I assume that copper will be slowly oxidized by oxygen, dissolved in water, forming copper(I) oxide (as stated here): $$\ce{4 \overset{\pm0}{Cu} + O2 -> 2 \overset{+I}{Cu}_2O}$$

Reaction rate might be affected by the hydrochloric acid (cf. here)

And some of the copper could actually be converted to $\ce{CuCl2}$ (cf. here): $$\ce{2HCl (aq) + \overset{\pm0}{Cu}(s) -> \overset{+II}{Cu}Cl2(aq) + H2(g)}$$

But what will happen to $\ce{Cu2O}$?

  1. Will it degrade to $\ce{CuO}$, which then forms $\ce{CuCl2}$? That would mean that one should find higher concentrations of $\ce{CuCl2}$ in the solution, as time goes on.
  2. Will it not react any further but instead get solved in the hydrochloric acid solution, because it is soluable in acids?
  3. Or will something else happen?

Note: The concentration of hydrochloric acid solution is around 10 %.

Related Questions:
Copper Doesn't React with Hydrochloric acid
Copper and hydrochloric acid

  • 1
    $\begingroup$ in presence of CuCl2 in excess of concentrated HCl copper will slowly dissolve, forming $\ce{[CuCl2]^{2-}}$ anion, which is readily oxidized by air. So, yes, copper will slowly dissolve in hydrochloric acid bath, especially if some 'seed' copper chloride is present. The process is slow, though. $\endgroup$
    – permeakra
    Commented Sep 28, 2015 at 16:30
  • $\begingroup$ also -- oxidation is typically a surface-reaction unless you have a very powerful oxidizer. Cu in air will form an oxide on the surface slowly. Water will speed it up some, HCl even more, and H2O2 (~30%) with HCl will oxidize very fast -- effectively dissolving the Cu and leaving CuCl2 dissolved in water. $\endgroup$
    – khaverim
    Commented Sep 29, 2015 at 3:09

1 Answer 1


In the presence of excess hydrochloric acid, copper oxides are not the immediate reaction product of copper metal and oxygen (e.g. from the air). The air does oxidize the copper...

$$\ce{Cu + 1/2 O2 + 2 H+ -> Cu^{2+} + H2O}$$

... but the final oxidation product is copper(II) ion, not copper(I). (Copper(I) is formed initially but is more prone to air oxidation than copper metal is, so it is relatively quickly converted to copper(II).) Additionally, in the presence of chloride anions, the copper(II) ions are preferentially complexed by chloride. The low concentration of hydroxide means that hydroxide complexes are not favored, and chloride is a better ligand than water for copper.

The actual complexes formed vary in their stoichiometry and their color. $\ce{CuCl2(aq)}$, $\ce{CuCl3^{-}(aq)}$ and $\ce{CuCl4^{2-}(aq)}$ are all possible:

$$\ce{Cu^{2+} + n Cl- -> CuCl_n^{(n-2)-}(aq)}$$

$$n \in \{2, 3, 4\}$$

Nurdrage made a nice video showing how the reaction of $\ce{HCl}$, air, and copper metal leads to copper chloride. It's the first of the three methods he presents for making copper chloride. As he notes, the reaction is very slow. $%edit$

  • $\begingroup$ Thanks! Do you think that this reaction could, after enough time, lead to a orange layer on the copper surface, when the same solution is exposed to additional "fresh" copper regularly? Because this is what I was observing after approximately one month. $\endgroup$
    – AstronAUT
    Commented Sep 28, 2015 at 17:22
  • $\begingroup$ Well, after some time, the $\ce{HCl}$ will be used up, so technically the solution would turn into water. It sounds possible. But I never tried. @AstronAUT $\endgroup$
    – Jan
    Commented Sep 28, 2015 at 17:24
  • $\begingroup$ That would mean, that this orange powder I observed could be Cu2O (red) mixed with CuO (black), which form in the absence of HCl, right? $\endgroup$
    – AstronAUT
    Commented Sep 28, 2015 at 17:34

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