The amount of Faraday required to discharge 4.5 moles of Al$^{3+}$ is

Analyze the question: We need the quantity of electricity (in Faradays) needed to discharge (deposit as metal) 4.5 moles of Al³⁺ ions.

Concept: One Faraday (1 F) is the charge carried by 1 mole of electrons (≈ 96,500 C).

The discharge reaction is:

Al³⁺ + 3e⁻ → Al

So 1 mole of Al³⁺ needs 3 moles of electrons, which is 3 Faradays.

Solution:

• 1 mole of Al³⁺ requires 3 F
• 4.5 moles of Al³⁺ require 4.5 × 3 = 13.5 F

Answer: Option A (13.5).

Tip: Faradays needed = moles of ion × charge on the ion. For example, 2 moles of Cu²⁺ need 2 × 2 = 4 F. Do not pick 4.5 (option B), which comes from forgetting to multiply by the charge of 3.