Chapter 8.13 - Half Reaction Method When Medium Is Basic

In the previous section, we saw the steps required for acidic medium. In this section, we will see the steps required for basic medium. 

Example 7
Permanganate ion reacts with bromide ion in basic medium to give manganese
dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Solution:
1. The reactants are: MnO₄^- (aq) and Br^- (aq). (See list of common polyatomic ions)
2. The products are: MnO₂ (s) and BrO₃^- (aq).
3. So the skeletal equation is:
MnO₄^- (aq) + Br^- (aq) → MnO₂ (s) + BrO₃^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Br is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Br^- (aq) → BrO₃^- (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Br atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are three excess O atoms on the product side. So we must add three H₂O molecules on the reactant side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq)
• Now there are six excess H atoms on the reactant side. So we must put six H⁺ ions on the product side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq) + 6H⁺
• To balance the six H⁺ ions, we add six OH^- ions on both sides. We get:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H⁺ + 6OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give six H₂O molecules. So the equation becomes:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ 6 No. × 1- (from the OH^- ion) = -6
   ♦ Total = -7
Product side:
   ♦ 1 No. × 1- (from the BrO₃^- ion) = -1
• So there is an excess charge of -6 on the reactant side. Therefore we must add 6e^- on the product side. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 6e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the reduction half reaction by '3'.
Also we keep the oxidation half reaction as such. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + Br^- (aq) + 6OH^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + BrO₃^- (aq) + 3H₂O (l) + 6e^-
• So the net equation is:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
Product side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ Total = -3
Product side:
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 1 No. × 1- (from the BrO3- ion) = -1
   ♦ Total = -3
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)

Example 8
Balance the following equation in basic medium:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
Solution:
The given skeletal equation is:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{
\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)
+\overset{-1}{Cl^{-}}(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cr is oxidized and Cl is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
CrO₂^- (aq) → CrO₄^2- (aq)
• The reduction half reaction is:
ClO^- (aq) → Cl^- (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Cr atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Cl atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the product side. So we must add two H₂O molecules on the reactant side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq)
• Now there are four excess H atoms on the reactant side. So we must put four H⁺ ions on the product side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq) + 4H⁺ (aq)
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H⁺ (aq) + 4OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give four H₂O molecules. So the equation becomes:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there is one excess O atom on the reactant side. So we must add one H₂O molecule on the product side. We get:
ClO^- (aq) → Cl^- (aq) + H₂O (l)
• Now there are two extra H atoms on the product side. So we must put two H⁺ ions on the reactant side. We get:
ClO^- (aq) + 2H⁺ → Cl^- (aq) + H₂O (l)
• To balance the two H⁺ ions, we add two OH^- ions on both sides. We get:
ClO^- (aq) + 2H⁺ + 2OH^- (aq) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give two H₂O molecules. So the equation becomes:
ClO^- (aq) + 2H₂O (l) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
ClO^- (aq) + H₂O (l) → Cl^- (aq) + 2OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the CrO₂^- ion) = -1
   ♦ 4 No. × 1- (from the OH^- ion) = -4
   ♦ Total = -5
Product side:
   ♦ 1 No. × 2- (from the CrO₄^2- ion) = -2
• So there is an excess charge of -3 on the reactant side. Therefore we must add 3e^- on the product side. We get:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l) + 3e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the ClO^- ion) = -1
Product side:
   ♦ 1 No. × 1- (from the Cl^- ion) = -1
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ Total = -3
• So there is an excess charge of -2 on the product side. Therefore we must add 2e^- on the reactant side. We get:
ClO^- (aq) + H₂O (l) + 2e^- → Cl^- (aq) + 2OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 3e^-.
    ♦ The reduction half reaction has 2e^-.
• So we multiply the oxidation half reaction by '2'.
Also we multiply the reduction half reaction by '3'. We get:
2CrO₂^- (aq) + 8OH^- (aq) → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^-
3ClO^- (aq) + 3H₂O (l) + 6e^- → 3Cl^- (aq) + 6OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2CrO₂^- (aq) + 8OH^- (aq) + 3ClO^- (aq) + 3H₂O (l) + 6e^- → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^- + 3Cl^- (aq) + 6OH^- (aq)
The net equation is:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
Product side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the CrO₂^- ion) = -2
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 3 No. × 1- (from the ClO^- ion) = -3
   ♦ Total = -7
Product side:
   ♦ 2 No. × 2- (from the CrO₄^2- ion) = -4
   ♦ 3 No. × 1- (from the 3Cl^- ion) = -3
   ♦ Total = -7
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)

Example 9
Permanganate (VII) ion MnO₄^- in basic solution oxidizes iodide ion I^- to produce molecular iodine (I₂) and manganese (iv) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:
The skeletal equation is:
MnO₄^- (aq) + I^- (aq) → MnO₂ (s) + I₂ (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{I^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{0}{I_2}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
I is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
I^- (aq) → I₂ (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of I^- in the reactant side. We get:
2I^- (aq) → I₂ (aq)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the I^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2I^- (aq) → I₂ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the oxidation half reaction by '3'.
• Also we multiply the reduction half reaction by '2'.
We get:
6I^- (aq) → 3I₂ (aq) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq) + 6e^-
So the net equation is:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
Product side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 6 No. × 1- (from the I^- ion) = -6
   ♦ Total = -8
Product side:
   ♦ 8 No. × 1- (from the OH^- ion) = -8
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)

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The following link gives three more examples:

Solved example 8.11 - Part (a), Part (b) and Part (c)

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• We have completed a discussion on half reaction method.
• In the next section, we will see redox reactions as the basis for titrations.

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Chapter 8.12 - Half Reaction Method When Medium Is Acidic

In the previous section, we saw the steps in half reaction method, related to neutral medium. We saw them while analyzing three examples. In this section, we will see the additional steps required for acidic medium.

Example 4:
Write the net ionic equation for the reaction of potassium dichromate(VI),
K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium(III) ion and the sulfate ion.
Solution:
1. The reactants are: K₂Cr₂O₇ and Na₂SO₃ 
2. K₂Cr₂O₇ is an ionic compound
   ♦ The ions are: K⁺ and Cr₂O₇^2-
• In the dissolved state, the two ions separate away from each other.
• Cr₂O₇^2- will not dissociate further. It is a polyatomic ion. (A list of such polyatomic ions can be seen here. It is better to remember their formula and names)
3. Na₂SO₃ is an ionic compound
   ♦ The ions are: Na⁺ and SO₃^2-
• In the dissolved state, the two ions separate away from each other.
• SO₃^2- will not dissociate further. It is a polyatomic ion.
4. K⁺ and Na⁺ are spectator ions. They do not take part in the reaction. In the reaction equation, they will appear as such on both sides, and thus will cancel out.
• So the actual reactants are: Cr₂O₇^2- and SO₃^2- 
5. We are given the products: chromium(III) ion and the sulfate ion.
• Chromium(III) ion means, the chromium ion with oxidation state +3. Obviously, it is the Cr³⁺ ion.
• Sulfate ion is known to us from the list. It is the SO₄^2- ion.
6. Steps (1) to (4) give us the reactants. Step (5) gives us the products.
• So now we can write the skeletal equation:
Cr₂O₇^2- + SO₃^2- → Cr³⁺ + SO₄^2-
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
S is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
SO₃^2- (aq) → SO₄^2- (aq)
• The reduction half reaction is:
Cr₂O₇^2- (aq) → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of S atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, SO₄^2- has one O more than SO₃^2-. So we must add one H₂O molecule on the reactant side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq)
• Now there are two extra H atoms on the reactant side. So we must put two H⁺ ions on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the SO₃^2- ion) = -2
Product side:
   ♦ 1 No. × 2- (from the SO₄^2- ion) = -2
   ♦ 2 No. × 1+ (from the H⁺ ion) = +2
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '3'. We get:
3SO₃^2- (aq) + 3H₂O (l) → 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- + 3SO₃^2- (aq) + 3H₂O (l) → 2Cr³⁺ (aq) + 7H₂O  + 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^- 
• The 6e^- on either sides, cancel each other.
• The 6H⁺ on the product side will take away 6 H⁺ from the reaction side. So 8H⁺ will remain on the reaction side.
• The 3H₂O on the reaction side will take away 3H₂O from the product side. So 4H₂O will remain on the product side.
• We get:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
Product side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 3 No. × 2- (from the SO₃^2- ions) = -6
   ♦ 8 No. × 1+ (from the H⁺ ions) = +8
   ♦ Total = 0
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ 3 No. × 2- (from the SO₄^2- ions) = -6
   ♦ Total = 0
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O

Example 5:
Balance the equation in acidic medium:
MnO₄^- (aq) + Cl^- (aq) → Mn²⁺ (aq) + Cl₂ (g)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cl is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Cl^- (aq) → Cl₂ (g)
• The reduction half reaction is:
MnO₄^- (aq) → Mn²⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of Cl^- on the reactant side. We get:
2Cl^- (aq) → Cl₂ (g)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are same on both sides. So it is already balanced
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are extra four O atoms on the reactant side. So we must add four H₂O molecules on the product side. We get:
MnO₄^- (aq) → Mn²⁺ (aq) + 4H₂O
• Now there are eight extra H atoms on the product side. So we must put eight H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ → Mn²⁺ (aq) + 4H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the Cl^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2Cl^- (aq) → Cl₂ (g) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
   ♦ 8 No. × 1+ (from the H⁺ ion) = +8
   ♦ Total = +7
Product side:
   ♦ 1 No. × 2+ (from the Mn²⁺ ion) = +2
• So there is an excess charge of +5 on the reactant side. Therefore we must add 5e^- on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ +5e^- → Mn²⁺ (aq) + 4H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 5e^-.
• So we multiply the oxidation half reaction by '5'.
• Also we multiply the reduction half reaction by '2'.
We get:
10Cl^- (aq) → 5Cl₂ (g) + 10e^-
2MnO₄^- (aq) + 16H⁺ + 10e^- → 2Mn²⁺ (aq) + 8H₂O
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 16H⁺ + 10e^- + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g) + 10e-
• The 10e^- on either sides, cancel each other. We get:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
Product side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 10 No. × 1- (from the SO₃^2- ions) = -10
   ♦ 16 No. × 1+ (from the H⁺ ions) = +16
   ♦ Total = +4
Product side:
   ♦ 2 No. × 2+ (from the Mn²⁺ ion) = +4
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)

Example 6:
Balance the equation in acidic medium:
Fe²⁺ (aq) + Cr₂O₇^2- → Fe³⁺ (aq) + Cr³⁺ (aq)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+2}{Fe^{2+}}\,(aq)+\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)\rightarrow\overset{+3}{Fe^{3+}}\,(aq)
+\overset{+3}{Cr^{3+}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Fe²⁺ (aq) → Fe³⁺ (aq)
• The reduction half reaction is:
Cr₂O₇^2- → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Fe atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2+ (from the Fe²⁺ ion) = +2
Product side:
   ♦ 1 No. × 3+ (from the Fe³⁺ ion) = +3
• So there is an excess charge of +1 on the product side. Therefore we must add 1e^- on the product side. We get:
Fe²⁺ (aq) → Fe³⁺ (aq) + 1e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 1e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '6'. We get:
6Fe²⁺ (aq) → 6Fe³⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 6Fe³⁺ (aq) + 6e^- + 2Cr³⁺ (aq) + 7H₂O
• The 6e^- on either sides, cancel each other.
• We get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
Product side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 6 No. × 2+ (from the Fe²⁺ ions) = +12
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ions) = +14
   ♦ Total = 24
Product side:
   ♦ 6 No. × 3+ (from the Fe³⁺ ions) = +18
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ Total = 24
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O

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The following link gives three more examples:

Solved example 8.10 - Part (a), Part (b) and Part (c)

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• We have seen the additional steps required in acidic medium when half reaction method is used.
• In the next section, we will see some examples, which take place in basic medium.
   ♦ There we will see the steps required for basic medium.


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Chapter 8.11 - Half Reaction Method When The medium Is Neutral

In the previous section, we completed the discussion on oxidation number method. In this section, we will see half reaction method.

• First we will see redox reactions taking place in neutral medium.
Example 1: We want to balance the equation:
Fe₂O₃ (s) + CO (g) → Fe (s) + CO₂ (g)
It can be written in 10 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(s)
+\overset{+2}{C}\,\overset{-2}{O}\,(g)\rightarrow\overset{0}{Fe}\,(s)
+\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is reduced and C is oxidized.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
CO (g) + O₂ (g) → CO₂ (g)
• The reduction half reaction is:
Fe₂O₃ (s) → Fe (s) + O₂ (g)
Step 4: Balance the oxidation half reaction.
• In our present case, we get:
2CO (g) + O₂ (g) → 2CO₂ (g)
Step 5: Balance the reduction half reaction.
• In our present case, we get:
2Fe₂O₃ (s) → 4Fe (s) + 3O₂ (g)
Step 6: Make the number of O₂ molecules the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has one O₂ molecule.
    ♦ The reduction half reaction has three O₂ molecules.
• So we multiply the oxidation half reaction by '3'. We get:
6CO (g) + 3O₂ (g) → 6CO₂ (g)
Step 7: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the oxidation half reaction do not have any charges. So we need not consider this step.
Step 8: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the reduction half reaction do not have any charges. So we need not consider this step.
Step 9: Add the two reactions together.
• In our present case, we get:
2Fe₂O₃ (s) + 6CO (g) + 3O₂ (g) → 4Fe (s) + 3O₂ (g) + 6CO₂ (g)
• The three O₂ molecules on either sides, cancel each other. We get:
2Fe₂O₃ (s) + 6CO (g) → 4Fe (s) + 6CO₂ (g)
• We see that, all the coefficients have a common factor '2'. So we can divide all the coefficients by '2'. We get:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos. 
Product side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos.
         ✰ So all atoms are balanced.
• In our present case, there are no ions. So we do not have to check charges.
◼  We see that, all atoms are balanced. So we can write the balanced equation as:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)

Example 2:
Consider the reaction that we saw at the beginning of this chapter. (Fig.1 of section 8.1)
• The skeletal equation can be written as:
Zn (s) + CuSO₄ (aq)  → ZnSO₄ (aq) + Cu (s)
• Since CuSO₄ and ZnSO₄ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Zn (s) + Cu²⁺ + SO₄^2- (aq)  → Zn²⁺ +SO₄^2- (aq) + Cu (s)
• We see that, SO₄^2- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
(Remember that, polyatomic ions like SO₄^2-, NO₃^- etc., do not split into individual atoms when dissolved in water. Some common polyatomic ions can be seen here. It is useful to remember their formula and names)
• So the net reaction is:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+\overset{+2}{Cu^{2+}}\,(aq)
\rightarrow \overset{+2}{Zn^{2+}}\,(aq)+\overset{0}{Cu}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Zn is oxidized and Cu is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Zn (s) → Zn²⁺ (aq)
• The reduction half reaction is:
Cu²⁺ (aq) → Cu (s)
Step 4: Balance the oxidation half reaction.
• In our present case, the number of Zn atoms are the same on both sides. So it is already balanced.
Step 5: Balance the reduction half reaction.
• In our present case, the number of Cu atoms are the same on both sides. So it is already balanced.
Step 6: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Zn²⁺ creates an excess charge of 2+ on the product side. So we must add 2e^- on the product side. We get:
Zn (s) → Zn²⁺ (aq) + 2e^-
Step 7: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Cu²⁺ creates an excess charge of 2+ on the reactant side. So we must add 2e^- on the reactant side. We get:
Cu²⁺ (aq) + 2e^- → Cu (s)
Step 8: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 2e^-.
• So they are already equal.
Step 9: Add the two half reactions together.
• In our present case, we get:
Zn (s) + Cu²⁺ (aq) + 2e^- → Zn²⁺ (aq) + 2e^- + Cu (s)
• The 2e^- on either sides, cancel each other. We get:
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Zn - 1 No, Cu - 1 No. 
Product side:
   ♦ Zn - 1 No, Cu - 1 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from the Cu²⁺ ion)
Product side:
   ♦ 2+ (from the Zn²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)

Example 3:
Consider the second reaction that we saw at the beginning of this chapter. (Fig.2 of section 8.1)
• The skeletal equation can be written as:
Cu (s) + AgNO₃ (aq)  → Cu(NO₃)₂ (aq) + Ag (s)
• Since AgNO₃ and Cu(NO₃)₂ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Cu (s) + Ag⁺ + NO₃^- (aq)  → Cu²⁺ + NO₃^- (aq) + Ag (s)
• We see that, NO₃^- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
• So the net reaction is:
Cu (s) + Ag⁺  → Cu²⁺ + Ag (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+\overset{+1}{Ag^{+}}\,(aq)
\rightarrow \overset{+2}{Cu^{2+}}\,(aq)+\overset{0}{Ag}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cu is oxidized and Ag is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Cu (s) → Cu²⁺ (aq)
• The reduction half reaction is:
Ag⁺ (aq) → Ag (s)
Step 4: Balance the oxidation half reaction.
• In our present case, the number of Cu atoms are the same on both sides. So it is already balanced.
Step 5: Balance the reduction half reaction.
• In our present case, the number of Ag atoms are the same on both sides. So it is already balanced.
Step 6: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Cu²⁺ creates an excess charge of 2+ on the product side. So we must add 2e^- on the product side. We get:
Cu (s) → Cu²⁺ (aq) + 2e^-
Step 7: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Ag⁺ creates an excess charge of 1+ on the reactant side. So we must add e^- on the reactant side. We get:
Ag⁺ (aq) + e^- → Ag (s)
Step 8: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 1e^-.
• So we multiply the reduction half reaction by '2'. We get:
2Ag⁺ (aq) + 2e^- → 2Ag (s)
Step 9: Add the two half reactions together.
• In our present case, we get:
Cu (s) + 2Ag⁺ (aq) + 2e^- → Cu²⁺ (aq) + 2Ag (s) + 2e^-
• The 2e^- on either sides, cancel each other. We get:
Cu (s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag (s)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cu - 1 No., Ag - 2 No. 
Product side:
   ♦ Cu - 1 No., Ag - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from two Ag⁺ ions)
Product side:
   ♦ 2+ (from one Cu²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Cu (s) + 2Ag⁺  → Cu²⁺ + 2Ag (s)

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• We have seen the steps for balancing a reaction using half reaction method.
   ♦ The examples that we saw, take place in neutral medium.
• In the next section, we will see some examples, which take place in acidic medium.
   ♦ There will be a few more steps in addition to those that we saw above.

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Chapter 8.10 - Oxidation Number Method When Medium Is Basic

In the previous section, we saw the steps required for acidic medium. In this section, we will see the steps required for basic medium.

Example 6
Permanganate ion reacts with bromide ion in basic medium to give manganese
dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Solution:
1. The reactants are: MnO₄^- (aq) and Br^- (aq). (See list of common polyatomic ions)
2. The products are: MnO₂ (s) and BrO₃^- (aq).
3. So the skeletal equation is:
MnO₄^- (aq) + Br^- (aq) → MnO₂ (s) + BrO₃^- (aq)
• We want to balance this equation. It can be written in 8 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Br is oxidized and Mn is reduced.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Mn decreases from +7 to +4
⇒ Each Mn atom gains (+7 - +4) = 3 electrons.
(ii) There is only one Mn atom on the reactant side.
⇒ There is a total gain of (1 No. × 3 electrons) = 3 electrons.
(iii) Oxidation number of Br increases from -1 to +5
⇒ Each Br atom loses (+5 - -1) = 6 electrons.
(iv) There is only one Br atom on the reactant side.
⇒ There is a total loss of (1 No. × 6 electrons) = 6 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 3 and loss of 6 do not tally. We can make them tally as follows:
If there are two Mn atoms on the reactant side, each one gaining three electrons, will result in a total gain of 6. Then they tally.
   ♦ So we must put '2' in front of MnO₄^- on the reactant side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, to balance the number of Mn atoms, we must put '2' in front of MnO₄ on the product side also. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow 2\,\overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
Step 6: Check the balancing of charges.
   ♦ Balance them using H⁺ if it is acidic medium.
   ♦ Balance them using OH^- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
   ♦ (2 No. × 1-) from the MnO₄^- ion = -2
   ♦ (1 No. × 1- ) from the Br^- ion = -1
   ♦ Total = -3
Product side:
   ♦ (1 No. × 1-) from the BrO₃^- ion = -1
   ♦ Total = -1
         ✰ So there is an excess of -2 charge on the reactant side.
• In the present case, since it is a basic medium, we must balance the charges using OH^- ions.
   ♦ We must add 2 Nos. of OH^- ions on the product side. We get:
2MnO₄^- (aq) + Br^- (aq) → 2MnO₂ (s) + BrO₃^- (aq) + 2OH^- (aq)
Step 7: Balance the number of H atoms by adding appropriate number of H₂O molecules.
• In our present case, there is an excess of 2 No. H atoms on the product side. So we must add 1 No. H₂O molecules on the reactant side. We get:
2MnO₄^- (aq) + Br^- (aq) + H₂O (l) → 2MnO₂ (s) + BrO₃^- (aq) + 2OH^- (aq)
Step 8: Due to the addition of H₂O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
   ♦ 8 + 1 = 9
Product side:
   ♦ 4 + 3 + 2 = 9
         ✰ So the O atoms are balanced
◼ The balanced equation is:
2MnO₄^- (aq) + Br^- (aq) + H₂O (l) → 2MnO₂ (s) + BrO₃^- (aq) + 2OH^- (aq)

Example 7
Balance the following equation in basic medium:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
Solution:
The given skeletal equation is:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
• We want to balance this equation. It can be written in 8 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{
\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)
+\overset{-1}{Cl^{-}}(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cr is oxidized and Cl is reduced.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Cl decreases from +1 to -1
⇒ Each Cl atom gains (+1 - -1) = 2 electrons.
(ii) There is only one Cl atom on the reactant side.
⇒ There is a total gain of (1 No. × 2 electrons) = 2 electrons.
(iii) Oxidation number of Cr increases from +3 to +6
⇒ Each Cr atom loses (+6 - +3) = 3 electrons.
(iv) There is only one Cr atom on the reactant side.
⇒ There is a total loss of (1 No. × 3 electrons) = 3 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 2 and loss of 3 do not tally. We can make them tally as follows:
If there are two Cr atoms on the reactant side, each one losing 3 electrons, will result in a total loss of 6.
If there are three Cl atoms on the reactant side, each one gaining 2 electrons, will result in a total gain of 6.
Then they tally.
   ♦ So we must put '2' in front of CrO₂^- on the reactant side.
   ♦ Also, we must put '3' in front of ClO^- on the reactant side.We get:
$\mathbf\small{\rm{2\,\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+3\,\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)+\overset{-1}{Cl^{-}}(aq)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, to balance the number of Cr atoms, we must put '2' in front of CrO₄^2-     on the product side.
Also, to balance the number of Cl atoms, we must put '3' in front of Cl^- on the product side. We get:
$\mathbf\small{\rm{2\,\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+3\,\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow 2\,\overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)+3\,\overset{-1}{Cl^{-}}(aq)}}$
Step 6: Check the balancing of charges.
   ♦ Balance them using H⁺ if it is acidic medium.
   ♦ Balance them using OH^- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
   ♦ (2 No. × 1-) from the CrO₂^- ion = -2
   ♦ (3 No. × 1- ) from the ClO^- ion = -3
   ♦ Total = -5
Product side:
   ♦ (2 No. × 2-) from the CrO₄^2- ion = -4
   ♦ (3 No. × 1-) from the Cl^-  ion = -3
   ♦ Total = -7
         ✰ So there is an excess of -2 charge on the product side.
• In the present case, since it is a basic medium, we must balance the charges using OH^- ions.
   ♦ We must add 2 Nos. of OH^- ions on the reactant side. We get:
2CrO₂^- (aq) + 3ClO^- (aq) + 2OH^- → 2CrO₄^2- (aq) + 3Cl^- (aq)
Step 7: Balance the number of H atoms by adding appropriate number of H₂O molecules.
• In our present case, there is an excess of 2 No. H atoms on the reactant side. So we must add 1 No. H₂O molecules on the product side. We get:
2CrO₂^- (aq) + 3ClO^- (aq) + 2OH^- → 2CrO₄^2- (aq) + 3Cl^- (aq) + H₂O
Step 8: Due to the addition of H₂O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
   ♦ 4 + 3+ 2 = 9
Product side:
   ♦ 8 + 1 = 9
         ✰ So the O atoms are balanced
◼ The balanced equation is:
2CrO₂^- (aq) + 3ClO^- (aq) + 2OH^- → 2CrO₄^2- (aq) + 3Cl^- (aq) + H₂O

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Now we will see some solved examples. The link is given below:

Solved example 8.9 Parts (a), (b), (c) and (d)

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• We have completed the discussion on oxidation number method. In the next section, we will see the half reaction method.


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Chapter 8.9 - Oxidation Number Method When Medium Is Acidic

In the previous section, we saw five steps related to neutral medium. We saw them while analyzing three examples. In this section, we will see the additional steps required for acidic medium.

Example 4:
Write the net ionic equation for the reaction of potassium dichromate(VI),
K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium(III) ion and the sulfate ion.
Solution:
1. The reactants are: K₂Cr₂O₇ and Na₂SO₃ 
2. K₂Cr₂O₇ is an ionic compound
   ♦ The ions are: K⁺ and Cr₂O₇^2-
• In the dissolved state, the two ions separate away from each other.
• Cr₂O₇^2- will not dissociate further. It is a polyatomic ion. (A list of such polyatomic ions can be seen here. It is better to remember their formula and names)
3. Na₂SO₃ is an ionic compound
   ♦ The ions are: Na⁺ and SO₃^2-
• In the dissolved state, the two ions separate away from each other.
• SO₃^2- will not dissociate further. It is a polyatomic ion.
4. K⁺ and Na⁺ are spectator ions. They do not take part in the reaction. In the reaction equation, they will appear as such on both sides, and thus will cancel out.
• So the actual reactants are: Cr₂O₇^2- and SO₃^2- 
5. We are given the products: chromium(III) ion and the sulfate ion.
• Chromium(III) ion means, the chromium ion with oxidation state +3. Obviously, it is the Cr³⁺ ion.
• Sulfate ion is known to us from the list. It is the SO₄^2- ion.
6. Steps (1) to (4) give us the reactants. Step (5) gives us the products.
• So now we can write the skeletal equation:
Cr₂O₇^2- + SO₃^2- → Cr³⁺ + SO₄^2-
• We want to balance this equation. It can be written in 8 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
S is oxidized and Cr is reduced.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Cr decreases from +6 to +3
⇒ Each Cr atom gains (+6 - +3) = 3 electrons.
(ii) There are two Cr atoms on the reactant side.
⇒ There is a total gain of (2 No. × 3 electrons) = 6 electrons.
(iii) Oxidation number of S increases from +4 to +6
⇒ Each S atom loses (+6 - +4) = 2 electrons.
(iv) There is only one S atom on the reactant side.
⇒ There is a total loss of (1 No. × 2 electrons) = 2 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 6 and loss of 2 do not tally. We can make them tally as follows:
If there are three S atoms on the reactant side, each one losing two electrons, will result in a total loss of 6. Then they tally.
   ♦ So we must put '3' in front of SO₃^2- on the reactant side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(g)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, to balance the number of S atoms, we must put '3' in front of SO₄^2- on the product side also.
• Also, to balance the number of Cr atoms, we must put 2 in front of Cr³⁺ in the product side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 6: Check the balancing of charges.
   ♦ Balance them using H⁺ if it is acidic medium.
   ♦ Balance them using OH^- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
   ♦ (1 No. × 2-) from the Cr₂O₇^2- ion = -2
   ♦ (3 No. × 2- ) from the SO₃^2- ion = -6
   ♦ Total = -8
Product side:
   ♦ (2 No. × 3+) from the Cr³⁺ ion = +6
   ♦ (3 No. × 2- ) from the SO₄^2- ion = -6
   ♦ Total = 0
         ✰ So there is an excess of -8 charge on the reactant side.
• In the present case, since it is an acidic medium, we must balance the charges using H⁺ ions.
   ♦ We must add 8 Nos. of H⁺ ions on the reactant side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)+8\,H^{+}\,(aq)+\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 7: Balance the number of H atoms by adding appropriate number of H₂O molecules.
• In our present case, there is an excess of 8 No. H atoms on the reactant side. So we must add 4 No. H₂O molecules on the product side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)+8\,H^{+}\,(aq)+\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)+4\,H_2O}}$
Step 8: Due to the addition of H₂O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
   ♦ 7 + 9 = 16
Product side:
   ♦ 12 + 4 = 16
         ✰ So the O atoms are balanced
◼ The balanced equation is:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8 H⁺ (aq) → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O (l)

Example 5:
Balance the equation in acidic medium:
MnO₄^- + Cl^- → Mn²⁺ + Cl₂
Solution:
We want to balance this equation. It can be written in 8 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cl is oxidized and Mn is reduced.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Mn decreases from +7 to +2
⇒ Each Mn atom gains (+7 - +2) = 5 electrons.
(ii) There is only one Mn atom on the reactant side.
⇒ There is a total gain of (1 No. × 5 electrons) = 5 electrons.
(iii) Oxidation number of Cl increases from -1 to 0
⇒ Each Cl atom loses (0 - -1) = 1 electron.
(iv) There is only one Cl atom on the reactant side.
⇒ There is a total loss of (1 No. × 1 electron) = 1 electron.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 5 and loss of 1 do not tally. We can make them tally using 2 steps:
(i) If there are 5 No. Cl^- ions on the reactant side, each one losing one electron, will result in a total loss of 5. Now they tally.
   ♦ So we must put '5' in front of Cl^- on the reactant side. We get:
$\mathbf\small{\rm{
\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+5\,\overset{-1}{Cl^{-}}\,(aq)
\rightarrow
\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, Mn is already balanced.
• Cl atoms are not balanced. 5 Cl atoms on the reactant side, cannot give 2 Cl atoms on the product side. So we use the least common multiple of 5 and 2, which is 10.
• 10 Cl atoms on the reactant side will give 5 Cl₂ molecules on the product side. We get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+5\,\overset{0}{Cl_2}\,(g)}}$
• But now, the tally of electrons that we obtained in step 4 is lost:
    ♦ One Mn atom gains 5 electrons.
    ♦ Ten Cl^- ions lose 10 electrons.
• So we must put '2' in front of MnO₄^- ion on the reactant side.
• To compensate, we must put '2' in front of Mn²⁺ ion on the product side also.
We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)
+5\,\overset{0}{Cl_2}\,(g)}}$
Step 6: Check the balancing of charges.
    ♦ Balance them using H⁺ if it is acidic medium.
    ♦ Balance them using OH^- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
   ♦ (2 No. × 1-) from the MnO₄^- ion = -2
   ♦ (10 No. × 1- ) from the Cl^- ion = -10
   ♦ Total = -12
Product side:
   ♦ (2 No. × 2+) from the Mn²⁺ ion = +4
   ♦ Total = +4
         ✰ So there is an excess of -8 charge on the reactant side.
• In the present case, since it is an acidic medium, we must balance the charges using H⁺ ions.
    ♦ We must add 8 Nos. of H⁺ ions on the reactant side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)+8\,H^{+}\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)+5\,\overset{0}{Cl_2}\,(g)}}$
Step 7: Balance the number of H atoms by adding appropriate number of H₂O molecules.
• In our present case, there is an excess of 8 No. H atoms on the reactant side. So we must add 4 No. H₂O molecules on the product side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)+8\,H^{+}\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)+5\,\overset{0}{Cl_2}\,(g)+4H_2O\,(l)}}$
Step 8: Due to the addition of H₂O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
    ♦ 8
Product side:
    ♦ 4
        ✰ So there is an excess of 4 Nos. O atoms on the reactant side.
• This can be balanced by providing a total of 8 Nos. H₂O molecules on the product side.
• To compensate, the H atoms, there must be a total of 16 H⁺ ions on the reactant side.
• The balanced equation is:
2MnO₄^- + 10Cl^- + 16H⁺ (aq) → 2Mn²⁺ + 5Cl₂ + 8H₂O (l)

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The following link gives three more examples:

Solved example 8.8 - Part (a), Part (b) and Part (c)

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• We have seen the additional steps required in acidic medium.
• In the next section, we will see some examples, which take place in basic medium.
   ♦ There we will see the steps required for basic medium.

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Chapter 8.8 - Balancing of Redox Reactions

In the previous section, we saw the paradox of fractional oxidation numbers. In this section, we will see balancing of redox reactions.

• We know how to balance the equations of simple chemical reactions. Balancing of redox reactions involves a few more steps.
• Redox reactions may occur in the following types of medium:
   ♦ neutral medium
   ♦ acidic medium
   ♦ basic medium
• These mediums can be explained in 4 steps.
1. Consider the various redox reactions.
• In some of those reactions, we will need to maintain the pH of the 'surrounding environment' less than 7.
    ♦ That means, we will need to maintain the environment acidic.
• In some others, we will need to maintain the pH greater than 7.
    ♦ That means, we will need to maintain the environment basic.
• In some others, we will need to maintain the pH equal to 7.
    ♦ That means, we will need to maintain the environment neutral.
2. Why do we want some environments to be acidic?
• Answer can be written in 3 steps:
(i) presence of ions
    ♦ When an environment is acidic, there will be H⁺ ions.
    ♦ When an environments is basic, there will be OH^- ions.
(ii) We will not want the presence of OH^- ions because, those ions will enter into the reaction
    ♦ If they enter into the reaction, we will not get the desired products.
(iii) We will want the presence of H⁺ ions because, they are necessary to get the desired products.
• So it will be important to maintain the environment acidic.
3. Why do we want some environments to be basic?
• Answer can be written in 3 steps:
(i) presence of ions
    ♦ When an environment is acidic, there will be H⁺ ions.
    ♦ When an environment is basic, there will be OH^- ions.
(ii) We will not want the presence of H⁺ ions because, those ions will enter into the reaction
    ♦ If they enter into the reaction, we will not get the desired products.
(iii) We will want the presence of OH^- ions because, they are necessary to get the desired products.
• So it will be important to maintain the environment basic.
4. If the presence of neither H⁺ nor OH^- is desirable, we will want to maintain the environment neutral.

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• There are two methods for balancing redox reactions.
(i) Method using oxidation numbers.
(ii) Method using half reactions.
• In this section, we will see the first method. We can learn this method by analyzing some examples. While analyzing, we will see the various steps also.
• First we will see redox reactions taking place in neutral medium.
Example 1: We want to balance the equation:
Fe₂O₃ (s) + CO (g) → Fe (s) + CO₂ (g)
It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(s)
+\overset{+2}{C}\,\overset{-2}{O}\,(g)\rightarrow\overset{0}{Fe}\,(s)
+\overset{+4}{C}\,\overset{-2}{O}_2\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is reduced and C is oxidized.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Fe decreases from +3 to 0
⇒ Each Fe atom gains (+3 - 0) = 3 electrons
(ii) There are two Fe atoms on the reactant side.
⇒ There is a total gain of (2 No. × 3 electrons) = 6 electrons.
(iii) Oxidation number of C increases from +2 to +4
⇒ Each C atom loses (+4 - +2) = 2 electrons.
(iv) There is one C atom on the reactant side.
⇒ There is a total loss of (1 No. × 2 electrons) = 2 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost
• In our present case, gain of 6 and loss of 2 do not tally. We can make them tally using 3 steps:
(i) If there are three C atoms on the reactant side, each one losing two electrons, will result in a total loss of 6. Now they tally.
   ♦ So we must put '3' in front of CO on the reactant side. So we get:
$\mathbf\small{\rm{ \overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(s)+3\,\overset{+2}{C}\,\overset{-2}{O}\,(g) \rightarrow \overset{0}{Fe}\,(s)+\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
(ii) But now, to balance the number of C atoms, we must put '3' in front of CO₂ on the product side also.
(iii) Also, to balance the number of Fe atoms, we must put '2' in front of Fe on the product side. We get:
$\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(s)
+3\,\overset{+2}{C}\,\overset{-2}{O}\,(g)\rightarrow
2\,\overset{0}{Fe}\,(s)+3\,\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
Step 5: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos. 
Product side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos.
         ✰ So all atoms are balanced.
• In our present case, there are no ions. So we do not have to check charges.
◼  We see that, all atoms are balanced. So we can write the balanced equation as:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)

Example 2:
Consider the reaction that we saw at the beginning of this chapter. (Fig.1 of section 8.1)
• The skeletal equation can be written as:
Zn (s) + CuSO₄ (aq)  → ZnSO₄ (aq) + Cu (s)
• Since CuSO₄ and ZnSO₄ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Zn (s) + Cu²⁺ + SO₄^2- (aq)  → Zn²⁺ +SO₄^2- (aq) + Cu (s)
• We see that, SO₄^2- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
(Remember that, polyatomic ions like SO₄^2-, NO₃^- etc., do not split into individual atoms when dissolved in water. Some common polyatomic ions can be seen here. It is useful to remember their formula and names)
• So the net reaction is:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+\overset{+2}{Cu^{2+}}\,(aq)
\rightarrow \overset{+2}{Zn^{2+}}\,(aq)+\overset{0}{Cu}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Zn is oxidized and Cu is reduced.
Step 3: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Cu²⁺ decreases from +2 to 0
⇒ Each Cu²⁺ ion gains (+2 - 0) = 2 electrons
(ii) There is only one Cu²⁺ ion on the reactant side.
⇒ There is a total gain of (1 No. × 2 electrons) = 2 electrons.
(iii) Oxidation number of Zn increases from 0 to +2
⇒ Each Zn atom loses (+2 - 0) = 2 electrons.
(iv) There is only one Zn atom on the reactant side.
⇒ There is a total loss of (1 No. × 2 electrons) = 2 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost
• In our present case, gain of 2 and loss of 2 tally.
Step 5: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Zn - 1 No., Cu - 1 No. 
Product side:
   ♦ Zn - 1 No., Cu - 1 No.
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from the Cu²⁺ ion)
Product side:
   ♦ 2+ (from the Zn²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)

Example 3:
Consider the second reaction that we saw at the beginning of this chapter. (Fig.2 of section 8.1)
• The skeletal equation can be written as:
Cu (s) + AgNO₃ (aq)  → Cu(NO₃)₂ (aq) + Ag (s)
• Since AgNO₃ and Cu(NO₃)₂ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Cu (s) + Ag⁺ + NO₃^- (aq)  → Cu²⁺ + NO₃^- (aq) + Ag (s)
• We see that, NO₃^- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
• So the net reaction is:
Cu (s) + Ag⁺  → Cu²⁺ + Ag (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+\overset{+1}{Ag^{+}}\,(aq)
\rightarrow \overset{+2}{Cu^{2+}}\,(aq)+\overset{0}{Ag}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cu is oxidized and Ag is reduced.
Step 3: Find the number of electrons which are transferred.
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Ag⁺ decreases from +1 to 0
⇒ Each Ag⁺ ion gains (+1 - 0) = 1 electron.
(ii) There is only one Ag⁺ ion on the reactant side.
⇒ There is a total gain of (1 No. × 1 electron) = 1 electron.
(iii) Oxidation number of Cu increases from 0 to +2
⇒ Each Cu atom loses (+2 - 0) = 2 electrons.
(iv) There is only one Cu atom on the reactant side.
⇒ There is a total loss of (1 No. × 2 electrons) = 2 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 1 and loss of 2 do not tally. We can make them tally using 2 steps:
(i) If there are two Ag atoms on the reactant side, each one gaining one electron will result in a total gain of 2. Now they tally.
   ♦ So we must put '2' in front of Ag⁺ on the reactant side. So we get:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+2\,\overset{+1}{Ag^{+}}\,(aq)
\rightarrow \overset{+2}{Cu^{2+}}\,(aq)+\overset{0}{Ag}\,(s)}}$
(ii) But now, to balance the number of Ag atoms, we must put '2' in front of Ag on the product side also. We get:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+2\,\overset{+1}{Ag^{+}}\,(aq)
\rightarrow \overset{+2}{Cu^{2+}}\,(aq)+2\,\overset{0}{Ag}\,(s)}}$
Step 5: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cu - 1 No., Ag - 2 No. 
Product side:
   ♦ Cu - 1 No., Ag - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from two Ag⁺ ions)
Product side:
   ♦ 2+ (from one Cu²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Cu (s) + 2Ag⁺  → Cu²⁺ + 2Ag (s)

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• We have seen five steps for balancing a reaction using oxidation number method.
   ♦ The examples that we saw, take place in neutral medium.
• In the next section, we will see some examples, which take place in acidic medium.
   ♦ There will be a few more steps in addition to the five that we saw above.

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