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 8.10 - 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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