$\begin{array}{l}\mathrm{Pb}\left(\mathrm{s}\right)+{\mathrm{Hg}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{PbSO}}_4\left(\mathrm{s}\right)+2\mathrm{Hg}\left(1\right){\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.92\mathrm{V}\\ {\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{PbSO}}_4\right)=2\times {10}^{-8},{\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Hg}}_2{\mathrm{SO}}_4\right)=1\times {10}^{-6}\end{array}$

Hence, E cell for a cell containing saturated solutions of the two salts in their respective electrodes would be:
Given [antilog (0.141) = 1.38]

${\mathrm{Pb}}_{\left(\mathrm{s}\right)}+{\mathrm{Hg}}_2{\mathrm{SO}}_{4\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{PbSO}}_{4\left(\mathrm{s}\right)}+2\mathrm{Hg}$

Can be visualized as 

$\begin{array}{l}{\mathrm{Pb}}_{\left(\mathrm{s}\right)}\to {\mathrm{Pb}}_{\mathrm{aq}}^{2+}+2{\mathrm{e}}^{-}\\ {\mathrm{SO}}_4^{2-}+{\mathrm{Pb}}^{2+}\to {\mathrm{PbSO}}_4\\ {\mathrm{Hg}}_2{\mathrm{SO}}_4\rightleftharpoons {\mathrm{Hg}}_2^{2+}+{\mathrm{SO}}_4^{2-}\\ {\mathrm{Hg}}_2{\mathrm{SO}}_4\rightleftharpoons {\mathrm{Hg}}_2^{2+}+{\mathrm{SO}}^{2-}\\ {\mathrm{Hg}}_2^{2+}+2{\mathrm{e}}^{-}\to 2\mathrm{Hg}\end{array}$

$\therefore$ same cell can be written as: 

$\begin{array}{l}\mathrm{Pb}+{\mathrm{Hg}}_2^{2+}\to {\mathrm{Pb}}^{2+}+2\mathrm{Hg}\\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}({\mathrm{E}}_{{\mathrm{Hg}}_2^{2+}|\mathrm{Hg}}^0-{\mathrm{E}}_{{\mathrm{Pb}}^{2+}|\mathrm{Pb}}^0)\\ -\frac{0.059}{2}\log \frac{\left({\mathrm{Pb}}^{2+}\right)}{\left({\mathrm{Hg}}^{2+}\right)}\\ =0.92+\frac{0.06}{4}\times 1.7\\ =0.92-0.0255\\ \cong 0.89\mathrm{V}\end{array}$

Now, ${\mathrm{Pb}}^{2+}=\frac{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{PbSO}}_4}{{\mathrm{SO}}_4^{2-}}$

$\begin{array}{l}{\mathrm{Hg}}_2^{2+}=\frac{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{Hg}}_2{\mathrm{SO}}_4}{{\mathrm{SO}}_4^{2-}}\\ \therefore {\mathrm{E}}_{\mathrm{cell}}^0=0.92+\frac{0.059}{2}\log \frac{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{PbSO}}_4}{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{Hg}}_2{\mathrm{SO}}_4}\end{array}$

$\therefore {\mathrm{SO}}_4^{2-}$ cancels for separate cell

$\begin{array}{l}{\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.059}{2}\log \frac{\left({\mathrm{Pb}}^{2+}\right)}{\left({\mathrm{Hg}}_2^{2+}\right)}\\ {\mathrm{Pb}}^{2+}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{PbSO}}_4}\\ {\mathrm{Hg}}^{2+}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{HgSO}}_4}\\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.059}{2}\log \sqrt{\frac{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{PbSO}}_4}{{\mathrm{K}}_{\mathrm{sp}}{\mathrm{Hg}}_2{\mathrm{SO}}_4}}\end{array}$

$\begin{array}{l}=0.92+\frac{0.059}{2}\log 2\times {10}^{-2}\\ -\frac{0.059}{4}\log 2\times {10}^{-2}\\ =0.92+\frac{0.059}{4}[2-\log 2]\end{array}$

=0.96V