Eored of the half-cell Hg2Cl2(s)+2e−→2Hg(s)+2Cl−(aq) is 0.27V. I – InterviewSolution

InterviewSolution

1.	Eored of the half-cell Hg2Cl2(s)+2e−→2Hg(s)+2Cl−(aq) is 0.27V. If pKsp corresponding to solubility product for Hg2Cl2(s) is 17 then find the E∘red for the half cell Hg2+2(aq)+2e→2Hg(s). (Given 2.303RTF=0.06)
Answer» $E_{r e d}^{o}$ of the half-cell $H g_{2} C l_{2} (s) + 2 e^{-} \to 2 H g (s) + 2 C l^{-} (a q)$ is $0.27 V$ . If $p K_{s p}$ corresponding to solubility product for $H g_{2} C l_{2} (s)$ is 17 then find the $E_{r e d}^{\circ}$ for the half cell $H g_{2}^{2 +} (a q) + 2 e \to 2 H g (s) .$ (Given $\frac{2.303 R T}{F} = 0.06)$

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