How many moles of copper would be deposited by passing 1 Faraday of electricity through a $CuCl_2$ solution?

Analyze the question: We need the moles of Cu deposited when 1 Faraday (1 mole of electrons, about 96,500 C) passes through a CuCl₂ solution.

Concept: Faraday's first law says the amount of substance deposited depends on the charge passed and on the number of electrons needed per ion.

1 Faraday = the charge carried by 1 mole of electrons.

Solution:

In CuCl₂ solution, copper exists as Cu²⁺. At the cathode:

Cu²⁺ + 2e⁻ → Cu

• 2 moles of electrons (2 F) deposit 1 mole of Cu
• So 1 mole of electrons (1 F) deposits 1/2 = 0.5 mole of Cu

(Check by mass: 0.5 mol × 63.5 g/mol ≈ 31.75 g of Cu deposited by 1 F.)

Answer: Option B, 0.5 moles.

Tip: Moles deposited by 1 F = 1 ÷ (charge on the ion).

• Na⁺ → 1 mole
• Cu²⁺ → 0.5 mole
• Al³⁺ → 1/3 mole

Option D (1 mole) would be correct only for a +1 ion, and 2 F would be needed for 1 mole of Cu.