Q61 / 240 · KCET 2026 · Chemistry
During the electrolysis of acidified water, $16$ g of $\text{O}_2$ gas is formed at anode. The volume of $\text{H}_2$ gas liberated at cathode under STP conditions is
- Correct answerA$22.4$ L
- B$11.2$ L
- C$2.24$ L
- D$1.12$ L
Solution
The overall reaction for the electrolysis of acidified water is:
$2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$
Number of moles of $\text{O}_2$ produced $= \dfrac{16}{32} = 0.5$ mol
From the balanced chemical equation, $1$ mole of $\text{O}_2$ is produced for every $2$ moles of $\text{H}_2$.
Number of moles of $\text{H}_2$ produced $= 2 \times 0.5 = 1$ mol
At STP, the volume occupied by $1$ mole of an ideal gas is $22.4$ L.
Volume of $\text{H}_2$ gas liberated $= 1 \times 22.4 = 22.4$ L
Answer: $22.4$ L
$2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$
Number of moles of $\text{O}_2$ produced $= \dfrac{16}{32} = 0.5$ mol
From the balanced chemical equation, $1$ mole of $\text{O}_2$ is produced for every $2$ moles of $\text{H}_2$.
Number of moles of $\text{H}_2$ produced $= 2 \times 0.5 = 1$ mol
At STP, the volume occupied by $1$ mole of an ideal gas is $22.4$ L.
Volume of $\text{H}_2$ gas liberated $= 1 \times 22.4 = 22.4$ L
Answer: $22.4$ L
Hence, the correct answer is option: 'A'.