# Cupric and cuprous copper [duplicate]

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.

• Regarding stability of Cu+ see: chemistry.stackexchange.com/questions/42855/… – Nilay Ghosh Sep 15 '19 at 13:44
• – Nilay Ghosh Sep 15 '19 at 13:44
• and who said copper(I) chloride is not formed? Wikipedia says that copper and chlorine directly reacts with each other to form copper(I) chloride at 450-900 C. – Nilay Ghosh Sep 15 '19 at 13:58

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 $$\ce{Cu -> Cu^2+}).$$