(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 $E^{\ominus}$ value undergoes reduction.
For example,in an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_{2}O$ are present near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
$(i) \ Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$(ii) \ H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\ominus} = 0.00 \ V$
Comparing the $E^{\ominus}$ values,reaction $(ii)$ has a higher value than reaction $(i)$. Therefore,$H^{+}$ ions from water are reduced at the cathode to produce $H_{2}$ gas.
The overall cathodic reaction is:
$H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
As $Na^{+}$ ions do not participate in the reaction,they remain in the solution as spectator ions,combining with $OH^{-}$ to form $NaOH$. The presence of $NaOH$ is confirmed by the pink color observed when phenolphthalein is added near the cathode.