A solution of Fe₂(SO₄)₃ is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]
Given : 1 F = 96500 C mol⁻¹
Atomic mass of Fe = 56 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

In Fe₂(SO₄)₃ iron is Fe³⁺, so every Fe atom deposited needs three electrons.

1. Find the charge. In Fe₂(SO₄)₃ the sulphate is SO₄²⁻, so iron is Fe³⁺. At the cathode Fe³⁺ + 3e⁻ → Fe, so 1 mol of Fe needs 3 F.

   Q = m/M × n × F = 0.3482/56 × 3 × 96500

2. Put in numbers. Moles of Fe = 0.3482/56 = 6.218 × 10⁻³ mol. Charge Q = 6.218 × 10⁻³ × 3 × 96500 ≈ 1800 C.
3. Find the time. t = Q/I = 1800/1.5 = 1200 s. Divide by 60 to get minutes: 1200/60 = 20 min.
4. So. The electrolysis runs for 20 minutes.

The common slip: 13. Taking iron as Fe²⁺ (2 electrons per atom) gives 800 s, which is about 13 min, not 20.

The idea to keep. Before using Faraday's law, read the metal's charge from the salt. That charge is the number of electrons per atom, and it multiplies straight into Q = m/M nF.

Answer from the NTA final answer key. Solution by PrepWiser.