0.059}{2}\log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
We should use the same number of electrons (n) for both the electrodes and thus for the following cell Ni(s) | Ni2+(aq) || Ag+(aq) | Ag. The cell reaction is Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s). The Nernst equation can be written as
\( \displaystyle E_{(cell)} = E^{\circ}_{(cell)} - \frac{RT}{2F}\ln\frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}} \)”