Pb(s)+Hg2SO4(s)⇌PbSO4(s)+2Hg(1)Ecello=0.92 VKsp(PbSO4)=2×10−8, Ksp(Hg2SO4)=1×10−6Hence, E cell for a cell containing saturated solutions of the two salts in their respective electrodes would be:Given [antilog (0.141) = 1.38]

Pb(s)+Hg2SO4(s)⇌PbSO4(s)+2HgCan be visualized as Pb(s)→Pbaq2++2e−SO42−+Pb2+→PbSO4Hg2SO4⇌Hg22++SO42−Hg2SO4⇌Hg22++SO2−Hg22++2e−→2Hg∴ same cell can be written as: Pb+Hg22+→Pb2++2HgEcell=Ecello(EHg22+|Hg0−EPb2+|Pb0)−0.0592log⁡(Pb2+)(Hg2+)=0.92+0.064×1.7=0.92−0.0255≅0.89 VNow, Pb2+=KspPbSO4SO42−Hg22+=KspHg2SO4SO42−∴ Ecell0=0.92+0.0592log⁡KspPbSO4KspHg2SO4∴ SO42− cancels for separate cellEcell=Ecell0−0.0592log⁡(Pb2+)(Hg22+)Pb2+=KspPbSO4Hg2+=KspHgSO4Ecell=Ecell0−0.0592log⁡KspPbSO4KspHg2SO4=0.92+0.0592log⁡2×10−2−0.0594log⁡2×10−2=0.92+0.0594[2−log⁡2]=0.96V