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# Balancing a Chemical Equation by Oxidation Number Method for JEE      LIVE
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## Chemical Equation: An Introduction

A chemical equation is the depiction of the chemical reaction that occurs between the reactants leading to the formation of the products. It is represented by the molecular formula of the compounds, a plus sign (+) to show what other compound is taken and/or formed, and a right-arrow ($\rightarrow$). For example, consider the reaction of formation of ammonia: $\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$ showing that nitrogen and hydrogen combine to form ammonia. Here, nitrogen and hydrogen are our reactants (left side of the arrow), and ammonia is the product (right side of the arrow).

Redox chemical reactions are the chemical equations in which oxidation and reduction occur simultaneously. Simply put, it transfers electrons from one species to another.

## Why Balance a Chemical Equation?

The chemical formulas for reacting and producing substances are represented in a chemical equation. The reactants' and products' atom counts are meant to be equal. The inclusion of stoichiometric coefficients to products and reactants is required to balance the chemical equations. This is important because any chemical equation must follow the law of constant proportions and conservation of mass, which states that the product and reactant sides of the equation must include the same amount of atoms of each element.

## Balancing by Oxidation Number Method

For balancing the redox reactions by oxidation number method, you find out the oxidation number of each of the participating atoms to keep track of the electron transfer. Balancing by oxidation number method requires certain steps which must be followed and are discussed below as steps of balancing a chemical equation by oxidation number method:

Step 1

Start by writing the unbalanced chemical equation. For example: $\mathrm{Fe}+\mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}$.

Step 2

Now, identify the oxidation states of each atom present.

Chemical Equation is -$\mathrm{Fe}+\mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}$

 Reactant Side Product Side Oxidation number of Fe = 0 Oxidation number of Fe = +3 Oxidation number of O = 0 Oxidation number of O = -2

Step 3

• List the change in oxidation number of each atom from reactant to product.

• So for Fe: 0 to +3 = Change of +3

• And for O: 0 to -2 = Change of -2

Step 4

• Now, the total increase and the total decrease in oxidation number is made equal by using the coefficient.

• That is, two atoms of Fe are needed for every three atoms of O, giving a total of +6 and -6.

Step 5

• These coefficients are then placed in front of the chemical formulas containing these atoms.

• However, in this case, one must note that from the unbalanced chemical equation, an oxygen molecule is involved, which contains two atoms of oxygen. Thus, the coefficient while writing a balanced chemical equation becomes $\dfrac{3}{2}$.

• So, the balanced chemical equation becomes: $2 \mathrm{Fe}+\dfrac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}$.

Step 6

All the other atoms were balanced if present, except for hydrogen and oxygen. In this case, oxygen is involved. Thus, it is balanced first with iron.

Step 7

Remaining oxygen atoms are balanced if present.

Step 8

Remaining hydrogen atoms are balanced if present.

The final balanced chemical equation becomes: $2 \mathrm{Fe}+\dfrac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}$.

Balancing a chemical equation examples:

Example 1

Step 1

Consider the chemical equation:  $\mathrm{Cu}+\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}$

Step 2

Chemical Equation is: $\mathrm{Cu}+\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}$

 Reactant Side Product Side Oxidation number of Cu = 0 Oxidation number of Cu = +2 Oxidation number of N = +5 Oxidation number of N = +4

Step 3

• List the change in oxidation number of each atom from reactant to product.

• So for Cu: 0 to +2 = Change of +2

• And for N: +5 to +4 = Change of -1

Step 4

• Now, the total increase and the total decrease in oxidation number is made equal by using the coefficient.

• That is, 1 atom of Cu is needed for every 2 atoms of N, giving a total of +2 and -2.

Step 5

• These coefficients are then placed in front of the chemical formulas containing these atoms.

• So, the balanced chemical equation becomes: $\mathrm{Cu}+2\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}$.

Step 6

• All the other atoms were balanced if present, except for hydrogen and oxygen, which is chlorine.

Step 7

• Remaining oxygen atoms are balanced. 6 oxygen atoms on the reactant side and 11 oxygen atoms on the product side.

• If we put 4 as a coefficient of HNO3 and 2 as a coefficient of H2O, we get 12 oxygen atoms on both sides. Thus, $\mathrm{Cu}+4\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2\mathrm{NO}_{2}+2\mathrm{H}_{2} \mathrm{O}$.

Step 8

Remaining hydrogen atoms are balanced if present.

The final balanced chemical equation becomes: $\mathrm{Cu}+4\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2\mathrm{NO}_{2}+2\mathrm{H}_{2} \mathrm{O}$.

## Conclusion

Chemical equations depict the utilization of reactant molecules and formation of the product molecules. Hence, balancing the chemical equations becomes necessary to understand the composition of the reaction happening and further use it to calculate other parameters of the reaction. The oxidation number method for balancing the chemical equations or redox chemical equations requires the calculation of the oxidation number, which helps us in deducing the number of electrons that are transferred. Calculating the coefficients on this basis gives us the balanced chemical equation.

Last updated date: 26th Sep 2023
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## FAQs on Balancing a Chemical Equation by Oxidation Number Method for JEE

1. Is the oxidation number method and half-reaction method the same?

No, both the methods are carried out differently but they both require the use of oxidation number of all the atoms to keep track of the electrons' change during oxidation and reduction. In the oxidation number method, we carry out everything in the given unbalanced equation. However, in the half-reaction method, we write 2 half reactions separately, balance them, add them again and balance them as a whole again if required. So, despite providing the same results, the 2 approaches are still different.

2. How to balance the reaction between ferric chloride and stannous chloride?

The unbalanced reaction between ferric chloride and stannous chloride is represented as:

$\mathrm{FeCl}_{3}+\mathrm{SnCl}_{2} \rightarrow \mathrm{FeCl}_{2}+\mathrm{SnCl}_{4}$.

Oxidation state of Fe changes from +3 to +2 which is a change of -1. Sn changes from +2 to +4 which is a change of +2. So, after equating both of them, 2 atoms of Fe are required for every 1 atom of Sn. Thus, the balanced chemical equation becomes:

$2\mathrm{FeCl}_{3}+\mathrm{SnCl}_{2} \rightarrow 2\mathrm{FeCl}_{2}+\mathrm{SnCl}_{4}$

(in this process, chlorine atoms end up getting balanced as well).