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

Options

A. `Na^(+)(aq) + e^(-) to Na(s), E_("cell")^(@)= -2.7 V`
B. `2H_(2)O(l) to O_(2)(g) + 4H^(+) (aq) + 4E_("cell")^(@) - 1.23 V`
C. `H^(+)(aq) + e^(-) to 1/2 H_(2), E_("cell")^(-) = 0.00 V`
D. `Cl^(-)(aq) to 1/2 Cl_(2)(g) + e^(-), E_("cell")^(@) = 1.36 V`

Answer

D

Explanation

In case of electrolysis of aqueous NaCl oxidation reaction occurs at anode as follows

But due to lower `E_("cell")^(@)` value water should get oxidised in preference of `Cl^(-)` (aq)
However, 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 result 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).