CUET · CHEMISTRY · PYQ PAPER 2025
Calculate the emf of the following cell at 298 K if \(E_{\text {Cell }}^{\circ}=1.10 \mathrm{~V}:\)
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.2 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.04 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})\)
- A 1.10 V
- B 1.07 V
- C 1.05 V
- D 0.030 V
Answer & Solution
Correct Answer
(A) 1.10 V
Step-by-step Solution
Detailed explanation
\(n = 2\) \(Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Ag}^{+}]^2} = \frac{0.04}{(0.2)^2} = 1\) \(E_{\text{Cell}} = E_{\text{Cell}}^{\circ} - \frac{0.0592}{n} \log Q\) \(E_{\text{Cell}} = 1.10 - \frac{0.0592}{2} \log (1)\) \(E_{\text{Cell}} = 1.10 - 0 = 1.10 \mathrm{~V}\)
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