\(0.04\, \ F\) of charge will liberate how much of oxygen gas (at STP\) from acidified water?

224 mL

The correct answer is option 3. 224 mL.

To determine the volume of oxygen gas liberated at STP from acidified water by \(0.04\, F\) of charge, we can follow these steps:

When acidified water undergoes electrolysis, oxygen gas is produced at the anode and hydrogen gas at the cathode. The half-reaction for oxygen gas formation is:

\(2H_2O \rightarrow O_2 + 4H^+ + 4e^-\)

This reaction shows that 4 moles of electrons (4 Faradays, \(4\, F\)) are required to liberate 1 mole of \(O_2\).

1 Faraday (\(1\, F\)) of charge corresponds to the liberation of \(\frac{1}{4}\) mole of \(O_2\) (since 4 Faradays are needed to liberate 1 mole of \(O_2\)).

For \(0.04\, F\) of charge, the moles of oxygen liberated are:

\(\text{Moles of } O_2 = 0.04 \, F \times \frac{1 \, \text{mole } O_2}{4 \, F} = 0.01 \, \text{mole of } O_2\)

At STP, 1 mole of any gas occupies 22.4 liters (22,400 mL).

Therefore, the volume of \(O_2\) liberated by \(0.01\) mole is:

\(\text{Volume of } O_2 = 0.01 \, \text{mole} \times 22,400 \, \text{mL/mole} = 224 \, \text{mL}\)

Conclusion: The volume of oxygen gas liberated by \(0.04 \, F\) of charge at STP is 224 mL.

The correct answer is: 3. 224 mL