Find the emf of the cell reaction Ni(s) + 2Ag+(0.001 M) -> Ni^2+(0.001 M) + 2Ag(s) (Given E^ cell = 1.05 V, 2.303RT/F = 0.059 at 298 K)

Correct answer: (C) 0.9615 V. Answer: (C) 0.9615 V Solution: The reaction transfers n = 2 electrons. Reaction quotient: Q = [Ni^2+]/[Ag+]^2 = 0.001/(0.001)^2 = 10-310-6 = 103 Applying the Nernst equation: Ecell = E^ cell - 0.059/n Q = 1.05 - 0.059/2 (103) Ecell = 1.05 - 0.0295 × 3 = 1.05 - 0.0885 = 0.9615 V

NEET 2022 · ChemistryNernst equationPrevious Year Question

Find the emf of the cell reaction $Ni (s) + 2 Ag X^{+} (0.001 M) Ni X^{2 +} (0.001 M) + 2 Ag (s)$ (Given $E_{ce l l}^{\circ} = 1.05 V$ , $\frac{2.303 R T}{F} = 0.059$ at $298 K$ )

Answer: (C) $0.9615 V$ . \textbf{Answer:} (C) $0.9615 V$ \textbf{Solution:} The reaction transfers $n = 2$ electrons.

A. $1.0385 V$
B. $1.385 V$
C. $0.9615 V$ ✓
D. $1.05 V$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $0.9615 V$ \textbf{Solution:} The reaction transfers $n = 2$ electrons. Reaction quotient: $Q = \frac{[ Ni X ^{2 +} ]}{[ Ag X ^{+} ] ^{2}} = \frac{0.001}{( 0.001 ) ^{2}} = \frac{1 0 ^{- 3}}{1 0 ^{- 6}} = 1 0^{3}$ Applying the Nernst equation: $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g Q = 1.05 - \frac{0.059}{2} lo g (1 0^{3})$ $E_{ce l l} = 1.05 - 0.0295 \times 3 = 1.05 - 0.0885 = 0.9615 V$

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Find the emf of the cell reaction Ni(s)+2AgX+(0.001M)​NiX2+(0.001M)+2Ag(s) (Given Ecell∘​=1.05 V, F2.303RT​=0.059 at 298 K)

Solution & Explanation

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Find the emf of the cell reaction $Ni (s) + 2 Ag X^{+} (0.001 M) Ni X^{2 +} (0.001 M) + 2 Ag (s)$ (Given $E_{ce l l}^{\circ} = 1.05 V$ , $\frac{2.303 R T}{F} = 0.059$ at $298 K$ )