Calculate the emf of the following cell at 25∘C : Ag(s)|Ag+(10−3M)||Cu2+(10−1M)|Cu(s) Given E∘cell=+0.46V and log10n=n.
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Solution
For the given cell, the redox reaction is 2Ag(s)+Cu2+(aq)⟶2Ag+(aq)+Cu(s) ∴n=2 Ecell=E∘cell−0.0591nlog[Ag+]2[Cu2+] =0.46−0.05912log(10−3)210−1 =0.46−0.02955log10−5 =0.46−0.02955×(−5) =0.46+0.14775=0.60775V.