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KCET · Chemistry · Electrochemistry

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

  1. A \(22.4\) L
  2. B \(11.2\) L
  3. C \(2.24\) L
  4. D \(1.12\) L
Verified Solution

Answer & Solution

Correct Answer

(A) \(22.4\) L

Step-by-step Solution

Detailed explanation

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