During electrolysis of aqueous 4, gNaOH of `O_(2)` gas is liberated at NTP at anode, `H_(2)` gas liberated at cathode is

Options

A. 2.8L
B. 5.6L
C. 11.2L
D. 22.4L

Explanation

To solve the problem of determining the volume of hydrogen gas liberated at the cathode during the electrolysis of aqueous NaOH when 4 g of O₂ gas is liberated at the anode, we can follow these steps:

### Step 1: Write the half-reaction at the anode
At the anode, hydroxide ions (OH⁻) are oxidized to produce oxygen gas (O₂). The balanced half-reaction is:
\[ 2 \text{OH}^- \rightarrow \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 + 2 e^- \]

### Step 2: Calculate the moles of O₂ produced
The molecular weight of O₂ is 32 g/mol. Given that 4 g of O₂ is liberated, we can calculate the number of moles of O₂ produced:
\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{32 \text{ g/mol}} = 0.125 \text{ mol} \]

### Step 3: Determine the number of moles of H₂ produced
From the balanced half-reaction, we see that for every 1 mole of O₂ produced, 2 moles of H₂ gas are produced at the cathode. Therefore, the moles of H₂ produced can be calculated as follows:
\[ \text{Moles of } H_2 = 2 \times \text{Moles of } O_2 = 2 \times 0.125 \text{ mol} = 0.25 \text{ mol} \]

### Step 4: Calculate the volume of H₂ produced at NTP
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of H₂ produced can be calculated as:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \text{ L/mol} = 0.25 \text{ mol} \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

### Conclusion
The volume of hydrogen gas liberated at the cathode during the electrolysis of aqueous NaOH, when 4 g of O₂ is liberated at the anode, is **5.6 liters**.

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