Explain: Reduction reaction is possible with higher $E^{\theta}$ value.

(N/A) Reduction reactions occur on the surface of the cathode. If more than one species are present near the cathode,the species with the higher $E^{\theta}$ value undergoes reduction.
For example,in an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_{2}O$ are available near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
The reduction half-reactions are:
$(i) Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\theta} = -2.71 \ V$
$(ii) H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\theta} = 0.00 \ V$
Comparing the $E^{\theta}$ values,reaction $(ii)$ has a higher value than reaction $(i)$. Therefore,reaction $(ii)$ occurs preferentially at the cathode. Consequently,$H^{+}$ ions from water are reduced to produce $H_{2}$ gas. The overall reaction at the cathode during the electrolysis of aqueous $NaCl$ is:
$(iii) H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
The $Na^{+}$ ions remain in the solution as spectator ions,forming $NaOH$ with the produced $OH^{-}$ ions. The presence of $NaOH$ is confirmed experimentally by the pink color observed upon adding phenolphthalein indicator near the cathode.