If $0.224 L$ of $H_{2}$ gas is formed at the cathode, the volume of $O_{2}$ gas formed at the anode under identical conditions is:

0.224 L

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0.448 L

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0.112 L

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1.12 L

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Solution

The correct option is D $0.112 L$
$H_{2} O \to H_{2} + \frac{1}{2} O_{2}$ .

From the above equation, it is clear that the volume of $O_{2}$ formed will be half of the formation of $H_{2}$ .

Since, volume of $H_{2}$ formed $= 0.224 L$

So, volume of $O_{2}$ formed $= \frac{0.224}{2} = 0.112 L$

Hence, the correct answer is an option $C$ .

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