Chemistry - Electrochemistry Question with Solution | TestHub

Given $E^{\circ }({\text{Cu}}^{+2}/\text{Cu})=+0.34\text{V}$ and $E^{\circ }({\text{H}}^{+}/{\text{H}}_2)=0.00\text{V}$.

Since the standard reduction potential of the copper electrode is greater than that of the standard hydrogen electrode, copper acts as the cathode and the hydrogen electrode acts as the anode.

In a galvanic cell, the cathode is the positive electrode where reduction takes place.

The reduction reaction at the positive electrode (copper cathode) is:

${\text{Cu}}_{(aq)}^{+2}+2e^{-}⟶{\text{Cu}}_{(s)}$