I did an experiment with copper sulfate where you place two copper rods in an aqueous solution of copper sulfate. You apply an electric current and copper metal builds up on the cathode. I noticed that the copper anode dissolved. Does the dissolved copper from the anode just turn into more copper sulfate? If not, what does it produce?

This is NOT a homework question. It’s an experiment that I performed out of pure curiosity.

  • $\begingroup$ What happened to the cathode? $\endgroup$ – DrMoishe Pippik Sep 17 '20 at 22:12
  • $\begingroup$ In my question, I said that copper built up on it. I probably should have added that it didn’t dissolve. $\endgroup$ – ElectronicsNoob Sep 17 '20 at 22:35

You have replicated electro-rafination of copper. Copper ions from the being dissolved anode are replacing the copper ions being deposited on the cathode.

Therefore overall electroneutrality of the solution is kept. The local neutrality is managed by electromigration of both ions $\ce{Cu^2+}$ and $\ce{SO4^2+}$.


For the record, to explain more generally some of the possible reactions around the electrolysis of aqueous CuSO4, I first cite this reference:

Whenever copper sulfate or CuSO4 is added to water, it gets dissolved in the water. As CuSO4 is an electrolyte, it splits into Cu+ + (cation) and SO4 − − (anion) ions and move freely in the solution. ... So SO4 − − ions will move towards anode where they give up two electrons and become SO4 radical.

where, more correctly, the sulfate radical is $\ce{.SO4-}$ (see, for example, this source) with the implied electrochemistry:

$\ce{ SO4(2-) = .SO4- + e- }$

[EDIT] And:

$\ce{ 2H2O -> 4H+ + O2 + 4e– }$

where both of the last two reactions are conditional on voltage level and the composition of the electrode (see discussion here) as in a copper anode versus a graphite anode, for example.


$\ce{ Cu = Cu+ + e- }$

[EDIT] where the attack on Cu metal itself forming cuprous (and further cupric) in a system with near-zero voltage at its onset (latter an equilibrium per the Bulter-Volmer equation) only requires a source of H+ and O2 (either from oxygen electrochemical generation here, or from air contact, in a spontaneous electrochemical cell reaction apart from any application of an external current). The alluded to spontaneous electrochemical reaction is given by:

$\ce{ O2 + 4 Cu/Cu(I) + 2 H+ → 4 Cu(I)/Cu(II) + 2 OH-}$

which accounts for the protective Cu2O formation on copper metal surfaces.

There is also a possible subsequent radical attack on cuprous with any created $\ce{.SO4-}$ radical anion (whose presence is apparently obviated in a two copper electrode scenario):

$\ce{Cu+ + .SO4- -> CuSO4 }$

Source: See this large file "Rate Constances for Reactions of Inorganic Radicals In Aqueous Solutions" published by the National Bureau of Standards, which is freely available here.

Interestingly, per the folllowing reactions [EDIT] assuming a sulfate radical presence:

$\ce{H+ + e- -> .H}$

$\ce{.H + .SO4- -> HSO4-}$

$\ce{H+ + HSO4- = H2SO4}$

implies the formation of H2SO4. Actually, the electrolysis of aqueous CuSO4 with only one copper electrode is a known path to Sulfuric acid, see, for example, How to make sulfuric acid by electrolysis of copper using an inert anode, where the employment of a non-inert anode (Cu), however, implies different possible chemistry at the anode.

[EDIT] An interesting related paper discussing nature of species involved, Understanding Persulfate Production at Boron Doped Diamond Film Anodes, to quote:

Density functional theory (DFT) modeling indicated that uncatalyzed oxidation of SO42− and HSO4− occurs at lower potentials than water oxidation, and that sulfate radical species (SO4−• and HSO4•) may be produced via direct electron transfer, or via reaction with hydroxyl radicals. The RDE experiments indicated that rates of persulfate generation were strongly dependent of the condition of the electrode surface, and that aged electrode surfaces favored water oxidation over direct SO42− and HSO4− oxidation. Combination of sulfate radical species in solution is the lowest energy pathway for persulfate production. Sulfate radical species may also react with radical sites on the electrode surface and produce chemisorbed intermediates that can stabilize sulfate radical species. Reaction of the chemisorbed intermediates with a bisulfate radical can produce persulfate via a surface catalyzed pathway. However, the activation barriers for this pathway are much higher than those for persulfate production via solution phase species.

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    $\begingroup$ Hmm, it seems to me Cu/Cu^2+ redox potential is too low to oxidate sulphates at potential where copper is being dissolved. So unless extreme current densities are applied, it would not happen. $\endgroup$ – Poutnik Sep 19 '20 at 5:47
  • $\begingroup$ Assuming sufficient voltage to form O2, in the presence of .H forms .HO2, which with Cu or Cu+ and H+ can lead to Copper-Fenton based hydroxyl radicals. Finally, .OH + SO4(2-) = OH- + .SO4- as a more likely path to the sulfate radical here, albeit, hardly simple chemistry. It does note, however, the significance/importance of a transition metal presence. $\endgroup$ – AJKOER Sep 19 '20 at 7:16
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    $\begingroup$ There is not the sufficient voltage. $\endgroup$ – Poutnik Sep 19 '20 at 7:19
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    $\begingroup$ Yes, it is,but not in context of Cu/CuSO4/Cu system. $\endgroup$ – Poutnik Sep 19 '20 at 7:38
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    $\begingroup$ I recommend withdrawing any mentioning of radicals, reserving it to inert anode scenarios, where it plays its role. As except of extreme case of diffusion limited copper dissolution, anodic potential never reaches high enough values needed for other oxidation reactions. $\endgroup$ – Poutnik Sep 19 '20 at 14:32

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