Question

Balance the following redox reactions by ion electron method.
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + S}}{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{(aq) + SO}}_{\text{4}}^{{\text{2 - }}}{\text{(aq)}} $ (in acidic solution).

Answer 1

Hint: In the above question, we have to balance the redox reaction. For this, we first have to divide the reaction into oxidation half and reduction half according to the oxidation state of transition metals. Then we will add that number of electrons gained or lost in the reaction. To balance this, we will add equivalent amount of $ {{\text{H}}^{\text{ + }}} $ and $ {{\text{H}}_{\text{2}}}{\text{O}} $ .

Formula Used:
Oxidation state= charge on compound- (charge on $ {{\text{i}}^{{\text{th}}}} $ ion $ {\text{ \times }} $ number of $ {{\text{i}}^{{\text{th}}}} $ ion)

Complete Step by step solution:
In the above question, since, we have to balance the below redox reaction:
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + S}}{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{(aq) + SO}}_{\text{4}}^{{\text{2 - }}}{\text{(aq)}} $
Let the oxidation state of Cr in $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}} $ be x.
 $ 2x - 2 \times 7 = - 2 $
 $ {\text{x = }}\frac{{{\text{12}}}}{{\text{2}}}{\text{ = 6}} $
Therefore, oxidation state of Cr in $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}} $ is 6.
Let the oxidation state of S in $ {\text{S}}{{\text{O}}_{\text{2}}} $ be x.
 $ x - 2 \times 2 = 0 $
 $ {\text{x = 4}} $
Therefore, the oxidation state of S in $ {\text{S}}{{\text{O}}_{\text{2}}} $ is 4.
The oxidation state of Cr in $ {\text{C}}{{\text{r}}^{{\text{3 + }}}} $ is 3.
Let the oxidation state of S in $ {\text{SO}}_{\text{4}}^{{\text{2 - }}} $ be x.
 $ x - 2 \times 4 = - 2 $
 $ {\text{x = 6}} $
Therefore, the oxidation state of S in $ {\text{SO}}_{\text{4}}^{{\text{2 - }}} $ is 6.
The oxidation state of Cr changes from $ {\text{ + 6}} $ to $ {\text{ + 3}} $ , so it is undergoing reduction.
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}} $
We will multiply 2 to the reactant side to balance the Cr atom. Since, The oxidation state of Cr changes from $ {\text{ + 6}} $ to $ {\text{ + 3}} $ and 2 Cr atoms are added, so that, we will add 6 electrons on the product side. So, we have:
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}} $
Let us now add 7 molecules of water to balance the oxygen atom.
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_{\text{2}}}{\text{O}} $
Let us now, balance the hydrogen atom:
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}}{\text{ + 14}}{{\text{H}}^{\text{ + }}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_{\text{2}}}{\text{O}} $ ....................(1)
The oxidation state of S changes from $ {\text{ + 4}} $ to $ {\text{ + 6}} $ , so it is undergoing oxidation.
 $ {\text{S}}{{\text{O}}_{\text{2}}} \to {\text{SO}}_{\text{4}}^{{\text{2 - }}} $
Since, the oxidation state of S changes from $ {\text{ + 4}} $ to $ {\text{ + 6}} $ , so we can add 2 electrons on the product side. And to balance, oxygen atoms will have 2 molecules of water. So, we get:
 $ {\text{S}}{{\text{O}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} $
Let us now balance the number of hydrogen atom:
 $ {\text{S}}{{\text{O}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}{\text{ + 4}}{{\text{H}}^{\text{ + }}} $ ....................(2)
Since, in equation (1), we have 6 electrons, we will multiply 3 to equation (2) and add it to equation (1). Hence, we get:
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}}{\text{ + 14}}{{\text{H}}^{\text{ + }}}{\text{ + 3S}}{{\text{O}}_{\text{2}}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_{\text{2}}}{\text{O + 3SO}}_{\text{4}}^{{\text{2 - }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}}{\text{ + 12}}{{\text{H}}^{\text{ + }}} $
Rearranging the above equation, we get:
 $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}{\text{ + 3S}}{{\text{O}}_{\text{2}}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + 3SO}}_{\text{4}}^{{\text{2 - }}} $
Hence, the balance equation is $ {\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{{\text{2 - }}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}{\text{ + 3S}}{{\text{O}}_{\text{2}}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + 3SO}}_{\text{4}}^{{\text{2 - }}} $ .

Note:
An oxidation-reduction reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is a type of chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.
Redox reactions are common and vital to basic functions of life such as photosynthesis, respiration, combustion, and corrosion or rusting.