Chemistry - Electrochemistry Question with Solution | TestHub

$\frac{{\displaystyle \stackrel{\mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-} \Delta {\mathrm{G}}_1^{\mathrm{o}}=-\mathrm{RTln}(2\times {10}^{-10})}{{\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Ag} {\mathrm{\Delta G}}_2^{\mathrm{o}}=-0.799\mathrm{F}}}}{{\displaystyle \mathrm{AgCl}+{\mathrm{e}}^{-}\to Ag+{\mathrm{Cl}}^{-} \Delta {\mathrm{G}}_3^{\mathrm{o}}=-\mathrm{FE}°}}$

Thus, $\Delta {\mathrm{G}}_3^{\mathrm{o}}=\Delta {\mathrm{G}}_1^{\mathrm{o}}+\Delta {\mathrm{G}}_2^{\mathrm{o}}$

So, $-{\mathrm{FE}}^{\circ }=-\mathrm{RT}\ln (2\times 10^{-10})-0.799\mathrm{F}$

$E^{\circ }=\frac{0.0591}{1}\log (2\times 10^{-10})+0.799=0.225\mathrm{V}$

$\mathrm{AgCl}+{\mathrm{e}}^{-}⟶\mathrm{Ag}+{\mathrm{Cl}}^{-}$, the Nernst equation is:

$E=E^{\circ }-\frac{0.0591}{1}\log [{\mathrm{Cl}}^{-}]$

$E=0.225-0.0591\log [{\mathrm{Cl}}^{-}]$

$E=0.225+0.0591\log \frac{K_{sp}}{\left[A{g}^{+}\right]}=0.68\mathrm{V}$