Given that $2 C l^{-} \to C l_{2} + 2 e^{-}, E^{\circ} = - 1.36 V;$
$H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-}, E^{\circ} = - 1.23 V .$ On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?

Cathode : Na and $H_{2}$

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Cathode : $N a, O_{2}$

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Anode : $O_{2}$

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Anode : $C l_{2}$

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Solution

The correct option is D
Anode : $C l_{2}$

Since $E_{\frac{N a^{+}}{N a}}^{\circ} = - 2.71 V$ < reduction potential of water (=- 0.83V), water will be reduced to $H_{2}$ and $O H^{-}$ at the cathode. Although
$E_{o x}^{\circ} (H_{2} O) = - 1.23 V > E_{\frac{C l^{-}}{C l_{2}}}^{\circ} = (- 1.36 V),$
Water will not be oxidised at the anode due to over potential of oxygen at platinum surface but $C l^{-}$ will oxidised to $C l_{2}$

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Similar questions

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{\circ} = 1.23 V

F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V

△ G^{\circ}

2 F e_{2} + + C l_{2} \to 2 F e^{3 +} + 2 C l^{-}

F e^{3 +} + e^{-} \to F e^{2 +}; E = 0.77 V

C l_{2} + 2 e^{-} \to 2 C l^{-}; E = 1.36 V

{2 H}^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2} ⟶ H_{2} O (l); E^{⊙} = + 1.23 V

F e^{2 +} + 2 e^{-} ⟶ F e (s); E^{⊙} = - 0.44 V

Δ G^{⊙}

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