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# $2CuI \to Cu + Cu{I_2}$, the reaction is:A) RedoxB) NeutralisationC) DisplacementD) None of the above

Last updated date: 15th Sep 2024
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Hint: You need to calculate the oxidation state of copper in CuI, Cu and $Cu{I_2}$ in the given reaction. Examine the relation between different calculated oxidation states. An increase in the oxidation number of an element in the given substance is termed as oxidation whereas a decrease in the oxidation number of an element in the given substance is termed as reduction.

Complete Solution :
Given reaction is-
$2CuI \to Cu + Cu{I_2}$
- Let us first look at the oxidation states of copper in the above reaction.
Oxidation state of copper (Cu) in CuI is +1.
Oxidation state of Cu as a product is 0.
Oxidation state of Cu in $Cu{I_2}$ is +2.
The idea of oxidation state has been applied to define some terms such as:

Oxidation: An increase in the oxidation number of an element in the given substance is termed as oxidation.
Reduction: A decrease in the oxidation number of an element in the given substance is termed as reduction.
Redox reaction: Reaction which involves change in the oxidation state of the interacting species or when oxidation and reduction occur simultaneously in a reaction, the reaction is redox reaction.

- We therefore conclude that in the given reaction, copper in first reduced from +1 oxidation (in CuI) to zero oxidation state (in Cu) and also oxidised from +1 oxidation (in CuI) to +2 oxidation state in $Cu{I_2}$. Thus, both oxidation and reduction reactions are occurring.
So, the correct answer is “Option A”.

Note: You must note that a given reaction can also be termed as a disproportionation reaction. Disproportionation reaction is defined as a redox reaction in which a compound undergoes oxidation as well as reduction. In our given reaction, copper is undergoing reduction as well as oxidation.