Oxidation Number Method

The composition and formulas for the reactants and products must be known when creating equations for oxidation-reduction processes. The oxidation number method is best demonstrated in the steps that follow:

While balancing redox reacction we need to be aware about the medium of the reaction. There are two mediums in which redox reaction are present, Acidic Medium and Basic Medium.

Steps to Balance Redox Reaction in Acidic Medium by Oxidation Number Method

Step 1: Assign the oxidation number to all elements in the reaction to identify atoms that change oxidation number during the reaction.

How to Find Oxidation Number?

The oxidation number for some common elements are given in the table below:

• Oxidation number of an atom or molecule is 0 (Zero). E.g., Cu, N₂
• Sum of oxidation number the n compound is 0. eg, CO: Oxidation number (O.N.) of Cu = +2 and O.N. of O is -2, i.e. +2 + (-2) = 0.
• More Electronegative atom = -ve O.S.
• Less Electronegative atom = +ve O.S.

Examples: Find oxidation numbers of,

• Cr₂O₇^2- 

2(x) + 7(-2) = -2, (Here, x is O.N. of Cu and -2 is O.N. of O)

2x – 14 = -2

2x = -2 + 14

x = 6 (Oxidation number of Cr)

• Na₃PO₃ 

3(+1) + (x) + 3(-2) = 0

3 + x – 6 = 0

x = +3 (Oxidation number of P)

Step 2: Identify the elements/atoms that undergo oxidation and reduction.

Step 3: After balancing atoms that have undergone oxidation and reduction, balance charge by cross multiplication.

Step 4: Balance all other atoms except oxygen and hydrogen.

Step 5: Add H₂O to balance the oxygen.

Step 6: By adding H⁺ molecules to the reactants or products, you can make the number of hydrogen atoms in the expression on both sides equal.

Let us use the example below to explain the steps involved in the method.

Example: Balance the Redox Reaction.

P  +  HNO₃   →   HPO₃  +  NO  +  H₂O

Solution:

Step 1: Assign the oxidation number to all elements in the reaction to identify atoms that change oxidation number during the reaction.

Step 2: Identify the elements/atoms that undergo oxidation and reduction. 

Oxidation: P  →  HPO₃

Reduction: HNO₃   →   NO

Step 3: After balancing atoms that have undergone oxidation and reduction, balance charge by cross multiplication.

P  +  HNO₃   →   HPO₃  +  NO  +  H₂O

There is an increase in the O.N. of P from 0 to +5 which is an increase of 5 in O.N. and there is a decrease in the O.N. of N from +5 to +2 which is a decrease of 3 in O.N.

Cross multiplication: 3P  +  5HNO₃   →   3HPO₃  +  5NO  +  H₂O

Step 4: Balance all other atoms except oxygen and hydrogen.

3P  +  5HNO₃   →   3HPO₃  +  5NO  +  H₂O 

Step 5: Add H₂O to balance the oxygen.

3P  +  5HNO₃   →   3HPO₃  +  5NO  +  H₂O 

Step 6: By adding H⁺ molecules to the reactants or products, you can make the number of hydrogen atoms in the expression on both sides equal.

3P  +  5HNO₃   →   3HPO₃  +  5NO  +  H₂O

Hence, the equation is balanced.

Steps to Balance the Redox Reaction in Basic Medium by Oxidation Number Method

Step 1: Assign the oxidation number to all elements in the reaction to identify atoms that change oxidation number during the reaction.

Step 2: Identify the elements/atoms that undergo oxidation and reduction.

Step 3: Balance all other atoms except oxygen and hydrogen.

Step 4: After balancing atoms that have undergone oxidation and reduction, balance charge by cross multiplication.

Step 5: Add H₂O to balance the oxygen.

Step 6: By adding H⁺ molecules to the reactants or products, you can make the number of hydrogen atoms in the expression on both sides equal.

Step 7: Add as many OH^– ions on both sides as a number of H⁺ ions on one side.

Let us use the example below to explain the steps involved in the method.

Example: Balance the Redox Reaction in Basic Medium by Oxidation Number Method.

Cl₂  +  IO₃^–  +  OH^–   →   IO₄^–  +  Cl^–  +  H₂O

Solution:

Step 1: Assign the oxidation number to all elements in the reaction to identify atoms that change oxidation number during the reaction.

Step 2: Identify the elements/atoms that undergo oxidation and reduction.

Oxidation: IO₃^–  →  IO₄^–

Reduction: Cl₂  →  Cl^–

Step 3: Balance all other atoms except oxygen and hydrogen.

Cl₂  +  IO₃^–  + OH   →   IO₄^–  +  2Cl^–  +  H₂O

Step 4: After balancing atoms that have undergone oxidation and reduction, balance charge by cross multiplication.

Cl₂  +  IO₃^–  +OH^–   →   IO₄^–  +  2Cl^–  +  H₂O

There is an increase in the O.N. of I from +5 to +7whicht is an increase of 2 in O.N. and there is a decrease in the O.N. of Cl from 0 to -2 that is a decrease of 2 in O.N.

Cancel difference of (2) from each other

Cl₂  +  IO₃^–  + OH^–   →   IO₄^–  +  2Cl^–  +  H₂O

Step 5: Add H₂O to balance the oxygen

Cl₂  +  IO₃^–   + H₂O   →   IO₄^–  +  2Cl^– 

Step 6: By adding H⁺ molecules to the reactants or products, you can make the number of hydrogen atoms in the expression on both sides equal.

Cl₂  +  IO₃^–  + H₂O   →   IO₄^–  +  2Cl^–  +  2H⁺

Step 7: Add as many OH- ions on both sides as the number of H+ ions on one side.

Cl₂  +  IO₃^–  + H₂O  +  2OH^–   →   IO₄^–  +  2Cl^–  +  2H⁺  +  2OH^–

Cl₂  +  IO₃^–  + H₂O  +  2OH^–   →   IO₄^–  +  2Cl^–  +  2H₂O

Cl₂  +  IO₃^–  +  2OH^–   →   IO₄^–  +  2Cl^–  +  (2H₂O – H₂O)

Cl₂  +  IO₃^–  +  2OH^–   →   IO₄^–  +  2Cl^–  +  H₂O

Hence, the equation is balanced.

Redox Reactions are the reaction in which oxidation and reduction occur in the same reaction. Balancing redox reactions is the process in which we balance redox reaction equations using various methods. There are generally two methods that are used to balance redox reactions that include,

• Oxidation Number Method
• Ion Electron Method

Balancing redox reactions is very important because redox reactions are observed all around us and having the knowledge of balancing redox reactions helps us to achieve various things in our daily life. Before starting with balancing the redox reaction we have to first know what Redox Reactions are, Oxidation and Reduction reactions, and Balancing Redox reactions using the Oxidation Number method and Ion-Electron method, their examples, and others in detail.