Reaction of Cu in hydrochloric acid bath with exposure to air over several months?
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...

$$\ce{Cu + 1/2 O2  + 2 H+ -&gt; Cu^{2+} + H2O}$$

... 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:

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

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

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.
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