When sulfuric acid reacts with copper, then $\ce{H2SO4}$ gives nascent oxygen $\ce{[O]}$. The reaction is: $$\ce{H2SO4 -> H2O + SO2 + [O]}$$

This nascent oxygen reacts with copper to create $\ce{CuO}$.

Therefore my question is that when copper reacts with $\ce{HCl}$ then why doesn’t $\ce{HCl}$ give nascent chlorine $\ce{[Cl]}$ so that it could react with copper and create copper(II) chloride $\ce{CuCl2}$?

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    – Jan
    Dec 21, 2015 at 14:32
  • $\begingroup$ This may be due to standard redox potentials between copper cations and chlorine anions, which can't give the driving force to generate nascent gas $Cl_{2}$, and the high solubility of $CuCl_{2}$ that doesn't allow this salt precipitation. $\endgroup$
    – Orr22
    Dec 21, 2015 at 14:37

1 Answer 1


When concentrated sulfuric acid is heated, then it decomposes via the reaction $$\ce{H2SO4 -> H2O + SO3}$$

I tried to determine under what circumstances the nascent oxygen (a free single atom of oxygen) is formed which would react with copper to create $\ce{CuO}$ but couldn't find the reaction in any modern source. I did find the proposed reaction in Essentials of Modern Chemistry By Charles Elwood Dull (1918) but I'm guessing that the reaction to give nascent oxygen is just wrong.

The gist here is that you can't piece reactions together from balanced equations but you must also define the conditions under which the reaction takes place. At a deeper level not only can the chemical equation be specified but also the mechanism of the reaction can be detailed.

Hydrochloric acid, $\ce{HCl}$, is typically in aqueous solution. Although copper won't react with pure hydrochloric acid, it can react with dissolved oxygen in hydrochloric acid. $$\ce{Cu (s) + 2HCl (aq) + 1/2O2 (aq) -> CuCl2 (aq) + H2O}$$

Oxygen can also be formed in chlorine water: $$\begin{align} \ce{Cl2 (aq) &-> HCl + HOCl}\\ \ce{HOCl &-> HCl + 1/2O2} \end{align}$$

But the concept of nacent oxygen from this reaction is wrong.

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    $\begingroup$ FYI: States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. I also adjusted the choice of maths environment and removed the blockquoting. $\endgroup$ Jan 4, 2016 at 11:05

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