Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11th Chemistry Online Class For 100% Result
About Lesson

Balancing of Redox reactions

There are two ways of balancing Redox reactions:

  • Oxidation number method
  • Half equation method

Oxidation method: The steps to be followed-

  1. Write the skeletal equation of reactants and products.
  2. Indicate the oxidation number of all the elements involved in the reaction.
  3. Calculate the increase or decrease in oxidation number per atom. Also, identify the oxidizing and reducing agents.
  4. Multiply the formula of oxidizing agent and reducing agent by suitable integers, so as to equalize the total increase or decrease in oxidation number as calculated in step c.
  5. Balance all atoms other than H and O.
  6. Finally balance H and O atoms by adding water molecules using hit and trial method.
  7. In case of Ionic reactions:
  • For acidic medium
  • First balance O atoms by adding water molecules to the deficient side.
  • Balance H+ ions to the side deficient in H atoms.
  • For Basic medium
  • First balance oxygen atom by adding water molecules to the deficient side.
  • Then to balance hydrogen, add water molecules equal to the number of deficiency of H atoms.
  • Also add equal number of OH ions to opposite side of the equation.

Example: Permagnate ion reacts with bromide ion in basic medium to give manganese dioxide and Bromate ion .

Step1: the skeletal ionic equation is :

MnO4 (aq) +Br (aq) —> MnO2 +BrO3

Step 2: assign oxidation numbers for Mn and Br

Redox Reactions Class 11th Chemistry Notes and NCERT Solution
Redox Reactions Class 11th

Half reaction method or Ion electron method

  1. Write the skeletal equation and indicate the oxidation number of all the elements which occur in skeletal equation
  2. Find out the species that are oxidized and reduced.
  3. Split the skeletal equation into two half reactions: oxidation half reaction and reduction half reaction
  4. Balance the two-half equation separately by rules described below:
  • In each half reaction first balance the atoms of element that has undergone a change in oxidation number.
  • Add electrons to whatever side is necessary to make up the difference in oxidation number in each half reaction.
  • Balance the charge by adding H+ ions, if the reaction occurs in acidic medium .For basic medium, add OH ions if the reaction occurs in basic medium.
  • Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms
  • In the acidic medium, H atoms are balanced by adding H + ions to the side deficient in H atoms.
  • However, in the basic medium H atoms are balanced by adding water molecules equal to number to H atoms deficient.
  • Add equal number of OH ions to opposite side of equation.
  • The two half reactions are then multiplied by suitable integers .so that the total number of electrons gained in half reaction becomes equal to total number of electrons lost in another half reaction.
  • Then the two half reactions are added up.
  • To verify the balancing, check whether the total charge on either is equal or not.

Example: Let us consider the skeletal equation:

Fe2+ + Cr2O72- –>  Fe3+ +Cr3+

Step 1: Separate the equation in to two halves:

Oxidation half reaction: Fe–>Fe3+

 Reduction half reaction: Cr2O72-  –>    Cr3+

Step 2: Balance the atoms other than hydrogen and oxygen in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms .For the reduction half reaction, we multiply the Cr3+ by 2 to balance Cr atoms.

Step 3: For reactions occurring in acidic medium, add water molecules to balance oxygen atoms and hydrogen ions are balanced by adding H atoms. Thus, we get:

Cr2O72- + 14 H+  +  6e –> 2 Cr3+ + 7H2O

Step 4: Add electrons to one side of the half reaction to balance the charges .if needed make the number of electrons equal in two half reactions by multiplying one or both half reaction by suitable coefficient.

The oxidation half reaction is thus written again to balance the charge .Now in the reduction half reaction there are 12 positive charges on the left hand side and only 6 positive charge on right hand side .Therefore, we add six electrons to left hand side .

Cr2O72- + 14 H+  +  6e –> 2 Cr3+ + 7H2O

To equalize the number of electrons in both reactions, we multiply oxidation half reaction by 6 and write as:

6Fe2+  –>  6Fe3+ +6e

Step 5: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side .This give us net ionic equation:

6Fe2+  + Cr2O72- + 14 H+ –>    2Cr3++6Fe3+ +7H2O

Step6: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number atoms and the charges.

Types of Redox reactions

  1. Combination reaction: When two or more reactants react to form single product.

i.e   a + b –> ab (in this single product is formed).

For the reaction to be Redox, both the elements should be in elemental form.

For example: All combustion reactions are Redox reactions.

C0        + O2 –>   CO2

Carbon  Oxygen  Carbondioxide

Redox Reactions Class 11th Chemistry Notes and NCERT Solution
Redox Reactions Class 11th Chemistry Notes and NCERT Solution
Redox Reactions Class 11th

There are two types of Redox reactions

  • Direct Redox reaction
  • Indirect Redox reaction

Direct Redox reaction: In which oxidation and reduction both occur in same beaker. In this electron so produced does not travel to large distance.

For example: A beaker containing Zinc rod dipped in Copper sulphate solution in this the following reaction occur:

Zn    +    CuSO4 –>       ZnSO4 + Cu

Zinc   CopperSulphate  Zinc Sulphate  Copper

In this zinc being more reactive displaces copper from copper sulphate and forms zinc sulphate and copper.

Indirect Redox reactions: In it oxidation and reduction occur in different beakers. The electron so produced has to travel a certain distance that leads to generation of current.

Example: Daniel cell: A cell containing Zn-Cu couple that we are going to study in detail now as given below but before that let us make you familiar with the general term used in redox reaction that is Redox couple.

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