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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$\begingroup$ Regarding stability of Cu+ see: chemistry.stackexchange.com/questions/42855/… $\endgroup$– Nilay GhoshCommented Sep 15, 2019 at 13:44
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$\begingroup$ and chemistry.stackexchange.com/questions/115674/… $\endgroup$– Nilay GhoshCommented Sep 15, 2019 at 13:44
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1$\begingroup$ 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. $\endgroup$– Nilay GhoshCommented Sep 15, 2019 at 13:58
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1 Answer
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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 $\ce{Cu -> Cu^2+}).$
answered Sep 15, 2019 at 12:54