(A) In the reaction $2CuI \to Cu + CuI_2$,the oxidation state of $Cu$ in $CuI$ is $+1$. In the product side,$Cu$ is in $0$ oxidation state and $Cu$ in

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(A) In the reaction $2CuI \to Cu + CuI_2$,the oxidation state of $Cu$ in $CuI$ is $+1$. In the product side,$Cu$ is in $0$ oxidation state and $Cu$ in $CuI_2$ is in $+2$ oxidation state. Since $Cu$ is simultaneously oxidized (from $+1$ to $+2$) and reduced (from $+1$ to $0$),this is a disproportionation reaction,which is a type of redox reaction.

Class 11 Chemistry Redox Reactions Auto oxidation and Disproportionation

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$2CuI \to Cu + CuI_2$,the reaction is

A
Redox
B
Neutralisation
C
Oxidation
D
Reduction

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$2CuI \to Cu + CuI_2$,the reaction is

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$3Br_2 + 6CO_3^{2-} + 3H_2O \to 5Br^- + BrO_3^- + 6HCO_3^-$
In the above reaction,