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Question

Given that 2ClCl2+2e,E=1.36V;
H2O2H++12O2+2e,E=1.23V. On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?


A

Cathode : Na and H2

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B

Cathode : Na,O2

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C

Anode : O2

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D

Anode : Cl2

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Solution

The correct option is D

Anode : Cl2


Since ENa+Na=2.71V < reduction potential of water (=- 0.83V), water will be reduced to H2 and OH at the cathode. Although
Eox(H2O)=1.23 V>EClCl2=(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


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