Given that 2Cl−→Cl2+2e−,E∘=−1.36V;
H2O→2H++12O2+2e−,E∘=−1.23V. On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?
Anode : Cl2
Since E∘Na+Na=−2.71V < reduction potential of water (=- 0.83V), water will be reduced to H2 and OH− at the cathode. Although
E∘ox(H2O)=−1.23 V>E∘Cl−Cl2=(−1.36 V),
Water will not be oxidised at the anode due to over potential of oxygen at platinum surface but Cl− will oxidised to Cl2