Sciencemadness Discussion Board

How to calculate time required for electrolysis until completion

Random - 27-7-2013 at 08:25

For example if I have battery 9V 200mA and electrolyse a saturated solution 250mL of CuSO4(aq), is it possible to calculate how much time I'll need?

For example I don't know if it's current or voltage that affects the speed as I'm fairly inexperienced here, but would connecting battreies speed up the process? Since I have a lot of rechargeable batteries.

[Edited on 27-7-2013 by Random]

plante1999 - 27-7-2013 at 08:43

Current, it is fairly easy to learn too, would like to explain but got no much time.

Learn coulomb etc...

DutchChemistryBox - 27-7-2013 at 09:30

Assuming that you know something about redox I think that these two formulas will answer allot of your question.

q=I*t
q= electric charge (C)
I = electric current (A)
t = time (s)

n(e)=q/F
n(e) = amount of electrons (mol)
F = faraday constant (96485 C/mol)

ElectroWin - 27-7-2013 at 12:51

Quote: Originally posted by Random  
For example if I have battery 9V 200mA and electrolyse a saturated solution 250mL of CuSO4(aq), is it possible to calculate how much time I'll need?

For example I don't know if it's current or voltage that affects the speed as I'm fairly inexperienced here, but would connecting battreies speed up the process? Since I have a lot of rechargeable batteries.


it's current that affects the speed.

the chemistry is affected by electrolyte, anode type, current density at each electrode, pH, temperature, and diffusion between the half cells. several redox reactions may be occurring in any circumstance. yields vary.

for example, if you have a copper anode, you will get SO2 gas evolved; hypothetically, as
2 Cu + SO4 2- --> 2 CuO + SO2 + 2 e-
or oxygen as
4 OH- --> 2 H2O + O2 + 4 e-

relative yields vary with circumstance.

as you can see, each mole of product will require x moles of electrons to flow, for some yield-dependent value of x.

1 mole = 6.022x10^23
1 coulomb = 6.241x10^18 electrons

1 ampere = 1 coulomb/second

so if x = 1, then you'd need 96506 s or 1.11 days of 1 ampere current to get 1 mole of yield.

probably x > 1. so at your 200 mA current youre looking at
weeks to get sensible yield.

But, note that increasing that current changes the chemistry as well: an example being, that if you expose copper or graphite anodes to more than 10 A/m^2 then they will corrode, whereas platinized titanium can handle much greater current density without corrosion.

so, what are you making?

[Edited on 2013-7-27 by ElectroWin]