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

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$2 H_{(aq)}^{+} + e^{-} \to \frac{1}{2} H_{2}; E_{cell}^{°} = + 0.00 volts$

${Na}_{(aq)}^{+} + e^{-} \to {Na}_{(s)}; E_{cell}^{°} = - 2.71 volts$

${Cl}_{(aq)}^{-} \to \frac{1}{2} {Cl}_{2} + e^{-}; E_{cell}^{°} = 1.36 volts$

$2 H_{2} O_{(l)} \to O_{2} + 4 H^{+} + 4 e^{-}; E_{cell}^{°} = + 1.23 volts$

answer is D.

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Detailed Solution

Chlorine gas is released at the anode during the electrolysis of aqueous sodium chloride solution.
Cl_(aq)+e⁻→1/2Cl₂+e−
E⁰_cell=1.36 V.
This is due to the fact that Cl^- ions have a lower discharge potential than OH^- ions.
Because oxidation of O₂ is a kinetically slow process that requires high voltage, the reaction at the anode with a lower value of E should be chosen, and so the above reaction occurs.

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