Copper sulfate is the electrolyte that is brought up most commonly when we talk about the refining of copper. Do any of the copper ions from the copper sulfate solution reduce at the anode like the copper ions that were oxidized from the cathode?

  • 2
    $\begingroup$ Oxidation is at the anode and reduction takes place at the cathode. Copper ions get reduced to copper metal at the cathode. $\endgroup$
    – Ed V
    Dec 9, 2022 at 15:46
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    $\begingroup$ At the cathode/anode is ongoing reduction/oxidation, what means the cathode/anode is kept at more negative/positive external potential in the cell electrolytic mode. While in the galvanic(active) mode, the cathode/anode has the more positive/negative potential (i.e. marking of electrodes switches places). $\endgroup$
    – Poutnik
    Dec 9, 2022 at 15:58
  • $\begingroup$ Anode = Oxydation $\endgroup$
    – Maurice
    Dec 9, 2022 at 16:43
  • $\begingroup$ Oh yeah sorry I got confused with the names of the electrodes,, Thanks for pointing it out $\endgroup$ Dec 10, 2022 at 1:32

2 Answers 2


Like commenters pointed out the direction of the process is the other way round.

Apart from that the answer is yes, copper ions from the sulfate will be reduced at the cathode but at the same time copper ions from the anode replace the now "missing" ions from the sulfate. So at the beginning of the process only ions from the original copper sulfate are reduced and soon after the amount of ions originating from the anode increases continuously until they all the ions in the solution have been replaced.

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    $\begingroup$ Such processes are usually stirred to retain current so it is diminishing returns on the original copper ions. $\endgroup$
    – jimchmst
    Dec 10, 2022 at 0:48
  • $\begingroup$ @jimchmst Could you explain a bit more please? $\endgroup$ Dec 11, 2022 at 23:49
  • $\begingroup$ @chem_student The net movement speed of the ions in the electrolyte is limited. By stirring you level out concentration differences along their path. This helps to increase the current flow and thus the transport rate of ions through towards the cathode. $\endgroup$
    – datenheim
    Dec 12, 2022 at 14:16

"Cathode is where the cations go" to be reduced.

In order to keep charge balance oxidation takes place at the anode.

With aqueous CuSO4 solution, if copper is not present in the anode, one can very conveniently use the reaction: $$\ce{2 H2O -> 4 H+ + 4 e- + O2 gas - 1.23 V}$$

With an impure copper anode, copper and any metal with a less negative oxidation potential, such as iron, will go into solution preferentially: $$\ce{Cu -> Cu++ + 2 e- - 0.34 V}$$ $$\ce{Fe -> Fe+++ + 3 e- + 0.04 V}$$ $$\ce{Al -> Al+++ + 3 e- + 1.66 V}$$

That goes with reduction at the cathode:

$$\ce{Cu++ + 2 e- -> Cu metal + 0.34 V}$$

Metal drops out, O2 gas goes away (or other metals stay in solution), H+ replaces Cu ++, SO4 does not react.

Note that, at the cathode, copper will "go first" because it has a higher positive reduction potential than H+:

$$\ce{2H+ aqueous + 2 e- -> H2 gas 0.00 V}$$

as compared with + 0.34 V for Cu ++.

Note: reactions are written in the order they proceed at anode and cathode. More positive V values signify order of reaction.

  • $\begingroup$ What you say, Robert DiGiovanni, is correct if the anode is not made in copper. The trouble is that ChemStudent speaks of the refining of copper. This happens when the anode is made of impure copper. Here copper gets dissolved at the anode giving $\ce{Cu^{2+}}$ ions which go into solution. The impurities of the impure copper block are not dissolved and fall down the electrolysis container. So the solution is not going to be more and more acidic, as you state. The copper ion concentration does not change much during electrolysis, and pure copper is deposited at the cathode. $\endgroup$
    – Maurice
    Jan 8, 2023 at 21:31
  • $\begingroup$ @Maurice yes, what an interesting way of refining copper. Thank you. "The copper ion concentration does not change much during electrolysis". You might lose your job if you threw any copper away. All can be plated out (and re-refined). I come from industry. We have to turn a dime. $\endgroup$ Jan 8, 2023 at 21:47

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