Electrolysis of Aqueous NaCl

To determine the volume of Cl2 gas produced in the time it takes to liberate 5 L of H2 gas during the electrolysis of aqueous NaCl, we need to consider the stoichiometry of the reaction and the molar volumes of the gases involved.

1. Electrolysis of Aqueous NaCl

During electrolysis, an electric current is passed through a solution to drive a chemical reaction. In the case of aqueous NaCl, the electrolysis process occurs as follows:

At the anode (positive electrode):
2 Cl- (aq) → Cl2 (g) + 2 e-

At the cathode (negative electrode):
2 H2O (l) + 2 e- → H2 (g) + 2 OH- (aq)

The overall reaction is:
2 NaCl (aq) + 2 H2O (l) → Cl2 (g) + H2 (g) + 2 NaOH (aq)

2. Molar Volume of Gases

At STP (standard temperature and pressure), 1 mole of any gas occupies a volume of 22.4 L. This value is derived from the ideal gas law and is applicable to gases at standard conditions.

3. Calculation

Since the reaction stoichiometry shows that the molar ratio between Cl2 and H2 is 1:1, we can use the molar volume of H2 gas to determine the volume of Cl2 gas produced.

Given:
Volume of H2 gas liberated = 5 L

Using the molar volume of gases at STP, we can determine the number of moles of H2 gas liberated:
n(H2) = V(H2) / Vm(H2)
n(H2) = 5 L / 22.4 L/mol = 0.2232 mol

Since the stoichiometry of the reaction indicates that the molar ratio of H2 to Cl2 is 1:1, the number of moles of Cl2 gas produced will be the same as the number of moles of H2 gas liberated.

To determine the volume of Cl2 gas, we can use the molar volume at STP:
V(Cl2) = n(Cl2) x Vm(Cl2)
V(Cl2) = 0.2232 mol x 22.4 L/mol = 5 L

Therefore, the volume of Cl2 gas produced in the time it takes to liberate 5 L of H2 gas during the electrolysis of aqueous NaCl is also 5 L.