Q28 · CSIR-NET Chemistry, December 2015

Paper: CSIR-NET December 2015 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

Given that $E^{o}\left(\mathrm{Cl_2}/\mathrm{Cl^{-}}\right)=1.35\,\mathrm{V}$ and $K_{sp}(\mathrm{AgCl})=10^{-10}$ at $25^{\circ}\mathrm{C}$, $E_0$ corresponding to the electrode reaction $\dfrac{1}{2}\mathrm{Cl_2}(g)+\mathrm{Ag^{+}}(\text{Soln.})+e^{-} \rightarrow \mathrm{AgCl}(s)$ is $\left[\dfrac{2.303RT}{F}=0.06\,V\right]$
(a)0.75 V
(b)1.05 V
(c)1.65 V
(d)1.95V
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
½Cl₂ + Ag⁺ + e⁻ → AgCl is ½Cl₂ + e⁻ → Cl⁻ (1.35 V) plus Ag⁺ + Cl⁻ → AgCl (K = 1/Ksp). E° = 1.35 + 0.06 log 10¹⁰ = 1.95 V (d).

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