During electrolysis of aqueous $N a O H, 4 g$ of $O_{2}$ gas is liberated at NTP at anode, $H_{2}$ gas liberated at cathode is:-

2.5

litres

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5.6

litres

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11.2

litres

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22.4

litres

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Solution

The correct option is B $5.6$ litres
Reaction for electrolysis of aq. NaOH is given as -

$2 N a O H \to 2 N a + O_{2} + H_{2}$

That means for every mole of $O_{2}$ liberated at anode same no of moles are liberated at cathode.

$n (O_{2}) = n (H_{2})$

$W (O_{2}) / M (O_{2}) = W (H_{2}) / M (H_{2})$

$4 / 32 = W (H_{2}) / 2$

$W (H_{2}) = 2 \times 4 / 32$

$W (H_{2}) = 0.25 g$

$H e n c e$ , 0.25 g of $H_{2}$ is liberated at cathode.
$H_{2} : O_{2}$
Ratio:- 2:1
∴ The volume of $H_{2}$
liberated will be twice of $O_{2}$ .

∴ Volume of $O_{2} \to$ 2.8litre

Volume of $H_{2} \to$ 5.6litre.

Hence, the correct option is B

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