Redox Reaction: Types, Applications, Examples, Questions

Redox Reactions: Chemistry deals with the study of composition, structure, and properties of varieties of matter and the change of one kind of matter. This transformation of one type of matter into another occurs through several different types of reactions. One such reaction is Redox Reaction, or simply an oxidation-reduction reaction.

What is Redox Reaction?

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

Oxidation-Reduction- A Classical Concept

Oxidation

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

For example,
\(2{\rm{Mg}} + {{\rm{O}}_2} \to 2{\rm{MgO}}\) (addition of oxygen)
\(2{{\rm{H}}_2}\;{\rm{S}} + {{\rm{O}}_2} \to 2\;{\rm{S}} + 2{{\rm{H}}_2}\) (removal of hydrogen)

Oxidizing agent or Oxidant-An oxidizing agent or oxidant is a substance that supplies oxygen or any other electronegative element or removes hydrogen or an electropositive element. An oxidizing agent, after carrying out oxygen itself, is reduced in a chemical reaction. Some important oxidizing agents are fluorine, manganese dioxide, hydrogen peroxide, potassium permanganate, etc.

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,
\({\rm{B}}{{\rm{r}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{S}} \to {\rm{2HBr + S}}\) (addition of hydrogen)
\({\rm{Cuo}} + {{\rm{H}}_2} \to {\rm{Cu}} + {{\rm{H}}_2}{\rm{O}}\) (removal of oxygen)

Reducing agent or Reductant– A reducing agent or a reductant may be defined as a substance which supplies hydrogen or any other electropositive element or removes oxygen or any other electronegative element. A reducing agent after reduction is itself oxidized in a chemical reaction. Some important reducing agents are carbon, carbon monoxide, aluminum, nitrous acid, etc.

Oxidation and Reduction are Complementary

Whenever any substance is oxidized, other substances are always reduced at the same time and vice versa. In other words, oxidation and reduction are complementary. For example,

i. The reaction between hydrogen sulfide and chlorine
\({{\rm{H}}_{\rm{2}}}{\rm{S + C}}{{\rm{l}}_{\rm{2}}} \to {\rm{2HCl + S}}\)
Here, \({{\rm{H}}_{\rm{2}}}{\rm{S}}\) is oxidized to \({\rm{S}}\) while \({\rm{C}}{{\rm{l}}_2}\) is reduced to \({\rm{HCl}}\).
ii. The reaction between manganese dioxide and hydrochloric acid-
\({\rm{Mn}}{{\rm{O}}_2} + {\rm{HCl}} \to {\rm{MnC}}{{\rm{l}}_2} + {\rm{C}}{{\rm{l}}_2} + {{\rm{H}}_2}{\rm{O}}\)
\({\rm{HCl}}\) is oxidized to \({\rm{C}}{{\rm{l}}_2}\) while \({\rm{Mn}}{{\rm{O}}_2}\) is reduced to \({\rm{MnC}}{{\rm{l}}_2}\).

Oxidation and Reduction- 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. For example,
i. Loss of electrons results in an increase in positive charg:
\({\rm{Na}} \to {\rm{N}}{{\rm{a}}^{\rm{ – }}}{\rm{ + }}{{\rm{e}}^{\rm{ – }}}\,\,:\,\,{\rm{Mg}} \to {\rm{M}}{{\rm{g}}^{{\rm{2 – }}}} + {\rm{2}}{{\rm{e}}^{\rm{ – }}}\)
ii. Loss of electrons results in a decrease in negative charge:
\({\rm{MnO}}_4^{2 – } \to {\rm{MnO}}_{\rm{4}}^{\rm{ – }}{\rm{ + }}{{\rm{e}}^{\rm{ – }}}\,\,:\,\,{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{4 – }} \to {\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 – }} + {{\rm{e}}^ – }\)

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,
i. Gain of electrons results in a decrease in positive charge:
\({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + }}{{\rm{e}}^{\rm{ – }}} \to {\rm{F}}{{\rm{e}}^{{\rm{2 + }}}}\,\,{\rm{:}}\,\,{\rm{2H}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{ + 2}}{{\rm{e}}^{\rm{ – }}} \to {\rm{Hg}}_2^{2 – }\)
ii. Gain of electrons results in an increase in negative charge:
\({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ + 2}}{{\rm{e}}^{\rm{ – }}} \to {\rm{2C}}{{\rm{l}}^{\rm{ – }}}\,\,{\rm{:}}\,\,{\rm{MnO}}_4^ –  + {{\rm{e}}^ – } \to {\rm{Mn}}{{\rm{O}}_4}^{2 – }\)

Oxidation-reduction as an Electron Transfer Process

In a chemical reaction, a substance can lose electrons only if there is another substance present that can gain electrons. This implies that oxidation can only take place only if the reduction also occurs at the same time and vice-versa. For example,

i. \(2{\rm{Na}} + {{\rm{O}}_2} \to {\rm{N}}{{\rm{a}}_2}{\rm{O}}\)

ii. \(2{\rm{Na}} + {\rm{C}}{{\rm{l}}_2} \to 2{\rm{NaCl}}\)

Each of the above reactions can be considered a sum of two half-reactions- one involving oxidation called oxidation half-reaction and the other involving reduction called reduction half-reaction. Reducing agents are electron donors, while oxidizing agents are electron acceptors.

To get the overall equation for a redox reaction, the two half-reactions are simply added-

i. \({\rm{Na}}\) is a reducing agent, while \({{\rm{O}}_2}\) is an oxidizing agent.

ii. \({\rm{Na}}\) is a reducing agent, while \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) is an oxidizing agent.

Thus, in a redox reaction, the oxidant is reduced by accepting electrons and the reductant is oxidized by losing electrons.

Oxidation Number

Oxidation number may be defined as the charge an atom of the element has in its ions or appears to have present in the combined state with other atoms. Thus, the oxidation number is also called oxidation states.

Calculation of an Oxidation Number

An oxidation number can be assigned to a given element or compound by following the following rules.

1. Any element will have zero as an oxidation number.
2. For ions, the oxidation number is always the net charge corresponding to the ion.
3. The hydrogen atom has an oxidation state of \({\rm{ + 1}}\). However, if it is bonded with an element with less electronegativity, it exhibits an oxidation number of \({\rm{ – 1}}\).
4. Oxygen has an oxidation of \({\rm{ – 2}}\) in most of its compounds.
5. In the case of peroxides, the oxidation number corresponding to oxygen is \({\rm{ – 1}}\).
6. All group \({\rm{1}}\) elements exhibit an oxidation state of \({\rm{ + 1}}\).
7. All group \({\rm{2}}\) elements exhibit an oxidation state of \({\rm{ + 2}}\).
8. A group of \({\rm{17}}\) elements has an oxidation number of \({\rm{ – 1}}\).
9. In the case of neutral compounds, the sum of all the oxidation numbers of the constituent atoms is always equal to zero.
10. In the case of polyatomic ions, the sum of all the oxidation numbers of the constituent atoms is equal to the net charge of the polyatomic ion.

Redox Reaction in terms of Oxidation Number

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, in the reaction between zinc and hydrochloric acid-

\({\rm{Zn}} + 2{\rm{HCl}} \to {\rm{ZnC}}{{\rm{l}}_2} + {{\rm{H}}_2}\)

The oxidizing agent will be the substance whose oxidation number decreases, and the reducing agent will be the substance whose oxidation number increases.

Types of Redox Reaction

Some important types of redox reactions are as follows:

1. Combination Reaction– A reaction in which two atoms or molecules combine to form a third molecule. For example,
\({\rm{A}} + {\rm{B}} \to {\rm{C}}\)

Some examples of combination reactions that are redox reaction are-

2. Decomposition Reaction- A reaction in which a molecule breaks down into two or more components is called a decomposition reaction. For a decomposition reaction to be a redox reaction, it is essential that one of the products of the reaction must be in the elemental state.
Some examples of decomposition reaction that are redox reaction-

3. Displacement Reaction- A reaction in which an atom or ion in a compound is replaced by an atom of some other element. In general, it is represented by-

\({\rm{X + YZ}} \to {\rm{XZ + Y}}\)

It can be further divided as-

a. Metal Displacement- In this reaction, a metal in the compound is displaced by other metals in the elemental state. For example,

b. Non-metal Displacement- In this reaction, a metal or a non-metal displaces another non-metal from its compound. For example,

4. Disproportionation Reactions– A reaction in which the same species is simultaneously oxidized and reduced. For such a reaction to occur, the reacting species must contain an element with at least three oxidation states. For example,

Balancing of Chemical Equations of Redox Reactions

Chemical equations of redox reactions can be balanced by using any one of the following methods:

Oxidation Number Method

The following steps are involved in this method-
i. Write the skeletal equation of all the reactants and products of the reaction.
ii. Indicate the oxidation number of each atom above its symbol.
iii. Calculate the increase and decrease in the oxidation number and identify oxidizing and reducing agents.
iv. Multiply the formulae of the oxidizing and reducing agent by a suitable integer to equalize the total increase or decrease in the oxidation number.
v. Balance all other atoms other than \({\rm{H}}\) and \({\rm{O}}\).
vi. Finally, balance \({\rm{H}}\) and \({\rm{O}}\) by adding \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) molecules by the hit and trial method.

Solved Examples on Balancing Redox Reactions by Oxidation Number Method

Let us understand the balancing of Redox reaction with the help of an example-

\({\rm{CuO + N}}{{\rm{H}}_{\rm{3}}} \to {\rm{Cu + }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

Step 1. Write the oxidation number of each atom.

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

Copper changes its \({\rm{O}}{\rm{.N}}\) from \({\rm{ + 2}}\) to 0 and Hydrogen changes its \({\rm{O}}{\rm{.N}}\) from \({\rm{ – 3}}\) to \({\rm{ 0}}\).

Step 3. Calculate the increase and decrease of \({\rm{O}}{\rm{.N}}\).

The decrease in \({\rm{O}}{\rm{.N}}\) can be observed in copper from \({\rm{ + 2}}\) to \({\rm{ 0}}\) and increase in \({\rm{O}}{\rm{.N}}\) can be observed in hydrogen from \({\rm{ – 3}}\) to \({\rm{ 0}}\).

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

\({\rm{3CuO + 2N}}{{\rm{H}}_{\rm{3}}} \to {\rm{Cu + }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

Step 5. Balance the number of copper and nitrogen on both sides.

\(3{\rm{CuO}} + 2{\rm{N}}{{\rm{H}}^{{{\rm{H}}_3}}} \to 3{\rm{Cu}} + {{\rm{N}}_2} + {{\rm{H}}_2}{\rm{O}}\)

Step 6. Now balance \({\rm{H}}\) and \({\rm{O}}\) atoms by hit and trial method.

\({\rm{3CuO + 2N}}{{\rm{H}}_{\rm{2}}} \to {\rm{3Cu + }}{{\rm{N}}_{\rm{2}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

Hence, the equation is balanced.

Half Reaction Method or Ion Electron Method

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- 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 half-reaction is equal to the total number of electrons lost in the other half-reaction.
vi. Verification.

Solved Examples on Balancing Redox Reactions by Ion Electron Method

Let us understand the balancing of Redox reaction with the help of an example-

\({{\rm{(C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}{\rm{)}}^{{\rm{ – 2}}}}{\rm{ + F}}{{\rm{e}}^{{\rm{ + 2}}}}{\rm{ + }}{{\rm{H}}^{\rm{ + }}} \to {\rm{C}}{{\rm{r}}^{{\rm{ + 3}}}}{\rm{ + F}}{{\rm{e}}^{{\rm{ + 3}}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

Step 1. Write the oxidation number of each atom.

Step 2. Find out the species which are oxidized and reduced.

Chromium changes its \({\rm{O}}{\rm{.N}}\) from \({\rm{ + 6}}\) to \({\rm{ + 3}}\)  thus it is reduced and Iron changes its \({\rm{O}}{\rm{.N}}\) from \({\rm{ + 2}}\) to \({\rm{ + 3}}\) and hence it is oxidised.

Step 3. Split the skeletal equation into two half-reactions- oxidation half-reaction and reduction half-reaction.

Step 4.

(i.) Balancing of half oxidation reaction-

\({\rm{F}}{{\rm{e}}^{{\rm{ + 2}}}} \to {\rm{F}}{{\rm{e}}^{{\rm{ + 3}}}}\)

As there is an increase in \(({\rm{1}}\,{\rm{O}}{\rm{.N}}\). Therefore, add one electron to the product side.

\({\rm{F}}{{\rm{e}}^{{\rm{ + 2}}}} \to {\rm{F}}{{\rm{e}}^{{\rm{ + 3}}}}{\rm{ + }}{{\rm{e}}^{\rm{ – }}}\)

Thus, the reaction is balanced.

(ii.) Balancing the half reduction reaction-

\({\left( {{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}} \right)^{{\rm{ – 2}}}} \to {\rm{C}}{{\rm{r}}^{{\rm{ + 3}}}}\)

There is a total decrease of \(({\rm{6}}\,{\rm{O}}{\rm{.N}}\) for \({\rm{Cr}}\) atoms. Therefore, add \({\rm{6}}\) electrons on the reactant side.

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

Balance \({{\rm{Cr}}}\) atoms on both sides.

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

In order to balance \({\rm{O}}\) atoms add seven \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) molecules to the product side and add fourteen \({{\rm{H}}^ + }\) to balance \({\rm{H}}\) atoms.

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

Step 5. Adding the two half reactions.

Step 6. Hence the balanced equation is-

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

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Applications of Redox Reaction

1. In electrometallurgy- For the isolation of reactive metals like sodium, potassium, magnesium, calcium, aluminum, etc., by the electrolysis of their fused salts.
2. In electrorefining of metals- The metals like copper, silver, gold, aluminum, etc., are purified by electrorefining.
3. In electroplating- It is coating a layer of superior metal over the surface of the inferior metal.
4. Preparation of chemicals- \({\rm{A}}\) large number of chemicals such as sodium hydroxide, chlorine, and potassium permanganate are commercially prepared by electrolysis methods that involve redox reactions.
5. Preparation of organic compounds- Kolbe’s electrolysis is very useful for the synthesis of hydrocarbons.
6. Redox reactions play a major role in some important processes like photosynthesis. The oxidation of water into molecular oxygen and reduction of carbon dioxide into sugar is a redox reaction.

Summary

This article teaches that oxidation and reduction are complementary reactions, and when oxidation and reduction occur simultaneously, it is a redox reaction. We studied the classical concept and the redox reaction as a concept of the transfer of electrons and oxidation numbers. We hope this detailed article on the Redox Reaction is helpful to you. If you have any questions about this article or in general about the Redox reactions, ping us through the comments section, and we will get back to you as soon as possible.

Frequently Asked Questions on Redox Reaction

Q.1. Give 2 examples of redox reactions.
Ans: The two examples of redox reactions are-
\({\rm{Zn}} + 2{\rm{HCl}} \to {\rm{ZnC}}{{\rm{l}}_2} + {{\rm{H}}_2}\)
\(2{\rm{Na}} + {{\rm{O}}_2} \to {\rm{N}}{{\rm{a}}_2}{\rm{O}}\)

Q.2. What is the importance of redox reactions?
Ans: The importance of redox reaction is that it is the principal source of energy in terms of natural, biological or artificial purpose. For example, The batteries used to produce DC current and storage cells used in vehicles work on redox reactions as well as the food we eat is converted into energy through redox reactions.

Q.3. What are redox reactions?
Ans: The reaction in which one substance gets oxidized and the other gets reduced simultaneously in a chemical reaction is called a redox reaction. In simple words, it means that a chemical reaction when oxidation and reduction are taking place at the same time is called redox reaction.

Q.4. What are oxidizing agents?
Ans: An oxidizing agent or oxidant is a substance that supplies oxygen or any other electronegative element or removes hydrogen or an electropositive element. An oxidizing agent, after adding oxygen itself, is reduced in a chemical reaction. Some important oxidizing agents are fluorine, manganese dioxide, hydrogen peroxide, potassium permanganate, etc.

Q.5. What are reducing agents?
Ans: A reducing agent or a reductant may be defined as a substance which supplies hydrogen or any other electropositive element or removes oxygen or any other electronegative element. A reducing agent after reduction is itself oxidised in a chemical reaction. Some important reducing agents are carbon, carbon monoxide, aluminium, nitrous acid, etc.

Q.6. What are the types of redox reactions?
Ans: There are four types of Redox reactions which are- Combination reaction, Decomposition reaction, Displacement reaction, and Disproportionation reaction.

Redox Reactions and Volumetric Analysis

Now that you are provided with all the necessary information on Redox Reactions and we hope this detailed article is helpful to you. If you have any queries on this page or in general about Redox reactions, ping us through the comment box below and we will get back to you as soon as possible.