Topic: EMF, Nernst equation and applications· Redox Reactions and Electrochemistry

Find the emf of the cell in which the following reaction takes place at 298 KNi(s) + 2Ag^+ (0.001\ M) → Ni^2+ (0.001\ M) + 2Ag(s)(Given that E^°_cell = 10.5 V, 2.303 RT/F = 0.059 at 298 K)

1. 1.05 V

2. 1.0385 V

3. 1.385 V

4. 0.9615 V

Doubtly Asked 30 Oct 2026 NEET UG 2022 neet Save WhatsApp Report

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Answer: Dropped in the official answer key: full marks to every candidate

Key idea: use the Nernst equation \(E=E^\circ-\dfrac{0.059}{n}\log Q\).

Reaction: \(\mathrm{Ni+2Ag^+\to Ni^{2+}+2Ag}\), n = 2, \(Q=\dfrac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^+}]^2}=\dfrac{10^{-3}}{10^{-6}}=10^3\).

\(E=E^\circ-\dfrac{0.059}{2}\times3=E^\circ-0.0885\) V.

The standard value is \(E^\circ=0.80-(-0.25)=1.05\) V, giving \(E=0.9615\) V (option 4). The paper prints \(E^\circ=10.5\) V, which gives 10.4115 V, not in any option.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Find the emf of the cell in which the following reaction takes place at 298 KNi(s) + 2Ag^+ (0.001\ M) → Ni^2+ (0.001\ M) + 2Ag(s)(Given that E^°_cell = 10.5 V, 2.303 RT/F = 0.059 at 298 K)

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