Redox Reactions And Electrochemistry Ques 736

Question: For the following cell reaction

Pb(s)+Hg2SO4(s)PbSO4(s)+2Hg(l) Ecello=0.92V Ksp(PbSO4)=2×108, Ksp(Hg2SO4)=1×106 Hence, Ecell is

Options:

A) 0.92 V

B) 0.89 V

C) 1.04 V

D) 0.95 V

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Q=[Pb2+][Hg22+]=Ksp(PbSO4)Ksp(Hg2SO4)

=2×1081×106

=2×102=0.14

Ecell=Ecello0.0592log0.14

=0.920.0295log0.14

=0.95V



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