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Question

Calculate the emf of the following cell at 25oC Ag(s)Ag+(103M)Cu2+(101M)Cu(s)
Given: Eocell=+0.46V and log10n=n

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Solution

CathodeCCl2++2eCCl
Anode2Ag2Ag++2e
Cell reaction Cu2++2Ag2Ag++Cu
E0cell=+0.46v
Ecell=0.46V0.05912log[Ag+]2[Cu2+]Ecell=0.46V0.05912log[103]2[101]Ecell=+0.61V
Thus the Emf of the reaction is +0.61V

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