Electrolysis of water is a decomposition reaction. The mole ratio of hydrogen and oxygen gases liberated during electrolysis of water is

Options

A. `1 : 1`
B. `2 : 1`
C. `4 : 1`
D. `1 : 2`

Explanation

To determine the mole ratio of hydrogen and oxygen gases liberated during the electrolysis of water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process of Electrolysis**:
- Electrolysis is a process where electrical energy is used to drive a non-spontaneous chemical reaction. In the case of water, when electricity is passed through water, it decomposes into hydrogen and oxygen gases.

2. **Write the Chemical Reaction**:
- The chemical reaction for the electrolysis of water can be represented as:
\[ 2H_2O \rightarrow 2H_2 + O_2 \]
- This equation shows that two molecules of water (H₂O) decompose to produce two molecules of hydrogen gas (H₂) and one molecule of oxygen gas (O₂).

3. **Balance the Reaction**:
- The equation is already balanced, as there are equal numbers of each type of atom on both sides of the equation:
- Left side: 2 H (from 2 H₂O) and 2 O (from 2 H₂O)
- Right side: 4 H (from 2 H₂) and 2 O (from O₂)

4. **Determine the Mole Ratio**:
- From the balanced equation, we can see that for every 2 moles of hydrogen gas produced, 1 mole of oxygen gas is produced.
- Therefore, the mole ratio of hydrogen to oxygen is:
\[ \text{Mole Ratio} = \frac{2 \text{ moles of } H_2}{1 \text{ mole of } O_2} = 2:1 \]

5. **Conclusion**:
- The mole ratio of hydrogen to oxygen gases liberated during the electrolysis of water is 2:1.

### Final Answer:
The mole ratio of hydrogen and oxygen gases liberated during the electrolysis of water is **2:1**.