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Question

The absolute value of standard reduction potential for CuS|Cu electrode in volts is
Given: Ksp of CuS=8.0×1036; EoCu|Cu2+=0.34 V.
(take log2=0.3)

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Solution

Half cell reactions are
at anode: CuCu2++2e
at cathode: CuS+2eCu+S2
so, overall cell reaction is: CuSCu2++S2
now, as Eocell=(Eoreduction)C(Eoreduction)A
=EoCuS|CuEoCu2+|Cu
=x0.34 V
Cell is in equilibrium, therefore, Ecell=0
So, Ecell=Eocell0.0592log[Cu2+][S2]
0=(x0.34)0.0592log[Cu2+][S2]
Eocell=0.0592logKsp
=0.03 log(8×1036)
=0.03[3log 236]
=0.03[3×0.336]
=0.03(35.1)=1.05 V
=x0.34=1.05 V
=0.71 V
hence, EoCuS|Cu=0.71 V

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