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

Easy

A

$\large 2{H_2}{O_{\left( l \right)}}\, \to \,{O_2} + \,4{H^ + } + \,4{e^ - };\,E_{cell}^0\, = \, + \,1.23\,volts$
B

$2{H^ + }_{\left( {aq} \right)} + \,{e^ - } \to \,\frac{1}{2}{H_2};\,E_{cell}^0\, = \, + 0.00\,volts$
C

$N{a^ + }_{\left( {aq} \right)}\, + \,{e^ - }\, \to \,N{a_{\left( s \right)}};\,E_{cell}^0 = \, - 2.71\,volts$
D

$C{l^ - }_{\left( {aq} \right)}\, \to \,\frac{1}{2}C{l_2}\, + \,{e^ - };\,E_{cell}^0\, = \,1.36\,volts$

Solution

At anode high SOP is oxidised first.
Options 2 and 3 are reduction reactions occurs at Cathode.
Between options 1 and 4 high SOP is for oxidation of Cl^- which occurs at anode.

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