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Question

Dissociation constant for Ag(NH3)+2 into Ag+ and NH3 is 6×1014, Eo for the half cell reaction is
Ag(NH3)+2+eAg+NH3
Given Ag++eAg;Eo=0.799V The answer is 0.0mn then m+n =

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Solution

AgAg++e;EoOP=0.799V
Ag(NH3)+2+eAg+2NH3;EoRP=?
------------------------------------------------
Ag(NH3)+2Ag+2NH3
Ecell=Eocell+0.0591log10[Ag(NH3)+2][Ag+][NH3]2
Ecell=0 at equilibrium and
Ecell=0.0591logKc=0.0591log(6×1014) (i)
=0.780V
Also, Eocell=EoOPAg/Ag++EoRPAg/Ag(NH3)+2/Ag (ii)
EoAg(NH3)+2/Ag=EoCellEoOPAg/Ag+
=0.780+0.799=+0.019V

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