In the electroysis of an aqueous solution of NaOH, 2.8 litre of oxygen gas at NTP was libreated at anode. How much of hydrogen gas was liberated at cathode ?

Options

A. 5.6 litre
B. 66 Paise
C. 560 ml
D. Rs.6.60

Answer

To solve the problem of how much hydrogen gas was liberated at the cathode during the electrolysis of an aqueous solution of NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electrolysis Process:**
- During the electrolysis of sodium hydroxide (NaOH), water is split into oxygen and hydrogen gas.
- At the anode (positive electrode), hydroxide ions (OH⁻) are oxidized to form oxygen gas (O₂).
- At the cathode (negative electrode), water is reduced to form hydrogen gas (H₂).

2. **Identifying the Reaction:**
- The half-reaction at the anode can be represented as:
\[ 4OH⁻ \rightarrow O₂ + 2H₂O + 4e⁻ \]
- The half-reaction at the cathode can be represented as:
\[ 2H₂O + 4e⁻ \rightarrow 2H₂ + 4OH⁻ \]

3. **Determining the Volume Ratio:**
- From the balanced equations, we can see that for every 1 mole of O₂ produced, 2 moles of H₂ are produced.
- This means the volume ratio of hydrogen to oxygen is 2:1.

4. **Calculating the Volume of Hydrogen:**
- We are given that 2.8 liters of oxygen gas (O₂) is liberated at the anode.
- Using the volume ratio, we can calculate the volume of hydrogen (H₂) produced:
\[ \text{Volume of H₂} = 2 \times \text{Volume of O₂} \]
\[ \text{Volume of H₂} = 2 \times 2.8 \, \text{liters} = 5.6 \, \text{liters} \]

5. **Conclusion:**
- Therefore, the volume of hydrogen gas liberated at the cathode is **5.6 liters**.

### Final Answer:
The amount of hydrogen gas liberated at the cathode is **5.6 liters**.

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