I was reading up on electroplating and I came across some things which really confused me:

Faraday's First Law says that the amount of metal plated onto the cathode is proportionate to the current, so it should be that a higher current= more metal plated onto the cathode (I assume). And from Ohm's Law, a higher current= higher voltage so (by syllogistic reasoning) a higher voltage should give you more metal plated onto the cathode.

However this isn't the case, because I did an experiment recently where the highest voltage didn't produce the highest amount of metal plated onto the cathode. I did some reading on why, and I read (on the internet) that it's because when you increase the current by a lot, the ions formed at the anode can't keep up with the rate of electron flow and electrons just end up combining with other things in solution and no metal forms.

I have some issues with this though; I used a solution with ions already in it, so shouldn't there be no problem for me? Even if the ions cannot be formed at the anode fast enough, I have ions already in solution, so I thought it would be fine. Plus, the experiment was run for only 25 minutes, so I doubt they would've all been used up or anything.

Also, even if it were true that the current was just too high for the ions to keep up; wouldn't the amount of metal formed be at least equal to or greater than the amount formed for lower voltages? It's just that you're forming more metal in less time, ie it gets faster (I think); so you should get the same amount at say 5 V that you do that 3 V, even if towards the end of the experiment in 5 V it gets inefficient and you stop plating. Because the amount of ions or anything doesn't change.

Sorry for the huge amount of writing, I just thought it'd be easier if I explained my thought process. I'd appreciate it a lot if anyone could help me! Thanks so much. :)

  • $\begingroup$ I added some clarification to my answer, is it good enough for you? $\endgroup$
    – permeakra
    Jun 12 '17 at 13:17

When you apply the required voltage for the desired cell, then your desired product is being produced, but if you provide an overpotential then you forced the other reaction to also occur(the undesired one).

For eg. If in a reaction two species can be oxidized then, of the required potential is applied then you will get the required product but if an over potential is applied then the other species will also get oxidized and you will have two oxidation products.


There are two types of control of what reaction may occur: thermodynamic control and kinetic control.

Thermodynamic control dominates in near-equilibrium conditions, which in case of electroplating means small voltages and relatively high concentrations.

When applying high voltage, metal ions in near-electrode space are quickly exhausted and we move from equilibrium conditions to kinetic conditions when the first reaction that may happen occurs. In water this is oxidation/reduction of water.

The main criteria for choice of the voltage in electroplating is so that diffusion of metal ions from solution happened fast enough, but electrolysis of water and/or oxidation of anions didn't happen. This means low voltages, preferably below 1 V when possible. Electrolysis of water may happen if voltage applied is above 1.23 eV, so exceeding this value is allowed only if the procedure and experimental setup is proven to work.

Why electrolysis of water in this setup bad? Leaving aside waste of energy, it creates bubbles of gas stick to the electrodes, resulting (at the very least) in inhomogeneous plating. Also, mind that the transport of ions in water is a diffusion process and electrode potential does not have that much effect on it.

  • $\begingroup$ I still dont understand why people describe blackened product when using higher voltages (say above 3V). Electrolysis of water should just produce hydrogen and oxygen gas. $\endgroup$
    – user391339
    Dec 30 '19 at 7:18
  • $\begingroup$ Also with an excess of the cupric sulfate we are driving the reaction forward by mass action so it must not be in equilibrium. Interesting about the kinetic aspect though is it as if we are raising the temperature higher than what would be possible in water? $\endgroup$
    – user391339
    Dec 30 '19 at 7:29
  • 1
    $\begingroup$ @user391339 most metal electrodes oxidize before water, so instead of oxygen we get metals entering the solution of producing oxides. If the electrodes contain impurities that oxidize harder than the main electrode components, the impurities form a precipitate. $\endgroup$
    – permeakra
    Dec 30 '19 at 17:23

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