If 5 amps of current are used in an electrolytic cell, how long would it take to produce;?

a) 50 grams of Na ________________
b) 50 grams of Mg ________________

2 thoughts on “If 5 amps of current are used in an electrolytic cell, how long would it take to produce;?”

1. Zor Prime says:

Lancenigo di Villorba (TV), Italy

The FIRST FARADAY’s LAW affirm that :”the electrical charge flown in an electrical circuit results proportional to mass of the chemical species processed at electrodes”.

CASE A
Well, 50 grams of Metallic Sodium results a mass corrensponding to

50 / 23 = 2.1 atom-grams

and it arrives from the following half-rection

Na+ + e —> Na

Since ONE ELECTRON LEADS TO ONE SODIUM ATOM, the Faraday’s Law is applied on the basis of this “1Electron : 1Sodium” Ratio

Sodium Atom-grams * 1 = Current * Time
2.1 * 1 = 5 * Time

and I get

Time = 2.1 / 5 = 0.43 s

CASE B
Well, 50 grams of Metallic Potassium results a mass corrensponding to

50 / 39 = 1.3 atom-grams

and it arrives from the following half-rection

K+ + e —> K

Since ONE ELECTRON LEADS TO ONE POTASSIUM ATOM, the Faraday’s Law is applied on the basis of this “1Electron : 1Potassium” Ratio

Potassium Atom-grams * 1 = Current * Time
1.3 * 1 = 5 * Time

and I get

Time = 1.3 / 5 = 0.26 s

I hope this helps you.

2. David B says:

Zor is forgetting to use the Faraday constant (F) 96.5 C/mol e-.

a) 50g of Na is 2.17 mol Na. To produce 2.17 mol of Na from Na+ will take 2.17×96.5 = 210 coulombs.

now #coulombs = current x time, so

time = 210/5 = 42.0 seconds

b) 50g of Mg is 2.06 mol Mg. To produce this from Mg2+ will take 2×2.06×96.5 = 397 coulombs (since it takes two mol e- to produce one mol Mg)

so

time = 397/5 = 79.4 s

edit: My mistake F is 96500 C/mol e-. So the answers are 1000 times larger a) is 43000 s or 11.9 hours and b) is 79400s or 66.2 hours.