Question

Balance the redox reaction by oxidation number method. It occurs in acidic medium.
${H_2}{O_2} + C{r_2}{O_7}^{ - 2} \to {O_2} + C{r^{ + 3}}$

Answer 1

Hint: As we know that the chemical equation given is an oxidation-reduction (redox) reaction. In this chemical equation $C{r_2}{O_7}^{2 - }$is an oxidising agent and ${H_2}{O_2}$is a reducing agent. In an acidic medium we will try to balance the chemical equation by using ${H_2}O$ and ${H^ + }$.

Complete step by step answer:
As we can see that in the given chemical equation , both oxidation and reduction is going on. The oxygen in the hydrogen peroxide in the reactant side has an oxidation number of $ - 1$. And the oxygen in the product side has an oxidation state of $0$ . So oxygen is undergoing an oxidation reaction.
$6{O^{ - 1}} \to 6{O^0} + 6{e^ - }$
The chromium has an oxidation state of $ + 6$ in the $C{r_2}{O_7}^{2 - }$ in the reactant side and it has an oxidation state of $ + 3$in the product side. So chromium is undergoing a reduction reaction. The reduction reaction can be written as $2C{r^{ + 6}} + 6{e^ - } \to 2C{r^{ + 3}}$ . The complete reaction can be written as : $C{r_2}{O_7}^{ - 2} + 3{H_2}{O_2} + 8{H^ + } \to 2C{r^{ + 3}} + 3{O_2} + 7{H_2}O$ . The ${H^ + }$ and ${H_2}O$ is used to balance the hydrogen and oxygen atoms in an acidic medium.
So the complete balanced equation in acidic medium is: $C{r_2}{O_7}^{ - 2} + 3{H_2}{O_2} + 8{H^ + } \to 2C{r^{ + 3}} + 3{O_2} + 7{H_2}O$
Additional information: Hydrogen peroxide ${H_2}{O_2}$ is also known as Dioidane, Oxidanyl and per hydroxic acid. In pure form it is a pale blue liquid. It is slightly more viscous than water.
 $C{r_2}{O_7}^{2 - }$ has an orange colour. The $C{r^{ + 3}}$has a green colour.

Note:
Always remember that $C{r_2}{O_7}^{2 - }$ is a very strong oxidising agent . ${H_2}{O_2}$ can act as both oxidising agent and reducing agent according to the situation. In an acidic medium we use ${H^ + }$ and ${H_2}O$to balance the oxygen and hydrogen atoms . The $C{r_2}{O_7}^{2 - }$is orange coloured and $C{r^{ + 3}}$ is green coloured.