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Chemistry · PYQ NEET UG 2022
हिन्दीFind the emf of the cell in which the following reaction takes place at \( 298\text{ K} \). \( \text{Ni(s)} + 2\text{Ag}^+ (0.001\text{ M}) \rightarrow \text{Ni}^{2+} (0.001\text{ M}) + 2\text{Ag(s)} \) (Given that \( E^\circ_{\text{cell}} = 10.5\text{ V} \), \( \frac{2.303RT}{F} = 0.059 \) at \( 298\text{ K} \))
Correct answer · Option 1
\( 0.9615\text{ V} \)
Explanation
Using Nernst equation: \( E = E^\circ - \frac{0.059}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} = 1.05 - \frac{0.059}{2} \log \frac{0.001}{(0.001)^2} = 1.05 - 0.0295 \log(1000) = 1.05 - 0.0885 = 0.9615\text{ V} \).
