CBSE Class 11 Chemistry Revision Notes Chapter 8

Class 11 Chemistry Revision Notes for Chapter 8 – Redox Reactions 

To make the chapter more easily understandable and help students prepare well for the exams, this  Class 11 Chemistry Chapter 8 Notes are easily accessible from Extramarks. By referring to these notes, students can make their notes and familiarise themselves with all the key concepts at once. Prepared by subject matter as per the latest CBSE Syllabus, students can leverage their exam preparation with Chemistry Class 11 Chapter 8 Notes to score high marks.

Access Revision Notes for Class 11 Chemistry Chapter 8 – Redox Reaction 

Introduction 

This chapter intends to ensure that students can identify redox reactions as a class

of reactions in which oxidation and reduction reactions occur simultaneously. Redox refers to an integral area of Chemistry. It is an essential reaction that concerns physical as well as biological phenomena. 

Oxidation 

The process that involves the addition of oxygen or any electronegative element or the removal of any electropositive element or hydrogen is called oxidation. 

Reduction 

The process that involves the addition of hydrogen or any electropositive element or the removal of any electronegative element or oxygen is called reduction.

Redox

Since oxidation and reduction always take place simultaneously, the term “redox” was developed to refer to this class of chemical events.

Oxidation Number 

• When an element transitions from its free elemental state to its combined form in molecules, it develops a seeming charge over each atom.
• The oxidation number is determined using a set of rules.
• A more practical way of determining the oxidation number is to keep track of shifts in electrons in a chemical reaction involving the synthesis of compounds that have been created.
• In this procedure, the complete transfer of electrons from a less electronegative atom to a more electronegative atom is always expected.

Rules Governing Oxidation Number 

The rules used to determine the oxidation number of elements in different compounds are based on the element’s electronegativity. You can have a better understanding and clarity of concepts by going through the notes provided by Extramarks.

Atom of Fluorine 

The most electronegative of all the known elements is Fluorine. It has an oxidation number of –1 in all of its compounds. 

Atom of Oxygen 

The atoms of oxygen and their oxides have an oxidation number of –2.

Hydrogen Atom 

The oxidation number of a hydrogen atom is +1 in general. However, it is –1 in the case of metallic hydrides.

Halogen Atom 

All halogen atoms have an oxidation number of –1. If a halogen atom is connected to an atom more electronegative than it, the oxidation number will be positive.

Metals 

• +1 is the oxidation number of alkali metals.
• +2 is the oxidation number of alkaline earth metals.
• +3 is the oxidation number of aluminium.
• The oxidation number of a metal can also be negative or 0.
• The number of oxidations in an element’s free state or allotropic forms is always zero.
• A molecule’s total number of oxidation states for all of its atoms is zero.
• The sum of the oxidation numbers of all the atoms in the ion is equal to the charge on an ion.
• The oxidation number of an element with group number n in the periodic table can range from (n – 10) to (n – 18). This, however, mostly applies to p-block components.

Calculation of Average Oxidation Number: Solved Examples 

Example 1: Q.Find out the oxidation number of underlined element: Na2S2

Ans: Let the oxidation number of S-atoms be y. Now work according to the rules stated before.

(+1)×2+(y)×2+(−2)×3=0     

y=+2

Individual Oxidation Number Calculation 

It is crucial to understand the structure of any compound before attempting to determine the particular oxidation number of an element inside it.

Individual oxidation number calculation formula:

Oxidation number = Number of electrons in the valence shell – Number of electrons taken up after bonding

Solved Examples 

Example  2:

Q.Calculate the individual oxidation number of each S-atom in Sodium Thiosulphate (Na2S2O3) with the help of its structure.

Ans: Number of electrons in the valence shell of the middle S-atom = 6

After bonding, the number of electrons left = 0

The oxidation number of central S-atom = 6 − 0 = +6

Number of electrons in the valence shell of terminal S-atom = 6

After bonding, the number of electrons left = 8

The oxidation number of terminal S-atom= 

6 − 8 = −2

Now, even the Average Oxidation number of S can be calculated, where, S=[6+(−2)]/2=+2 (as we have calculated before). 

Miscellaneous Examples 

Q.To calculate the individual oxidation number we need to take help from the structures of the molecules. Find out the oxidation of Cr in CrO5.

Ans: It is evident that there are two peroxide linkages and one double bond in CrO5 from its structure. The contribution of each peroxide linkage is −2. Now, let the oxidation number of Cr be y.

∴y+(−2)2+(−2)=0 or y=6

∴ The oxidation number of Cr=+6 

The paradox of Fractional Oxidation Number 

The structural parameters show that the fractional oxidation number represents the average oxidation state of all atoms of the element under study. It also shows that the atoms of the element for which the fractional oxidation state is realised are actually present in different oxidation states. The structure of C3O2, for instance, exhibits similar bonding possibilities. Each element’s element with an asterisk (*) has an oxidation number that is distinct from the rest of the element’s atoms.

Oxidising and Reducing Agent 

Oxidising Agent or Oxidant : 

Oxidants are substances that oxidise others while reducing themselves during a chemical process. They cause an element’s oxidation number to decrease or cause an element to gain electrons in a redox process.

Reducing Agent or Reductant : 

Reductants are substances that can both oxidise and reduce other molecules during a chemical reaction. They increase the oxidation number or cause the element to lose electrons in a redox reaction.

Redox Reaction 

Redox reactions are chemical processes that involve simultaneous oxidation and reduction.

Disproportionation Reaction 

A disproportionation reaction is a redox process in which the same element that is present in a molecule in a particular oxidation state is oxidised and reduced simultaneously. 

List of Some Important Disproportionation Reactions 

1. H2O2 → H2O + O2 
2. P4 + OH− → PH3 + H2PO2– 
3. HNO2 → NO + HNO3

Balancing of Redox Reactions 

Two conditions must be met by any balanced equation.

• Atom balancing (mass balance): There should be an equal amount of atoms of each species on the reactant and product sides.
• Charge balance: The total of the real charges in the equation must be equal on both sides.

A redox equation can be balanced in the following two ways:

• Number change technique of oxidation
• Half-cell method or ion-electron method

Students are recommended to try the second method (Ion electron method) to balance redox reactions because the first strategy is not especially effective in doing so.

Method of Ion Electrons 

Using this technique, redox equations in two different mediums are changed.

• Acidic Medium 
• Basic Medium

Balancing in Acidic Medium 

For balancing redox reactions, students are advised to use the ion-electron technique in an acidic medium. Refer to Chapter 8 Chemistry Class 11 Notes for understanding this in detail. 

Balancing in Basic Medium 

Bases dissolve into OH− ions in solution; therefore, balancing redox reactions in basic conditions requires OH−. Refer to Chemistry Chapter 8 Class 11 Notes for understanding this in detail. 

Concept of Equivalents:

Equivalent Mass of Element 

The number of parts by mass of an element that displaces or reacts from a compound containing 8 parts by mass of oxygen, 1.008 parts by mass of hydrogen, and 35.5 parts by mass of chlorine is known as the equivalent weight of that element.

Equivalent Weight (E) 

An equivalent weight, also known as gram equivalent, is the mass of one equivalent, i.e. the mass of a given substance that will combine with or displace a fixed quantity of another substance. 

• For a solution, the number of equivalents is equal to N1V1, where N denotes normality and V denotes volume in litres.
• When expressed in grammes, equivalent mass—a pure number—becomes known as gramme equivalent mass.
• The comparable mass of a substance may vary depending on the circumstances.
• The molecular mass will not always be less than the corresponding weight, however, this is not an absolute rule.
• Number of Equivalents = mass of specieseq. wt. of that species

Valency Factor Calculation 

The valency factor equals the element’s valence for elements. For acids, the number of hydrogen ions that can be replaced per acid molecule is the valence factor.

Acid – Base Reaction 

The actual number of hydrogen or hydroxide ion-exchanged in the reaction is the valence factor in an acid-base reaction. More hydrogen or hydroxide ions may be replaceable in the acid or base than what is actually replaced in the reaction. The number of hydrogen ions replaced by each molecule of the base from the acid is denoted by v. f.

Salts: 

ln Non-Reacting Condition 

The total amount of negative or positive charges present in the chemical is known as the valency factor.

Normality 

Normality is the number of equivalents of solute present in one litre or 1000 millilitres of a solution. 

Law of Equivalence

According to the law, one equivalent of one element must combine with one equivalent of another. A chemical reaction results in the same number of equivalents or milliequivalents of products when equivalents or milliequivalents of reactants react in the same proportion.

Titrations 

The process of estimating a solution’s concentration by allowing a precisely measured volume to react with a known-concentration standard solution of another substance is called titrations. The standard solution so used is known as Titrant. 

Titrants are of two types: 

• Primary Titrants/Standards 
• Secondary Titrants/Standards 

Indicators 

Indicators are auxiliary materials added to allow physical detection of titration completion. When the titration is finished, the colour of the indicator usually changes.

Hydrogen Peroxide 

In both acid and base mediums, Hydrogen Peroxide can act as a reducing agent as well as an oxidising agent.

Hardness of water (Hard water does not give lather with soap)

Calcium and Magnesium bicarbonates cause temporary hardness while calcium and Magnesium chlorides and sulphates cause permanent hardness. The water sample can be softened by boiling or by using washing soda.

Measurement of Hardness 

Hardness is measured in terms of Parts Per Million (ppm) of Calcium Carbonate (CaCO3)

or its equivalent. 

Solved Examples 

Q1. A fresh H2O2 solution is labelled 11.2V. It has the same concentration as a solution which is:

• 3.4%(w/w)
• 3.4%(v/v)
• 3.4%(w/v)
• None of these

Explanation: Molarity of H2O2  = vol. strength 11.2=11.211.2=1

Now, %(w/v)= wt. of solute in g wt. of solution in mL×100

= Molarity × Mol. wt. of solute ×110

=1×34×110=3.4%

Ans. c. 

Q2. Find out the normality of a solution containing 13.4g of sodium oxalate in a 100mL solution. 

Ans: Normality = (wt. in g/eq. wt)/vol of solution in a litre 

Here, eq. wt. of Na2C2O4=134/2=67

so N=(13.4/67)/(100/1000)=2N 

Class 11 Redox Reactions Revision Notes 

Extramarks, one of the leading e-learning platforms, is known for providing the best educational materials for students of all classes. The review notes offered by Extramarks for CBSE Class 11 Redox Reactions are a great option for students to comprehend this chapter in great detail. These notes are written by the subject-matter experts, ensuring the reliability and quality of the chapter in complete detail.

Sub-Topics Covered Under Redox Reactions 

• Balance Redox Reactions 
• Classical Idea of Redox Reactions
• Oxidation Number 
• Redox Reactions & Electrode Potential 
• Types of Redox Reactions 
• Redox Reactions as the Basis of Titrations 
• Redox Reactions 

Some Important Points to Determine the Oxidation Number 

• In an uncharged compound, the algebraic sum of the oxidation numbers of the atoms is zero. The algebraic sum of an ion’s charge is equal to its charge.
• All of the elements’ oxidation numbers in their elementary states are zero.
• Metal in amalgams has no oxidation at all.

Importance of Redox Reactions Revision Notes 

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Tips to Use Notes of Chapter 8 Chemistry Class 11 

• Students are advised to read the chapter first from the NCERT Chemistry book and highlight the sections they have trouble understanding after having a basic idea of the chapter. 
• They must carefully go through the Class 11 Chemistry Chapter 8 Notes to understand the challenging sections identified earlier to avoid any silly mistakes and be confident.
• Students can apply what they have learned by practising the questions that are provided at the end of the chapter in the textbook.