I am a high school student and for my chemistry lab, I did an electrolyis experiment on how the changing concentration of electrolyte (copper sulfate) will change the rate at which the copper is deposited on the cathode.

My hypothesis was that as the concentration is increased, so will the rate of electrolysis. However, my results are completely opposite. When I use 0.1 M solution, loads of copper is deposited on the cathode but as I increase the concentration, the amount of copper deposited becomes progressively less and less. With 1 M solution, there was almost no copper being deposited.

This makes no sense to me. Could somebody explain what is going on? Am I doing something wrong?

I used both copper electrodes. I did not use distilled water, just normal tap water. I used 12 V power supply.


I tried adding a little sulfuric acid (making the pH around 3) to the copper sulfate but it did not make any difference on electrolysis.

I tried to observe the electrolysis very closely today and noticed that there was some sort of oily liquid falling off from the anode. Also, I noticed that initially some flaky brown/black substance fell off from the anode (which may or may not be copper)and soon there was a silvery layer formed on the anode.

Also, I checked using Hanna checker that the overall pH of the solution was around 3 but the pH near the anode read between 9 and 14.

Also, I noticed that when I used a higher concentration, the copper deposition on the cathode was brown and clearly copper. But when I used lower concentration, even though the amount of deposit was a lot more, the substance was probably not pure copper as I had assumed earlier. It was black and flaky. It was certainly different to the deposit I observed when I used higher concentration of copper sulfate.

Does anybody know now what is happening? Is there anything that I can do to fix the problem?


I'm thinking maybe at the higher concentrations your tap water is making a film of $\ce{Cu(OH)2}$ on the surface of your electrode, and that is what's shutting down your electroplating. Your tap water might be mildly alkaline (tap water pH's can range from about $6.5$ to as high as $9.5$).

Try adding a little sulfuric acid to your solutions. If alkaline films are the problem, that will fix it.

It will also increase the conductivity of your solutions tremendously, because hydrogen ions are extremely mobile.

The solubility of copper sulfate decreases with increasing sulfuric acid concentration, so don't add too much acid.

  • $\begingroup$ Thanks. Today I have chemistry class and will definitely give sulphuric acid a go. Do you think around 5 ml of diluted acid to 200ml CuS04 will be enough? $\endgroup$ – Joe Slater Jan 29 '15 at 6:07
  • $\begingroup$ I don't know, it depends on the concentration of the acid. If you bring the pH down to around 3 to 5, it should work. $\endgroup$ – Fred Senese Jan 29 '15 at 6:11
  • $\begingroup$ Can you please look at the updates in the question and help me. $\endgroup$ – Joe Slater Jan 29 '15 at 18:05

Two things are very important in electrochemistry.

1- Cleanliness: Your tap water contains all sorts of impurities that are probably electroactive. My recommendation, take some tap water, don't add any $\ce{CuSO4}$ and repeat the experiment. If you see any deposit, that is not copper. I won't even attempt to make any guesses as to what that deposit may be.

2- Potential control: This is very tricky in the type of setup you are using. In two electrode experiments like you are doing, the solution resistance, the distance between the electrodes and the current you are drawing all effect the actual potential your electrode experiences. With the constant 12V you are applying, the current that is drawn is also a function of the above parameters, so it is very hard to determine what the actual electrochemical potential will be.

Gamry has an excellent application note about 2,3 and 4 electrode experiments here: http://www.gamry.com/application-notes/two-three-and-four-electrode-experiments/ if you would like to read up on the details.

  • $\begingroup$ I think the cleanliness of the tap water was probably the reason for the wierd results. In the end, I just decided to do another experiment seeing this one did not work. If I get a chance, I will definitely try electrolysis on only tap water and see what happens. Thanks very much. $\endgroup$ – Joe Slater Feb 24 '15 at 12:43

Kohlrausch law states that $\Lambda_m=\Lambda_m^{\circ}-\kappa \sqrt{c}$ with $\Lambda_m = \frac{k}{c}$ (in which $k$ is the electrical conductivity) for strong electrolytes and low concentrations, which is due to the strong interaction of ions. In that light your findings are not inconsistent. I am not sure though about the magnitude of $\kappa$ and whether for the conditions (copper sulfate as electrolyte, 0.1M to 1.0M concentration) you listed the law holds. At least the point I want to make is that it is not obvious or even correct to assume that conductivity depends linearly on the concentration of the electrolyte or that there is even a positive relation at high concentrations.

This is definitely not intended as a definite answer (nice tautology).

  • $\begingroup$ Is copper sulphate a strong electrolyte? Also, I read on this website: antoine.frostburg.edu/chem/senese/101/redox/faq/… $\endgroup$ – Joe Slater Jan 29 '15 at 6:12
  • $\begingroup$ that increasing copper sulphate concentration upto 1.0 M should increase deposition of copper as per Nernst equation. $\endgroup$ – Joe Slater Jan 29 '15 at 6:13
  • $\begingroup$ @Joe Slater, that's correct, it should (but of course I'd say that, that's me in the link). $\endgroup$ – Fred Senese Jan 29 '15 at 6:24
  • $\begingroup$ @FredSenese wow. What a coincidence. Thanks for your website. It is quite helpful. So kohlrausch law does hold true for 1 M copper sulphate? Are Nernst equation and kohlrausc law derived from one another? $\endgroup$ – Joe Slater Jan 29 '15 at 6:27
  • $\begingroup$ No, they're different things. Kohlrausch's law only works for strong electrolytes that are fairly dilute, as Jori says. Behavior in concentrated solutions is complicated; the concentration is different from the activity (an effective thermodynamic concentration) in concentrated solutions, and you'll also get ion pair formation (direct hookups of $\rm Cu^{+2}$ and $\rm SO_4^{-2}$, in this case). $\endgroup$ – Fred Senese Jan 29 '15 at 6:38

The solute must ionize, and as the concentration of ions in increased, further ionization is inhibited. The non-ionized molecules are inert, and interfere with reaction rate.


Whatever contamination in your tap water should be a great deal less than the concentration of CuSO4, but might be a factor in your experiment.

Conductivity is measured using a non-depositing AC current. The DC resistance or its inverse will depend on many factors, but all will require a measurement of voltage and current simultaneously.

Measuring the rate of electrolysis by visual means is inadequate, since it is obvious to you that the character of the deposit of copper changes. A better measure is determination of mass of copper deposited - but the accuracy required for a short experiment requires a very sensitive balance - and, unfortunately, since copper is sometimes flaking off, measuring the cathode is going to be rough. Measuring the loss of copper at the anode could be a better measurement.

But even better (easily more accurate) is measuring the current. When current flows, it will be carried by Cu++ and SO4-- ions (unless you add H2SO4, in which case, H+ will also be carrying current). Adding H2SO4 complicates the experiment, because you are electroplating copper and electrolyzing water at the same time.

So here is my suggestion: measure the voltage between the electrodes AND the current flowing, simultaneously, at one concentration of CuSO4 (no H2SO4) in water (how about using distilled water instead of tap water to make your results more universal?). Then dump in some more CuSO4.5H2O and follow the change in current and voltage. Both will change. You may adjust the voltage OR current to equal the original voltage or current. In either case, the apparent DC resistance (R= E/I https://en.wikipedia.org/wiki/Ohm%27s_law) of the solution will decrease.


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