CHEMICAL EQUATION

 CHEMICAL EQUATION

 

The equation used to express an actual chemical reaction in terms of symbols and chemical formulae of various substances is called chemical equation. The substances present on the left hand side of the equation are called reactants whereas those on the right hand side are called products. The equation which has equal number of atoms on its both sides is called balanced chemical equation. The equation which is not balanced is known as skeleton equation. The coefficients of balanced chemical equation are called stoichiometric coefficients. E.g.

 

SIGNIFICANCE OF CHEMICAL EQUATION

 

1) It tells us about names of reactants and products.

2) Number of moles of reactants and products.

3) It tells us about the relative number of molecules of reactants and products

4) It tells us about the relative amount of reactants and products in grams.

5) It tells us about the volumes of gaseous reactants and products.
 

 BALANCING OF CHEMICAL EQUATION

 

1) Hit and trial method

2) Partial equation method

3) Ion-electron method

 

4) Oxidation Number Method

 

1) Hit and trial method

 

 1) KClO₃     →  KCl   + O₂

 2) C₆H₆   +   O₂     →   CO₂   +   H₂O

 3) Fe₃O₄ + H₂   →  Fe   +   H₂O

 4) FeS₂ + O₂   →  Fe₂O₃ + SO₂

 5) As₂O₃ + SnCl₂ + HCl     →  SnCl₄ +   As   + H₂O

6) Na₂CO₃ + HCl    →  NaCl   + CO₂   + H₂O

7) KMnO₄   + H₂SO₄    →   K₂SO₄   + MnSO₄   + H₂O   + O₂

8) K₂Cr₂O₇   +   H₂SO₄    →  K₂SO₄   + Cr₂ (SO₄)₃   + H₂O   +   O₂

9) C₄H₁₀   +   O₂     →   CO₂   +   H₂O

 

10) Ca₂CO₃ + HCl    →  CaCl₂  + CO₂   + H₂O

 

11) NH₄Cl  +  Ca(OH)₂    →  CaCl₂  + NH₃  + H₂O 

 

12) Fe   +    H₂SO₄    →  Fe₂ (SO₄)₃    +   H₂

 

13)  Al   +   O₂     →   Al₂O₃

 

2.PARTIAL EQUATION METHOD

This method is simpler than Hit & Trial method.In this method the reaction is supposed to take place in two or more steps.Each step is written in the form of a partial equation.This method involves the following steps:

1. Write Partial equation for each step.

2. Balance each partial equation by Hit & Trial method.

3. Then add the partial equations to get the final equation.

4. While adding the partial equations cancel out the products that do not appear finally.

5. Lastly convert the atomic equation in the molecular form.

PRACTICE SESSION 

Balance the following skeleton equations using partial equation method:

1.)  KMnO₄   + H₂SO₄    + (COOH)₂   →   K₂SO₄   + MnSO₄   + H₂O   + CO₂

2.)  KMnO₄    +  H₂SO₄  +  FeSO₄  →  K₂SO₄  + MnSO₄  + Fe₂ (SO₄)₃  + H₂O  

3.)  KMnO₄   +  H₂SO₄  +  H₂S  →  K₂SO₄  + MnSO₄  +  S  + H₂O  

4.)  KMnO₄   +   H₂SO₄  +  KI   →  K₂SO₄   + MnSO₄   + H₂O   +  I₂

5.)  K₂Cr₂O₇   +   H₂SO₄  +  KI   →  K₂SO₄   + Cr₂ (SO₄)₃   + H₂O   +  I₂

6.)  K₂Cr₂O₇   +  H₂SO₄  +  FeSO₄  →  K₂SO₄  + Cr₂ (SO₄)₃   + Fe₂ (SO₄)₃  + H₂O  

 

7.)  K₂Cr₂O₇   +  H₂SO₄  +   H₂S  →  K₂SO₄  + Cr₂ (SO₄)₃   +   S  + H₂O  

 

8.) K₂Cr₂O₇   +  H₂SO₄  +   HBr →  K₂SO₄  + Cr₂ (SO₄)₃   +   Br₂ + H₂O  

 

3.Ion-Electron Method (Redox Reactions)

 

The balancing of a chemical equations by ion-electron method (using half reactions)for redox reactions is done according to the following steps:

1.Find the elements whose oxidation numbers are changed. 

 

2.Choose the substance, which acts as an oxidizing agent and  as a reducing agent.

3. Write Separate half equations for oxidizing and reducing agents. 

4.Balance oxygen atoms on both sides by adding water

a) Balance hydrogen atoms by adding hydrogen ions on  both sides.

b) Calculate the oxidation number on both sides of the equation.

 

c) add electrons to whichever side is necessary, to make up the difference.

 

d) Balance the half equation so that both sides get the same charge.

e) Add two balanced half equations to get net balanced equation

 

f) Multiply one or both half equations by suitable numbers if needed  so that on adding the two equations, the electrons are balanced.

Redox reactions take place in all the three media acidic or basic or neutral. If H+ ions appear on either side of the equation, the reaction takes place in acidic medium. If OH- ions appear on either side of the equation, the solution is basic. If neither H+ nor OH- ions are present, the reaction occurs in neutral solution. For balancing redox reactions involving acidic and basic media, the method has to be modified slightly.

 

 

PRACTICE SESSION 

Q:-Balance the following  ionic equation for what happens when pot dichromate reacts with dilute sulfuric acid sol ? 

         Cr₂O₇ ^2-   →   Cr^3-

       
        Cr₂O₇ ^2-   +  14H⁺ + 6e^-    →  Cr^3-   +  7H₂O   

    

Q:-Balance the following  ionic equation for what happens when pot permanganate reacts with dilute sulfuric acid sol ?

      MnO₄ ⁺  +  8H⁺ + 5e^-    →     Mn²⁺+  4H₂O 

 

4. OXIDATION NUMBER METHOD 

 The very first thing that we need to understand is the concept of oxidation number.It is the charge which an atom has in its ionic form or appears to have when present in the combined state with other atoms.This number is also known by the term Oxidation state.

1. All the atoms in the elemental form have zero oxidation number.

2. It may be positive or negative as sodium has +ve oxidation number and chlorine as -ve. 

3.In neutral molecules sum of  all oxidation numbers of all the atoms is Zero.

4.In poly atomic ions sum of oxidation numbers of all atoms = the charge on the ion.

5.d and f block elements exhibit variable oxidation states.

example : Find out the oxidation number of Cr and Mn in the following compounds and ions  

 K₂Cr₂O₇

 Cr = O.N : +6

KMnO₄  

 

Mn = O.N :+7  

 

Cr₂O₇ ^2-

  Cr = O.N : +6

 MnO₄ ⁺

 

 Mn = O.N :+7    

 

Balancing Redox Reactions

Oxidation number method is based on the difference in oxidation number of oxidizing agent and the reducing agent. Half-reaction method depends on the division of the redox reactions into oxidation half and reduction half. It depends on the individual which method to choose and use.

Step 1

Correctly write the formula for the reactants and the products of the chemical reaction.

Step 2

Determine correctly the atoms that undergo oxidation number change in the given reaction by allocating the oxidation number of the individual elements present in the reaction.

Step 3

Calculate the oxidation number on the basis of each atom for the given molecule or ion of the chemical reactions.

Step 4

Keep in mind the involvement of the ions if the reaction occurs in water. Accordingly, add H⁺ or OH^– ions in the appropriate side of the reaction. Overall, the ionic charges of reactant and products will be equal. However, if the reaction takes place in acidic solution then add H⁺ ions in the chemical equation. Similarly, if the reaction takes place in the basic solution add OH^– ions in the chemical equation.

Step 5

It is very important to equate the number of hydrogen atoms on each side of the equation by adding water molecules or H₂O molecules. Additionally, it is necessary to check the oxygen atoms present in the equation. It will be a balance reaction if there are equal numbers of oxygen atoms present in both the reactant as well as the product the side.

For instance, a reaction is given where Fe²⁺ ions are converted to Fe³⁺ ions by dichromate ions in an acidic solution. The dichromate ions (Cr₂O₇^2–) are reduced to Cr³⁺ ions in the reaction. We have to balance the above redox reaction. The steps to balance this equation is as follows.

Step 1

First, we have to write the basic ionic form of the equation.

Step 2

Divide the equation into two separate half reaction-oxidation half and reduction half.

• Oxidation half

• Reduction half

Step 3

In the third step of balancing redox reactions by half-reaction method, we will balance the atoms present in each half of the reaction except O and H atoms. In the example question, the oxidation part of the reaction in terms of Fe atoms is already balanced. Therefore, we will just balance the reduction part of the reaction. In this case, we will multiply Cr³⁺ by 2 in order to balance Chromium atoms.

 

Step 4

We know that the reaction takes place in an acidic solution. Therefore, we have to add water molecules (H₂O) for balancing the O atoms of the equation and H⁺ for balancing the H atoms in the equation. Now the equation is

 

Step 5

Now, we will need to balance the charges in both the half reactions. Therefore, we need to multiply the appropriate number to one or both the half reaction and make the number of electrons same. Balancing the oxidation half of the reaction

For the reduction half, there are 12 positive charges on the left side of the equation and 6 positive charges on the right side of the equation. Therefore, we need to add 6 more electrons on the left side of the equation to balance the reduction half.

 

Now, to equate the electrons in two halves of the reactions, we will multiply 6 in the oxidation half reaction. Thus, we get

Step 6

The two halves of the equations are added to complete the overall reactions. After the addition of two reaction halves, cancel the electrons on both sides. The net ionic equation can be written as:

 

Step 7

Finally, we have to verify whether the equation consists of the same type, number, and charges on both sides of the equation. Moreover, the equation is completely balanced in terms of atoms and charges.

In case of a basic solution, we have to balance the atom similar to acidic solution. Then the equal number of OH^– ions addition is done for each H⁺ ion, in both the halves of the equation. When the H⁺ ions and OH^– ions will be present on the same side of the equation, we have to combine the ions and write H₂O.

Solved Exercise

Q. Solve the net ionic equation where potassium dichromate(VI) (K₂Cr₂O₇) reacts with sodium sulphite (Na₂SO₃) in an acidic medium to form sulphate ion and chromium(III) ion.

Solution:

Step 1

The basic equation or the skeletal form of the reaction is

Step 2

Correctly assign oxidation numbers to Cr and S

Therefore, from the reaction we can decipher that dichromate ion is the oxidant in the reaction and the sulphite ion is the reductant in the reaction

Step 3

Calculation of the increase and the decrease in the oxidation number for making each side of the equation equal

Step 4

We know from the equation, the reaction occurs in the acidic medium. Moreover, we can see that the ionic charges in both the sides of the equation are not same. Therefore, we will add 8H⁺ I order to make the ionic charges equal.

Step 5

In the final step, we will calculate the required amount of water molecules and add it on the right side of the equation to make the equation a balanced redox reaction. In the given equation, we will need to add 4 water molecules on the right side to balance the equation.

Q2. Write the balanced redox equation by half-reaction method when Permanganate(VII) ion (MnO₄) produces iodine molecule (I₂)  and manganese (IV) oxide (MnO₂) in a basic medium.

Solution

Step 1

Firstly, we will write the base form of the equation

Step 2

In this step, we will find the two half-reactions and write them

Oxidation Half

Reduction Half

Step 3

We will balance the iodine atoms present in the oxidation half of the reaction. Therefore, the equation will become

Step 4

We will balance the oxygen atoms in the reduction half of the reaction by the addition of two water molecules. Refer to the equation below

Now, we will balance the H atoms by the addition of four H+ ions on the left half of the reduction half-reaction.

We know that the reaction occurs in a basic medium. Thus, to balance four H+ ions, we need to add four OH^– ions to each side of the equation.

Finally, we will interchange the H+ ions and OH^– ions with the water molecule. The final equation of the fourth step will be

Step 5

The charges on the two half-reactions are balanced

Finally, we have to equalize the electrons in the above reactions. Thus, we will multiply the oxidation half of the equation by 3 and the reduction half of the reaction by 2.

Step 6

We have to determine the net reaction by the addition of two halves of the reaction and by the cancellation of electrons on each side.

Step 7

Finally, we have to verify the equation in terms of the number of atoms and charges on both sides. Additionally, verification needs to be done about the equations written on both sides.

 

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2

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