Calculate the emf of the following cell at 298 K if \( E_{cell}^o= 1.10 V \):
\( Ni(s) + 2Ag^+(0.2M) → Ni^{2+} (0.04M) + 2Ag(s) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.10 V
The emf of a cell at concentrations other than 1 M is given by the Nernst equation. At 298 K, for a cell reaction in which n electrons are transferred:
\( \displaystyle E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\frac{[\text{products}]}{[\text{reactants}]} \)
Solids are left out of the ratio. In the reaction
\( \displaystyle \ce{Ni(s) + 2Ag+(aq) -> Ni^2+(aq) + 2Ag(s)} \)
nickel loses two electrons (\( \ce{Ni -> Ni^2+ + 2e-} \)) and two silver ions gain them, so n = 2. The concentration of \( \ce{Ag+} \) is squared, as in the equation:
\( \displaystyle E_{\text{cell}} = 1.10 - \frac{0.059}{2}\log\frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}} \)
\( \displaystyle \frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^{2}} = \frac{0.04}{(0.2)^{2}} = \frac{0.04}{0.04} = 1 \)
\( \displaystyle E_{\text{cell}} = 1.10 - \frac{0.059}{2}\log 1 = 1.10 - 0 = 1.10\ \mathrm{V} \)
So the emf of the cell is 1.10 V, the same as the standard emf.
Shortcut: Before using the log, work out the reaction quotient. Here \( (0.2)^2 = 0.04 \) equals \( [\ce{Ni^2+}] \), so the quotient is 1 and \( E_{\text{cell}} = E^\circ_{\text{cell}} \).
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag(s) Ecello = 1.05 V”