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CUET · CHEMISTRY · PYQ PAPER 2025

Calculate the emf of the following cell at 298 K if \(E_{\text {Cell }}^{\circ}=1.10 \mathrm{~V}:\)
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.2 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.04 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})\)

  1. A 1.10 V
  2. B 1.07 V
  3. C 1.05 V
  4. D 0.030 V
Verified Solution

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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