In the electrolysis of aqueous sodium chloride solution, which of the half cell reactions will occur at anode?

Options

A. `CI_(aq) to (1)/(2)CI_(2)+ e^(-), E_("cell")^(0)= 1.36 "volts"`
B. `2H_(2)O_(l) to O_(2)+ 4H^(+)+ 4e^(-), E_("cell")^(0)= 1.23 "volts"`
C. `Na_(aq)^(+) e^(-) to Na_(s), E^(0)= -2.71 "volts"`
D. `H_(aq)^(+)+ e^(-) to (1)/(2)H_(2), E_("cell")^(0)= 0.00"volts"`

Answer

A

Explanation

During electrolysis of aqueous NaCl following oxidation reaction are possible:
`Cl_(aq) to (1)/(2) Cl_(2(g))+e^(-)` , `E_("cell")^(o)=1.36 V` ...(i)
`2H_(2)O_(l) to O_(2(g))+4H_(aq)^(+)+4e^(-)`, `E_("cell")^(o)=1.23 V` ...(ii)
The reaction at anode with lower value of `E^(o)` is preferred and therefore, water should get oxidised in preference to `Cl_(aq)^(-)`. However on account of overpotential of oxygen, reaction (i) is preferred.
Thus,
Cathode: `H_(2)O_(l)+e^(-) to (1)/(2) H_(2(g))+OH_(aq)^(-)`
Anode: `Cl_(aq)^(-) ti (1)/(2)Cl_(2(g))+e^(-)`