Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.
6OH⁻ + Cl⁻ → ClO₃⁻ + 3H₂O + 6e⁻
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO₃ = 122.6 g mol⁻¹, F = 96500 C)

Solution

Chemistry ma’am · 3 steps

Six electrons per chlorate ion: get the moles of KClO₃, find the charge, then divide by time.

1. Moles of chlorate. n(KClO₃) = 10.0/122.6 = 0.0816 mol. Each ClO₃⁻ ion needs 6 electrons, so the moles of electrons are 6 × 0.0816 = 0.489 mol.
2. Charge needed. Each mole of electrons carries one faraday, so Q = 0.489 × 96500 = 4.72 × 10⁴ C.

   I = Q/t = (6 × 10.0/122.6 × 96500)/(10 × 3600) = (4.72 × 10⁴)/36000

3. So. I ≈ 1.31 A, which rounds to 1.

The common slip: 4723. Leaving the time as 10 instead of 36000 s gives about 4723, not 1. Convert hours to seconds before dividing.

The idea to keep. In any electrolysis problem, read the electrons per formula unit straight off the half-equation. Then charge = moles × n × F, and always put time in seconds.

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