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 $\ce{Cu2O}$?
- 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.
- Will it not react any further but instead get solved in the hydrochloric acid solution, because it is soluable in acids?
- Or will something else happen?
Note: The concentration of hydrochloric acid solution is around 10 %.
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