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

---

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

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

---

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


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com

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)

---

The following link gives three more examples:

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

---

• 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.

Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com

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.

---

• 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)

---

• 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.

---

Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com