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 8.10 - 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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