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# Balance the equation using Oxidation number method.${{\text{S}}^{ - 2}}{\text{ + N}}{{\text{O}}_3}^ - \to {\text{ N}}{{\text{O}}_2}{\text{ + S}}{{\text{O}}_4}$

Last updated date: 16th Jun 2024
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Hint: Oxidation number method involves splitting the given equation into two separate halves and balancing the two equations separately. It is required to identify the reduction half and the oxidation half.

As mentioned, the equation must be divided into two halves. The first half will be the reaction that denotes the reactant that is undergoing reaction. The other half represents the reactant that is undergoing oxidation. For this we must find out the change in the OS states of both Sulphur and Nitrogen.
Oxidation half: Sulphur is undergoing change from $- 2 \to + 6$ and therefore it has undergone oxidation. The reaction will look like this:
${{\text{S}}^{ - 2}} \to {\text{S}}{{\text{O}}_4}^{ - 2}$
Reduction half: Nitrogen is undergoing change from $+ 5 \to + 4$. Therefore, the reaction will look like this:
${\text{ N}}{{\text{O}}_3}^ - \to {\text{ N}}{{\text{O}}_2}$
Now it is required to add the electrons so as to make sure that the number of electrons lost in the oxidation half is equal to the number of electrons gained in the reduction half on both the sides of the two reactions.
O: ${{\text{S}}^{ - 2}} \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - }$
R: ${\text{ N}}{{\text{O}}_3}^ - + {e^ - } \to {\text{ N}}{{\text{O}}_2}$
Now we must balance all the elements in both reactions except for Hydrogen and oxygen.
O: ${{\text{S}}^{ - 2}} \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - }$
R: ${\text{ N}}{{\text{O}}_3}^ - + {e^ - } \to {\text{ N}}{{\text{O}}_2}$
We must now balance the hydrogen and oxygen atoms.
O: ${{\text{S}}^{ - 2}} \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - } + 8{H^ + }$
R: ${\text{ N}}{{\text{O}}_3}^ - + {e^ - } + 2{H^ + } \to {\text{ N}}{{\text{O}}_2}$
The reason why the hydrogen ions are added is to balance the charges on both the sides of the reactions.
The next step is to balance the $O$atoms by adding oxygen.
O: ${{\text{S}}^{ - 2}} + 4{H_2}O \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - } + 8{H^ + }$
R: ${\text{ N}}{{\text{O}}_3}^ - + {e^ - } + 2{H^ + } \to {\text{ N}}{{\text{O}}_2} + {H_2}O$
Now we must make sure that the same number of electrons are present in both halves. Therefore, we will multiply by 8 because the oxidation half has eight electrons and the reduction half has only one electron.
The new reactions will be:
O: ${{\text{S}}^{ - 2}} + 4{H_2}O \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - } + 8{H^ + }$
R: ${\text{8 N}}{{\text{O}}_3}^ - + 8{e^ - } + 16{H^ + } \to 8{\text{ N}}{{\text{O}}_2} + 8{H_2}O$
Now we can finally add both the equations together. The reaction will be:
${{\text{S}}^{ - 2}} + {\text{8 N}}{{\text{O}}_3}^ - + 8{e^ - } + 16{H^ + } + 4{H_2}O \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{e^ - } + 8{H^ + } + 8{\text{ N}}{{\text{O}}_2} + 8{H_2}O$
Subtract the reactants that are common on both sides and add the reactants that are common on the same side.
The new reaction will be: ${{\text{S}}^{ - 2}} + {\text{8 N}}{{\text{O}}_3}^ - + 8{H^ + } \to {\text{S}}{{\text{O}}_4}^{ - 2} + 8{\text{ N}}{{\text{O}}_2} + 4{H_2}O$

So, the correct answer is Option A.

Note: Make sure that the sum of signs on the left-hand side is equal to the sum of the signs on the right-hand side. If both are equal then we can say that the reaction is balanced.
Also, it is important to understand if the reaction takes place in the acidic or basic medium. As the ions to be added will differ on the product side. that is, if it is acidic then we will see hydrogen ions on the reactant side and if it is basic then we will see hydroxide ions on the reactant side.