CUET · CHEMISTRY · PYQ PAPER 2025
A cell is written as
\(Mg\left|Mg^{2+}(0.0001 M) \| Ag^{+}(0.01 M)\right| Ag\)
Calculate its \(E_{\text {cell }}\) if \(E_{\text {cell }}^{\ominus}=1.25 V\) at 298 K
- A 1 .28 V
- B 1.25 V
- C 1.19 V
- D 1.30 V
Answer & Solution
Correct Answer
(B) 1.25 V
Step-by-step Solution
Detailed explanation
\(Q = \frac{[Mg^{2+}]}{[Ag^{+}]^2} = \frac{0.0001}{(0.01)^2} = \frac{10^{-4}}{10^{-4}} = 1\) \(E_{\text {cell }} = E_{\text {cell }}^{\\ominus} - \frac{0.0592}{n} \log Q\) \(E_{\text {cell }} = 1.25 - \frac{0.0592}{2} \log(1)\) \(E_{\text {cell }} = 1.25 - 0 = 1.25 V\)
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