In the extraction of $\ce{Sn}$ from $\ce{SnO2}$ by carbon reduction method, why is $\ce{CO}$ formed instead of $\ce{CO2}$ as per the following reaction?

$\ce{SnO2 + C -> Sn + CO}$


1 Answer 1


Above a $1000\ \mathrm K$, the $\Delta_\mathrm fG$ of $\ce {CO}$ from $\ce C$ is more negative than that of $\ce{CO2}$ formation from $\ce C$. Therefore, during smelting, when coke ($\ce C$) reacts with $\ce {SnO2}$, the formation of $\ce {CO}$ rather than $\ce {CO2}$ is thermodynamically preferable.

Below that temperature, the reduction with carbon is not possible since $\Delta_\mathrm fG(\ce{SnO2})\lt\Delta_\mathrm fG(\ce{CO2})\lt\Delta_\mathrm fG(\ce{CO})$. Therefore at all temperatures where this reduction is possible, thermodynamically carbon monoxide formation would be more favourable as compared to carbon dioxide formation. To see for yourself, the variation of $\Delta_\mathrm fG$ of various compounds, have a look at Ellingham Diagrams, a link to a comprehensive set of which is Given Here.

$\Delta_\mathrm fG$ is the Gibbs free energy of formation.

  • 3
    $\begingroup$ Nice answer! I wonder if we can talk of Boudard reaction for this case too.... $\endgroup$
    – G M
    Oct 3, 2013 at 13:36

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