Reduction of water is carried out near the cathode in the presence of aqueous $Na^{+}$ ions. Explain why.

(N/A) Reduction reactions occur on the surface of the cathode. If more than one species is present near the cathode,the species with the higher standard reduction potential $(E^{\ominus})$ value undergoes reduction.
For an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_2O$ are present near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
The possible reduction reactions are:
$(i) \ Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$(ii) \ H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\ominus} = 0.00 \ V$
Since the $E^{\ominus}$ value for the reduction of $H^{+}$ (or $H_2O$) is higher than that of $Na^{+}$,the reduction of water (or $H^{+}$) is thermodynamically favored.
Therefore,the reaction occurring at the cathode is:
$H_2O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
As a result,$H_2$ gas is evolved at the cathode,and $OH^{-}$ ions accumulate,making the solution basic,which can be confirmed by the pink color produced with phenolphthalein.