Chemistry Question. PLEASE ANSWER

The Flying Sloth · May 16, 2016

So, we did an experiment using an electrolytic cell to test the electrodeposition of copper on a stainless steel electrode. we are modifying the voltages applied and the concentration of CuSO4. when we raise the voltage (until 8 V when the copper starts to 'burn') deposition occurs at a faster rate. This is expected, unfortunately we have some very puzzling data to do with the concentrations, namely that when we lower the concentration of CuSO4 more copper is deposited. This should not happen and goes against every chemistry principal I know of in that at lower concentrations the liquid is far less conductive and has less ions in the solution in which case less copper should be dissolved and deposited. We have over 200 data points on this experiment and are finding it impossible to explain why this is happening. Any help would be greatly appreciated.

And yes we did use a photospectrometer to test before and after concentrations with no unexpected results.

sorry if this is hard to understand for most of you.

Thankyou for any information you may have.

EDIT, we are using 1, .5 and .25 M CuSO4

Edited May 16, 2016 by The Flying Sloth
Clarification

FRN · May 26, 2016

tl;dr..

Since I'm not a chemist or anything related to it, I'd like to make a chemistry joke, but all the good ones ARGON.

You_are_a_cunt · May 26, 2016

My Chemistry is a bit iffy, but I'll give it some though.

I do, however, remember using 0.5M CuSO4.

Faraday's Law says you will ideally deposit one gram equivalent weight of metal for every 96,487 ampere-seconds (coulombs) you apply.

When you double the voltage from, say 3 volts to 6 volts you'll probably more or less double the current because of Ohm's Law, I = E / R, so you more or less plate twice as fast.

At 8 volts you will deplete the solution of copper ions in the area of the cathode rather quickly. The electrons you are making available have to go somewhere, with the result that the electrons will reduce the hydrogen in the water to nascent hydrogen gas instead of depositing copper which isn't there. This causes the "burning".

After a bit of digging, I found this: http://antoine.frostburg.edu/chem/senese/101/redox/faq/cu-deposition.shtml

Quote

The concentration of the Cu2+ ion will influence the amount of copper you see deposited. For the reduction of Cu2+ ion at the cathode, Cu2+ (aq) + 2e- = Cu(s). The Nernst equation implies that the cathode's potential is proportional to the log of the molarity of Cu2+:

E = Eo + (RT/nF) ln [Cu2+]

where Eo is the electrode potential under standard conditions (about +0.34 V for this reaction), n is the number of moles of electrons transferred (2 moles of electrons per mole of Cu deposited), F is the charge per mole of electrons transferred (F = 96487 Coulombs/mol), R is the gas law constant (8.31451 J/mol K; isn't it interesting how this constant crops up in places that have nothing to do with ideal gases?) and [Cu2+] is the molarity of the copper(II) ions in the solution.

When [Cu2+ ] is less than 1 M, the log term will be negative, and the electrode potential will be more negative. That means that the lower the concentration of Cu2+, the more negative potential (applied voltage) will be required. For a given applied voltage then, the more dilute the CuSO4 solution is, the less copper will be deposited on the cathode.

This isn't a bad thing, if your objective is purifying the copper. Using a highly concentrated copper(II) ion solution will deposit more copper on the cathode, but the deposit may be spongy or coarse and will not adhere well to the cathode. Commercial electroplating operations keep metal cation concentrations (and current density) low.

Not sure if any of this helps, but it's something.

I could try to talk to my mom and see if she has any idea what's going on, but I can't guarantee anything

The Flying Sloth · May 27, 2016

12 hours ago, revsilverspine said:

My Chemistry is a bit iffy, but I'll give it some though.

I do, however, remember using 0.5M CuSO4.

Faraday's Law says you will ideally deposit one gram equivalent weight of metal for every 96,487 ampere-seconds (coulombs) you apply.

When you double the voltage from, say 3 volts to 6 volts you'll probably more or less double the current because of Ohm's Law, I = E / R, so you more or less plate twice as fast.

At 8 volts you will deplete the solution of copper ions in the area of the cathode rather quickly. The electrons you are making available have to go somewhere, with the result that the electrons will reduce the hydrogen in the water to nascent hydrogen gas instead of depositing copper which isn't there. This causes the "burning".

After a bit of digging, I found this: http://antoine.frostburg.edu/chem/senese/101/redox/faq/cu-deposition.shtml

Not sure if any of this helps, but it's something.

I could try to talk to my mom and see if she has any idea what's going on, but I can't guarantee anything

Thanks, but we understand the voltage part, it is the "more electrodeposition as the concentration goes down" part that we are stuck on. I have been able to devise my own chemical theory for this but it would always be good to not have to disprove Ohm's laws and other related laws.