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Question: In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur a...

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

A

Na+(aq)+e−→Na(s);EºCell=2.71V]N{a^{+}}_{(aq)} + e^{-} \rightarrow Na_{(s)};Eº_{Cell} = 2.71V\rbrack

B

2H2O(l)→O2(g)+4H+(aq)+4e−;Eºcell=1.23V2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4{H^{+}}_{(aq)} + 4e^{-};Eº_{cell} = 1.23V

C

H+(aq)+e−→12H2(g);Eºcell=0.00V{H^{+}}_{(aq)} + e^{-} \rightarrow \frac{1}{2}H_{2(g)};Eº_{cell} = 0.00V

D

Cl−(aq)→12Cl2(g)+e−;Eºcell=1.36VC{l^{-}}_{(aq)} \rightarrow \frac{1}{2}Cl_{2(g)} + e^{-};Eº_{cell} = 1.36V

Answer

Cl−(aq)→12Cl2(g)+e−;Eºcell=1.36VC{l^{-}}_{(aq)} \rightarrow \frac{1}{2}Cl_{2(g)} + e^{-};Eº_{cell} = 1.36V

Explanation

Solution

At the anode, the following oxidation reactions are possible:

(4) Cl−(aq)→1/2Cl2(g)+e−Eºcell=1.36VC{l^{-}}_{(aq)} \rightarrow 1/2Cl_{2(g)} + e^{-}Eº_{cell} = 1.36V

(2) 2H2O(l)→O2(g)+4H+(aq)+4e−Eºcell=1.23V2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4{H^{+}}_{(aq)} + 4e^{-}Eº_{cell} = 1.23V

The reaction at anode with lower value of Eŗ is preferred and therefore water should get oxidized in preference to Cl−(aq)C{l^{-}}_{(aq)} however, on account of over potential of oxygen, reaction (4) is preferred.