In the electrolysis of aqueous sodium chloride solution which of the hall cell reaction will occur at anode?

Options

A. `Na^(+)(aq)+e^(-)rarrNa(s):E_("cell")^(@)=-2.71V`
B. `2H_(2)O(l)rarrO_(2)(g)+4H^(+)(aq) +4e^(-) E_("cell")^(@)-123V`
C. `H^(+)(aq)+e^(-)rarr(1)/(2)H_(2)(g):E_("cell")^(@)=0.00V`
D. `Cl^(-)(aq)rarr(1)/(2)Cl_(2)(g)+e^(-),E_("cell")^(@)=1.36V`

Explanation

In case of electrolysis of aqueous NaCl oxidation reaction occurs at anode as follows
`Cl^(-)(aq)rarr(1)/(2)Cl_(2)(g)+e^(-) E_(@)=1.36V`
`2H_(2)O(l)rarrO_(2)(g)+4H^(+)(aq)+4e^(-) E_("cell")^(@)=1.23V`
but due to lower `E_("cell")^(@)` value water should get oxidised in preference of `C l^(-)(aq)`.
Howe ver, the actual reaction taking place in the concentrated solution of NaCl is (d) and not (b) i.e., `Cl_(2)` is produced and not `O_(2)`.
This unexpected resuylt is explained on the basis of the concept of overvoltage. i.e., water needs greater voltage for oxidation to `O_(2)` (as it is kinetically slow process) than that needed for oxidation of `Cl^(-)` ions to `Cl_(2)`. thus, the correct option is (d) not (b).