Balance Redox Reactions

Everything in life requires balance. Too much of chocolate is not good as well. Even reactions need balancing. An important part of chemistry is balancing redox reactions. What are redox reactions? The rusting of iron is a redox reaction. Many other redox reactions take place around us. Let’s learn about them and study the steps of balancing redox reactions.

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Redox Reactions

The reduction is the gain of electrons whereas oxidation is the loss of electrons. The combination of reduction and oxidation reaction together refers to redox reaction/redox process. As discussed, it is very important to understand “balancing redox reactions”.

There are generally two methods for balancing redox reactions (chemical equations) in a redox process. The two methods are- Oxidation Number Method & Half-Reaction Method.

Browse more Topics under Redox Reactions

• Classical Idea of Redox Reactions
• Oxidation Number
• Redox Reactions and Electrode Potential
• Types of Redox Reactions
• Redox Reactions as the Basis of Titrations
• Redox Reactions – Electron Transfer Reactions

Balancing Redox Reactions

Oxidation number method is based on the difference in oxidation number of oxidizing agent and the reducing agent. Half-reaction method depends on the division of the redox reactions into oxidation half and reduction half. It depends on the individual which method to choose and use.

Balancing Redox Reactions by Oxidation Number Method

As with every other reaction, it is very important to write the correct compositions and formulas. A very important thing to keep in mind while writing oxidation-reduction reactions is to correctly write the compositions and formulas of the substances and products present in the chemical reaction. The steps of the oxidation number method are as follows:

Step 1

Correctly write the formula for the reactants and the products of the chemical reaction.

Step 2

Determine correctly the atoms that undergo oxidation number change in the given reaction by allocating the oxidation number of the individual elements present in the reaction.

Step 3

Calculate the oxidation number on the basis of each atom for the given molecule or ion of the chemical reactions. If the numbers are not equal then multiply it to such a number that overall these numbers become equal. If in a case, two substances are either only oxidized or only reduced then this signifies that something is wrong with the chemical reaction. This signifies that either the formulas of the reactant or the products are incorrect. This can also mean that the allocation of oxidation numbers is incorrect.

Step 4

Keep in mind the involvement of the ions if the reaction occurs in water. Accordingly, add H⁺ or OH^– ions in the appropriate side of the reaction. Overall, the ionic charges of reactant and products will be equal. However, if the reaction takes place in acidic solution then add H⁺ ions in the chemical equation. Similarly, if the reaction takes place in the basic solution add OH^– ions in the chemical equation.

Step 5

It is very important to equate the number of hydrogen atoms on each side of the equation by adding water molecules or H₂O molecules. Additionally, it is necessary to check the oxygen atoms present in the equation. It will be a balance reaction if there are equal numbers of oxygen atoms present in both the reactant as well as the product the side.

We will further understand the steps of balancing redox reactions by solving a problem on the basis of oxidation number method.

Balancing Redox Reactions by Half-Reaction Method

In this procedure, we split the equation into two halves. Thereafter, we balance both the parts of the reaction separately. Finally, we add them together to give a balanced equation. We will demonstrate this method with an example so as to understand the steps of balancing redox reactions by half-reaction method.

For instance, a reaction is given where Fe²⁺ ions are converted to Fe³⁺ ions by dichromate ions in an acidic solution. The dichromate ions (Cr₂O₇^2–) are reduced to Cr³⁺ ions in the reaction. We have to balance the above redox reaction. The steps to balance this equation is as follows.

Step 1

First, we have to write the basic ionic form of the equation.

Step 2

Divide the equation into two separate half reaction-oxidation half and reduction half.

• Oxidation half

• Reduction half

Step 3

In the third step of balancing redox reactions by half-reaction method, we will balance the atoms present in each half of the reaction except O and H atoms. In the example question, the oxidation part of the reaction in terms of Fe atoms is already balanced. Therefore, we will just balance the reduction part of the reaction. In this case, we will multiply Cr³⁺ by 2 in order to balance Chromium atoms.

Step 4

We know that the reaction takes place in an acidic solution. Therefore, we have to add water molecules (H₂O) for balancing the O atoms of the equation and H⁺ for balancing the H atoms in the equation. Now the equation is

Step 5

Now, we will need to balance the charges in both the half reactions. Therefore, we need to multiply the appropriate number to one or both the half reaction and make the number of electrons same. Balancing the oxidation half of the reaction

For the reduction half, there are 12 positive charges on the left side of the equation and 6 positive charges on the right side of the equation. Therefore, we need to add 6 more electrons on the left side of the equation to balance the reduction half.

Now, to equate the electrons in two halves of the reactions, we will multiply 6 in the oxidation half reaction. Thus, we get

Step 6

The two halves of the equations are added to complete the overall reactions. After the addition of two reaction halves, cancel the electrons on both sides. The net ionic equation can be written as:

Step 7

Finally, we have to verify whether the equation consists of the same type, number, and charges on both sides of the equation. Moreover, the equation is completely balanced in terms of atoms and charges.

In case of a basic solution, we have to balance the atom similar to acidic solution. Then the equal number of OH^– ions addition is done for each H⁺ ion, in both the halves of the equation. When the H⁺ ions and OH^– ions will be present on the same side of the equation, we have to combine the ions and write H₂O.

Solved Question for You

Q. 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.

Q2. 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.