Question

For a cell, $Cu ( s ) \mid Cu ^{2+}\left(0.001 M || Ag ^{+}(0.01 M ) \mid\right.$ $Ag ( s )$ the cell potential is found to be $0.43 V$ at $298 K$. The magnitude of standard electrode potential for $Cu ^{2+} / Cu$ is $-\times 10^{-2} V$.
$\left[\text { Given }: E _{ Ag ^{\dagger} / Ag }^{\Theta}=0.80 V \text { and } \frac{2.303 RT }{ F }=0.06 V \right]$

Answer 1

At anode
$Cu \rightarrow Cu ^{2+}+2 e ^{-}$
At cathode
$ 2 Ag ^{+}+2 e ^{-} \rightarrow 2 Ag$
$ \text { Cell reaction } \rightarrow Cu +2 Ag ^{+} \rightarrow Cu ^{2+}+2 Ag$
$E _{\text {cell }}= E _{\text {cell }}^0-\frac{0.06}{2} \log \frac{\left[ Cu ^{2+}\right]}{\left[ Ag ^{+}\right]^2}$
$0.43= E _{\text {cell }}^0-\frac{0.06}{2} \log \frac{(0.001)}{(0.01)^2}$
$ E _{ cell }^0=0.46 $
$ E _{ cell ^0}^0= E _{ Ag ^{+} / Ag }^0- E _{ Cu ^{2+} / Cu }^0 $
$0.46=0.80- E _{ Cu ^{2+} / Cu }^0 $
$ E _{ Cu ^{2+} / Cu }^0=0.34$ volt
$ E _{ Cu ^{2+/ Cu }}^0=34 \times 10^{-2}$