Potassium chlorate is prepared by the electrolysis of KCl in basic solution
6OH⁻ + Cl⁻ → ClO₃⁻ + 3H₂O + 6e⁻
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO₃ using a current of 2 A is_________.
(Given : F = 96,500 C mol –1 ; molar mass of KCIO₃ = 122 g mol –1 )

Solution

Chemistry ma’am · 4 steps

Find the charge the reaction needs, then remember only 60% of the current is doing that work.

1. Moles of product. n(KClO₃) = 10/122 = 0.0820 mol. The half-reaction shows 6 electrons released for every ClO₃⁻ formed, so you need 6 faradays per mole.
2. Charge needed. Q = 6 × 10/122 × 96500 = 4.746 × 10⁴ C. This is the charge that must actually go into the reaction.
3. Useful current. Only 60% of the 2 A drives this reaction, so the effective current is 0.6 × 2 = 1.2 A.

   t = Q/I_(eff) = (4.746 × 10⁴ C)/(1.2 A) = 3.955 × 10⁴ s

4. So. t = 39549/3600 = 10.99 h, which rounds to 11 hours.

The common slip: 7. Ignoring the 60% efficiency and using the full 2 A gives about 6.6 h, which rounds to 7, not 11.

The idea to keep. Faraday's law gives the charge the reaction needs. Current efficiency only lowers the useful current, so divide by it and the time goes up.

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