So my book sates that in the extraction of cuprous oxide, the following reaction takes place.

$$\ce{Cu2O + C -> 2 Cu + CO}$$

This makes sense to me, if it is seen as the following two reactions coupled.

$$ \begin{align} \ce{Cu2O &-> Cu + 1/2 O2} \\ \ce{C + 1/2 O2 &-> CO} \end{align} $$

But as per the Boudouard reaction, carbon monoxide is the dominant form of oxide of carbon at higher temperatures, however using carbon for reduction is also possible at lower temperatures, which can be seen from Ellingham diagram.

So, are both $\ce{CO}$ and $\ce{CO2}$ possible as side products depending on temperature at which the extraction of $\ce{Cu}$ is done?


1 Answer 1


@ Paras Khosla. Your first equation is correct, but not your second one : $\ce{Cu_2O}$ reacts with $\ce{C}$ without getting decomposed previously into $\ce{Cu + O_2}$. $\ce{Cu_2O}$ is more stable than $\ce{CuO}$ above $1000$°C

Furthermore, $\ce{Cu_2O}$ does not react easily with Carbon in the solid state. The contact between the reacting solids is not good. The contact is better if $\ce{Cu_2O}$ is liquid. But $\ce{Cu_2O}$ melts at $1026$°C. So, the reaction starts at this temperature, forming $\ce{CO}$ which is more stable than $\ce{CO_2}$ at such a high temperature.


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