Chemistry - Electrochemistry Question with Solution | TestHub

$$\begin{array}{rl}& \mathrm{Pb}⟶{\mathrm{Pb}}^{+2}+2{\mathrm{e}}^{-}[{\mathrm{PbSO}}_4]\underline{\dots (1)}\\ & {\mathrm{Pb}}^{+2}+2{\mathrm{e}}^{-}⟶\mathrm{Pb}[{\mathrm{PbCl}}_2]\underline{\dots (2)}\\ & Addingequation(1)\&(2)\\ & {\mathrm{Pb}}^{+2}(fromcathode)⟹{\mathrm{Pb}}^{+2}\\ & {\mathrm{E}}_{cell}={\mathrm{E}}_{cell}^0-\frac{0.059}{2}\log \left[\frac{[{\mathrm{Pb}}^{+2}{]}_{anode}}{[{\mathrm{Pb}}^{+2}{]}_{cathode}}\right]\\ & {\mathrm{E}}_{cell}=0-\frac{0.059}{2}\log \left[\frac{[{\mathrm{Pb}}^{+2}{]}_{anode}}{[{\mathrm{Pb}}^{+2}{]}_{cathode}}\right]\left\{\therefore {\mathrm{E}}_{cell}^0=0\right\}\\ & {\mathrm{PbCl}}_2\rightleftharpoons {\mathrm{Pb}}_{{\mathrm{S}}_1}^{+2}+2{\mathrm{Cl}}^{-}\\ & (\mathrm{KSP}{)}_{{\mathrm{PbCl}}_2}={\mathrm{S}}_1\cdot (2{\mathrm{S}}_1{)}^2=4{\mathrm{S}}_1^3\\ & {\mathrm{PbSO}}_4\rightleftharpoons {\mathrm{Pb}}_{{\mathrm{S}}_2}^{+2}+{\mathrm{SO}}_4^{2-}\\ & {\mathrm{S}}_1=\left(\frac{(\mathrm{KSP}{)}_{{\mathrm{PbSO}}_4}}{4}\right)=({\mathrm{S}}_2{)}^2\\ & {\mathrm{E}}_{cell}=\frac{-0.059}{2}\log \left[\frac{[{\mathrm{Pb}}^{+2}{]}_{anode}}{[{\mathrm{Pb}}^{+2}{]}_{cathode}}\right]\end{array}$$