Concept:
Electrolysis of acidified water involves the decomposition of water into its constituent gases, hydrogen and oxygen, by passing an electric current through it. The amount of hydrogen and oxygen liberated during electrolysis can be calculated using Faraday's laws of electrolysis.

Faraday's laws of electrolysis:
1. The amount of any substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through it.
2. The amount of any substance liberated during electrolysis is directly proportional to its chemical equivalent weight.

Calculation:
1. Chemical equation for electrolysis of water:
2H2O(l) → 2H2(g) + O2(g)

2. From the equation, it can be seen that for every 2 moles of hydrogen gas liberated, 1 mole of oxygen gas is liberated.

3. The volume of gas liberated can be converted to its corresponding number of moles using the ideal gas law:
PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant, and T is the temperature of the gas.

4. Assuming the temperature and pressure are constant, the number of moles of hydrogen gas liberated can be calculated using the volume of hydrogen gas liberated:
n(H2) = PV/RT

5. Using the stoichiometry of the reaction, the number of moles of oxygen gas liberated can be calculated:
n(O2) = n(H2)/2

6. The volume of oxygen gas liberated can then be calculated using the ideal gas law:
V(O2) = n(O2)RT/P

7. Substituting the given values:
V(H2) = 5.6 cm^3
P = 1 atm
T = 298 K
R = 0.0821 L atm mol^-1 K^-1

n(H2) = (1 atm x 5.6 cm^3)/(0.0821 L atm mol^-1 K^-1 x 298 K) = 0.000233 mol

n(O2) = 0.000233 mol/2 = 0.0001165 mol

V(O2) = (0.0001165 mol x 0.0821 L atm mol^-1 K^-1 x 298 K)/(1 atm) = 0.0029 L = 2.9 cm^3

Answer:
The volume of oxygen gas liberated is 2.9 cm^3.

Volume of oxygen librated = 1/2 of v H2.
Bole to 2.8 cm^3