In the electrolysis of `H_(2)O`, 11.2 litres of `H_(2)` was liberated at cathode at NTP. How much `O_(2)` will be liberated at anode under the same conditions ?

Options

A. 11.2 litres
B. 22.4 litres
C. 32 g
D. 5.6 litres

Answer

To solve the problem of how much oxygen will be liberated at the anode during the electrolysis of water when 11.2 liters of hydrogen is produced at the cathode, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electrolysis Reaction**:
The electrolysis of water can be represented by the following reaction:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]
This means that for every 2 moles of water, 2 moles of hydrogen gas and 1 mole of oxygen gas are produced.

2. **Determine the Moles of Hydrogen Produced**:
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Given that 11.2 liters of hydrogen is produced, we can calculate the number of moles of hydrogen:
\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar Volume}} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \]

3. **Use Stoichiometry to Find Moles of Oxygen**:
From the balanced equation, we see that 2 moles of hydrogen produce 1 mole of oxygen. Therefore, the moles of oxygen produced can be calculated as:
\[ \text{Moles of } O_2 = \frac{0.5 \text{ moles of } H_2}{2} = 0.25 \text{ moles} \]

4. **Calculate the Volume of Oxygen Produced**:
Now, we can find the volume of oxygen produced using the molar volume:
\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times \text{Molar Volume} = 0.25 \text{ moles} \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

### Final Answer:
The volume of oxygen liberated at the anode is **5.6 liters**.