Chapter 8.11 - Half Reaction Method When The medium Is Neutral

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

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

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

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

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

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