Chemistry - Electrochemistry Question with Solution | TestHub

Given $E_{{\text{Cd}}^{2+}|\text{Cd}}^{\circ }=-0.403\text{V}$ and $E_{{\text{H}}^{+}|{\text{H}}_2}^{\circ }=0.00\text{V}$.

Since the standard reduction potential of cadmium is lower than that of hydrogen, cadmium has a higher tendency to get oxidized. Therefore, cadmium will act as the anode (oxidation) and the standard hydrogen electrode will act as the cathode (reduction).

At anode: ${\text{Cd}}_{(s)}⟶{\text{Cd}}_{(aq)}^{2+}+2{\text{e}}^{-}$

At cathode: $2{\text{H}}_{(aq)}^{+}+2{\text{e}}^{-}⟶{\text{H}}_{2(g)}$

Adding the two half-cell reactions, the net cell reaction is:

${\text{Cd}}_{(s)}+2{\text{H}}_{(aq)}^{+}⟶{\text{Cd}}_{(aq)}^{2+}+{\text{H}}_{2(g)}$

Answer: ${\text{Cd}}_{(s)}+2{\text{H}}_{(aq)}^{+}⟶{\text{Cd}}_{(aq)}^{2+}+{\text{H}}_{2(g)}$