I have the equation:

$$\ce{2 Cu2O + Cu2S -> 6 Cu + SO2}$$

My doubt is that since copper is reduced (it goes from $+1$ to a $0$ O.S.), then which of the two reactants is reduced? Or why would it be fine to say that only copper is reduced and not an entire reactant? Because from my experience, I've seen that an entire reactant is considered to be reduced or oxidized. So is it right to tell only a part of the compound is oxidized?

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    $\begingroup$ Compounds are not reduced; elements are. To say that a compound is reduced is an abuse of language. Sometimes it is forgivable, sometimes it isn't. $\endgroup$ Feb 4, 2019 at 7:23
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    $\begingroup$ So is it more correct to refer to a reactant as a reducing/oxidizing agent rather than saying "a compound is reduced/oxidized"? $\endgroup$ Feb 4, 2019 at 7:29
  • $\begingroup$ Exactly. But then again, is can so happen that one element in a compound is oxidized and another reduced; what would you call that? $\endgroup$ Feb 4, 2019 at 7:33
  • $\begingroup$ So that would then be referred to as a disproportionation reaction, I believe? And to determine which type of reagent it is, we would have to look at the particulars of the mole ratio and the individual changes in oxidation number of that particular reactant, right? $\endgroup$ Feb 4, 2019 at 7:39
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    $\begingroup$ No and no. Disproportionation is a reaction where the same element reduces and oxidizes at the same time. As for the type of reagent, this is not something worth caring about. Call it a reducing and oxidizing agent at once, if you wish, or don't call it anything special at all. $\endgroup$ Feb 4, 2019 at 7:56

2 Answers 2


For accounting purposes, we can assign oxidation numbers to all the atoms. Oxygen could be -II in all species here, then sulfur would be +IV in $\ce{SO2}$ and -II in $\ce{Cu2S}$ (oxygen and sulfur are in the same group and both are more electronegative than copper, so assigning the same oxidation number in $\ce{Cu2S}$ and $\ce{Cu2O}$ makes sense). Oxidation numbers of atoms in elements are zero, and the copper in both compounds would be +I to balance out the -II of the chalcogens.

Once you are in agreement with your discussion partner on these oxidation states, you can make a statement about which atoms undergo oxidation or reduction: Sulfur gets oxidized, and copper in both copper compounds gets reduced.

In cases where only one atom in a compound changes oxidation states, or when all changes are in the same direction, we do talk about the compound being reduced or oxidized. For example, it is common to say that when you oxidize an aldehyde, you get a carboxylic acid (and when you reduce it, you get an alcohol or an alkane).

On the other hand, if you hydrate an alkene, some folks don't consider that a redox reaction.


I am restating the question below (to slightly adjust the formatting and make it easier to address):

I have the equation:

$$2\ce{{Cu_2O} + {Cu_2S}⟶{6Cu} + {SO_2}}$$

Here we have compounds $\ce{{Cu}_2O}$, $\ce{{Cu}_2S}$ for the reactants, and the products of metallic $\ce{Cu}$ and gas $\ce{S{O_2}}$; compounds share electrons and so it difficult to say that they are "reduced" or "reduce" since there is a shared bond in any case to start with.

Getting beyond that, however, we see that for $\ce{Cu_2O}$ that just like water that is $\ce{H_2O}$, Copper shares two of its electrons with Oxygen. In the case of $\ce{Cu}$ Copper, these electrons are not shared anymore, so in this case $\ce{Cu}$ is reduced (the opposite of oxidation where electrons are shared), and the $\ce{O}$ is oxidized, since the copper is no longer sharing negative charge, increasing its positive charge.

My doubt is that since copper is reduced (it goes from +1 to a 0 O.S.), then which of the two reactants is reduced? Or why would it be fine to say that only copper is reduced and not an entire reactant? Because from my experience, I've seen that an entire reactant is considered to be reduced or oxidized. So is it right to tell only a part of the compound is oxidized?

Oxidation and reduction is usually formally defined in the sense as in this link https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/oxred_3.php but I am trying to address the problem from a conceptual perception to make progress in answering the question at hand.

Also, for $\ce{{Cu}_2S}$, we have that the reactant $\ce{Cu}$ Copper is sharing two of its electrons with $\ce{S}$ Sulphur and in the product that metallic $\ce{Cu}$ Copper is no longer sharing these electrons, receives them back in the reaction, and therefore is also reduced in the forward reaction.

For Copper, it is a a metal at the end, and individual atoms are "covalent reduced" here, where we are taking oxidation and reduction to be defined in a sense of negative charge being moved (reduction) to the Copper atom in the case of this metal $\ce{Cu}$; but something to be careful about is that this charge that moves back to the Copper atom is not charge of the Oxygen atom or charge of the Sulfur atom. Rather, this is shared charge that belongs originally to the neutral Copper atoms.

Now let us consider the compounds. All the reactant compounds are neutral to start with. But in the course of the reaction, the shared charge that belonged to the Copper atoms is returned to those Copper metal atoms. And the movement of the charge away from the Oxygen $\ce{2 O}$ and from $\ce{2{Cu}_2O}$ results in the lowering of negative charge shared there on the $\ce{2 O}$ and is therefore oxidation, though not in an ionic sense since the charge was only shared to begin with.

Similarly, the reactant compound $\ce{{Cu}_2S}$ has a movement of shared charge to the Copper $\ce{{Cu}_2}$ and away from the $\ce{S}$ Sulfur, a covalent oxidation of the Sulfur. The usual application of oxidation and reduction is for ionic compounds and all compounds of the reactants and the products are covalent or metallic bonds.

But to conceptually address the question, the shared metallic electrons are returned to metallic copper $\ce{Cu}$ (a non-ionic reduction characteristic) and in the course of the reaction the shared electrons are taken back from the final Sulfur Dioxide Compound product $\ce{SO_2}$ (a non-ionic oxidation characteristic).

If it were not enough to consider the movement of negative charge for Copper reactants towards the metal in the course of the reaction -- there is an additional bonding process!

Interesting enough also, is that the Copper atoms have completely received their covalent-bonded charge back - but that means that it had been taken from somewhere. And the only product left is $\ce{SO_2}$; so that implies that the move of negative charge in the reaction towards the Copper metal product (considered here like a covalent reduction) must result in a covalent-like oxidation (movement of negative charge away) of the $\ce{SO_2}$ and this effect probably influences the resulting strength of the resulting neutral metallic and neutral covalent bonds at the end of the reaction.

The chemical Sulfur Dioxide $\ce{SO_2}$ is covalent, meaning that the Oxygen atoms $\ce{O_2}$ newly share their charge with the Sulfur atom $\ce{S}$ in the course of the reaction. Although this bond is covalent, charge still moves away from the Oxygen atoms towards the Sulfur atoms, having a reduction effect on the Sulfur and an Oxidation effect on the Oxygen atoms (within the covalent compound)!

I appreciate an upvote if this answer addresses the question as well as comments from any angle!

  • $\begingroup$ Cu++ ion is reduced to Cu]0]. The reducing agent is sulfide Whether the mechanism is stepwise decomposition of CuS then reduction of CuO by elemental S or some concerted mechanism requires some chemical study. Since oxidation numbers are assigned by fiat there is little need to be concerned about bond covalency. $\endgroup$
    – jimchmst
    Aug 16, 2023 at 6:03
  • $\begingroup$ @jimchmst For oxidation and reduction I am going from starting reactants to final products. (What is Copper(I) Oxide?) byjus.com/chemistry/copper-oxide notes that "Copper(I) Oxide is also called as cuprous oxide, an inorganic compound with the chemical formula $\ce{Cu2O}$. It is covalent in nature. " The statement "Cu++ ion is reduced to Cu]0]" is in the opposite direction of the question: (the product Copper metal to a reactant covalent bond), The forward reaction is the question $\ce{2Cu2O + Cu2S⟶6Cu + SO2}$; that oxidized in a reverse reaction is reduced in a forward reaction. $\endgroup$ Aug 16, 2023 at 7:28

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