Question

Find out the oxidation states.
${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}} + {\text{ }}{{\text{I}}^{\text{ - }}} \to {\text{ C}}{{\text{r}}^{{\text{3 + }}}} + {\text{I}}{{\text{O}}_{\text{3}}}^{\text{ - }}$

Answer 1

Oxidation state is the number of electrons that the atom of an element loses or gains during chemical bond formation. The oxidation number can be zero, negative, or positive. Use the oxidation number rules to calculate the oxidation state of all atoms.

Complete answer
The reaction given to us is
${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}} + {\text{ }}{{\text{I}}^{\text{ - }}} \to {\text{ C}}{{\text{r}}^{{\text{3 + }}}} + {\text{I}}{{\text{O}}_{\text{3}}}^{\text{ - }}$
We have to determine the oxidation states of all the atoms given in the reaction.
According to the oxidation number rules the oxidation state of monatomic ions is nothing but the charge on it. Thus the oxidation state of ${{\text{I}}^{\text{ - }}}$ and ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ ions is nothing but their respective charges.

Oxidation state of ${{\text{I}}^{\text{-}}}$ = -1
The oxidation state of ${\text{C}}{{\text{r}}^{{\text{3+}}}}$ = +3

Now using the oxidation number rule for the oxygen we can determine the oxidation number of ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2-}}}$ and ${\text{I}}{{\text{O}}_{\text{3}}}^{\text{-}}$ as follows:

The oxidation state of oxygen is always -2 except in peroxide. The peroxide oxidation state of oxygen is -1. Since ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2-}}}$ is not peroxide we will consider the oxidation state of oxygen as-2 and will calculate the oxidation state of ${\text{Cr}}$ as follows:
(Number of ${\text{Cr}}$ atoms) (Oxidation state of ${\text{Cr}}$) + (Number of oxygen atom) (Oxidation state of oxygen) = charge on ion

(Oxidation state of ${\text{Cr}}$) + (7) (-2) = -2
Oxidation state of ${\text{Cr}}$ = +6

Similarly, as ${\text{I}}{{\text{O}}_{\text{3}}}^{\text{-}}$ is not peroxide we will consider oxidation state of oxygen as -2 and will calculate the oxidation state of ${\text{I}}$ as follows:

(Number of ${\text{I}}$ atom) (Oxidation state of ${\text{I}}$) + (Number of oxygen atom) (Oxidation state of oxygen) = charge on ion

(Oxidation state of ${\text{I}}$) + (3) (-2) = -1
Oxidation state of ${\text{I}}$ = +5
Thus, the oxidation state of ${\text{Cr}}$ in ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}$ is +6 and oxidation state of oxygen is -2.

The oxidation state of ${{\text{I}}^{\text{-}}}$ is -1.
The oxidation state of ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ is +3
The oxidation state of ${\text{I}}$ in ${\text{I}}{{\text{O}}_{\text{3}}}^{\text{-}}$ is +5 and oxidation state of oxygen is -2.

Note: Oxidation state is the only apparent charge which represents the positive or negative character of the atom. It is necessary to denote the charge of oxidation state even if it is positive. Oxidation is the loss of electrons while reduction is the gain of electrons. During oxidation, the oxidation number of an atom increases while during reduction the oxidation number of an atom decreases.