# Balancing Redox Reactions

In balancing redox reaction Reduction is defined as the gain of electrons and oxidation is defined as loss of electrons. In other words, oxidation is loss whereas reduction is the gain. The reaction which consists of both oxidation and reduction is termed as a redox reaction. Now we need to learn how to balance these redox reactions.

There are two ways of balancing a redox reaction. One method is by using the change in oxidation number of oxidizing agent and the reducing agent and the other method is based on dividing the redox reaction into two half reactions-one of reduction and other of oxidation.

### Balancing by oxidation number

This method can be best explained with the help of the following steps:

• Correct formula should be written for both reactant and product.
• Assign the oxidation number to all the elements present in the reaction and hence identify the atoms that will undergo a change in oxidation number.
• Increase or decrease of oxidation number should be calculated.
• Make the total ionic charges of reactant and product equal by adding H+ or OH when the reaction is taking place in water.
• Hydrogen atoms in the reaction are made equal by adding H2O molecule either to reactant side or product side.

### Balancing by half-reaction method

In this method, the redox reaction is split into two half reactions-one of reduction and other of oxidation and then combined back to give a balanced reaction.

This method can be best explained with the help of the following steps:

• Put the unbalanced equation in the ionic form for the given reaction.
• Divide the equations into half reactions: Oxidation half and reduction half
• For reactions occurring in acidic medium, add H2O to balance O atoms and H+ to balance H atoms.
• Balancing of charges should be done by adding electrons
• Add the two half-reactions when the number of electrons exchanged is equal to form a redox reaction.

Let’s explain these steps with the help of an example.

1. $$Fe (OH)_2 + H_2O_2 \rightarrow Fe(OH)_3 + H_2O$$
2. We balance the half reactions:$$(Fe(OH)_2 + OH^- \rightarrow Fe(OH)_3 + e^-) x 2$$
$$H_2O_2 + 2e^- \rightarrow 2OH^-$$
After balancing we can find that, there is no presence of $$H_2O$$.
3. The balanced equation is:
$$2 Fe(OH)_2 + H_2O_2 \rightarrow 2 Fe(OH)_3$$

#### Practise This Question

How many H2O  should be added to the RHS of the following equation to balance number of oxygen atoms on the left ?

6Fe2++Cr2O27+14H+6Fe3++2Cr3++ ___________