How To Balance Redox Reaction: Exhaustive Medium And Methods

In this article, “how to balance redox reaction” different types of balancing method of redox reactions are discussed briefly.

Redox reaction is basically an electron transfer reaction. Electrons are donated by reducing agent to oxidizing agent to drive a redox reaction towards forward direction. Balancing of a redox reaction is essential to determine the number of electrons involved in the redox reaction.

Some frequently asked questions with answers on how to balance the redox reaction is described in this article.

How to balance a redox reaction in basic medium?

To balance any redox reaction in basic medium, these following steps must be maintained.

Oxidation and reduction half reaction will be determined at the very first step.
Then the elements participate in those half reactions will be balanced keeping aside oxygen and hydrogen (O,H)
One H₂O is added to the same side and 2OH^– to the other side for each oxygen to balance the number of oxygens in redox reaction.

Let’s take an example of a redox reaction balanced in basic medium.

Ag (s) + Zn²⁺ (aq)→ Ag₂O (aq) + Zn (s)

Oxidation half reaction: 2Ag +2OH^–→ Ag₂O +H₂O +2e^–——-1 no equation
Reduction half reaction: Zn²⁺ → Zn +2e^– ————–2 no equation
(1no equation ×1)——- 2Ag + 2OH^– → Ag₂O + H₂O + 2e^–
(2no equation× 1)——- Zn²⁺→ Zn +2e^–
Net balanced equation——- 2Ag + Zn²⁺ + 2OH^– → Ag₂O +Zn +H₂O

To know more please follow: Peptide Bond Formation: How, Why, Where, Exhaustive Facts Around It

How to balance a redox reaction in acidic medium?

The steps to balance the redox reaction in acidic medium are written below-

Like the balancing method in basic medium, at the first step the oxidation and reduction half reaction should be identified.
The number of all the participating elements will be balanced without oxygen and hydrogen.
To balance each of the oxygen on one side of the equation one water molecule (H₂O) is added on the same side and on the other side H⁺ is added.

Let’s take an example of an unbalanced redox equation and see the steps to balance the equation.

Fe²⁺ +MnO₄^– → Fe³⁺ + Mn²⁺

Oxidation half reaction: Fe²⁺→Fe³⁺ + e^– ——–1 no equation
Reduction half reaction: MnO₄^– + 8H⁺→Mn²⁺+ 4H₂O + 5e^–——2 no equation
1no equation × 5)——— 5Fe²⁺ → 5Fe³⁺ + 5e^–
2no equation ×1)———MnO₄^– + 8H⁺→ Mn²⁺ + 4H₂O + 5e^–
Net balanced equation is—— 5Fe²⁺ + MnO₄^– + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

How to balance redox reaction by half reaction method?

Another method of balancing redox reaction is “half reaction method” or “half equation method”. Any redox reaction consists of two half reaction, oxidation and reduction half reaction. Research says that it is better and suitable method than oxidation number method of balancing redox reaction.

In the half reaction, atoms are balanced separately and after the balancing of all the participating atoms net balanced redox reaction is developed by adding the two half reactions. Not only the atoms, the electrons involved in redox reaction should be balanced before developing the net balanced equation.

Let’s take an example to clear the balancing method.

Fe²⁺ (aq) + Cr₂O₇^2- (aq) → Fe³⁺ (aq)+ Cr³⁺ (aq)

Oxidation half reaction: Fe²⁺ → Fe³⁺ + e^– ———1 no equation
Reduction half reaction: Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O + 6e^– —————- 2 no equation

14 numbers of H⁺7 water molecules are added to balance the total number of oxygen and hydrogens.

(1no equation × 6) ———– 6Fe²⁺ → 6Fe³⁺ + 6e^–
(2no equation ×1) ———– Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O + 6e^–
Net balanced equation is: Cr₂O₇^2- + 6Fe²⁺+ 14H⁺→ 2Cr³⁺ + 7H₂O + 6Fe³⁺

To know more please check: 11+ First Order Reaction Example: Detailed Explanations

How to balance redox reaction by oxidation number method?

One of the methods of balancing redox reaction is oxidation number change method. As we all know that redox reaction is none other than electron transfer reaction and oxidation number must be changed due to the electron transfer.

Let’s account an example to make it clear.

Fe₂O₃ (s) + CO (g) → Fe (s) + CO₂ (g)

In the above example, Fe is reduced to its +3 oxidation state to 0 oxidation state and carbon is oxidized by Fe₂O₃ and change of oxidation state takes place is from +2 to +4.

how to balance redox reaction — **Oxidation- Reduction Reaction**

Thus, oxidation number increases for the oxidation half reaction is 2 and for reduction half reaction oxidation number is decreased by 3. Least common number (LCM) of these two numbers (2 and 3) is 6. So, oxidation and reduction half reaction are multiplied by 3 and 2 respectively to balance the equation.

Net balanced equation is:

3Fe₂O₃ (s) + 2CO (g) → 3Fe (s) + 2CO₂ (g)

To know more please check: 12+ Exothermic Reaction Examples: Detailed Explanations

How to balance a redox reaction by oxidation number method in acidic medium?

Let’s consider an example to clear the balancing method of redox reaction.

HNO₃ + H₃AsO₃ (aq) → NO (g) + H₃AsO₄ (aq)+H₂O (l)

Oxidation number of nitrogen is decreased from +5 to +2 and increase of oxidation state of As is +2. The increase of oxidation number must be equal to the decrease of oxidation number.

These two integers are used as coefficients of the redox reaction and placed in front of the substances.

2HNO₃ + 3H₃AsO₃ (aq) → 2NO (g) + 3H₃AsO₄ (aq) +H₂O (l)

How to balance a redox reaction with only one reactant?

Formation of more than one product from one reactant in redox reaction is called disproportionation reaction. The one reactant is oxidized and reduced simultaneously.

These following steps should be followed to balance any disproportionation reaction-

The oxidation number of both the reactant and products should be determined.
Identify the atoms for which oxidation numbers are changed from reactant to product.
At the end integers are placed as coefficients before the substances to balance the redox reaction.

Let’s take an example to make it clear.

2CuCl → CuCl₂ + Cu.

CuCl is only one reactant in this redox reaction. It is oxidized to CuCl₂ and reduced to molecular Cu at a time. Oxidation number is changed for oxidation is +1 and -1 for reduction respectively. Thus, the coefficients “2” is multiplied with reactant and Cu to balance this equation.

To know more please go through: 10+ Redox Reaction Examples: Detailed Explanations

How to balance electrons in a redox reaction?

Balancing of electrons that are gained and lost during a redox reaction is very important part of a redox reaction. All the methods are used to balance the redox reaction (ion electron method, oxidation number method, half equation method) must involve the balancing of electrons step.

To balance the electrons in a redox reaction these steps are followed-

The first step of balancing electrons is to identify the oxidation and reduction half reaction.
Balance those two reactions separately by using any suitable method.
Calculate the electrons that are lost in oxidation half reaction and gained in reduction half reaction.
At the end two balanced half equations are added with each other to obtain the net balanced redox reaction equation.

Let’s account an example-

Mg + Al³⁺→ Mg²⁺ + Al

Oxidation half reaction: Mg → Mg²⁺ + 2e^– ———1 no equation
Reduction half reaction: Al³⁺+ 3e^– → Al ———-2 no equation
(1 no equation × 3)——— 3Mg → 3Mg²⁺ + 6e^–
(2 no equation × 2)——— 2Al³⁺+ 6e^– → 2Al
Net balanced equation: 3Mg + 2Al³⁺→ 3Mg²⁺ + 2Al

Relationship between redox reactions and oxidation reaction

In a redox reaction both the oxidation and reduction take place simultaneously in a single reaction. If one reactant is oxidized then other reaction must be reduced in that reaction. But in oxidation reaction any substance is only oxidized by oxygen or any other oxidizing agent.

For example-

Magnesium is oxidized in presence of oxygen and form magnesium oxide (MgO) as product. In this reaction only oxidation takes place not reduction in the same reaction. But for the following reaction- Ce⁴⁺ + Fe²⁺ → Ce³⁺ + Fe³⁺. Ce⁴⁺is reduced to Ce³⁺ and Fe²⁺ is oxidized to Fe³⁺ simultaneously. So it is a redox reaction.

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Aditi Roy

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I am Aditi Ray, a chemistry SME on this platform. I have completed graduation in Chemistry from the University of Calcutta and post graduation from Techno India University with a specialization in Inorganic Chemistry. I am very happy to be a part of the Lambdageeks family and I would like to explain the subject in a simplistic way.
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