Explain: Oxidation of $Cl^{-}$ ion is carried out near the anode when electrolysis of aqueous (concentrated) $NaCl$ solution is carried out using an inert electrode.

(N/A) $NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
At the anode,both $Cl^{-}$ ions and $H_{2}O$ molecules are present.
Two possible oxidation reactions are:
$(i) \ Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2(g)} + e^{-} \quad E^{\circ} = 1.36 \ V$
$(ii) \ 2 H_{2}O_{(l)} \rightarrow O_{2(g)} + 4 H^{+}_{(aq)} + 4 e^{-} \quad E^{\circ} = 1.23 \ V$
Generally,the reaction with the lower $E^{\circ}$ value is preferred. However,due to the overpotential of oxygen,the oxidation of water is kinetically hindered. Consequently,the oxidation of $Cl^{-}$ ions occurs at the anode,resulting in the evolution of $Cl_{2}$ gas.