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I recently did a lab at school surrounding the decomposition of copper(II) carbonate into copper(II) oxide, and subsequently into pure copper, when sufficient heat is added. What I don't understand is why the copper(II) carbonate can't decompose into copper(I) oxide, rather than copper(II) oxide.

For example, copper(II) carbonate decomposing into copper(II) oxide would be:

$$\ce{CuCO3 ->CuO + CO2}$$

While copper(II) carbonate decomposing into copper(I) oxide would be:

$$\ce{4CuCO3 -> 2Cu2O + 4CO2 + O2}$$

If anyone could help me understand why copper(II) carbonate decomposes into copper(II) oxide rather than copper(I) oxide, I would greatly appreciate it.

Edit: I've been nosing around Wikipedia a bit and discovered that copper(I) oxide reacts with atmospheric moisture to become copper(II) oxide. This would explain why copper (I) oxide is not present; it has degraded into copper(II) oxide. I'm still interested in how copper(I) oxide and water react with respect to reactants and products, so if anyone knows what the equation between the products and reactants is, I would like to know.

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  • $\begingroup$ Hello and welcome to Chemistry.SE. A good way to familiarize yourself with the way this site works is by taking the short tour. Also, as this is a homework-type of question, you can find a good discussion of the homework policy here. Regarding your question, do you have any ideas about why copper(I) was not formed in the first step? How hot did you have to heat the CuO to form pure Cu? $\endgroup$ – airhuff Mar 1 '17 at 23:20
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It comes to stability of products, in this case interaction of metal ion with ligands (here oxygen). In water solutions $\ce{Cu^2+}$ is usually more stable than $\ce{Cu^1+}$, it's quite different story in non-aqueous solvents like acetonitrile. In solid: $\ce{CuCl2}$ is more stable than $\ce{CuCl}$, but iodide $\ce{CuI}$ is more stable than $\ce{CuI2}$.

The simplest explanation would be hard-soft acid-base theory: $\ce{Cu^2+}$ is a hard acid, and $\ce{Cu+}$ is a soft one. Oxygen in copper oxide or $\ce{Cl-}$ in chloride are hard bases and $\ce{I-}$ is a soft one. The same to the same: soft acids goes to soft bases, and hard acids likes the hard bases.

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Since @Kris_R already answered that Cu(II) is much more stable than Cu(I) according to HSAB theory and thus form $\ce{CuO}$ rather than $\ce{Cu2O}$ on decomposition but it is further noted that copper(II) oxide can be further decomposed to copper(I) oxide on further application of heat.

$$\ce{4CuO ->[1026-1100°C] 2Cu2O + O2}$$

(reaction source)

So, complete reaction would be:

$$\ce{CuCO3 ⟶[\Delta 1] CuO + CO2}$$ $$\ce{4CuO ->[\Delta 2 ] 2Cu2O + O2}$$

Note that you can't write both reactions in one single equation(as mentioned in the question) as the two reactions proceed at different temperatures.

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