Chemistry - Electrochemistry Question with Solution | TestHub

During the electrolysis of aqueous sodium chloride, the following reactions take place:

At the cathode (reduction): Water is reduced in preference to ${\text{Na}}^{+}$ ions because the standard reduction potential of water is higher than that of ${\text{Na}}^{+}$.
$2{\text{H}}_2\text{O}(l)+2{\text{e}}^{-}\to {\text{H}}_2(g)+2{\text{OH}}^{-}(aq)$
Thus, ${\text{H}}_2$ gas is liberated at the cathode, not at the anode.

At the anode (oxidation): Chloride ions are oxidized in preference to water due to the overpotential of oxygen.
$2{\text{Cl}}^{-}(aq)\to {\text{Cl}}_2(g)+2{\text{e}}^{-}$
Thus, ${\text{Cl}}_2$ gas is liberated at the anode, and ${\text{O}}_2$ gas is not liberated.

In the solution, ${\text{Na}}^{+}$ and ${\text{OH}}^{-}$ ions remain, converting the aqueous $\text{NaCl}$ solution into sodium hydroxide ($\text{NaOH}$).

Therefore, the statement that ${\text{H}}_2$ gas is liberated at the anode is incorrect.

Answer: ${\text{H}}_2$ gas is liberated at anode.