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This question already has an answer here:

Copper has two chlorides: $\ce{CuCl2}$ and $\ce{CuCl}.$ Copper reacts directly with chlorine to form a copper(II) chloride. Why doesn't it form copper(I) chloride? Is it because $\ce{CuCl2}$ is more stable than $\ce{CuCl}?$ I want a more foundational approach to answer this question.

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marked as duplicate by Mithoron, Mathew Mahindaratne, Buck Thorn, Tyberius, Todd Minehardt Sep 19 at 3:13

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There are several ways to look at this question.

On one hand, the heat of formation of $\ce{CuCl2}$ $(\pu{49.2 kcal/mol})$ is greater than the heat of formation of $\ce{CuCl}$ $(\pu{32.5 kcal/mol}),$ so if there is excess $\ce{Cl2},$ the product would naturally be the dichloride. On a kinetic basis, even with just a stoichiometric amount of chlorine $(\ce{1/2 Cl2}$ plus $\ce{1 Cu}),$ gaseous $\ce{Cl2}$ might produce a little cuprous chloride from bulk metal, but the product would be more finely divided and more likely to react, so the reaction would tend toward $\ce{CuCl2},$ leaving half the copper unreacted.

The electrochemical potentials reflect the same thing: in water, if you have enough oxidizing power to produce $\ce{Cu+}$ $(\pu{-0.52 V}),$ there is enough to go the rest of the way to $\ce{Cu^2+}$ $(\pu{-0.1 V}$ for $\ce{Cu+ -> Cu^2+},$ or $\pu{-0.34 V}$ for $\ce{Cu -> Cu^2+}).$

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