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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?

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

2 Answers 2

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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.

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  • $\begingroup$ 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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Somebody mistyped something. We do require a higher temperature than that quoted in the question.

Zinc is a chalcophilic element, generally available as a sulfide ore which is first roasted to the oxide. The oxide may then be reduced to zinc metal by coke (carbon), but it requires a higher temperature than $400°\text{C}$. From Wikipedia:

For further processing two basic methods are used: pyrometallurgy or electrowinning. Pyrometallurgy reduces zinc oxide with carbon or carbon monoxide at 950 °C (1,740 °F) into the metal, which is distilled as zinc vapor to separate it from other metals, which are not volatile at those temperatures.[1] The zinc vapor is collected in a condenser.[2] The equations below describe this process:[2]

${\displaystyle {\ce {ZnO + C ->[950°C] Zn + CO}}}$

${\displaystyle {\ce {ZnO + CO ->[950°C] Zn + CO2}}}$

The Ellingham diagram then shows that this temperature is sufficient to give carbon monoxide formation up to one full atmosphere carbon monoxide pressure; the carbon/carbon monoxide line is clearly below the zinc/zinc oxide line. We do not oxidize coke to carbon dioxide at that partial pressure (the carbon/carbon dioxide line is slightly above the zinc/zinc oxlide line), but because the free energy difference is small we do get a significant partial pressure below one atmosphere which is made up by the carbon monoxide.

The carbon monoxide can be burned to remove the toxic gas and provide some of the heat energy needed to run the process.

Cited References

  1. Bodsworth, Colin (1994). The Extraction and Refining of Metals. CRC Press. p. 148. ISBN 978-0-8493-4433-6.

  2. Porter, Frank C. (1991). Zinc Handbook. CRC Press. ISBN 978-0-8247-8340-2.

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