Chemistry - Electrochemistry Question with Solution | TestHub

During the electrolysis of aqueous sodium chloride, the ions present in the solution are ${\mathrm{Na}}^{+}$, ${\mathrm{Cl}}^{-}$, ${\mathrm{H}}^{+}$, and ${\mathrm{OH}}^{-}$.

At the cathode, reduction takes place. The two possible reduction reactions are:
${\mathrm{Na}}_{(aq)}^{+}+e^{-}\to {\mathrm{Na}}_{(s)}$ ($E^{\circ }=-2.71$ V)
$2{\mathrm{H}}_2{\mathrm{O}}_{(l)}+2e^{-}\to {\mathrm{H}}_{2(g)}+2{\mathrm{OH}}_{(aq)}^{-}$ ($E^{\circ }=-0.83$ V)

Since the standard reduction potential of water is higher than that of sodium ions, water is preferentially reduced at the cathode. Thus, hydrogen gas (${\mathrm{H}}_{2(g)}$) is released at the cathode.

At the anode, oxidation of chloride ions takes place due to the overpotential of oxygen, releasing chlorine gas (${\mathrm{Cl}}_{2(g)}$).

Answer: ${\mathrm{H}}_{2(\mathrm{g})}$