Question

The following reaction ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}+\text{S}{{\text{O}}_{2}}+\text{N}{{\text{a}}_{2}}\text{Cr}{{\text{O}}_{4}}\to \text{C}{{\text{r}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}+\text{N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{4}}$
Balance the equation with the oxidation method.

Answer 1

To solve this question, we need to identify the oxidation part and the reduction part of the given chemical reaction. Then we will have to balance the given chemical reaction. We shall balance the charge, the number of atoms of chromium and sodium and then the number of oxygen and hydrogen atoms.

Complete Step-by-Step Solution
Let us write the oxidation number of each of the molecules of the chemical compound taking part in the chemical reaction
$\overset{+1}{\mathop{{{\text{H}}_{2}}}}\,\overset{+6}{\mathop{\text{S}}}\,\overset{-2}{\mathop{{{\text{O}}_{4}}}}\,+\overset{+4}{\mathop{\text{S}}}\,\overset{-2}{\mathop{{{\text{O}}_{2}}}}\,+\overset{+1}{\mathop{\text{N}}}\,{{\text{a}}_{2}}\overset{+6}{\mathop{\text{Cr}}}\,\overset{-2}{\mathop{{{\text{O}}_{4}}}}\,\to \overset{+3}{\mathop{\text{C}{{\text{r}}_{2}}}}\,{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}+\overset{+1}{\mathop{\text{N}{{\text{a}}_{2}}}}\,\overset{+6}{\mathop{\text{S}}}\,\overset{-2}{\mathop{{{\text{O}}_{4}}}}\,$
Let us first identify the oxidation part of the reaction
Oxidation: $2N{{a}^{+}}+S{{O}_{2}}+2{{H}_{2}}O\to N{{a}_{2}}S{{O}_{4}}+4{{H}^{+}}+2{{e}^{-}}$
Let us now identify the reduction part of the reaction
Reduction: $3S{{O}_{4}}^{2-}+2N{{a}_{2}}Cr{{O}_{4}}+16{{H}^{+}}+6{{e}^{-}}\to C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+4N{{a}^{+}}+8{{H}_{2}}O$
Now, we have to write the overall reaction and balance it
But first we will have to multiply the oxidation part of the chemical reaction by $3$ for easy calculations
Now, the oxidation part becomes
Oxidation: $6N{{a}^{+}}+3S{{O}_{2}}+6{{H}_{2}}O\to 3N{{a}_{2}}S{{O}_{4}}+12{{H}^{+}}+6{{e}^{-}}$
So, we get the overall reaction as:
$6N{{a}^{+}}+3S{{O}_{2}}+6{{H}_{2}}O+3S{{O}_{4}}^{2-}+16{{H}^{+}}+2N{{a}_{2}}Cr{{O}_{4}}\to 3N{{a}_{2}}S{{O}_{4}}+12{{H}^{+}}+C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+4N{{a}^{+}}+8{{H}_{2}}O$
Upon further calculations, we get
$2N{{a}^{+}}+4{{H}^{+}}+3S{{O}_{2}}+3S{{O}_{4}}^{2-}+2N{{a}_{2}}Cr{{O}_{4}}\to 3N{{a}_{2}}S{{O}_{4}}+C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+2{{H}_{2}}O$
Now, we know that
Sodium ion and sulphate ion will form sodium sulphate
$2N{{a}^{+}}+S{{O}_{4}}^{2-}\to N{{a}_{2}}S{{O}_{4}}$
And
Hydrogen ion and sulphate ion will form sulphuric acid
$4{{H}^{+}}+2S{{O}_{4}}^{2-}\to 2{{H}_{2}}S{{O}_{4}}$
So, the total reaction will become:
$N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}+2N{{a}_{2}}Cr{{O}_{4}}\to 3N{{a}_{2}}S{{O}_{4}}+C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+2{{H}_{2}}O$
Upon balancing the chemical equation, we get
$\text{2}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}+3\text{S}{{\text{O}}_{2}}+2\text{N}{{\text{a}}_{2}}\text{Cr}{{\text{O}}_{4}}\to \text{C}{{\text{r}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}+2\text{N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{4}}+2{{H}_{2}}O$

Additional Information
The total number of molecules of a chemical species involved in a chemical reaction is described by a stoichiometric coefficient. The ratio between the reacting species and the products formed in the reaction is established. The total number of atoms of an element present in a species (in a balanced chemical equation) is equal to the product of the stoichiometric coefficient and the number of element atoms present in the species in a single molecule.

Note
The addition of stoichiometric coefficients to the reactants and products includes balancing chemical equations. This is important because the law of conservation of mass and the law of constant proportions must be complied with by a chemical equation, that is, the same number of atoms of each element must exist on the reactant side and the product side of the equation.