Question

96.5 amperes current is passed through the molten AlCl$_3$ for 100 seconds. The mass of aluminum deposited at the cathode is (Atomic weight of Al = 27 u)

• A. 0.90 g
• B. 0.45 g
• C. 1.35 g
• D. 1.8 g

Hint:

Faraday's first law relates the mass deposited during electrolysis to the charge passed: \(m = \frac{Q \cdot M}{n \cdot F}\), where \(n\) is the number of electrons per ion.

Answer 1

The Correct Option is A

During the electrolysis of molten AlCl$_3$, aluminum is deposited at the cathode: \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\). Thus, 1 mole of Al (27 g) requires 3 moles of electrons. The charge passed is: \[ Q = I \times t = 96.5 \times 100 = 9650 \, \text{C} \] Using Faraday's constant (\(F = 96500 \, \text{C mol}^{-1}\)), the moles of electrons are: \[ \text{Moles of electrons} = \frac{Q}{F} = \frac{9650}{96500} = 0.1 \, \text{mol} \] Moles of Al deposited: \[ \text{Moles of Al} = \frac{\text{Moles of electrons}}{3} = \frac{0.1}{3} = \frac{1}{30} \, \text{mol} \] Mass of Al: \[ \text{Mass} = \frac{1}{30} \times 27 = 0.9 \, \text{g} \] This matches option 1. Thus, the correct answer is 0.90 g.