Chemistry - Electrochemistry Question with Solution | TestHub

The cell representation is ${\text{Cu}}_{(s)}\mid {\text{Cu}}_{(1M)}^{2+}\Vert {\text{Ag}}_{(1M)}^{+}\mid \text{Ag}$.

Since the concentrations of the ions are $1\text{M}$, the given emf is the standard emf of the cell, $E_{\text{cell}}^{\circ }=0.463\text{V}$.

The standard reduction potential of the anode (Cu) is $E_{{\text{Cu}}^{2+}/\text{Cu}}^{\circ }=0.337\text{V}$.

Using the relation for standard cell potential:

$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$

$E_{\text{cell}}^{\circ }=E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }-E_{{\text{Cu}}^{2+}/\text{Cu}}^{\circ }$

Substituting the given values:

$0.463=E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }-0.337$

$E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }=0.463+0.337=0.800\text{V}$

The standard potential of the Ag electrode is $0.8\text{V}$.

Answer: $0.8\text{V}$