Question Analysis: Identification of the feasible reaction at the anode during the electrolysis of aqueous NaCl.

Step 1 (Governing Principle & Theoretical Basis): [1 Mark]

During electrolysis, the reaction with the lower oxidation potential (or higher reduction potential) is theoretically preferred. However, for the oxidation reaction at the anode, we must consider the overpotential. The oxidation of water to oxygen requires an extra potential (overpotential) to overcome the kinetic barriers. Therefore, the reaction with the lower electrode potential is not always the one that occurs.

Step 2 (Step-by-Step Comparative Analysis): [2 Marks]

• Reaction (I): $\text{Cl}^- (\text{aq}) \longrightarrow \frac{1}{2} \text{Cl}_2 \, (g) + \text{e}^- \quad E^\circ_{\text{ox}} = -1.36 \text{ V}$
• Reaction (II): $2\text{H}_2\text{O} \, (l) \longrightarrow \text{O}_2 \, (g) + 4\text{H}^+ (\text{aq}) + 4\text{e}^- \quad E^\circ_{\text{ox}} = -1.23 \text{ V}$

Mathematically, Reaction (II) has a higher oxidation potential. However, due to the overpotential of oxygen, the actual potential required for the oxidation of water is significantly higher than $1.23 \text{ V}$. Consequently, the discharge of $\text{Cl}^-$ ions becomes kinetically more favorable than the discharge of $\text{OH}^-$ or $\text{H}_2\text{O}$.

Step 3 (Final Concluded Answer): [1 Mark]

Conclusion: Reaction (I) is feasible at the anode.

Reason: Although the oxidation potential of water is lower, the overpotential required for the evolution of oxygen gas makes the oxidation of water kinetically difficult. Thus, $\text{Cl}^-$ ions are preferentially oxidized to $\text{Cl}_2$ gas at the anode.