Question

A current of 0.5 ampere is passed through molten AlCl\(_3\) for 96.5 seconds. The mass of aluminium deposited at cathode is x mg and volume of chlorine liberated (at STP) at anode is y mL. x and y are respectively

• A. 18.0, 22.4
• B. 13.5, 16.8
• C. 9.0, 11.2
• D. 4.5, 5.6

Hint:

Faraday's laws problems can be solved systematically: 1. Calculate total charge: \(Q = It\). 2. Calculate moles of electrons: \(n_e = Q/F\). 3. Use the stoichiometry of the half-reaction to find moles of product. 4. Convert moles of product to mass (using molar mass) or volume (using molar volume at STP).

Answer 1

The Correct Option is D

Step 1: Calculate the total charge passed through the electrolyte.
Charge (Q) = Current (I) \(\times\) Time (t).
\( Q = 0.5 \text{ A} \times 96.5 \text{ s} = 48.25 \) C.
Step 2: Calculate the number of moles of electrons transferred.
One Faraday (F) is the charge of one mole of electrons, which is approximately 96500 C/mol.
Moles of electrons = \( \frac{Q}{F} = \frac{48.25}{96500} = \frac{4825}{9650000} = \frac{1}{2000} = 0.0005 \) moles.
Step 3: Calculate the mass of Aluminium deposited (x).
The reaction at the cathode is: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \).
This shows that 3 moles of electrons are required to deposit 1 mole of Aluminium.
Moles of Al deposited = \( \frac{\text{moles of electrons}}{3} = \frac{0.0005}{3} \) moles.
Molar mass of Al is 27 g/mol.
Mass of Al = Moles \(\times\) Molar mass = \( \frac{0.0005}{3} \times 27 = 0.0005 \times 9 = 0.0045 \) g.
The question asks for the mass x in milligrams (mg).
\( x = 0.0045 \text{ g} = 4.5 \text{ mg} \).
Step 4: Calculate the volume of Chlorine liberated (y).
The reaction at the anode is: \( 2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^- \).
This shows that 2 moles of electrons are produced for every 1 mole of Cl\(_2\) gas.
Moles of Cl\(_2\) liberated = \( \frac{\text{moles of electrons}}{2} = \frac{0.0005}{2} = 0.00025 \) moles.
At STP, 1 mole of any gas occupies 22400 mL.
Volume of Cl\(_2\) (y) = Moles \(\times\) Molar volume at STP.
\( y = 0.00025 \times 22400 = \frac{1}{4000} \times 22400 = \frac{22.4}{4} = 5.6 \) mL.
Therefore, x = 4.5 and y = 5.6.