I have always read that the one having higher discharge potential undergoes the reaction( oxidation or reduction). But In this example why is there also oxidation of water?sch

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    $\begingroup$ What were the alternatives? (Hint: there were none.) $\endgroup$ Apr 13, 2020 at 17:39
  • $\begingroup$ But i dont get it. Why not just Cu2+ -->Cu. Why is there the reaction H2O---> H2 if it have lower discharge potential ? $\endgroup$ Apr 13, 2020 at 17:56
  • $\begingroup$ Why not? Because you can't have your hamburger for free. The electrons didn't just appear out of nowhere and reduce copper. They have to be coming from somewhere. Something is going to be losing these electrons. Something has to be oxidized. Just what could that be? $\endgroup$ Apr 13, 2020 at 18:18
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    $\begingroup$ It will. That is, until all copper is reduced and we have nothing more to reduce but hydrogen. $\endgroup$ Apr 13, 2020 at 19:25
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    $\begingroup$ Copper cannot be oxidized. It can be reduced. And it will be reduced. H2O could be reduced in H2 if the over voltage is high enough, which is difficult to decide, as it depends on the surface of the cathode.. $\endgroup$
    – Maurice
    Apr 13, 2020 at 19:29

1 Answer 1


The question is not fully complete (not your mistake but someone who wrote this exercise). I am not surprised that you are confused. The question is indeed silent about the electrodes. You should have asked the question: What is the cathode made of? What is the anode made of before starting the question. The second trick in your question in your question is, which I liked, is that it is an example of exhaustive electrolyte. You passed a huge current which consumed all free metal ions and the solvent can also be reduced or oxidized.

So let us look at your problem:

At the cathode you have two explicit choices: (i) Reduction of Cu ions which are floating in the solution, (ii) Reduction of the solvent, which is water in your case to hydrogen! There is nothing else which can be reduced here as per your question.

At the anode you still have two implicit (hidden) choices: (i) Oxidation of water and (ii) Oxidation of the anode material.

Your solved questions bluntly assumes that the electrode is inert, say it is made of carbon or Pt or gold, whatever but inert. So are left with only one choice. Oxidation of water. However, let us say they actually told you that the anode is made of Copper:

Oxidation potential of Cu

Cu(s) ---------> Cu$^{2+}$(aq) + 2e$^{-1}$ -0.34 V

Oxidation potential of water

2 H$_2$O(l)---------> O$_2$(g) + 4 H$^+$(aq) + 4e$^{-1}$ -1.23 V

Which one do you think will get oxidized? The Cu electrode instead of water!

This is the process and principle behind copper plating.

When your examiners are quiet about electrode materials, assume the solvent is going to oxidize, i.e., water.


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