Balancing of Redox Reactions: Introduction, Methods, Examples

An important guiding principle in science is the law of conservation of mass, which states that in an isolated system, the mass can neither be created nor be destroyed but can be transformed from one form to another. This law is followed for the balancing of equations.

What is Redox Reaction?

The reaction in which one substance gets oxidized and the other gets reduced simultaneously is called a redox reaction. In simple words, it means that in a chemical reaction, when oxidation and reduction are taking place simultaneously, it is called a redox reaction.

A Classical Concept

Oxidation

Oxidation may be defined as a process that involves the addition of oxygen or any other high electronegative element or as a process that involves the removal of hydrogen or any other electropositive element. For example,

\(2{\text{Mg}} + {{\text{O}}_2} \to 2{\text{MgO}}\) (addition of oxygen)
\(2{{\text{H}}_2}~{\text{S}} + {{\text{O}}_2} \to 2~{\text{S}} + 2{{\text{H}}_2}{\text{O}}\) (removal of hydrogen)

Reduction

Reduction may be defined as a process that involves the addition of hydrogen or an electropositive element or a process that involves the removal of oxygen or an electronegative element. For example,
\({\text{B}}{{\text{r}}_2} + {{\text{H}}_2}~{\text{S}} \to 2{\text{HBr}} + {\text{S}}\)(addition of hydrogen)
\({\text{CuO}} + {{\text{H}}_2} \to {\text{Cu}} + {{\text{H}}_2}{\text{O}}\) (removal of oxygen)

Electron Transfer Concept

Oxidation may be defined as a process in which an atom or an ion loses one or more electrons. That is why oxidation is also called de-electronation. This loss of electrons either increases the positive charge or decreases the negative charge of an atom or ion.

Reduction may be defined as a process in which an atom or ion gains one or more electrons. That is why reduction is also called electronation. This gain of electrons either decreases the positive charge or increases the negative charge of an atom or ion.

For example, \({\text{2Na + }}{{\text{O}}_{\text{2}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}\) (an unbalanced reaction)

Oxidation Number Method

Oxidation is a chemical change in which there occurs an increase in the oxidation number of an atom or atoms, while reduction is a chemical change in which there occurs a decrease in the oxidation number of an atom or atoms. Thus, a redox reaction may be defined as a reaction in which the oxidation number of atoms changes.

For example, \({\text{Zn}} + 2{\text{HCl}} \to {\text{ZnC}}{{\text{l}}_2} + {{\text{H}}_2}\)

Balancing of Redox Reaction

The need to balance a redox equation is to keep the atoms/electrons/oxidation number on the reactant side equal with the atoms/electrons/oxidation number on the product side. Chemical equations of redox reactions can be balanced by using any one of the following methods:

I. Oxidation Number Method

The following steps are involved in this method:

1. Write the skeletal equation of all the reactants and products of the reaction.
2. Indicate the oxidation number of each atom above its symbol.
3. Calculate the increase and decrease in the oxidation number and identify oxidizing and reducing agents.
4. Multiply the formulae of the oxidizing and reducing agent by a suitable integer to equalize the total increase or decrease in the oxidation number.
5. Balance all other atoms other than \({\text{H}}\) and \({\text{O.}}\)
6. Finally, balance \({\text{H}}\) and \({\text{O}}\) by adding \({{\text{H}}_2}{\text{O}}\) molecules by the hit and trial method
7. In the reactions taking place \({\text{n}}\) the acidic medium, balance the \({\text{O}}\) atoms by adding required \({{\text{H}}_2}{\text{O}}\) molecules to the side deficient in \({\text{O}}\) atoms. Then balance the \({\text{H}}\) atoms by adding to \({{\text{H}}^ + }\) the side deficient in \({\text{H}}\) atoms.
8. In the basic medium, first, balance the number of negative charges by adding the required number of \({\text{O}}{{\text{H}}^ – }\) ions to the side deficient in the magnitude of the charges. Then add \({{\text{H}}_2}{\text{O}}\) molecules on the other side in order to balance the \({\text{O}}{{\text{H}}^ – }\) ions added.
9. Let us see some examples of how to balance a redox reaction by the Oxidation number method.

Oxidation Number Method Examples

Example 1: Balance the Equation

\({{\text{C}}_6}{{\text{H}}_6} + {{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\)

Solution:

Step 1. Write the \({\text{O}}.{\text{N}}.\) of each atom in the skeleton equation.

Step 2. Identify the atoms which undergo a change in \({\text{O}}.{\text{N}}.\)

The \({\text{O}}.{\text{N}}.\) of carbon changes from \( – 1\) to \( + 4,\) and the \({\text{O}}.{\text{N}}.\) of oxygen changes from \( 0\) to \( – 2.\)

Step 3. Calculate the total increase in \({\text{O}}.{\text{N}}.\)

The \({\text{O}}.{\text{N}}.\) increase for carbon from \( – 1\) to \( +4\) and decrease for oxygen from \( 0\) to \( – 2.\) For one carbon atom, the increase is \( 5,\) so for \( 6\) carbon atoms, it will be \( 30,\) and for \( 1\) oxygen atom, the decrease is \( 2,\) so for \( 2\) oxygen atoms, it will be \( 4.\)

Step 4. Equate the increase and decrease in \({\text{O}}.{\text{N}}.\) on the reactant side after taking out a common factor of \( 2.\)

\(2{{\text{C}}_6}{{\text{H}}_6} + 15{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\)

Step 5. Balance the number of \({\text{C}}\) and \({\text{O}}\) atoms on both sides of the equation.

\(2{{\text{C}}_6}{{\text{H}}_6} + 15{{\text{O}}_2} \to 12{\text{C}}{{\text{O}}_2} + 6{{\text{H}}_2}{\text{O}}\)

Hence, the equation is balanced.

Example 2. Balance the Equation in an Acidic Medium

\({\text{Cu}} + {\text{NO}}_3^ – \to {\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{u}}^{2 + }}\)

Solution: Step 1. Write the \({\text{O}}.{\text{N}}.\) of each atom of the skeleton equation.

Step 2. Identify the atoms which undergo change in \({\text{O}}.{\text{N}}.\)

The oxidation number of copper is changing from \(0\) to \(+2,\) and the oxidation number of nitrogen is changing from \(+5\) to \(+4.\)

Step 3. Calculate the increase and decrease in \({\text{O}}.{\text{N}}.\) with respect to reactant atoms.

There is an increase in the \({\text{O}}.{\text{N}}.\) of copper from \(0\) to \(+2\) that is an increase of \(2\) in \({\text{O}}.{\text{N}}.,\) and there is a decrease in the \({\text{O}}.{\text{N}}.\) of nitrogen from \(+5\) to \(+4\) that is a decrease of \(1\) in \({\text{O}}.{\text{N}}.\)

Step 4. Equate the increase and decrease in \({\text{O}}.{\text{N}}.\) on the reactant side \({\text{Cu}} + 2{\text{NO}}_3^ – \to {\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{u}}^{2 + }}\)

Step 5. Balance the number of \({\text{Cu}}\) and \({\text{N}}\) atoms on both sides of the equation.

\({\text{Cu}} + 2{\text{NO}}_3^ – \to 2{\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{u}}^{2 + }}\)

Step 6. As the reaction is carried in the acidic medium, balance the number of \({\text{O}}\) atoms by adding two \({{\text{H}}_2}{\text{O}}\) molecules on the product side.

\({\text{Cu}} + 2{\text{NO}}_3^ – \to 2{\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{u}}^{2 + }} + 2{{\text{H}}_2}{\text{O}}\)

Step 7. To balance the number of \({\text{H}}\) atoms, add \(4{{\text{H}}^ + }\) on the reactant side.

\({\text{Cu}} + 2{\text{NO}}_3^ – + 4{{\text{H}}^ + } \to 2{\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{u}}^{2 + }} + 2{{\text{H}}_2}{\text{O}}\)

Hence, the equation is balanced.

Example 3. Balance the Equation in Basic Medium

\(\left[{{\text{C}}r{{\left({{\text{OH}}} \right)}_4}} \right] + {{\text{H}}_2}{{\text{O}}_2} \to {\text{CrO}}_4^{2 – } + {{\text{H}}_2}{\text{O}}\)

Solution: Step 1. Write the \({\text{O}}.{\text{N}}.\) of each atom in the skeleton equation

Step 2. Identify the atoms which undergo change in \({\text{O}}.{\text{N}}.\)

The \({\text{O}}.{\text{N}}.\) of chromium changes from \( + 3\) to \( + 6\) and the \({\text{O}}.{\text{N}}.\) of oxygen atoms in peroxide changes from \( -1\) to \( -2\)

Step3. Calculate the increase and decrease in \({\text{O}}.{\text{N}}.\) with respect to the reactant atoms.

There is an increase in the \({\text{O}}.{\text{N}}.\) of chromium from \( + 3\) to \( + 6\) that is an increase of \( 3\) in \({\text{O}}.{\text{N}}.\) and there is a decrease in the \({\text{O}}.{\text{N}}.\) of one oxygen atom in hydrogen peroxide from \( -1\) to \( -2\) that is a decrease of \( 1\) in \({\text{O}}.{\text{N}}.\)

Step 4. Equate the increase and decrease in \({\text{O}}.{\text{N}}.\) in the reactant side.

\(2{\left[{{\text{Cr}}{{\left({{\text{OH}}} \right)}_4}} \right]^ – } + 3{{\text{H}}_2}{\text{OC}}{{\text{r}}_2} \to {\text{Cr}}{{\text{O}}_4}^{2 – } + {{\text{H}}_2}{\text{O}}\)

Step 5. Balance the number of \({\text{Cr}}\) atoms in the equation.

\(2{\left[{{\text{Cr}}{{\left({{\text{OH}}} \right)}_4}} \right]^ – } + 3{{\text{H}}_2}{{\text{O}}_2} \to 2{\text{CrO}}_4^{2 – } + {{\text{H}}_2}{\text{O}}\)

Step 6. In order to balance the number of oxygen atoms, add five H_2 O molecules on the product side. \(2{\left[{{\text{Cr}}{{\left({{\text{OH}}} \right)}_4}} \right]^ – } + 3{{\text{H}}_2}{{\text{O}}_2} \to 2{\text{CrO}}_4^{2 – } + 6{{\text{H}}_2}{\text{O}}\)

Step 7. As the reaction is carried in the basic medium, in order to balance the number of negative charges add two \({\text{O}}{{\text{H}}^ – }\) ions on the reactant side and two \({{\text{H}}_2}{\text{O}}\) molecules on the product side.

\(2{\left[{{\text{Cr}}{{\left({{\text{OH}}} \right)}_4}} \right]^ – } + 3{{\text{H}}_2}{{\text{O}}_2} + 2{\text{O}}{{\text{H}}^ – } \to 2{\text{CrO}}_4^{2 – } + 8{{\text{H}}_2}{\text{O}}\)

Hence the equation is balanced.

Half Reaction Method or Ion Electron Method

The following steps are involved in this method:

1. Write the skeletal equation and indicate the oxidation number of all the  elements above their symbol.
2. Find out the species which are oxidized and reduced.
3. Split the skeletal equation into two half-reactions- oxidation half-reaction and reduction half-reaction.
4. Balance the two half-reactions separately.
5. The two half-reactions are then multiplied by suitable integers so that the total number of electrons gained in the reduction half-reaction is equal to the total number of electrons lost in the oxidation half-reaction.
6. Verification.

Half Reaction Method Examples

Example 1. Balance the following equation by the Ion-electron method. \({\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 – } + {\text{S}}{{\text{n}}^{2 + }} + {{\text{H}}^ + } \to {\text{C}}{{\text{r}}^{3 + }} + {\text{S}}{{\text{n}}^{4 + }} + {{\text{H}}_2}{\text{O}}\)

Solution: Step 1. Write the \({\text{O}}.{\text{N}}.\) of the atoms involved in the equation:

Step 2. Identify the atoms which undergo a change in the \({\text{O}}.{\text{N}}.\)

There is a change in the \({\text{O}}.{\text{N}}.\) of \({\text{Cr}}\) from \( + 6\) to \( + 3\) and Sn from \( + 2\) to \( + 4.\)

Step 3. Find out the species involved in the oxidation and reduction half reactions.

Sn undergoes oxidation as there is an increase in the \({\text{O}}.{\text{N}}.\) from \( + 2\) to \( + 4\) and \({\text{Cr}}\) undergoes reduction as there is a decrease in \({\text{O}}.{\text{N}}.\) from \( + 6\) to \( + 3.\)

Oxidation half reaction: \({\text{S}}{{\text{n}}^{2 + }} \to {\text{S}}{{\text{n}}^{4 + }}\)

Reduction half reaction: \({\left({{\text{C}}{{\text{r}}_2}{{\text{O}}_7}} \right)^{2 – }} \to {\text{C}}{{\text{r}}^{3 + }}\)

Step 4. Balancing the oxidation half reaction

As the increase in \({\text{O}}.{\text{N}}.\) as a result of oxidation is \(2,\) so add two electrons on the product side to balance the change in \({\text{O}}.{\text{N}}.\)

\({\text{S}}{{\text{n}}^{2 + }} \to {\text{S}}{{\text{n}}^{4 + }} + 2{e^ – }\)

Since the charge is already balanced, therefore the equation is balanced.

Step 5. Balancing of reduction half reaction:

The decrease in \({\text{O}}.{\text{N}}.\) per \({\text{Cr}}\) atom is \(3\) and the total decrease in \({\text{O}}.{\text{N}}.\) for two \({\text{Cr}}\) atoms is \(6.\) Therefore, add 6 electrons on the reactant side.

\({\left({{\text{C}}{{\text{r}}_2}{{\text{O}}_7}} \right)^{2 – }} + 6{{\text{e}}^ – } \to {\text{C}}{{\text{r}}^{3 + }}\)

Balance \({\text{Cr}}\) atoms on both sides of the equation

\({\left( {{\text{C}}{{\text{r}}_2}{{\text{O}}_7}} \right)^{2 – }} + 6{{\text{e}}^ – } \to 2{\text{C}}{{\text{r}}^{3 + }}\)

In order to balance O atoms add seven \({{\text{H}}_2}{\text{O}}\) molecules on the product side and then to balance \({\text{H}}\) atoms add \({\text{14}}\,{{\text{H}}^ + }\) on the reactant side.

\({\left({{\text{C}}{{\text{r}}_2}{{\text{O}}_7}} \right)^{2 – }} + 6{{\text{e}}^ – } + 14{{\text{H}}^ + } \to 2{\text{C}}{{\text{r}}^{3 + }} + 7{{\text{H}}_2}{\text{O}}\)

Step 6. Adding the two half reactions:

i. In order to equate the electrons, multiply the oxidation half reaction by \(3\) and add to the reduction half reaction.

\(\frac{{\frac{{\left[{{\text{S}}{{\text{n}}^{2 + }} \to {\text{S}}{{\text{n}}^{4 + }} + 2{{\text{e}}^ – }} \right] \times 3}}{{{{\left({{\text{Cr}}{{\text{O}}_7}} \right)}^{2 – }}6{{\text{e}}^ – } + 14\,{{\text{H}}^ + } \to 2\,{\text{C}}{{\text{r}}^{3 + }} + 7\,{{\text{H}}_2}{\text{O}}}}}}{{3\,{\text{S}}{{\text{n}}^{2 + }} + {{\left({{\text{Cr}}{{\text{O}}_7}} \right)}^{2 – }} + 14\,{{\text{H}}^ + } \to 3\,{\text{S}}{{\text{n}}^{4 + }} + 2\,{\text{C}}{{\text{r}}^{3 + }} + 7\,{{\text{H}}_2}{\text{O}}}}\)

Summary

In this article, we studied what a redox reaction is and why it should be balanced. We also studied the two different methods to balance a redox reaction. We learnt how to balance the redox reaction step-wise by the oxidation number method and by the ion-electron method. We also studied that the reaction in the acidic medium and basic medium have different methods of balancing.

FAQs on Balancing of Redox Reactions

Q.1. How do you balance redox reactions in acidic and basic solutions?
Ans: In the reactions taking place in the acidic medium, balance the \({\text{O}}\) atoms by adding required \({{\text{H}}_2}{\text{O}}\) molecules to the side deficient in \({\text{O}}\) atoms. Then balance the \({\text{H}}\) atoms by adding \({{\text{H}}^ + }\) to the side deficient in \({\text{H}}\) atoms.
In the basic medium, first, balance the number of negative charges by adding the required number of \({\text{O}}{{\text{H}}^ – }\) ions to the side deficient in the magnitude of the charges. Then add \({{\text{H}}_2}{\text{O}}\) molecules on the other side in order to balance the \({\text{O}}{{\text{H}}^ – }\) ions added. 

Q.2. What are the three methods for balancing redox equations?
Ans: The three methods for balancing redox reactions are:
1. Balancing redox reaction by oxidation number method in acidic medium
2. Balancing redox reaction by oxidation number method in basic medium.
3. Balancing redox reaction by the ion-electron method.

Q.3. How do you identify redox reactions?
Ans: Redox reactions can be identified by the change in oxidation number from the reactant side to the product side. Any reaction in which there is a change in the oxidation number is not a redox reaction. For a reaction to be redox, there should be an increase in oxidation number, as well as there should be a decrease in oxidation number. 

Q.4. How do you balance the redox reactions by the ion-electron method?
Ans: The following steps are involved in this method-
i. Write the skeletal equation and indicate the oxidation number of all the elements above their symbol.
ii. Find out the species which are oxidized and reduced.
iii. Split the skeletal equation into two half-reactions as oxidation half-reaction and reduction half-reaction.
iv. Balance the two half-reactions separately.
v. The two half-reactions are then multiplied by suitable integers so that the total number of electrons gained in one half-reaction is equal to the total number of electrons lost in the other half-reaction.
vi. Verification.