Why is ZnO reduced in this reaction? Consider the following Ellingham diagram.



At 673K, the Gibbs energy value of oxidation of Zn is more negative than the Gibbs energy value of oxidation of C (coke). So, Zn should be oxidised to ZnO. But in various sources like the textbooks of class 12 of CBSE (India), it is mentioned that ZnO is reduced to Zn and C (coke) is oxidised to CO at 673K in the reaction between ZnO and C. Looking at the Ellingham diagram, coke should not reduce ZnO to Zn at 673K.



So why is ZnO is reduced to Zn in this case?
This method works because one of the reaction products, $\ce{CO}$, is gaseous (at the stated reaction temperature) and is continually removed. Following Le Chatelier's Principle, the equilibrium:

$$\ce{ZnO(s) + C(s) &lt;=&gt; Zn(l) + CO(g)}$$

is pushed to the right and the reaction proceeds.

There are numerous of these reactions where Ellingham data don't tell the full story, notably also the reduction of magnesia ($\ce{MgO}$) by coke, is in accordance with the same principle.

One of the most interesting cases is a lab preparation of caesium metal that takes place according to:

$$\ce{CsCl(s) + Li(l) &lt;=&gt; Cs(g) + LiCl(s)}$$

The reaction is carried out under vacuum and at about $970\ \mathrm{K}$, conditions under which $\ce{Cs}$ is much more volatile than $\ce{Li}$. The former thus distils off and is caught in a condenser, pushing the reaction equilibrium to the right.