Consider the following Ellingham diagram.

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.

Extraction of Zn from ZnO

So why is ZnO is reduced to Zn in this case?

  • $\begingroup$ NCERT contains incorrect information. The temperature is 1673 K. Verified from earlier versions and other sources $\endgroup$
    – user87597
    Jan 5, 2020 at 8:45

1 Answer 1


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:

$$\ce{ZnO(s) + C(s) <=> Zn(l) + CO(g)}$$

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:

$$\ce{CsCl(s) + Li(l) <=> Cs(g) + LiCl(s)}$$

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.

  • $\begingroup$ For us to remove CO on formation, shouldn't ZnO get reduced first? But the oxidising gibbs energy change of ZnO is greater than coke, so how can the reaction move in the forward direction? Also why is it an equilibrium reaction? $\endgroup$ Dec 22, 2017 at 2:54
  • $\begingroup$ ALL reactions are in essence equilibrium reactions, this one is no exception. By removing one of the reaction products continually, its concentration (chemical activity) is lowered, thus pushing the equilibrium to the right. Have you read the linked page on Le Chatelier's Principle? $\endgroup$
    – Gert
    Dec 22, 2017 at 15:56

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