CBSE Class 11 Chemistry Notes For Chapter 8 Redox Reactions

CBSE Class 11 Chemistry Notes For Chapter 8 Redox Reactions Introduction

Chemical reactions, according to their nature, are divided into different classes, such as combination reactions, decomposition reactions, elimination reactions, polymerisation reactions and many other types.

Oxidation-reduction or redox-reaction is a class of chemical reactions in which oxidation and reduction occur simultaneously. A large number of chemical and biological reactions belong to this class. Some processes that are associated with oxidation-reduction reactions are the rusting of iron, the production of caustic soda and chlorine by electrochemical method, the production of glucose by photosynthesis, the generation of electric power by battery, fuel cell, etc.

Redox Concept Reactions According To Electronic

Oxidation reaction:

A chemical reaction in which an atom, an ion or a molecule loses one or more electrons is called an oxidation reaction. An atom, ion or molecule is oxidised by loss of electron (s).

Examples: Oxidation reactions involving—

Loss of electron(s) by an atom:

Generally atoms of metallic elements such as Na, K, Ca, etc., undergo oxidation by losing electron(s), thereby producing positive ions.

⇒ \(\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+e ; \mathrm{K} \longrightarrow \mathrm{K}^{+}+e ; \mathrm{Ca} \longrightarrow \mathrm{Ca}^{2+}+2 e\)

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Loss of electron(s) by a cation:

Some cations such as Fe²⁺, Sn²⁺, Cu⁺, etc., undergo oxidation by losing one or more electrons, thereby forming higher charges.

Fe²⁺→ Fe³⁺ + e; Sn²⁺ → Sn⁴⁺+ 2e

Cu²⁺ → Cu²⁺ + e

Loss of electron(s) by an anion:

Anions such as I^– and Br^– ions oxidise to neutral atoms or molecules by losing electron(s).

I₂^– →I₂ + 2e;2Br^– — Br₂ + 2e

Loss of electron(s) by a molecule:

Neutral molecules such as H₂, H₂O₂ and H₂O oxidise to cations by losing one or more electrons

H₂ → 2H+ + 2e;  H₂O₂→ O₂ + 2H⁺ + 2e

H₂O→ ½ O₂ + 2H⁺+ 2e

Reduction reaction:

A chemical reaction in which an atom, an ion or a molecule gains one or more electrons is called a reduction reaction. An atom, ion or molecule is reduced by the gain of electron(s).

Examples: Reduction reactions involving—

Gain of electron(s) by an atom:

Atoms of different elements in particular, such as, atoms of chlorine, bromine, oxygen and other non-metals are reduced to anions by gaining electron(s).

Cl + e → Cl^–

Br+ e→ Br^–

O+ 2e — O^2-

Gain of electron(s) by a cation:

Cations such as H⁺, Fe²⁺, Fe³⁺, Cu²⁺ etc., are reduced to neutral atoms or cations with lower charges by accepting electron(s)

H⁺+ e → H;  Fe²⁺+2e → Fe

Fe³⁺+ e → Fe²⁺ ; Fe³⁺+ 3e  → Fe

Cu²⁺+ 2e →Cu ; Cu²⁺+ e  → Cu⁺

Gain of electron(s) by a molecule:

Neutral molecules such as Cl₂, O₂, H₂O₂ etc., are reduced by gaining one or more electrons.

Cl₂ + 2e → 2CF ; O₂  + 4H⁺ + 4e→2H₂ O

H₂O₂ + 2H+ + 2e →  2H₂ O

Oxidant and reductant in light of electronic concept

Oxidant:

In a redox reaction, the species that itself gets reduced by accepting electron(s) but oxidises other substances is called an oxidant or oxidising agent. So, an oxidising agent is an electron acceptor. The greater the tendency of a substance to accept electrons, the greater the oxidising power it possesses.

Examples:

Oxygen (O₂), hydrogen peroxide (H₂O₂), halogens (F₂, Cl₂, Br₂, I₂), nitric acid (HNO₃), potassium permanganate (KMnO₄), potassium dichromate (K₂Cr₂O₇), sulphuric acid (H₂SO₄), etc.

Reductant:

In a redox reaction, the species that itself gets oxidised by losing electron(s) but reduces other substances is called a reductant or a reducing agent. So, a reducing agent is an electron donor.

The substance has a high tendency to lose electrons and acts as a strong reducing agent. Alkali metals (Na, K, Rb, Cs, etc.,) of group-IA of the periodic table show a strong tendency to lose electrons and hence behave as powerful reducing agents.

Examples: Hydrogen (H₂), hydrogen sulphide (H₂S), carbon (C), some metals (Na. K, Fe. Al, etc.), stannous chloride (SnCl₂), sulphur dioxide (SO₂), oxalic add (H₂C₂O₄), sodium thiosulphate (Na₂S₂O₂), etc.

According to an electronic concept;

• Oxidation involves the loss of one or more electrons. Reduction involves the gain of one or more electrons.
• Oxidants are electron acceptors. Reductants are electron donors.

Identification of oxidants and reductants with the help of electronic concept

Reaction: 2KI(aq) + Br₂(l)→2KBr(aq) + l₂(s)

The reaction can be represented in ionic form as

2K⁺ (aq) + 2I (aq) + Br₂(l)→ 2K^+- (aq)+ 2Br^–(aq) + I₂(s)……………………..(1)

This equation shows that in the reaction, the K⁺ ion does not undergo any change. The only function that it does In the reaction is to balance the charge. So, the K⁺ ion only acts as a spectator ion in the reaction.

Hence, the net ionic equation of the reaction is-

2I(aq) + Br₂(l)→ I₂(s) + 2Br^–(aq)

Equation(2) shows that the I^– ion produces I₂ by losing electrons, while Br₂ forms Br^– ions by gaining electrons. 1 lenco, In this reaction, the conversion of I^– into

[2l^–(aq) → l₂(s) + 2e] is an oxidation reaction. On the other hand, the conversion of Br₂ into Br^– →2Br (l)] is a reduction reaction. Thus, in this reaction, ion, i.e., Kl is the reductant and Br₂ is the oxidant.

Oxidation-reduction occur simultaneously

Neither an oxidation reaction nor a reduction reaction can occur alone. Oxidation and reduction reactions are complementary to each other. In a reaction system, if a reactant loses an electron, then there must be another reactant In the system that will gain the lost electron. Thus, oxidation and reduction reactions must occur together.

In a redox reaction, the reducing agent gets oxidised by losing electron(s), while the oxidising agent gets reduced by accepting the lost electron(s). For example, Metallic zinc reacts with copper sulphate in a solution to form metallic copper and zinc sulphate.

⇒ \(\mathrm{Zn}(s)+\mathrm{CuSO}_4(a q) \rightarrow \mathrm{ZnSO}_4(a q)+\mathrm{Cu}(s) \downarrow\)

In an aqueous solution, CuSO₄ and ZnSO₄ exist almost completely In dissociated state. So, the reaction can be expressed in the form of an ionic equation as

⇒ \(\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)\)

In this reaction, the Zn atom loses two electrons, forming the Zn^2- ion (oxidation). On the other hand, the Cu²⁺ ion accepts these two lost electrons to produce Cu^– atom (reduction)

In this reaction, the Zn-atom can lose electrons only because the Cu²⁺ ion present in the reaction system accepts those lost electrons. Alternatively, the Cu²⁺ ion can get reduced by accepting electrons only because the Zn-atom present in the reaction system lose electrons.

So, in a reaction, when a substance undergoes oxidation, another substance present in the reaction system undergoes reduction. Tints it can be said that oxidation and reduction occur simultaneously.

Half-reaction

Any redox reaction consists of two half-reactions, one is for an oxidation reaction and the other is for a reduction reaction Both of these reactions are called a half-reaction.

In a redox reaction, the half-reaction Involving oxidation Is called oxidation half-reaction, and the half-reaction involving reduction is called reduction half-reaction.

Redox reaction = Oxidation half-reaction Reduction half-reaction

Reaction 1:

Reaction 2:

Reaction 3:

Oxidation State And Oxidation Number

According to electronic theory, redox reactions can be explained in terms of electron transfer. The electronic theory can be applied in the case of ionic compounds because, in the formation of ionic compounds, one reactant gives up the electron(s) and another reactant accepts those electron(s).

However, this theory cannot be applied in the case of a redox reaction involving covalent compounds due to the absence of cations or anions in such compounds. Thus this theory is unable to Identify the oxidant and reductant in such types of reactions.

To overcome this problem, the concept of oxidation number has been introduced. All types of redox reactions can be explained based on oxidation number. Each constituent element of any compound has a definite valency. Similarly, it can be assumed that each atom of any element has a definite oxidation number.

Oxidation state

An atom of an element converts into an ion when it loses or gains an electron. The loss of one or more electrons by an atom results in the formation of a cation, while the gain of one or more electrons produces an anion.

Cation is the oxidised state and anion is the reduced state of an element For example, the Na+ ion is the oxidised state of the Na atom and the Cl- ion represents the reduced state of the Cl -atom. The state of oxidation or reduction of an element presenting a compound is called the oxidation state of that element.

Oxidation state Definition:

The state of oxidation or reduction in which an atom of an element exists in a compound is called the oxidation state of the element in that compound.

From the element charge of the compound ion of an element present in an ionic compound, the oxidation state of that element can be easily determined. If the element exists as a cation in the compound, the element is said to be in the oxidised state. On the other hand, if the element exists as an anion in the compound, then the element is said to be in a reduced state.

Example:

In the formation of the compound ZnCl₂, the Zn atom loses two electrons to produce a Zn²⁺ ion and two CIatoms accept two electrons, one electron each to yield two Cl- ions. Zn²⁺ ion combines with two Cl- ions to form a ZnCl₂ molecule.

Thus, in ZnCl₂, the Zn -atom exists in the oxidised state, while the Cl -atom exists in the reduced state. Cations or anions derived from the same element may exist in different oxidation states in different compounds. For example, in CuCl₂, Cu exists as Cu²⁺, while in CuCl, it exists as Cu⁺ ion. Thus, the oxidation state of Cu in CuCl₂ is higher than that in CuCl.

Oxidation number

Oxidation number Definition:

The oxidation number of an element in a compound is a definite number, which indicates the extent of oxidation or reduction to convert an atom of the element from its free state to its bonded state in the compound.

If oxidation is necessary to effect such a change, then the oxidation number will be positive. The oxidation number will be negative when such a change requires reduction. The oxidation number of an atom or molecule in its free state is considered to be zero (0).

The number expressing the die oxidation state of the atom of an element in a compound denotes the oxidation number of the elements in the compound.

The oxidation number of elements in electrovalent compounds:

The charge that an atom of an element In the molecule of an electrovalent compound carries, is equal to the oxidation number of the element In a compound, According to the nature (positive or negative) of the charge, the oxidation number may be positive or negative. The number of electrons (s) lost or gained by an atom during the formation of an Ionic compound determines the oxidation number of the clement in that compound.

Examples:

ln NaCl, sodium and chlorine exist ns Na⁺ and (11- ions, respectively. So the oxidation numbers of sodium and chlorine are +1 and – 1, respectively. In FeCI₂> Iron and chlorine are present as Fe²⁺ and Cl^– Ions, so the oxidation numbers of iron and chlorine are +2 and -1, respectively.

The oxidation number of elements in covalent compounds:

The formation of a covalent compound does not involve the direct transfer of electron(s) between the participating atoms; instead, a covalent bond Is formed by the sharing of electrons.

When two atoms of different electronegativities form a covalent bond(s) through the sharing of one or more electron pairs, they do not get an equal share of the electrons.

The more electronegative atom acquires a greater possession of the shared electron pair (s) than the less electronegative atom.

As a result, the more electronegative atom acquires a partial negative charge. It Is assumed that the atom of the more electronegative element has gained electron(s) i.e. it has been reduced and the less electronegative atom has lost electron(s) i.e., It has been oxidised.

The oxidation number of an atom In a covalent compound is considered to be equal to the number of electron pair(s) the atom shares with one or more atoms of different electronegativities in the compound, If the atom concerned is of higher electronegativity, then Its oxidation number is taken as negative, and if It is of lower electronegativity, its oxidation number is taken as positive.

In the case of a molecule of an element, such as H₂, N₂, O₂, Cl₂ etc., the two atoms of the same electronegativity are covalently linked by sharing electron pair (s) between them. Hence, these two atoms use the electron pair(s) equally and none of the atoms acquire a positive or negative charge. Therefore, the oxidation number of the atoms In the molecule of these elements is regarded as zero.

Example:

1. In hydrogen chloride molecule (HCl) one electron pair exists between H and Cl-atoms, As the electronegativity of chlorine is more than that of hydrogen, the oxidation number of H^–  +1 and that of Cl =-1,

2. In water (H₂O) molecule, the O -atom shares two electron palms, one each with two separate H -atoms. As oxygen is more electronegative than hydrogen, the oxidation number of H -atom +l and that of O -atom =-2.

3. In a carbon dioxide molecule (CO₂), a carbon atom shares four electron pairs, to each with two separate O atoms. As oxygen has higher electronegativity than carbon, in CO₂ molecule, the oxidation number of C = + 4 and that of O =-2

4. In the ethylene (C₂H₄) molecule, two equivalent carbon atoms share two C=C electron pairs between themselves, Since these two electron pairs are equally shared by two C -atoms, they have no role in determining the oxidation number of C.

Again, each carbon atom shares two electron pairs with two separate H -atoms. As carbon is more electronegative than hydrogen, the oxidation number of each H -atom = + 1 and the oxidation number of each C -atom =-2.

Rules For Calculating the Oxidation Number Of An Element

The following rules are to be followed in determining the oxidation number of an element in a compound.

The oxidation number of an element in its free or elementary state is taken as zero (0).

Example:

⇒ \(\stackrel{0}{\mathrm{Z}} \mathrm{n}, \stackrel{0}{\mathrm{Cu}}, \stackrel{0}{\mathrm{C}} \mathrm{L}_2, \stackrel{\ominus}{\mathrm{P}}_4, \stackrel{0}{\mathrm{~S}}_8 \text {, etc. }\) Etc.,

The oxidation number of a monoatomic ion in an ionic compound is equal to its charge.

Example:

In FeCl₂, iron and chlorine exist as Fe²⁺ and Cl^–. So, in i.e., Cl₂, the oxidation number of Fe and Cl are +2 and -1 respectively.

In the case of a polyatomic ion, the sum of the oxidation numbers of all the atoms present in it is equal to the charge of the Ion.

Examples:

The sum of oxidation numbers of all the atoms presenting [Fe(CN)₈]4- =-1.0 The sum of the oxidation numbers of the atoms Cr₂O₇ ^2- ion =-2.

Determination of the oxidation numbers of the atoms in covalent compounds has been discussed in

The algebraic sum ofthe oxidation numbers of all atoms in a neutral molecule is zero(O).

Example: In the FeCl₃ molecule, the oxidation number of Fe is +3 and the oxidation number of Cl is -1. So, the total oxidation number of the atoms in the FeCl₃ molecule =(+3) + 3 × (- 1) = 0.

The oxidation number of hydrogen:

In metallic hydride, it is always

In all hydrogen-containing compounds except for metallic hydrides, it is +1.

Example:

The oxidation number of oxygen in its compounds:

The oxidation number of oxygen in most compounds=-2

Example:

• In peroxide compounds (H₂O₂ Na₂O₂), the oxidation number of oxygen =-1.
• In superoxides (Example; KO₂), the oxidation number of oxygen =-1/2.
• Since fluorine is more electronegative than oxygen, the oxidation number of oxygen in F₂O = +2.
• In all fluorine-containing compounds, the oxidation number of fluorine =- 1.
• There are some elements which always show, fixed oxidation numbers in their compounds.

Example:

Alkali metals (Li, Na, K, Rb, Cs) always show a +1 oxidation state in their compounds. The oxidation number of alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra ) is +2 in their compounds. The oxidation number of Zn in its compounds is +2 and the oxidation number of Al in its compounds is +3.

The maxim oxidation1 number of an element cannot exceed its group numbering in the periodic table.

The following always have definite oxidation numbers in their compound

Calculation of oxidation number in some compounds

The oxidation number of an element in a compound can be calculated if the oxidation numbers of other elements in the compound are known.

The oxidation number of S in H₂SO₄:

Suppose, the oxidation number of S in H₂SO₄ = x.

The total oxidation number of two H-atoms in the H₂SO₄ molecule = 2 × (+1) = +2.

Total oxidation number of four O -atoms in H₂SO₄ molecule = 4 × (-2) =-8 .% Total oxidation number of all atoms in a H₂SO₄ molecule = +2 + x + (-8) = x- 6.

Now, the sum of the oxidation numbers of all atoms in a molecule = 0.

Therefore, x – 6 = 0 or, x = +6

∴ The oxidation number of S in H₂SO₄ = +6

The oxidation number of Cl in KClO₄:

If the oxidation number of Cl = x, then the total oxidation number of all the atoms in the KClO₄ molecule

=+1+x+ 4 × (-2) = x- 7

The sum of oxidation numbers for all the atoms present in a molecule = 0.

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of Cl in KClO₄ =

The oxidation number of N in NH₄NO₃:

NH₄NO₃ is an ionic compound in which NH₄ and NO-₃ are the cation and anion, respectively.

Let the oxidation no. of in NH⁺₄ ion be x. Then the total oxidation no. of the atoms present in this ion = x + 4.

For a polyatomic ion, the oxidation no. of the ion is equal to its charge i.e., the oxidation number of the NH₄ ion

= +1. x + 4

= +1 or, x = -3.

Again, if the oxidation number of N in NO₃ is y, then y + 3 × (-2) = -1 or, y = +5.

Hence, in NH₄NO₃ the oxidation number of one Natom is -3 and that of another N -atom is +5.

The oxidation number of Cl in Ca(OCl)Cl:

In this compound. The Cl -atom in OCl^– is linked with the O-atom, and another Cl atom exists as the Cl^– ion. The oxidation number of the Cl -atom that exists as Cl^– ion =-1.

Let the oxidation number of the Cl atom in OCP be x.

∴ – 2+ x =-1 or, x =+l

So, in Ca(.OCDCl, the oxidation number of one Cl atom is 1 and that of another Cl^– atom is +1.

The oxidation number of Mn in KMnO₄: Suppose, the oxidation number ofMn.in KMn04 = x.

Total oxidation number ofthe atoms in KMnO₄ =+1 + x+ 4 × (-2) = x-7

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of in KMnO₄ =+7

The oxidation number of P in H₄P₂O₇:

Let the oxidation number of P in H₄P₂O₇ be x. Hence, the total oxidation number of P-atom in the H₄P₂O₇ molecule

= 4 × (+1) -+2 × x + 7(-2)

= 4-+2x -14

= 2x- 10

2x- 10 = 0

Or, x = 5

Therefore, the oxidation number of P in H₄P₂O₇ = +5

The oxidation number of Fe in Fe(CO)₅:

CO is a neutral ligand (molecule) and its oxidation number = 0. So, the oxidation number of Fein Fe(CO)- is zero (0).

The oxidation number of Fe in K₄[Fe(CN)₆]:

Suppose, the oxidation number of Fe in K4[Fe(CN)₆] =x. Hence, 4 × (+1) + x+ 6 × (-1) = 0

The oxidation number of an element in 3 compounds may be zero(0):

In compounds like C₆H₁₂O₆, HCHO, CH₂C₁₂ etc., the oxidation number of carbon is zero (0). Suppose, the oxidation number ofCin C₆H₁₂O₆ = x. So, 6x+ 12x(+1) + 6x(-2) = 0

∴ x = 0

Some exceptions regarding the determination of oxidation number

The anomaly fractional oxidation state:

Since electrons can not be transferred fractionally, the fractional oxidation state of an element seems to be a hypothetical case. But in compounds like Fe₃O_{4, and} NaS₄O6 the oxidation states of Fe, and S are \(+\frac{8}{3}\) and +2.5, respectively.

The fractional oxidation state is only the average oxidation state of an element when two or more of its atoms with different oxidation states are present in a compound. For such compounds, the actual oxidation state can be determined by knowing the structure of the compound.

1. The oxidation number of Cr in CrO₅:

According to the usual method, the oxidation number of Cr in the CrO₅ molecule would be +10.

However, the oxidation number of Cr can never exceed 6 because the total number of electrons in its 3d and 4s orbitals is 6. From the chemical structure of CrO₅, it can be shown that the oxidation number of Cr in CrO₅ is, in fact, +6.

Let the oxidation number of Cr in the CrO₅ molecule be x.

x +  1 × (-2) (for O )+ 4 × (-1)(for O-atoms linked in O – O bond ) = 0

x = +6.

Hence, the oxidation number of Cr in CrO₅ = +6.

2. The oxidation number of S in H₂SO₅:

According to the tyre .usual method, the oxidation number of sulphur in the H₂SO₆ molecule is +8

The oxidation number of sulphur can never exceed + 6. The chemical structure of H₉SO₆ shows that the oxidation number of sulphur in H₂SO₅ is

Suppose, the oxidation number of S in H₂SO₅ = x

2 × (+1)( For H-atoms) + x + 2 × (-1) ( (For O-atoms held by O—O bond + 3 × (-2) ( For other- O-atoms)

x = + 6

Hence, the oxidation number of the S -atom in H₂SO₅ = +6

3. The oxidation number of S in Na₂S₂O₃:

According to the usual method, the average value of oxidation numbers of S in Na₂S₂O₃ molecule = +2

However the reaction of Na₂S₂O₃ with dilute H₂SO₄, one of the two S -atoms in Na₂S₂O₃ is precipitated as sulphur and the other is oxidised to SO₂.

Therefore, the two S -atoms in the Na₂S₂O₃ molecule are not identical. Consequently, their oxidation numbers cannot be the same. The chemical structure of the Na₂S₂O₃ molecule indicates that the two sulphur atoms in it are linked by a coordinate bond.

The oxidation number of the S -atom accepting the electron pair in the coordinate bond is considered to have an oxidation number of -2. If the oxidation number ofthe other S -atom is taken as .x, then.

2 × (+1)( For Na-atoms)  + 3 × (-2) ( (For O-atoms) + x × 1 + 1 × (-2) = 0 ( (For S-atom by coordinate bond)

∴ x = +6

Therefore, in the Na₂S₂O₃ molecule, the oxidation number of one S -atom is -2 and that of the other is +6

4. The oxidation number of S in Na₂S₄O₆:

According to the usual method, the average value of oxidation numbers of S in Na₂S₄O₆ molecule would be + 2.5

In atoms, this molecule is covalently oxidationlinkednumberis zero. If of the two oxidation sulphur numbers of each ofthe remaining two S -atoms is x, the

x × 2 (For S)+ 2× 0 (For S-S)+ 6 ×(-2)(For O) ×  2×  (+1) (For Na)= 0

Or 2x- 12 +12 = 0, x =+5

Therefore, the oxidation number of each of the two remaining S -atoms in Na₂S₄O₆ is +5

Explanation Of Oxidation-Reduction In Terms Of Oxidation Number

According to the concept of oxidation number, oxidation is a chemical reaction in which the oxidation number of atoms increases and reduction is a chemical reaction in which the oxidation number of an atom decreases.

So, oxidation means an increase in oxidation number, whereas reduction means a decrease in oxidation number. Examples of oxidation and reduction are as follows.

Explanation of oxidation-reduction reaction

Consider the reaction of HNO₃ with H₂S, forming nitric oxide (NO) and sulphur. In this reaction, the oxidation number of N decreases from +5 in HNO₂ to +2 in NO and the oxidation number of S increases from -2 in H₂S to 0 in S. So, the reaction brings about a reduction of HN03 and oxidation of H₂S.

In a redox reaction, all the atoms of a participating reactant do not change their oxidation number. Only one atom of the reactant changes oxidation number. This element is called an effective or reactive element.

In the previous example, only the N -atom of HNO₃ changes oxidation number. The oxidation numbers of hydrogen or oxygen remain the same before and after the reaction.

The reaction of FeSO₄ with KMnO4 acidified with dilute H₂SO₄ results in K₂SO₄, MnSO₄, FeSO₄ and H₂O.

In this reaction, the oxidation number of Mn decreases (+ 7→ + 2) and the oxidation number of Fe increases (+2 → + 3). Therefore, in the reaction, KMnO₄ is reduced and FeSO₄ is oxidised

Identification of oxidant and reductant based on oxidation number

In redox reactions, the substance that gets oxidised is the reducing agent and the substance that gets reduced is the oxidising agent. Based on oxidation number, it can be stated that in a redox reaction, the substance in which the oxidation number of an atom increases is the producing agent and the substance in which the oxidation number of an atom decreases is the oxidising agent.

Example:

In presence of H₂SO₄, K₂Cr₂O₇ reacts with KI to form I₂ and chromic sulphate [Cr₂(SO₄)₃]

Here, the oxidation number of Cr decreases from +6 to +3 and the oxidation number of 1- increases from -1 to 0. Therefore, K₂Cr₂O₇ acts as an oxidising agent and KI acts as a reducing agent.

How a redox reaction is identified:

At first oxidation number of each of the constituent elements of the participating substances is assigned. If the oxidation numbers of the elements change, then the reaction is identified as a redox reaction. If none of the elements shows any change in oxidation number, then the reaction is not a redox reaction.

Example: Identify whether the given two reactions are redox reactions or not

In this reaction, the oxidation numbers of all the atoms of the participating substances remain the same. Hence, it is not a redox reaction.

In this reaction, the oxidation number of N increases 0) and the oxidation number of oxygen decreases (0 →2). Hence, it is a redox reaction.

Auto Oxidation-Reduction Reactions

There are some redox reactions in which the same substance gets partially oxidised and reduced. This type of reaction is termed an auto oxidation-reduction reaction.

Example: Potassium chlorate (KCLO₃) on heating decomposes to produce KCl and O₂ gas:

In this reaction, the oxidation number of Cl decreases from +5 to -1 and the oxidation number of oxygen increases from- 2 to 0. So, in this reaction of KClO₃, one atom (O) is oxidised and the other (Cl) is reduced.

Lead nitrate undergoes thermal decomposition to produce PbO, NO gas and 02 gas:

In this reaction, the oxidation number of the N -atom reduces from +5 to +4 and the oxidation number of the O -atom increases from -2 to zero (0 ). Hence, in the reaction, the N atom is reduced and the O -atom is oxidised. So, Pb(NO₃)₃ in its thermal decomposition undergoes oxidation and reduction simultaneously.

Ammonium nitrate (NH₄NO₃) on heating decomposes to produce water vapour, N₂ gas and O₂ gas

NFI₄NO₃ is an ionic compound, consisting of ammonium cation (NH⁺₄) and nitrate anion (NO₃). The oxidation numbers of the N -atom in NH₄⁺  and NO₂ ions are -3 and 5 respectively.

In this reaction, the oxidation number of N in the NH₄ ion increases (-3 → 0) and the oxidation number of N in NO₃ decreases (+ 5 → 0). Therefore, NH₄NO₃ undergoes oxidation and t In this reaction, reduction at die same time.

Disproportionation And Comproportionation Reactions

Disproportionation reaction in which an element of a reactant undergoes oxidation and reduction simultaneously, resulting in two substances in which one of the elements exists in a higher oxidation state and the other exists in a lower oxidation state

Examples:

In the reaction of chlorine with cold and dilute NaOH solution, sodium hypochlorite (NaOCl) and sodium chloride are formed. In this reaction oxidation number of chlorine decreases (0→ -1) and increases (0→ +1) at the same time. thus, chlorine is simultaneously oxidised and reduced in the reaction.

Therefore, this reaction is an example of a disproportionation reaction.

When white phosphorus is heated with a caustic soda solution, phosphine (PH₄) and sodium hypophosphite (NaH₂PO₂) and produced.

Here P. undergoes oxidation and reduction simultaneously. In one of the products (PH₄), P exists in a lower oxidation state (-3) and the other product (NaH₄PO₂), it exists in a higher oxidation state (+1). Hence, the tills reaction is an example of a disproportionation reaction.

Comproportionation Reaction:

It is a reaction in which two reactants containing a particular element but in two different oxidation states react with each other to produce a substance in which the said element exists in an intermediate oxidation state.

Therefore, a comproportionation reaction is the opposite of a disproportionation reaction.

Example: KBrO₃ reacts with KBr in an acidic medium to produce Br₂.

In this reaction, the oxidation number of one Br-torn decreases (from +5 to. 0) and that of another Br-atom increases (from -1 to 0 ). Br₂ is formed by the oxidation of KBrO₃ and the reduction of KBr. The oxidation number of Br₂ is zero, which is intermediate between the oxidation numbers of Br atoms in KBrO, (+5 ) and KBr (-1 ). Therefore, it is a comproportionation reaction.

Equivalent Mass Of Oxidant And Reductant

The equivalent mass of oxidants and reductants is calculated by following two different methods. In one method, the calculation is done in terms of several electron (s) gained by an oxidant or the number of electrons (s) lost by a reductant. The other method takes into account the change in the oxidation number of an element present in the oxidising and reducing agents.

Oxidation number method:

The equivalent mass of an oxidant or reductant denotes the number obtained by dividing the molecular mass of the oxidant or reductant with the change in oxidation number of an element in the oxidant or reductant in their respective reduction or oxidation reaction.

⇒ \(\begin{aligned}
& \text { Equivalent mass of the oxidant }
\end{aligned}=\frac{\text { Molecular or formula mass of oxidant }}{\begin{array}{c}
\text { Total change in oxidation number } \\
\text { of an element present in a molecule } \\
\text { of the oxidant during its reduction }
\end{array}}\)

⇒ \(\begin{aligned}
& \text { Equivalent mass of the reductant }
\end{aligned}=\frac{\text { Molecular or formula mass of reductant }}{\begin{array}{c}
\text { Total change in oxidation number } \\
\text { of an element present in a molecule } \\
\text { of the reductant during its reduction }
\end{array}}\)

Determination of equivalent mass of oxidants 

Determination of equivalent mass of reductants

Electronic method: Equivalent mass of an oxidant or reductant is a number obtained by dividing the molecuslar mass or formula mass of an oxidant or reductant by the number ofelectron(s) gained or lost by a molecule of that oxidant or reductant during reduction or oxidation ofthe respective compound.

The equivalent mass of an oxidant or a reductant is formulated  as:

Equivalent mass ofthe oxidant = \(\frac{\text { Molecular or formula mass of oxidant }}{\text { Number of electron(s) gained by each molecule of oxidant during reduction }}\)

Equivalent mass bf the reductant = \(\frac{\text { Molecular mass or formula mass of a reductant }}{\text { Number of electron(s) lost by each molecule of reductant during oxidation }}\)

Determination of equivalent mass of oxidants

Determination of equivalent mass of reductants