If an electrolysis plant operates its electrolytic cells at a total current of 1.0*10⁶ amp, how long will it take to produce one metric ton (one million grams) of Mg(s) from seawater containing Mg²⁺? (1 faraday = 96,485 coulombs)

Question

Quizplus · Accepted Answer

The stoichiometry of the electrolytic process is:

Mg2+(aq) + 2e- --> Mg(s)

Therefore, one faraday of electricity (96,485 coulombs) can produce 1/2 mol of magnesium:

(1 mol Mg)/(2 mol e-) x (2 F) x (96,485 C/F) = 48,242.5 C/mol Mg

The mass of Mg produced by one faraday of electricity is:

(1 mol Mg) x (24.31 g/mol) = 24.31 g

Therefore, the number of faradays required to produce one metric ton (1,000,000 g) of Mg is:

(1,000,000 g)/(24.31 g/F) = 41,136.5 F

The time required to produce this amount of Mg at a current of 1 x 1010 A is:

t = (41,136.5 F)/(1.0 x 1010 A) = 4.11365 x 10-3 s

This is equivalent to 4.11365 ms, or 4.11365 x 10-6 s.

Therefore, the total time required to produce one metric ton of Mg is:

4.11365 x 10-6 s x 41,136.5 F = 169.5 s = 2.825 h

The answer closest to this value is C (2.2 h).

If an Electrolysis Plant Operates Its Electrolytic Cells at a Total