Question

How much charge in coulombs is required for the reduction of one mole of Al³⁺ to Al?

Options

• 1.930 × 10⁵ C

• 2.895 × 10⁵ C

• 2.895 × 10⁴ C

• 1.930 × 10⁴ C

Answer 1

2.895 × 10⁵ C

Explanation:

\[\ce{Al^{3+} + 3e^- -> Al}\]

1 mole of electron = 1 F = 96500 coulombs

∴ 3e⁻ = 3F = 3 × 96500 = 2,89,500 = 2.895 × 10⁵ coulombs