For the reduction of metal oxide to metal, suggest a reducing agent cheaper than aluminium

Dear student,

It is easier to reduce metal oxides in the liquid state as it means that the temperature of the substance is higher. According to the Ellingham diagram, the value of ΔG for the formation of metal oxides increases as temperature increases. At these higher temperatures, many metal graphs can be observed below this one (except for very high reactive metals of course) and therefore the reduction of the metal oxide is easier.

For reduction of metal oxide the Gibbs energy should be negative, this is a mandatory condition. In a solid state, the entropy or randomness will be less due to the high intermolecular force of attraction while in a liquid state, the entropy will be more comparatively, therefore Gibbs energy will be more negative.

Thus, how displacement can be used.

Carbon is a good reducing agent. It reacts with the metal oxide-forming metal and carbon dioxide. Coke is a readily and cheaply available form of carbon and is therefore used in reduction. 


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Element which is cheaper than aluminium and can be used for the reduction purposes is carbon. Carbon reacts with metallic oxides to form carbon dioxide and metal is obtained.
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Element which is?cheaper than aluminium?and can be used?for the reduction?purposes is carbon. Carbon reacts with?metallic oxides?to form carbon dioxide and?metal?is obtained.
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