Chemistry - Electrochemistry Question with Solution | TestHub

Cell reaction, (i) ${\mathrm{Ag}}^{+}+e^{-}⟶\mathrm{Ag}(\mathrm{s}),E_{\mathrm{red}}^{\circ }=0.8\mathrm{V}$ (ii) ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_7+2{\mathrm{H}}^{+}+2e^{-},E_{\text{oxid }}=-0.05\mathrm{V}$ Hence, for reaction $$\begin{array}{rl}2{\mathrm{Ag}}^{+}+{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}& ⟶2\mathrm{Ag}(s)+{\mathrm{C}}_6{\mathrm{H}}_{12}\mathrm{O}7+2{\mathrm{H}}^{+}\\ E_{\text{cell }}^{\circ }& =(0.8-0.05)=0.75\mathrm{V}\\ E_{\text{cell }}^{\circ }& =E_{\text{anode }}^{\circ }-E_{\text{cathode }}^{\circ }=[-0.05-(-0.8){]}_{\mathrm{V}}\end{array}$$ On the basis of oxidised electrode potential $=0.75\mathrm{V}$ By Nernst equation, $E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{2.303RT}{nF}{\log }_{10}K$ At equilibrium, $E_{\text{cell }}=0$ $$\begin{array}{rl}{E}_{\text{cell }}^{\circ }& =\frac{2.303RT}{nF}{\log }_{10}K\\ 0.75& =\frac{0.0592}{2}{\log }_{10}K\\ {\log }_{10}K& =\frac{2\times 0.75}{0.0592}=25.33\end{array}$$ or In $K=2.303\times {\log }_{10}K=2.303\times 25.33=58.35$