On carrying out the electrolysis of acidified water, the volume of hydrogen liberated at `STP` condition is `22.4L`. The volume of oxygen liberated is
a.22.4 L
b.44.8 L
c.11.2 L
d.2.24 L

Options

A. `22.4L`
B. `44.8L`
C. `11.2L`
D. `2.24L`

Explanation

To solve the problem of determining the volume of oxygen liberated during the electrolysis of acidified water when 22.4 L of hydrogen is produced, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electrolysis Reaction**:
The electrolysis of water can be represented by the following half-reactions:
- At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
- At the anode (oxidation):
\[ 4OH^- \rightarrow 2H_2O + O_2(g) + 4e^- \]

2. **Combine the Half-Reactions**:
The overall reaction for the electrolysis of water can be written as:
\[ 2H_2O \rightarrow 2H_2(g) + O_2(g) \]
This shows that for every 2 moles of hydrogen gas produced, 1 mole of oxygen gas is produced.

3. **Determine the Mole Ratio**:
From the balanced equation, we can see that the mole ratio of hydrogen to oxygen is:
\[ 2:1 \]
This means that for every 2 volumes of hydrogen produced, 1 volume of oxygen is produced.

4. **Calculate the Volume of Oxygen**:
Given that the volume of hydrogen produced is 22.4 L, we can use the mole ratio to find the volume of oxygen:
\[ \text{Volume of } O_2 = \frac{22.4 \, \text{L}}{2} = 11.2 \, \text{L} \]

5. **Conclusion**:
Therefore, the volume of oxygen liberated during the electrolysis of acidified water is:
\[ \text{Volume of } O_2 = 11.2 \, \text{L} \]

### Final Answer:
The volume of oxygen liberated is **11.2 L** (Option c).

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