When refining copper through electrolysis, do the copper ions from the electrolyte copper sulfate get reduced at the anode?

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?




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


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