The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution  is represented by : 

1. Na+(aq) + e- ⟶ Na(s) ; \(E_{cell}^{o} \ = \ -2.71 \ V \)

2. 2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e; \(E_{cell}^{o} \) = 1.23 V

3. H+(aq) + e-\(\frac{1}{2}\)H2(g) ; \(E_{cell}^{o} \) = 0.00 V

4. Cl-(aq) ⟶ \(\frac{1}{2}\)Cl2(g) + e- ; \(E_{cell}^{o}\) 1.36 V

HINT: Higher the value of oxidation potential, the higher will be the tendency to oxidize.
Explanation:
In the case of electrolysis of aqueous NaCl, an oxidation reaction occurs at the anode as follows
Cl-aq12Cl2g+e-   E°=1·36V2H2OlO2g+4H+aq+4e-   E°cell =1.23V
But due to lower ECell° value water should get oxidized in preference of Cl- (aq).
However, the actual reaction taking place in the concentrated solution of NaCl is (d) and not (b) i.e., Clis produced and not O2.
This unexpected result is explained on the basis of the concept of ‘overvoltage’, i.e., water needs greater voltage for oxidation to O2 (as it is a kinetically slow process) than that needed for oxidation of Cl- ions to Cl2. Thus, the correct option is (4) not (2).