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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