Question

$Pb\left(s\right)+H{g}_{2}S{O}_4\rightleftharpoons PbS{O}_4\left(s\right)+2Hg\left(l\right)$ design the cell if both electrolytes are present in their saturated solution state. Given $E_{Pb/P{b}^{2+}}^o$ and $E_{Hg/H{g}_2^{2+}}^o$ are 0.126 and -0.789 V respectively and K${}_{sp}$ of PbSO${}_4$ and Hg${}_2$SO${}_4$ are $2.43\times {10}^{-8}$ and $1.46\times {10}^{-6}$ respectively
The e.m.f. of given cell reaction to nearest integer

Answer 1

$E_{OP}^o$ is more for Pb and thus,

Anode : $Pb\to P{b}^{2+}+2e$

Cathode : $H{g}_2^{2+}+2e\to 2Hg$

Cell is $Pb\left|\underset{saturated}{PbS{O}_4}\right|\left|\underset{saturated}{H{g}_{2}S{O}_4}\right|Hg$

Also, $E_{cell}=E_{O{P}_{Pb}}^o+E_{R{P}_{Hg}}^o+\frac{0.059}{2}log\frac{\left[H{g}_2^{2+}\right]}{\left[P{b}^{2+}\right]}$

$\therefore E_{O{P}_{Pb}}^o=0.126V$

and $E_{R{P}_{Hg}}^o=+0.789V$

and for $PbS{O}_4\rightleftharpoons P{b}^{2+}+S{O}_4^{2-}$

$\therefore K_{sp}=\left[P{b}^{2+}\right]\left[S{O}_4^{2-}\right]=[P{b}^{2+}{]}^2$

or $\left[P{b}^{2+}\right]=\sqrt{\left(K_{sp}\right)}=\sqrt{(2.43\times {10}^{-8})}$

For $H{g}_{2}S{O}_4\rightleftharpoons H{g}_2^{2+}+S{O}_4^{2-}$

$\left[H{g}_2^{2+}\right]=\sqrt{\left(K_{sp}\right)}=\sqrt{(1.46\times {10}^{-6})}$

$\therefore E_{cell}=0.126+0.789+\frac{0.059}{2}log\frac{\sqrt{1.46\times {10}^{-6}}}{\sqrt{2.43\times {10}^{-8}}}$
$=0.941V$