Types of Redox Reactions - Important Topic of JEE

VSAT 2022

What are Redox Reactions?

Redox reactions are  reactions in which the oxidation and reduction occurs at the same time in a single reaction. Oxidation is defined as the loss of electrons. This simply signifies that the species being oxidized undergoes a positive change in oxidation state. Reduction is defined as the loss of electrons. The species being reduced gains electrons and undergoes a negative change in oxidation state. Redox reaction can also be defined as the reaction in which the loss and gain of electrons both takes place in the same reaction. The oxidizing agent is the substance that is being reduced in a chemical reaction, while the reducing agent is the substance that is being oxidized.


Atoms


Atoms showing redox reaction


Types of Redox Reaction

A redox reaction is a chemical reaction in which reactants lose and gain electrons. Changes in the oxidation states of the reacting species can be used to detect this electron transfer.


Decomposition Reaction

The breakdown of a substance into various compounds is what this type of reaction involves. In the first reaction sodium is being reduced from +1 to 0 while hydrogen oxidizes from -1 to 0. In the second reaction hydrogen reduces from +1 to 0 while oxygen goes from -2 to 0. Both these reactions are examples of  decomposition redox reactions.

$\begin{align} &2 \mathrm{NaH} \rightarrow 2 \mathrm{Na}+\mathrm{H}_{2} \\ \\ &2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}_{2}+\mathrm{O}_{2} \end{align}$


Combination Reaction

These reactions are the inverse of decomposition processes, in these reactions two compounds combine  to generate a single compound in the form

$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}$.

$\begin{align} &H_{2}+Cl_{2} \rightarrow 2 HCl \\ \\ &C+O_{2} \rightarrow CO_{2} \\ \\ &4 Fe+3 O_{2} \rightarrow 2 Fe_{2} O_{3} \end{align}$

In the first reaction chlorine is being reduced from 0 to -1 while hydrogen oxidizes from 0 to +1. In the second reaction oxygen reduces from 0 to -2 while carbon oxidizes from 0 to +4. In the third reaction iron is oxidized from 0 to +3 and oxygen reduces from 0 to -2. Hence all these are examples of redox combination reactions.


Displacement Reaction

An atom or an ion of a compound is replaced by an atom or an ion of another compound in this reaction. It can be represented as $\mathrm{X}+\mathrm{YZ} \rightarrow \mathrm{XY}+\mathrm{Z}$

$\mathrm{CuSO}_{4}+\mathrm{Zn} \rightarrow \mathrm{Cu}+\mathrm{ZnSO}_{4}$


Copper is reduced from +2 to 0 and zinc is oxidized from 0 to +2. This reaction is a displacement fresox reaction.


Disproportionation Reactions

Disproportionation reactions are those that involve the same reactant being oxidized and reduced.

$\mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{NaH}_{2} \mathrm{PO}_{2}+\mathrm{PH}_{3}$

Phosphorus is oxidized and reduced from 0 to +1 in NaH2PO3 and -3 in PH3 hence this is an example of redox disproportionation reaction. Hence these are different types of redox reactions.

Redox reaction type examples with answers are as follows:-

$\begin{align} &~~~0 \quad\quad 0 \quad~~~~+3~~~~-2 \\ &4 \mathrm{Al}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} ~\mathrm{O}_{3} \\ \\ &+4 \quad~~ 0 ~~\quad~~+6~-2 \\ &~\mathrm{SO}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{S}~~~ \mathrm{O}_{3} \end{align}$

These are two examples of redox reactions.


Oxidation Number

Oxidation number or oxidation state is the charge on an atom if all its bonds were assumed to be completely ionic. The element with higher electronegativity gets a negative charge and the atom with less electronegativity gets a positive charge. In a molecule one atom makes bonds with many other elements. The oxidation state of an element is proportional to the number of electrons involved in bond formation. Oxidation number can be zero, positive or negative. Depending on how many electrons are gained or lost, the atom might have different oxidation states. Oxidation number helps us to understand which element is being oxidized and which is being reduced in a chemical reaction.

  • An element in a free state has oxidation number  zero.

  • The charge of a monatomic ion is equal to its oxidation number.

  • H has an oxidation number of +1, but when coupled with less electronegative elements, it becomes -1.

  • In most compounds, the oxidation number of O is -2, while in peroxides, it is -1.

  • A Group 1 element  has an oxidation number of +1.

  • The oxidation number of a Group 2 element in a compound is +2.

  • The oxidation numbers of all the atoms in a neutral substance sums up to zero.

  • The charge of a polyatomic ion is equal to the sum of respective charges on all the elements present in the ion.


Oxidation and Reduction

Oxidation is defined as the loss of electrons and the resulting increase in the oxidation state of a given reactant. For example  

$2 \mathrm{~S}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{2}(\mathrm{~g})$ Sulfur is being oxidized from 0 to +4 here.

Reduction is the process of gaining electrons and decreasing the oxidation state of a reactant. For example 

$2 \mathrm{FeCl}_{3}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{FeCl}_{2}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq})$ iron is reduced from +3 to +2 in this reaction.

Oxidizing agents are electron-accepting species that tend to undergo a reduction in redox processes.

Some examples of oxidizing agent are 

$\mathrm{KMnO}_{4}, \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{HNO}_{3}, \mathrm{KClO}_{3}$

A reducing agent is an electron-donating substance that has a tendency to lose  electrons.

Some examples of reducing agents are

FeCl2, FeSO4, SnCl2, Hg2Cl2


Identification of oxidizing and Reducing Agent 

In a compound, if an element is in its highest possible oxidation state. It has the ability to act as an oxidizing agent. KMnO4, K2Cr2O7, HNO3, H2SO4, HClO4 are some examples.


When an element in a compound has the lowest oxidation state, it can act as a reducing agent. H2S, H2C2O4, FeSO4, SnCl2 are some examples.


The compound will function as an oxidizing agent if a highly electronegative element is in its highest oxidation state.


The compound functions as a reducing agent when a highly electronegative element is in its lowest oxidation state.


Balancing of Redox Reactions

The redox reaction can be balanced in two ways. The change in oxidation number of the oxidizing and reducing agents is one technique, while the other is based on separating the redox reaction into two half processes, one of reduction and the other of oxidation.

Method 1:Balancing of Redox Reaction by Oxidation State Method

The number of electrons gained during the reduction reaction equals the number of electrons lost during the oxidation reaction, according to this technique.

Oxidation of Fe+2 to Fe+3

$2 \mathrm{Fe}^{+2} \rightarrow 2 \mathrm{Fe}^{+3}+2 \mathrm{e}^{-}$

Multiply the reaction with 5

Reduction half reaction $\mathrm{Mn}^{+7}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}$

Multiply the reaction by 2

By adding coefficients reaction becomes

2Mn+7 + 10Fe+2 → 10Fe+3 + Mn+2

$2 \mathrm{KMnO}_{4}+10 \mathrm{FeSO}_{4}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+2 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O}$

In the above reaction we have also added some other counter ions and water molecules to write the reaction properly. How to add these counter ions and water molecules can be explained with another example.


Method 2:Ion Electron method 

  • Divide the entire reaction into two halves, one for oxidation and the other for reduction.

  • Other than the 'O' and 'H' atoms, balance first.

  • In an acidic or neutral media, add H2O molecules to balance oxygen atoms, and H+ ions to balance hydrogen atoms.

  • The oxygen atom is balanced in an alkaline media by adding an H2O molecule and an equal number of hydrogen atoms.

  • H+ atoms are still imbalanced, thus OH is added on the other side.

  • The addition of electrons balances the charges.

  • Multiply by a large enough integer to cancel the number of electrons.

  • Add both half-reactions together, subtract similar terms, and write the final equation.

For example balance by ion electron method:

$\mathrm{Ag}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})$

Step 1: Write Separate Half Reactions 

$\begin{align} 2 \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{aq}) \\ \\ \mathrm{Zn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{s}) \end{align}$

Step 2: Add H2O Add Balance H+ Ions 

$H_2O(l)+2Ag(s)\longrightarrow Ag_2O(aq)+2H^+(aq)$

$Zn^{2+}(aq)\longrightarrow Zn(s)$

Step 3: Balance Electrons 

$\begin{align} &\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{aq})+2 \mathrm{H}^+(\mathrm{aq})+2 \mathrm{e}^{-} \\ \\ &\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) \end{align}$

Step 4: Add Both Equations

$\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Ag}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{aq})+2 \mathrm{H}^{+}$


Summary

Redox reactions are those in which the number of valence electrons in the reactant atom/ion differs from the number of valence electrons in the product atom/ion. During the reaction, the atom/ion may have received or lost electrons. As a result, an atom/ion is either oxidized or reduced. There are four types of redox reaction displacement,decomposition,combination and disproportionation reaction.


The charge denotes the number of electrons that the atom/ion has in comparison to the neutral atom. In comparison to the neutral atom, the charge also has a positive sign if the electron is lost and a negative sign if the electron is acquired.


Oxidation number or oxidation state is the charge on an atom if all its bonds were assumed to be completely ionic. Depending on how many electrons are gained or lost, the atom might have different oxidation states. The oxidation state of neutral atoms is zero. The reaction can be balanced by any of the two methods.

FAQs on Types of Redox Reactions - Important Topic of JEE

1. When KMnO4 acts as an oxidizing agent and ultimately forms  MnO24-, MnO2, Mn2O3 and Mn+2. Then what is the number of electrons trans­ferred in each case?

Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron is known as a redox reaction.


Mn has oxidation numbers of 7, 6, 4, 3 and 2 in KMnO4, MnO42-, MnO2, Mn2O3, and Mn+2. The change in oxidation number correlates to the number of electrons exchanged. The number of electrons transported in each case when KMnO4 acts as an oxidizing agent and forms MnO42-, MnO2, Mn2O3, Mn2+ are 1,3,4,5, respectively.

2. What is redox titration?

A redox reaction between the titrant and the analyte is used to determine the concentration of a given analyte in a laboratory setting. A potentiometer or a redox indicator may be required for these types of titrations.


The titrant and the analyte undergo an oxidation-reduction reaction, which is used in redox titration. One of the most common laboratory methods for determining the quantity of unknown analytes is redox titration. The titration of potassium permanganate (KMnO4) against oxalic acid (C2H2O4) is an example of a redox titration.


$2 \mathrm{KMnO}_{4}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{MnSO}_{4}+10 \mathrm{CO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}$

Comment