Q101 · CSIR NET Chemistry, December 2015
Paper: CSIR NET December 2015 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium
Given that $\mathrm{E}^{\circ}\left(\mathrm{Cl}_{2} / \mathrm{Cl}^{-}\right)=1.35 \mathrm{V}$ and $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=10^{-10}$ at $25^{\circ} \mathrm{C}$, $E^{0}$ corresponding to the electrode reaction $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{Ag}^{+}($Soln. $)+\mathrm{e}^{-} \rightarrow \mathrm{AgCl}(\mathrm{s})$
(a)0.75 V
(b)1.05 V
(c)1.65 V
(d)1.95 V
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
½Cl₂ + e⁻ → Cl⁻ (1.35 V) plus Ag⁺ + Cl⁻ → AgCl (K = 10¹⁰): E° = 1.35 + 0.059 × 10 ≈ 1.95 V (d).
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