Problem#39Exercise on Electrolytic Cells[Solved]

Questions

1-How long it takes to deposit 0.635 g of Cu at the cathode during the electrolysis of CuSO4(aq) solution, if the current used is of 0.200A?  2-      An industrial cell electrolyzes an aqueous solution of NaCl using a current of 100 kiloamps for 90,000 seconds. How many kilograms of chlorine does it produce?

Answers:

1- Cu²⁺(aq)  + 2e  →  Cu(s)

2 moles of electrons are required to produce 1 mole of Cu.

In other words, 2 x 96320 C produce 63.5 g of Cu.

Then the production 0.635 g requires: (2 x 96320 C/63.5g) x 0.635 = 1,926.4C

t = 1,926.4C/0.200A = 9632 s = 160.5 min = 2.68 h

N.B: If you apply directly the formula: t = (m x 2F)/(i x M) = (0.635 x 2 x 96320)/ (0.200 x 63.5) = 9632 s.

2- 2Cl^-(aq)   → Cl₂(g)  + 2e

2 moles of e are required to produce 1 mole of Cl₂; in other words, 2 x 96320 or 192,640 C are required to produce 1 mole or 70.90g of Cl₂.

Quantity of electricity used in the process: It = 100,000 A x 90,000 s = 9 x 10⁹ C

The amount of Cl₂ produced: (70.90 g/192,640 C) x 9 x 10⁹ C = 3,312,396 g = 3,312.396 Kg = 3.312 Metric tonne.