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

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${Na}_{(aq)}^{+} + e^{-} \to {Na}_{(s)}; E_{cell}^{0} = - 2 .71 volts$

${Cl}^{-}_{(aq)} \to {Cl}_{2} + e^{-}; E_{cell}^{0} = + 1 .36 volts$

$2 H_{(aq)}^{+} + e^{-} \to \frac{1}{2} H_{2}; = + 0 .00 volts$

$2 H_{2} O_{(a q)} ⟶ O_{2} + 4 H^{+} + 4 e^{-}; E_{cell}^{0} = + 1 .23 volts$

answer is D.

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Detailed Solution

When aqueous solution of NaCl is electrolysed, the reaction takes place are

Cathode: $2 H^{+} + 2 e^{-} \to H_{2} (reduction)$

Anode: $2 C {l^{-}}_{(aq)} \to {Cl}_{2 (g)} + 2 e^{-} (oxidation)$

The E⁰_cell for oxidation of Cl^-= +1.36V which is higher than oxidation potential of H₂O = +1.23V

Hence Cl^- is oxidised to form Cl_2.

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