Electroplating with copper sulphate for science project

A. Hi Tiffany,
I would suggest rechecking the voltage and current because A = V / R and you should not be able to hold both the current and the voltage constant with solutions of varying conductivity unless you are introducing another variable.

Your voltage is high for a beaker [beakers on eBay or Amazon [affil links] plating experiment. The "efficiency" of acid copper sulphate plating should be close to 100% at reasonable currents and voltages. But there should be sulfuric acid in the solution for proper plating (maybe that's not allowed in school). What may be happening is that you are plating at a faster rate than the low concentration solutions can handle, so most of your Faraday's Law current is going to the liberation of hydrogen rather than the deposition of copper, and as you increase the concentration you are moving towards a more reasonable plating rate, and plating more while generating less hydrogen gas (you should be able to see the bubbles if you're generating lots of it).

As for the lower deposition rate at higher concentration it may well be "anode polarization" -- what happens in this case is the copper sulphate concentration gets too high very close to the anode and it interferes with normal ion transport and conductivity.
Luck & Regards,


Ted Mooney, P.E. RET
Striving to live Aloha
finishing.com - Pine Beach, New Jersey

A. Hi again. Based on your questions, here is what I'd like you to think about --

Current (amperage) is actually the measure of how many electrons you are pumping through your power supply and copper wiring per unit of time. One way of viewing Faraday's Law is that each of those electrons you are pumping must be matched by an atom getting stripped of its electrons, migrating through the solution over to the cathode, and recapturing its electrons. So for each two electrons you move, you reduce one atom of Copper on the anode to a Cu++ ion, it migrates to the cathode, re-captures those two electrons and becomes an atom of copper on the cathode.

So, if you hold the current and time constant, the amount of copper you deposit should remain constant regardless of concentration, voltage, temperature, or anything else ... EXCEPT ... Faraday's Law doesn't actually say how much copper must deposit, it says how much ionic transfer must occur. Any difference is accounted for by some other ionic activity, and the one that matters in electroplating is the liberation of hydrogen gas from water. If there are not sufficient Cu++ ions at the cathode to accept all the electrons you are pumping there, what will happen is that water will split into hydrogen gas and hydroxide ions to make up the deficit:

2 H2O + 2e- ⇨ H2(gas) + 2 OH-

i.e., 2 'excess' electrons will split two molecules of water into one atom of hydrogen gas and two ions of hydroxide.

Usually, operating under proper conditions, copper plating is nearly 100% efficient (nearly 100% of the electricity deposits copper, and very little is 'wasted' generating hydrogen gas. But under poor deposition conditions, a significant amount can be wasted, and your experiment seems to be doing that: demonstrating the waste or inefficiency.

The above should make you realize that voltage is not actually a control factor here. What an electroplater actually tries to do is to keep the current low enough that s/he gets essentially 100% efficiency. What I can tell you about your voltage is that 5 v is higher than appropriate for a 250 ml beaker. Even at 1-1/2 volts I think you generate significant hydrogen. My guess is that your Vernier power supply ⇦this on eBay] is only putting out something like 2 volts to maintain the 0.6A.

Usually available on eBay;

sometimes on
AbeBooks or Amazon

(affil links)
free pdf is currently available from academia.edu

A. Hi Colleen,
It's unfortunate that students do this particular experiment -- correlating copper deposition quantity to copper concentration -- because there is actually no dependable relationship :-(
Demonstrating the truth of Faraday's Law would probably be a better experiment.

But yes, the main half call reaction at the cathode is reduction per your equation. The undesired half-cell reaction which occurs to a more limited extent is 2H2O + 2e- --> H2^ + 20H- And the main half cell reaction at the anode is the reverse, i.e., Cu - 2e- --> Cu2+. The undesired half-cell reaction which occurs to a more limited extent is 2H20 - 4e- --> 02^ + 4H+
Luck & Regards,


Ted Mooney, P.E. RET
Striving to live Aloha
finishing.com - Pine Beach, New Jersey

A. Not a stupid question at all, Melissa. You are very industrious to plate 100 samples, good for you, and your results are exactly correct! Here is what happens--

The copper ions are sitting in solution, doing essentially nothing until you apply electricity. Then, when you apply electricity and a copper ion is touching your nickel cathode, two electrons flow into it reducing it from a Cu⁺⁺ ion to a Cu⁰ metal atom. So the amount of copper deposited is proportional to the number of electrons that flowed, i.e., the coulombs or ampere-seconds, NOT the copper concentration. This is Faraday's Law, and we're very happy you demonstrated it so well.

There is, however, a second reaction that occurs to a much lesser degree. Those two electrons can sometimes instead be wasted in the liberation of hydrogen from water,
2H₂O + 2e- => H₂ + 2OH-.

If 100 percent of your electricity goes into depositing copper, the process is said to be operating at 100 percent efficiency. If 10 percent of the electricity goes into liberating hydrogen from water, the process is operating at 90 percent efficiency. At high copper concentrations and reasonable current, like you used, copper plating is close to 100 percent efficient because it is much easier to reduce copper than hydrogen; but if you lower the copper concentration way down, the efficiency will drop off. You can understand this by remembering that I started by saying "a copper ion is touching your nickel cathode". Suppose that the concentration is so low that at a given instant no copper ions are near or touching the cathode? The electricity still needs to flow, and will do so by liberating hydrogen from the water.

You did a great experiment and demonstrated Faraday's Law -- with the great 'control' of showing that copper concentration within reasonable limits has no effect on plating rate! (but had you run another trial with a greatly reduced copper concentration, you might have seen the reduced efficiency that results from too little copper in solution).

A. Hi, Dan. The black deposit you are seeing is called "burning" or "smut", and I can tell you basically what it is, but it's hard to put a fine point on it in terms of percentages, because it is the result of a process that is out of control and therefore erratic and unpredictable. One component of it is a very finely deposited copper powder (if powders are fine enough they tend to look black because the tiny balls bounce light away at skew angles instead of reflecting it back.

It is caused, as previously mentioned, by trying to plate at a higher current than the solution can sustain. The electricity must take two paths to complete the circuit, electrons flow through the copper wire, while positively charged copper ions flow through the solution. If there isn't enough copper in the solution to sustain the current you are passing through the wire, the instant a copper ion reaches the cathode it is instantly reduced to a spec of powder instead of slowly growing a proper grain structure. Additionally, water is hydrolyzed into hydrogen and oxygen to try to carry the current.

It is possible to successfully plate copper with a low concentration of copper in the solution, but you must plate at a very low voltage and consequently slower so the small concentration of copper ions is sufficient to match the electrons you are passing through the wire. Good luck.

Regards,

A. Hi Aditi. Sorry, but your experiment will not be successful because electroplating copper onto a steel rod is a problem. Copper is more noble than steel, so copper will deposit onto the rod without any electricity applied. As for the reaction involved, the copper sulphate ionizes into Cu++ and SO4- -, and the Cu++ ions migrate over to the steel surface. When they reach it, they "steal" electrons from the steel because they are more noble than the steel is. Thus the copper ions become neutralized and deposit as copper metal onto the rod, while simultaneously the iron (steel) atoms lose their electrons and ionize into solution. When the steel surface is completely covered with copper metal, the reaction will stop.

However, if you apply electricity at that point, you can make it continue. You'll pump electrons to the cathode, where they will continue to reduce Cu++ ions to copper metal, and the electrons you pulled from the anode will cause the copper atoms of the anode to dissolve into solution as Cu++ ions.

Regards,

A. Hi Robert. 3500 parts per year is only 15-20 per working day, so your production needs are not far beyond an experimental laboratory. But the only practical way to get the copper on them is probably electroplating, and these thicknesses are in the range of usual electroforming thicknesses and will likely require a plating time of almost a full day.

Before plating, the parts will require electrocleaning, rinsing, acid dip neutralization, rinsing, Wood's Nickel Striking, and rinsing before the copper plating. A possible alternative for less space is sandblasting followed by rinsing and Woods's Nickel Striking.

The plating is thick enough that maybe, depending on the shape of the parts, you can rely on a shrink-wrap cohesion effect more than adhesion to keep the copper attached to the parts. So maybe you can just sandblast the parts, rinse off the dust, and quickly start copper plating without the strike. (Must the nickel plating be retained?)

Years ago I was involved with the re-plating of the ironwork of Philadelphia city hall's William Penn clock tower; the original plating (acid copper plating directly onto iron & steel) was in the same general thickness range as you are contemplating and worked for many decades ... but once it became perforated there was rapid horrible undercutting corrosion because there was zero adhesion.

Regards,

A. Hi again Robert. If at all possible you should be buying your plating solutions from a reputable plating process provider. If that is impossible, then you need to at least build a good plating library so you can research the matter.

Old nickel plating is difficult to activate for good adhesion even with a Wood's Nickel Strike. And chemically removing it without proprietaries or creating an environmental mess with nitric acid is difficult; blasting is better. Personally I would guess the best procedure for good adhesion on springs with old nickel plating would be to blast the old nickel off, then either nickel strike or cyanide copper strike, then acid copper plate.

But you don't build a plating plant based on theory. You need to try these things before you build a plant to do them.

Regards,

A. Hi Anand. Sorry to have to say this but, unfortunately, your experiment is a poor one. First, the variable that you are varying (acid concentration) is far removed from the heart of the matter of what electroplating is really about as a science. Secondly your constants (like 9V battery) are poorly chosen (9V is WAY too much; 3 V is more than enough, probably too much). Thirdly, a scale which is incapable of measuring what you are need to measure really isn't much use. Apologies :-(

Please see our paper "Faraday's Law of Electrolysis for Plating" which explains what the science of electroplating is really about. If you read and understand it, I think you will immediately rethink your experiment and figure out a better way to structure it. Good luck.

Regards,

A. Hi Kiran, that's an ambitious science project! I didn't carefully check your numbers, but they look about 2X too high. Remember that in acid copper sulphate, the oxidation state/valence of the copper is +2, which means you need to move 2 electrons through the power supply and wiring, not 1, for each ion of copper you want to transfer from anode to cathode through the solution.

But don't use an iron cathode because copper, being more noble than iron, will spontaneously deposit a thin layer on it without electricity, slightly changing the data. I would suggest just using two copper electrodes and carefully measuring the weight of both, and the weight of the solution before and after the experiment (if the solution & beaker aren't too heavy for your scale), which will offer checks and balances against measurement errors. In the unlikely event that the total of the three changes, he'll have another thing to consider & report -- that some of the weight might have evaporated as water or turned to hydrogen and floated off :-)

I don't know the surface area of those 4 inch long bars, and you probably only want to submerge 3 to 3-1/2 inches in the solution, because you won't want the connection submerged in the solution and possibly corrupting the situation. But I wouldn't suggest trying to plate at more than about 10 amps/square foot maximum, so I think the 1.25 Amps is probably high. Shooting for less copper deposited or plating for 12 or 24 hours might be better.

A beaker [beakers on eBay or Amazon [affil links] of 300-500 mL would be ideal because he can also record and track the solution level, to see if some was lost to evaporation or hydrogen. I'd get the beaker first and see how much solution corresponds to a 3 to 3-1/2" level (but obviously less than 300-500 mL).

With the before & after weight of the anode, the before & after weight of the cathode, the before & after weight of the solution, the ampere-hours of electricity, and the before & after solution level he should have outstanding data for his terrific experiment.

Luck & Regards,

A. Hi again Kiran. I have never actually done the experiment, but:

1. When people talk about buying or using copper sulphate, they usually actually mean copper sulphate pentahydrate, CuSO4·5H20. This is the blue crystals or powders you generally see. It is quite soluble, so he should be able to readily dissolve it in water. The advantage of using crystals or powder over buying a liquid is that you'll know how much actual copper sulphate you have, whereas if you buy a liquid then you'll have to trust the percentage from the label to know how much copper sulphate you have versus water.
A standard copper plating bath is about 210 grams per liter but you can operate at much lower concentration, and you may not be able to dissolve quite that much anyway, so buying 100 grams for a 300-500 mL beaker should be more than enough. You're also supposed to have about 30 grams per liter of sulfuric acid in your plating bath, but you didn't tell us how old your son is so I don't know if that's appropriate. You may want to just use vinegar ⇦in bulk on eBay or Amazon [affil links] instead. In that case, I'd probably use about 1 part vinegar to 4 parts water (200 grams per liter).

2. 0.5 Amp sounds okay to me.

3. Your understanding is correct, but you don't really even have to worry about the sulphate to understand it: when your power supply strips two electrons from the anode and pushes them through the copper wiring to the cathode, what used to be an atom of Cu⁰ on the anode has now been turned into a Cu⁺² ion in solution. It will migrate over to the cathode where it re-meets those 2 electrons and becomes an atom of copper again.

4. The anode weight and solution weight don't factor into his calculation for the weight gain at the cathode; that's a free-standing measurement; sorry for confusing/misleading you ...
But, just as one atom of copper should deposit on the cathode for each 2 electrons transferred, one atom of copper should be removed from the anode. So the weight loss from the anode should match the weight gain on the cathode -- just a built-in check.
My suggestion about weighing the solution and watching for evaporation was probably overkill, I only mentioned it because if the weight gain of the cathode doesn't match the weight loss from the anode, the only explanation is that there is either more or less copper dissolved in the solution at the end of the test than at the beginning; probably just skip that and weigh the anode and cathode before and after rather than going nuts with data :-)

Luck & Regards,

A. Please do not forget the 2 most important ingredients of the experiment: safety glasses ⇦this on eBay or Amazon [affil links] and gloves ⇦this on eBay or Amazon [affil links] .

Thank you Mr. Mooney so very much for such detailed guidance!

I think my son now has ample clarity to embark on his experiment.

Per Leon's advice, I will be sure to ensure he keeps his safety gloves and glasses at all times ;-)