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Matter
OBJECTIVES By the end of this topic, the trainee should be able to 1.Define matter 2.Explain state of matter 3.Distinguish between physical and chemical changes 4.Explain the gas laws
0/4
Introduction
States of Matter
Physical and Chemical Properties and Changes
The Gas Laws
Atoms , Elements and Compounds
OBJECTIVES By the end of this topic , the trainee should be able to; 1.Define Elements, Compounds and Mixtures 2.Describe the structure of an atom 3.Describe how to determine the Atomic number ,Mass number and Isotopes
0/3
Introduction
Atomic Structure
Atomic Structure of Isotopes
The Periodic Table
OBECTIVES By the end of this topic, the trainee should be able to : 1.State the historical contribution on development of the periodic table 2.Explain the periodic trends of elements and their compounds 3.State the diagonal relationships of the periodic table
0/3
Introduction
Families of the Periodic Table
The Periodic Trends
The S-Block Element
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Explain the chemistry of group I and II elements 2.State the application of group I and two elements and their compounds
0/4
Introduction
Electronic Configuration of S-Block Elements
Physical Properties of S-Block Elements
Chemical Properties of S-Block Elements
Chemical Bonds
OBJECTIVES By the end of these topic, the trainee should be able to 1.Identify different types of bonds 2.Describe their properties
0/2
Introduction
Types and Properties of Chemical Bonds
Chemical Equilibrium
OBJECTIVES By the end of this topic , the trainee should be able to : 1.Define chemical equilibria 2.Explain types of equilibria 3.Determine equilibrium constant 4.Describe factors affecting chemical equilibrium
0/6
Introduction
Measuring Equilibrium Rates
Equilibrium Basics
Types of equilibria
Predicting The Direction of a Reaction
Factors Affecting equilibrium
Introduction To Organic Chemistry
By the end of this topic , the trainee should be able to : 1.Explain the aspects of organic chemistry 2.Describe hydrocarbons 3.Classify organic molecules explain chemical reactions of simple organic molecules 4.Explain the properties , synthesis and uses of simple organic molecules
0/10
Introduction
Functional Groups
Alkanes
Alkenes and Alkynes
Alkanols(Alcohols)
HaloAlkanes
Alkanals and Alkanones
Carboxylic Acids
Esters
Amines
Acids, Bases and Salts
OBJECTIVES By the end of this session , the trainee should be able to : 1.State properties of acids and bases 2.Differentiate between strong and weak acids 3.Explain types and properties of salts
0/2
Introduction
Properties Of Acids And Bases
PH Analysis
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define the term PH 2.Explain the basic theory of PH 3.State the relationship between PH and color change in indicators 4.Explain the term buffer solution 5.Describe the preparation of buffer solutions 6.State the application of buffer solutions
0/5
Introduction
Neutral, acidic and basic solutions
PH and Its Importance
Determination Of PH
Buffer Solutions
Sampling and Sample Preparation
OBJECTIVE By the end of this topic, the trainee should be able to : 1.Define the terms used in sample preparation 2.State the importance of sampling 3.Describe the techniques of sampling 4.Describe the procedure for sample pre-treatment 5.State sample storage methods
0/5
Introduction
Types of Samples and Sampling
Sample preparation
Sample Losses During Preparation
Sample Identification and Storage
Separation Techniques
OBJECTIVES By the end of this topic , the trainee should be able to : 1.Define separation, extraction and purification 2.Describe the separation , extraction and purification techniques 3.Explain the methods of determining purity of substances
0/2
Introduction
Separation Techniques
Heating and Cooling Techniques
OBJECTIVES To identify various techniques used for heating and cooling substances in the laboratory
Heating and Cooling Techniques
OBJECTIVES To identify various techniques used for heating and cooling substances in the laboratory
0/1
Heating and Cooling Techniques used in the lab
Distillation Techniques
By end of this topic, Trainee should be able to : 1. Define distilation 2. State and explain various distillation techniques 3. Outline Various distillation techniques 4. Outline the applications of Distillation techniques
0/3
Introduction
Types of Distillation Techniques 
Distillation and Reflux Condensers
Crystallization Techniques
OBJECTIVES By the end of the topic, the learner should be able to: 1.To define crystallization 2.To describe crystallization process 3.To carry out crystallization procedure
0/1
Crystallization Technique
Solvent Extraction Techniques
OBJECTIVES By the end of the topic, the learner should be able to 1.Define solvent extraction 2.Explain terms used in solvent extraction 3.Describe methods of solvent extraction 4.Describe selection of appropriate solvents for solvent extraction 5.Determine distribution ration 6.Outline factors actors influencing the extraction efficiency 7.Describe Soxhlet extraction
0/1
Introduction
Chromatography Techniques
OBJECTIVES By the end of this topic, the learner should be able to: 1.Define chromatography techniques 2.Explain terms used in chromatography techniques 3.Describe principles of chromatography techniques 4.Explain types of chromatography techniques 5.Carry out chromatography experiments 6.Determine RF factor 7.Outline electrophoresis
0/6
Introduction to Chromatography Techniques
Types of Chromatography Techniques
Thin Layer Chromatography
Column Chromatography
Retardation Factor (Rf)
Gas-Liquid Chromatography
Titrimetric Analysis
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define terms used in titrimetric analysis 2.Describe types of titrimetric analysis 3.Balance chemical reactions 4.Work out calculations involved in titrimetric analysis
0/6
Introduction
Requirements For Titrimetric Analysis
Titration procedure:
Molarity
Apparatus for titrimetric analysis
Classification Of Reactions In Titrimetrlc Analysis
Redox Titration
Redox Titration is a laboratory method of determining the concentration of a given analyte by causing a redox reaction between the titrant and the analyte. Redox titration is based on an oxidation-reduction reaction between the titrant and the analyte. It is one of the most common laboratory methods used to identify the concentration of unknown analytes. Redox reactions involve both oxidation and reduction. The key features of reduction and oxidation are discussed below.
0/5
Introduction
The oxidation number 
Balancing Redox Reactions
Examples Of Redox Titration
Standardization of Redox solutions
Complexiometric Titration
omplexometric Titration or chelatometry is a type of volumetric analysis wherein the colored complex is used to determine the endpoint of the titration. The method is particularly useful for determination of the exact number of a mixture of different metal ions, especially calcium and magnesium ions present in water in solution .
0/5
Introduction to Complexiometric Titration
EDTA titration
Types of EDTA titrations
EDTA titration experiment 
Mechanism of action of indicator:
Gravimetric Analysis
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define gravimetric analysis 2.Describe the principles of gravimetric analysis 3.Describe the steps involved in gravimetric analysis 4.Explain factors affecting gravimetric analysis 5.Describe the equipments and apparatus used in gravimetric analysis 6.Carry out gravimetric analysis
0/8
Introduction
Principles Of Gravimetric Analysis
Determination of Mass in gravimetric analysis
Factors Affecting Separation Of Metals And Compounds
Preparation of pure precipitates
Example of gravimetric Analysis
Gravimetric Calculations
Draft Lesson
Calorimetric Analysis
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define terms and units used in thermochemistry 2.Determine enthalpy changes in chemical reactions 3.Determine heat capacity and specific heat capacity 4.Compare calorific values of different materials 5.Determine different heat reactions 6.Apply law of conservation of energy and Hess law in thermochemical calculations
0/4
Introduction
Calorimetry
Heat Capacity
Enthalpy of Phase Transitions
Chemistry Techniques for Science Laboratory Technicians
About Lesson

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Balancing Redox Reactions

Oxidation-Reduction or “redox” reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers.

The Half Equation Method is used to balance these reactions.

In a redox reaction, one or more element becomes oxidized, and one or more element becomes reduced. Oxidation is the loss of electrons whereas reduction is the gain of electrons. An easy way to remember this is to think of the charges: an element’s charge is reduced if it gains electrons (an acronym to remember the difference is LEO = Lose Electron Oxidation & GER = Gain Electron Reduction).

 Redox reactions usually occur in one of two environments: acidic or basic. In order to balance Redox equations, understanding oxidation states is necessary.

Some points to remember when balancing redox reactions

  • The equation is separated into two half-equations, one for oxidation, and one for reduction.
  • The equation is balanced by adjusting coefficients and adding H2O, H+, and e–in this order:
    1. Balance the atoms in the equation, apart from O and H.
    2. To balance the Oxygen atoms, add the appropriate number of water (H2O) molecules to the other side.
    3. To balance the Hydrogen atoms (including those added in step 2), add H+
    4. Add up the charges on each side. They must be made equal by adding enough electrons (e–) to the more positive side.
  • The e–on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same.
  • The half-equations are added together, cancelling out the electrons to form one balanced equation. Cancel out as much as possible.

These reactions can take place in either acidic or basic solutions.

  • (If the equation is being balanced in a basic solution, the appropriate number of OH– must be added to turn the remaining H+ into water molecules)
  • The equation can now be checked to make sure it is balanced.
  1. In Acidic Aqueous Solution

Balance this reaction

               MnO−4+I−⟶I2+Mn2+

Solution

Steps to balance:

Step 1: Separate the half-reactions that undergo oxidation and reduction.

Oxidation:

                I−⟶I2

This is the oxidation half because the oxidation state changes from -1 on the left side to 0 on the right side. This indicates a gain in electrons.

Reduction:

                MnO−4⟶Mn2+

This is the reduction half because the oxidation state changes from +7 on the left side to +2 on the right side. This indicates a reduction in electrons.

Step 2: In order to balance this half reaction we must start by balancing all atoms other than any Hydrogen or Oxygen atoms.

Oxidation:

               2I−⟶I2

In order to balance the oxidation half of the reaction you must first add a 2 in front of the I on the left hand side so there is an equal number of atoms on both sides.

Reduction:

            MnO−4⟶Mn2+

For the reduction half of the reaction, you can notice that all atoms other than Hydrogen and Oxygen are already balanced because there is one manganese atom on both sides of the half reaction.

Step 3: Balance Oxygen atoms by adding H2O to the side of the equation that needs Oxygen. Once you have completed this step add H+ to the side of the equation that lacks HH atoms necessary to be balanced.

Oxidation:

                      2I−⟶I2

Because there are no Oxygen or Hydrogen atoms in this half of the reaction, it is not required to perform any kind of balancing.

 

Reduction:

                 MnO−4⟶Mn2+ +  4H2O

The first step in balancing this reaction using step 3 is to add4 H2O atoms in order to balance the Oxygen atoms with the 4 on the other side of MnO4–

Reduction:

                 MnO−4+8H+⟶Mn2++4H2O

Now that the Oxygen atoms have been balanced you can see that there are 8 H atoms on the right hand side of the equation and none on the left. Therefore, you must add 8 H+ atoms to the left hand side of the equation to make it balanced.

Step 4: Now that the two half reactions have been balanced correctly one must balance the charges in each half reaction so that both the reduction and oxidation halves of the reaction consume the same number of electrons.

Oxidation:

           2I−⟶I2+2e−

Because of the fact that there are two I’s on the left hand side of the equation which a charge of -1 we can state that the left hand side has an overall charge of -2. The I on the left side of the equation has an overall charge of 0. Therefore to balance the charges of this reaction we must add 2 electrons to the right side of the equation so that both sides of the equation have equal charges of -2.

Reduction:

           5e−+8H++MnO−4⟶Mn2++4H2O

Looking at the left hand side of the equation you can notice that there are 8 Hydrogen atoms with a +1 charge. There is also a MnO−4 ion that has a charge of -1. When we add these two charges up we can calculate that the left hand side of the equation has an overall charge of +7. The right hand side has an Mn atom with a charge of +2 and then 4 water molecules that have charges of 0. Therefore, the overall charge of the right side is +2. We must add 5 electrons to the left side of the equation to make sure that both sides of the equation have equal charges of +2.

Step 5: Multiply both sides of both reactions by the least common multiple that will allow the half-reactions to have the same number of electrons and cancel each other out.

Oxidation:

                    10I−→5I2+10e−

We multiply this half reaction by 5 to come up with the following result above.

Reduction: 10e−+16H++2MnO−4→2Mn2++8H2O  

We multiply the reduction half of the reaction by 2 and arrive at the answer above.

By multiplying the oxidation half by 5 and the reduction half by 2 we are able to observe that both half-reactions have 10 electrons and are therefore are able to cancel each other out.

Step 6: Add the two half reactions in order to obtain the overall equation by canceling out the electrons and any H2O and H+ ions that exist on both sides of the equation.

Overall:

          10I−+16H++2MnO−4⟶5I2+2Mn2++8H2O

In this problem, there is not anything that exists on both halves of the equation that can be cancelled out other than the electrons. Finally, double check your work to make sure that the mass and charge are both balanced. To double check this equation you can notice that everything is balanced because both sides of the equation have an overall charge of +4.

  1. In Basic Aqueous Solutions

The balancing procedure in basic solution differs slightly because OH− ions must be used instead of H+ ions when balancing hydrogen atoms. To give the previous reaction under basic conditions, sixteen OH− ions can be added to both sides. on the left the OH− and the H+ ions will react to form water, which will cancel out with some of the H2O on the right.

10I−(aq)+2MnO−4(aq)+16H+(aq)+16OH−(aq)⟶5I2(s)+2Mn2+(aq)+8H2O(l)+16OH−(aq)

On the left side the OH- and the H+ ions will react to form water, which will cancel out with some of the H2O on the right:

10I−(aq)+2MnO−4(aq)+16H2O(l)⟶5I2(s)+2Mn2+(aq)+8H2O(l)+16OH− (aq)

Eight water molecules can be canceled, leaving eight on the reactant side:

10I−(aq)+2MnO−4(aq)+8H2O(l)⟶5I2(s)+2Mn2+(aq)+16OH−(aq)

This is the balanced reaction in basic solution.

 Example

Balance the following in an acidic solution.

             SO32−(aq)+MnO−4(aq)→SO2−4(aq)+Mn2+(aq)

Solution

To balance a redox reaction, first take an equation and separate into two half reaction equations specifically oxidation and reduction, and balance them.

Step 1: Split into two half reaction equations: Oxidation and Reduction

Oxidation: 

                   SO32− (aq)⟶SO42− (aq)

[ oxidation because oxidation state of sulfur increase from +4 to +6]

Reduction:

             MnO+4(aq)⟶Mn2+(aq)

[ Reduction because oxidation state of Mn decreases from +7 to +2]

Step 2: Balance each of the half equations in this order:

  • Atoms other than H and O
  • O atoms by adding H2Omolecules with proper coefficients
  • H atoms by adding H+with proper coefficients

The S and Mn atoms are already balanced,

Balancing O atoms

Oxidation:SO32− (aq)+H2O(l)→SO42− (aq)

Reduction:MnO4(−aq→Mn2+(aq)+4H2O(l)

Then balance out H atoms on each side

Oxidation:SO2−3(aq)+H2O(l)→SO42− (aq)+2H+(aq)

Reduction:MnO4− (aq)+8H+→Mn2+(aq)+4H2O(l)

Step 3: Balance the charges of the half reactions by adding electrons

Oxidation:SO32− (aq)+H2O(l)→SO42− (aq)+2H+(aq)+2e−

Reduction:MnO4− (aq)+8H++5e−→Mn2+(aq)+4H2O(l)

Step 4: Obtain the overall redox equation by combining the half reaction, but multiply entire equation by number of electrons in oxidation with reduction equation, and number of electrons in reduction with oxidation equation.

Oxidation:5×[SO32− (aq)+H2O(l)→SO42− (aq)+2H+(aq)+2e−]

Reduction:2×[MnO4− (aq)+8H++5e−→Mn2+(aq)+4H2O(l)]

Overall Reaction:

Oxidation:5SO32− (aq)+5H2O(l)→5SO42− (aq)+10H+(aq)+10e−

Reduction:2MnO4− (aq)+16H++10e−→2Mn2+(aq)+8H2O(l)

total:5SO32− (aq)+5H2O(l)+2MnO4− (aq)+16H++10e−→5SO42− (aq)+10H+(aq)+2Mn2+(aq)+8H2O(l)+10e−

Step 5: Simplify and cancel out similar terms on both sides

To get

5SO32−(aq)+2MnO4−(aq)+6H+→5SO42−(aq)+2Mn2+(aq)+3H2O(l)

Balance this reaction in both acidic and basic aqueous solutions

         MnO4−(aq)+SO32− (aq)⟶MnO2 (s) +SO42− (aq)

Solution

First, they are separated into the half-equations:

            MnO4− (aq)⟶MnO2(s)

This is the reduction half-reaction because oxygen is LOST)

And                 SO32− (aq)⟶SO42− (aq))

(the oxidation, because oxygen is GAINED)

Now, to balance the oxygen atoms, we must add two water molecules to the right side of the first equation, and one water molecule to the left side of the second equation:

           MnO4− (aq)⟶MnO2(s)+2H2O(l)

           H2O(l)+SO32− (aq)⟶SO42− (aq)

To balance the hydrogen atoms (those the original equation as well as those added in the last step), we must add four H+ ions to the left side of the first equation, and two H+ ions to the right side of the second equation.

            4H++MnO4− (aq)⟶MnO2(s)+2H2O(l)

            H2O(l)+SO32− (aq)⟶SO42− (aq)+2H+

Now we must balance the charges. In the first equation, the charge is +3 on the left and 0 on the right, so we must add three electrons to the left side to make the charges the same. In the second equation, the charge is -2 on the left and 0 on the right, so we must add two electrons to the right.

             3e−+4H++MnO4− (aq)⟶MnO2(s)+2H2O(l)

               H2O(l)+SO32− (aq)−⟶SO42− (aq)+2H++2e−

Now we must make the electrons equal each other, so we multiply each equation by the appropriate number to get the common multiple (in this case, by 2 for the first equation, and by 3 for the second).

                2(3e−+4H++MnO4− (aq)⟶MnO2(s)+2H2O(l))

                 3(H2O(l)+SO32− (aq)⟶SO42− (aq)+2H++2e−)

With the result:

6e−+8H++2MnO4−(aq)⟶2MnO2(s)+4H2O(l)

3H2O(l)+3SO32− (aq)⟶3SO42− (aq)+6H++6e−

Now we cancel and add the equations together. We can cancel the 6e– because they are on both sides. We can get rid of the 6H+ on both sides as well, turning the 8H+ in the first equation to 2H+. The same method gets rid of the 3H2O(l) on the bottom, leaving us with just one H2O(l) on the top. In the end, the overall reaction should have no electrons remaining. Now we can write one balanced equation:

2MnO4− (aq)+2H++3SO32− (aq)⟶H2O(l)+2MnO2(s)+3SO42− (aq) 

The equation is now balanced in an acidic environment.

To balance in a basic environment add OH− to each side to neutralize the H+ into water molecules:

2MnO4−(aq)+2H2O+3SO32−(aq)⟶H2O(l)+2MnO2(s)+3SO42− (aq)+2OH−

and then cancel the water molecules

2MnO4−(aq)+H2O+3SO32−(aq)⟶+2MnO2(s)+3SO42− (aq)+2OH−

The equation is now balanced in a basic environment.

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