Balancing Redox Equations | Chemistry Class 11 - NEET PDF Download

Balancing of Redox Chemical Equations

Every chemical equation must be balanced according to the law of conservation of mass. In a balanced chemical equation, the atoms of various species involved in the reactants and products must be equal in number. 

Redox reaction can be balanced through 

1. Oxidation number method
2. Half Reaction method

Difference between Oxidation Number Method & Half Reaction Method

Oxidation Number Method

When writing equations for oxidation-reduction reactions, similar to other reactions, you need to know the compositions and formulas of the substances involved in the reaction and those that are produced. 

Let's understand it by the following example, 

Mg + HNO₃ → Mg(NO₃)₂ + N₂O + H₂O

It involves the following steps.

Step I: Write the skeleton equation (if not given).

Step II: Assign the oxidation number of each atom.

Step III: Balance atoms other than H and O in two processes.

Step IV: Equalize the total increase or decrease in oxidation number.

4Mg + 2HNO₃ → 4Mg(NO₃)₂ + NO₂O

Step V: Balance H and O

8H⁺ + 4 Mg + 2HNO₃ + 8NO₃^– → 4 Mg (NO₃)₂ + N₂O + 5H₂O

4 Mg + 10 HNO₃ → 4 Mg (NO₃)₂ + N₂O + 5H₂O

Half Reaction Method

In this approach, you balance each half of the equation separately and then combine them to create a balanced overall equation. This method of balancing was developed by Jette and Lamer in 1927.

Let's understand this with an example:

Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O

It involves the following steps:

Step I: Write the redox reaction in ionic form.

Cu + H⁺ + NO^–₃ → Cu²⁺ + NO + H₂O

Step II: Split the redox reaction into its oxidation half and reduction half-reaction.

Step III: Balance atoms of each half-reaction (except H and O) by using simple multiples.

Cu → Cu²⁺ and NO^–₃ → NO

(Except for H and O, all atoms are balanced)

Step IV: Balance H and O as

(i) For acidic and neutral solutions- Add the H₂O molecule to the side deficient in oxygen and H+ to the side deficient in hydrogen.

(ii) For alkaline solutions- For each excess of oxygen, add one water molecule to the same side and OH^– ion to the other side to balance H.

Step V: Add electrons to the side deficient in electrons.

Step VI: Equalize the number of electrons in both reactions by multiplying a suitable number.

Step VII: Add the two balanced half-reactions and cancel common terms of opposite sides.

Step VIII: Convert the ionic reaction into molecular form by adding spectator ions.

(Ions that are present in solution but do not take part in the redox reaction are omitted while writing the net ionic equation of a reaction and are known as spectator ions.)

Solved Examples

Example: Balance the following Reaction
 VO²⁺ + MnO₄¯ ---> V(OH)₄⁺ + Mn²⁺

Solution:

1) Half reactions:

VO²⁺ ---> V(OH)₄⁺
MnO₄¯ ---> Mn²⁺

2) Balance:

3H₂O + VO²⁺ ---> V(OH)₄⁺ + 2H⁺ + e¯
5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O

3) Equalize electrons:

15H₂O + 5VO²⁺ ---> 5V(OH)₄⁺ + 10H⁺ + 5e¯
5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O

4) Add:

11H₂O + 5VO²⁺ + MnO₄¯ ---> 5V(OH)₄⁺ + Mn²⁺ + 2H⁺

Example: Balance the equation for the reaction of a stannous ion with pertechnetate in an acidic solution. Products are stannic ions, Sn^4+, and technetium(IV), and Tc⁴⁺ ions.

Solution:

1) Net ionic:

TcO₄¯ + Sn²⁺ ---> Tc⁴⁺ + Sn⁴⁺

2) Half-reactions:

TcO₄¯ ---> Tc⁴⁺
Sn²⁺ ---> Sn⁴⁺

3) Balance:

3e¯ + 8H⁺ + TcO₄¯ ---> Tc⁴⁺ + 4H₂O
Sn²⁺ ---> Sn⁴⁺ + 2e¯

4) Equalize electrons:

6e¯ + 16H⁺ + 2TcO₄¯ ---> 2Tc⁴⁺ + 8H₂O
3Sn²⁺ ---> 3Sn⁴⁺ + 6e¯

5) Add:

16H⁺ + 2TcO₄¯ + 3Sn²⁺ ---> 2Tc⁴⁺ + 3Sn⁴⁺ + 8H₂O

Example: Solve the net ionic equation where potassium dichromate(VI) (K₂Cr₂O₇) reacts with sodium sulphite (Na₂SO₃) in an acidic medium to form sulphate ion and chromium(III) ion.

Solution:

Step 1-

The basic equation or the skeletal form of the reaction is

Step 2-

Correctly assign oxidation numbers to Cr and S

Therefore, from the reaction we can decipher that dichromate ion is the oxidant in the reaction and the sulphite ion is the reductant in the reaction

Step 3-

Calculation of the increase and the decrease in the oxidation number for making each side of the equation equal

Step 4-

We know from the equation, the reaction occurs in the acidic medium. Moreover, we can see that the ionic charges in both the sides of the equation are not same. Therefore, we will add 8H⁺ I order to make the ionic charges equal.

Step 5-

In the final step, we will calculate the required amount of water molecules and add it on the right side of the equation to make the equation a balanced redox reaction. In the given equation, we will need to add 4 water molecules on the right side to balance the equation.

Example: Write the balanced redox equation by half-reaction method when Permanganate(VII) ion (MnO₄) produces iodine molecule (I₂)  and manganese (IV) oxide (MnO₂) in a basic medium.

Solution:

Step 1-

Firstly, we will write the base form of the equation

Step 2-

In this step, we will find the two half-reactions and write them

Oxidation Half

Reduction Half

Step 3-

We will balance the iodine atoms present in the oxidation half of the reaction. Therefore, the equation will become

Step 4-

We will balance the oxygen atoms in the reduction half of the reaction by the addition of two water molecules. Refer to the equation below

Now, we will balance the H atoms by the addition of four H+ ions on the left half of the reduction half-reaction.

We know that the reaction occurs in a basic medium. Thus, to balance four H+ ions, we need to add four OH^– ions to each side of the equation.

Finally, we will interchange the H+ ions and OH^– ions with the water molecule. The final equation of the fourth step will be

Step 5-

The charges on the two half-reactions are balanced

Finally, we have to equalize the electrons in the above reactions. Thus, we will multiply the oxidation half of the equation by 3 and the reduction half of the reaction by 2.

Step 6-

We have to determine the net reaction by the addition of two halves of the reaction and by the cancellation of electrons on each side.

Step 7-

Finally, we have to verify the equation in terms of the number of atoms and charges on both sides. Additionally, verification needs to be done about the equations written on both sides.