Question

\[\text{SO}_{2}\] reduces \[\text{Cr}_{2}\text{O}_{7}^{2-}\]$ to $\[\text{Cr}^{2+}\] . The change in oxidation number of \[\text{Cr}\] is:

Answer 1

Hint: We can see that it is given in the reduction of \[\text{Cr}_{2}\text{O}_{7}^{2-}\] takes place due to the presence of \[\text{SO}_{2}\] and leads to the formation of \[\text{Cr}^{2+}\] . Therefore, we can say that in this reaction, \[\text{SO}_{2}\] act as a reducing agent.

Complete step by step answer:
We can assume that the oxidation number of \[\text{Cr}\] in \[\text{Cr}_{2}\text{O}_{7}^{2-}\] be \[x\]. The value of \[x\] is calculated as shown below.
$ 2 x+7(-2) =-2$
x =$\dfrac{-2+14}{2}$
=+6
We can say that in the case of \[{Cr}^{2+}\] the oxidation number of \[{Cr}\] in \[Cr^{2+}\] is equal to the charge on \[{Cr}\] atom that is \[+2\].
We can write the reduction reaction in which \[t{Cr}_{2}{O}_{7}^{2-}\] reduces to \[{Cr}^{2+}\] can be written as follows.
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow 2 \mathrm{Cr}^{2+}$
We can see that in the above reaction the change in the oxidation number per \[\text{Cr}\] atom is \[-4\]. Thus we can conclude that the oxidation number of \[\text{Cr}\] decreases by four.

Note:
We know that the decrease in the oxidation number of species in a redox reaction is regarded as reduction. However, the increase in the oxidation number is considered as oxidation.