Using good quality reagents, during the electrolysis of $\ce{CuSO4}$ (at $\mathrm{15V ~ \& ~ 2A}$), I have noticed a significant amount of Hydrogen generated at the negative electrode in addition to copper deposition.

This generation occurred at all voltages, which surprised me. I expected a only the reduction of copper in the reaction. The solution was blue, so I know that copper sulfate was still in solution. At first, I thought that the copper might be reacting to the sulfuric acid being generated, but the copper that has fallen off the electrode is not bubbling.

I am using a platinum and copper electrodes.

Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Can you calculate the products of simultaneous electrochemistry reactions?

  • $\begingroup$ which electrodes are used. either platinum/ inert or copper one. $\endgroup$
    – solanki...
    Commented Dec 31, 2015 at 20:24
  • $\begingroup$ Hydrogen is not released. at cathode - copper is deposited and at anode - O2 gas released. $\endgroup$
    – solanki...
    Commented Dec 31, 2015 at 20:33
  • 3
    $\begingroup$ Hydrogen evolution is common at high current densities. The reason is that hydrogen cation and to slightly lower extent hydroxyde anion are highly mobile in water solutions, so hydrogen transport (and hydrolysis of water) is kinetically favoured over salt ions transport (and salt electrolysis). This is a constant nuisance in electronchemical processes and can be avoided only by use of voltages below water electrolysis voltage or use of special cathodes (such as carbon), as hydrogen evolution is kinetically hindered on some cathodes. $\endgroup$
    – permeakra
    Commented Dec 31, 2015 at 22:32
  • $\begingroup$ What books would you recommend for me to learn about this process, permeakra7? -Honestly, I am more interested in understanding the mathematical relationship that predicts the mechanics of this hydrogen transport behavior. All my books predict only Cu and O2 as products. I tried lower voltages and it did seem to diminish the Hydrogen production, but not remove it entirely. I am going to try some different electrodes in addition to my copper/platinum pair tomorrow and see what that does. $\endgroup$
    – Mcruggs
    Commented Jan 1, 2016 at 13:37
  • 3
    $\begingroup$ @Mcruggs: If you're interested in a good fundamental understanding, I would recommend Electrochemical Methods by Bard and Faulkner. There's a whole chapter on electrochemical kinetics (chapter 3 I think) that derives general rate expressions which take mass transport into account. $\endgroup$
    – Sean Doris
    Commented Jan 3, 2016 at 3:17

2 Answers 2


There are a few key factors that I expect are contributing here:

  • $\ce{CuSO4}$ concentration – If your $\left[\ce{Cu^2+}\!\right]$ in solution is low, then that will disfavor deposition, as permeakra noted in a comment. You want at least $25$ to $100~\mathrm{g\over L}$ of copper in the electrolyte.

  • Cell voltage – $15~\mathrm V$ is an exorbitantly high voltage for copper electrodeposition. In order to avoid gassing at the cathode, you'll want to operate down in the $1.5$ to $2.0~\mathrm V$ range, tops.

  • Electrolyte conductivity – you don't mention anywhere that you've added other constituents to try to increase the conductivity.

    • If this is a home experiment, I would try dissolving the $\ce{CuSO4}$ in table or pickling vinegar instead of plain water. That will give you more charge carriers in solution, and also decrease the $\mathrm{pH}$, both of which should help.

    • If this is a school or work experiment with access to proper lab equipment and chemicals disposal services, I would use $1\%$ to $10\%~\ce{H2SO4}$ as the base electrolyte. That will really get your conductivity up, and your cell voltage down.

Finally, on the methods side, I would recommend choosing your voltage in order to initially target a current density in the vicinity of $\sim\!10~\mathrm{mA\over cm^2}$, as a value where I would expect to see little to no $\ce{H2}$ evolution. You can always nudge the voltage up from there as needed.

Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Almost certainly not.

Can you calculate the products of simultaneous electrochemistry reactions?

Not with much certainty. Most of the time the best one can do is make qualitative arguments about what reactions might be happening. Detailed modeling of all reactions occurring at an electrode is still prohibitively difficult, as far as I know.


Hydrogen evolution will be favored by increased hydrogen concentration (lower pH). As the copper plates out, the Cu++ ion is replaced by H+. The very first plating out should be with the least hydrogen, and then hydrogen gas evolved should increase as the electrolysis continues. How can we reduce the pH without precipitating the copper?

I have not done the experiment, but I think that adding a bit of ammonia would convert the H+ to NH4+ and raise the pH enough to reduce hydrogen evolution without precipitating copper (Cu++ --> soluble Cu(NH3)4++). You might get more creative and use other amines.

Cell voltage is important also, as suggested in the comments. High current forces something to deposit; copper is preferred, but if the current demands, hydrogen will deposit also.

If you want to get really inventive, you might consider a cathodic poison that is not incorporated into the copper deposit. Mercury would work, but is a no-no for many reasons; sulfur and arsenic compounds also probably not good; maybe a long chain water-soluble polymer, or maybe even a sugar would coat the copper cathode and interfere with hydrogen deposition without becoming incorporated. Good luck!

  • $\begingroup$ Platinum is expensive. How about using a copper anode? No H2SO4 generated - only more CuSO4! $\endgroup$ Commented Feb 9, 2018 at 15:34

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