Question

please explain oxidation/reduction inthe following chemical reactions:
1. Cu + I2 -------> CuI2

Answer 1

$\large\mathsf{Ahoy!!}$
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$\mathsf{\underline{ \ Given \ equation \ : \ }}$

$\mathsf{Cu \ + \ I_2 \ \rightarrow \ CuI_2}$



${\large{\boxed{\mathsf{\red{ \ 1st \ Method \ : \ }}}}}$


${\large{\boxed{\mathsf{ \ Oxidation \ - \ }}}}$

Addition of Oxygen or an electronegative element OR Removal of Hydrogen or an electropositive element.

${\large{\boxed{\mathsf{ \ Reduction \ - \ }}}}$

Removal of Oxygen or an electronegative element OR Addition of Hydrogen or an electropositive element.

In the given equation, there is a addition of electronegative element I to Cu, therefore, Cu is oxidised to $\mathsf{CuI_2}$.

Also, there is a addition of electropositive element Cu to $\mathsf{I_2}$, therefore, $\mathsf{I_2}$ is reduced to $\mathsf{CuI_2}$.



${\large{\boxed{\mathsf{\red{ \ 2nd \ Method \ : \ }}}}}$


${\large{\boxed{\mathsf{ \ Oxidation \ - \ }}}}$

Loss of electrons.

${\large{\boxed{\mathsf{ \ Reduction \ - \ }}}}$

Gain of electrons.

Now,

Copper is a metal, so it has a tendency to lose electrons.

$\mathsf{Cu \ \rightarrow \ {Cu}^{2+} \ + \ 2{e}^{-}}$

Iodine is a non - metal, so it has a tendency to gain electrons.

$\mathsf{I_2 \ + \ 2e^- \ \rightarrow \ 2I^-}$

Therefore, Copper is oxidized and Iodine is reduced.



${\large{\boxed{\mathsf{\red{ \ 3rd \ Method \ : \ }}}}}$

${\large{\boxed{\mathsf{ \ Oxidation \ - \ }}}}$

Increase in oxidation number.

${\large{\boxed{\mathsf{ \ Reduction \ - \ }}}}$

Decrease in Oxidation number

First, we have to write the oxidation number of all elements and compounds involved in the equation.

Oxidation number of Cu = 0

Oxidation number of $\mathsf{I_2}$ = 0

Oxidation number of Cu and $\mathsf{I_2}$ in $\mathsf{CuI_2}$ = +2 and -1

From these, we can see that the oxidation number of Cu is increases from 0 to +2, therefore, Cu is oxidised.

Also, the oxidation number of $\mathsf{I_2}$ is decreased from 0 to -1.

$\therefore$
At last, we arrived at the result that Cu is oxidised and $\mathsf{I_2}$ is reduced.



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$\implies$ BeBrainly

Answer 2

First, we have to write the oxidation number of all elements and compounds involved in the equation.

Oxidation number of Cu = 0

Oxidation number of  = 0

Oxidation number of Cu and  in  = +2 and -1

From these, we can see that the oxidation number of Cu is increases from 0 to +2, therefore, Cu is oxidised.

Also, the oxidation number of  is decreased from 0 to -1.

we get  the result that Cu is oxidised and  is reduced.