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Q.

Fused NaOH is subjected to electrolysis using inert electrodes. The reaction taking place at anode would be

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a

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

b

$2 H_{2} O \to 4 H^{+} + O_{2} + 4 e^{-}$

c

$2 H_{2} O + O_{2} + 4 e^{-} \to 4 O H^{-}$

d

$4 O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$

answer is A.

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Detailed Solution

Fused NaOH:

$2 NaOH ⇌ 2 {Na}^{+} + 2 {OH}^{-}$

At anode:

$4 O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$

At cathode:

$2 {Na}^{+} + 2 e^{-} ⇌ 2 Na$

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