Find the emf of the cell in which the following reaction takes place at 298 K $Ni(s) + 2Ag^+ (0.001\ M) \rightarrow Ni^{2+} (0.001\ M) + 2Ag(s)$ (Given that $E^\circ_{cell} = 10.5\ V$, $\frac{2.303RT}{F} = 0.059$ at 298 K)

Question

Dr. Abubacker · Accepted Answer

[Solution] $E = E^\circ - \frac{0.059}{2}\log\frac{[Ni^{2+}]}{[Ag^+]^2} = E^\circ - 0.0295 	imes 3 = E^\circ - 0.0885$. With $E^\circ = 10.5$ V as printed, E = 10.41 V, which matches no option (with 1.05 V it would be 0.9615 V). No option is correct (bonus question as per source key).

Q. Find the emf of the cell in which the following reaction takes place at 298 K $Ni(s) + 2Ag^+ (0.001\ M) \rightarrow Ni^{2+} (0.001\ M) + 2Ag(s)$ (Given that $E^\circ_{cell} = 10.5\ V$, $\frac{2.303RT}{F} = 0.059$ at 298 K)

Explanation

Reviewed by Dr. Abubacker

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