Find the emf of the cell: Ni(s) + 2Ag^+ (0.001 M) → Ni^2+ (0.001 M) + 2Ag(s) [Given that E^_cell = 1.05 V , 2.303RT/F = 0.059 at 298 K ] - Chemistry MCQ NEET 2022 2026 | ExamTest.live

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Find the emf of the cell: Ni (s) + 2 Ag^{+} (0.001 M) \to Ni^{2 +} (0.001 M) + 2 Ag (s) [Given that E_{ce ll}^{\circ} = 1.05 V, \frac{2.303 RT}{F} = 0.059 at 298 K]

Solution & Step-by-step Explanation

Use Nernst Equation: E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.059}{n} lo g \frac{[ Ni ^{2 +} ]}{[ Ag ^{+} ] ^{2}} n = 2 (since Ni \to Ni^{2 +} + 2 e^{-}). E_{ce ll} = 1.05 - \frac{0.059}{2} lo g \frac{1 0 ^{- 3}}{( 1 0 ^{- 3} ) ^{2}} E_{ce ll} = 1.05 - 0.0295 lo g (\frac{1 0 ^{- 3}}{1 0 ^{- 6}}) E_{ce ll} = 1.05 - 0.0295 lo g (1 0^{3}) E_{ce ll} = 1.05 - 0.0295 (3) = 1.05 - 0.0885 = 0.9615 V

Try it yourself before checking the explanation above.

Find the emf of the cell: Ni (s) + 2 Ag^{+} (0.001 M) \to Ni^{2 +} (0.001 M) + 2 Ag (s) [Given that E_{ce ll}^{\circ} = 1.05 V, \frac{2.303 RT}{F} = 0.059 at 298 K]

1.0385 V

1.385 V

0.9615 V

1.05 V

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