Question

$4.5\, g$ of aluminium (at. mass $27$ amu) is deposited at cathode from a molten electrolyte containing $Al^{3+}$ ions by a certain quantity of electric charge.The volume of hydrogen produced at STP from $H^+$ ions in a solution by the same quantity of electric charge will be

• A. 44.8 L
• B. 11.2 L
• C. 22.4 L
• D. 5.6 L

Answer 1

The Correct Option is D

To solve the problem of determining the volume of hydrogen produced using the same quantity of electric charge that deposits 4.5 g of aluminum, we need to use Faraday's laws of electrolysis and the stoichiometry of the reactions involved.

1. First, determine the moles of Al deposited. The atomic mass of aluminum is 27 g/mol. Thus, the number of moles of aluminum is: \text{moles of } Al = \frac{4.5 \, \text{g}}{27 \, \text{g/mol}} = 0.1667 \, \text{mol}.
2. According to Faraday's laws, the amount of substance deposited or liberated is directly proportional to the quantity of charge passed through the electrolyte. The reaction at the cathode for aluminum depositing is: Al^{3+} + 3e^- \rightarrow Al.
3. Thus, 1 mole of Al requires 3 moles of electrons. Therefore, 0.1667 mol of Al requires: 0.1667 \times 3 = 0.5001 \, \text{mol of electrons}.
4. The same quantity of electric charge is used to produce hydrogen. The reaction at the cathode for hydrogen gas evolution from H^+ ions is: 2H^{+} + 2e^- \rightarrow H_2.
5. This means 2 moles of electrons will produce 1 mole of H_2. Therefore, 0.5001 mol of electrons will produce: \frac{0.5001}{2} = 0.25005 \, \text{mol of } H_2.
6. Lastly, calculate the volume of H_2 at STP. At STP, 1 mole of gas occupies 22.4 L. So, 0.25005 mol of H_2 will occupy: 0.25005 \times 22.4 \, \text{L} = 5.6 \, \text{L}.

Therefore, the volume of hydrogen produced at STP from H^+ ions with the same quantity of electric charge is 5.6 L.