Ag→Ag++e;EoOP=−0.799V
Ag(NH3)+2+e→Ag+2NH3;EoRP=?
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∴Ag(NH3)+2⇌Ag+2NH3
∴Ecell=Eocell+0.0591log10[Ag(NH3)+2][Ag+][NH3]2
∵Ecell=0 at equilibrium and
∴Ecell=0.0591logKc=0.0591log(6×10−14) (i)
=−0.780V
Also, Eocell=EoOPAg/Ag++EoRPAg/Ag(NH3)+2/Ag (ii)
∴EoAg(NH3)+2/Ag=EoCell−EoOPAg/Ag+
=−0.780+0.799=+0.019V