Statement A: The electrolysis of aqueous NaCI solution gives hydrogen gas at the cathode and chlorine gas at the anode.

Statement B: Chlorine has higher oxidation potential than water.

In aqueous solution, water is available in plenty. Water can undergo both Oxidation and reduction.

$$\begin{gathered} {\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}^{+1}+2{\mathrm{e}}^{-1}; {\mathrm{E}}^0=-1.23\mathrm{V} \\ 2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-1}\to {\mathrm{H}}_{2\left(\mathrm{g}\right)}+2{\mathrm{OH}}^{-1}; {\mathrm{E}}^0=-0.83\mathrm{V} \\ 2{\mathrm{Cl}}^{-1}\to {\mathrm{Cl}}_{2\left(\mathrm{g}\right)}+2{\mathrm{e}}^{-1};{\mathrm{E}}^0=-1.36\mathrm{V} \end{gathered}$$

At Anode, Water competes with Chloride ion to undergo Oxidation. But, due to higher over voltage, the Oxidation of Water becomes energetically unfavourable and the Chloride ion is Oxidised to Chlorine gas.

At Cathode, Water competes with ${\mathrm{Na}}^{+1}$ ion to undergo reduction. But, the reduction of ${\mathrm{Na}}^{+1}$is energetically more difficult than the reduction of water as SRP of ${\mathrm{Na}}^{+1}$is -2.71V whereas the SRP of Water is -0.83V. So, Water is readily reduced in to Hydrogen gas.

Chlorine has higher reduction potential (+1.36V) than water (+1.23V).