Chapter 8.4 - Types Of Redox Reactions

In the previous section we saw how oxidation number can be used to determine whether a reaction is a redox reaction or not. In this section, we will see the different types of redox reactions.

There are four types of redox reactions:
A. Combination reactions
B. Decomposition reactions
C. Displacement reactions
D. Disproportionation reactions

A. Combination reactions
This can be explained in 5 steps:
1. The combination reaction is the reaction in which two or more substances react to form a single new substance.
    ♦ They are also known as synthesis reactions.
2. The general form of a combination reaction is:
A + B → C
3. In order that, a combination reaction is a redox reaction, at least one of the reactants must be in the elemental form. That means, both A and B cannot be compounds.
4. All combustion reactions
    ♦ are combination reactions
    ♦ and at the same time
    ♦ redox reactions also.
• Recall that, in combustion, a substance reacts with elemental oxygen.
    ♦ An example is: $\mathbf\small{\rm{\overset{0}{C}\,(s)+\overset{0}{O}_2\,(g)\rightarrow \overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
    ♦ Another example is: $\mathbf\small{\rm{2\,\overset{+2}{C}\,\overset{-2}{O}\,(s)+\overset{0}{O}_2\,(g)\rightarrow 2\,\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
• Here, O gets reduced. It is the oxidant.
5. Combination reactions can be redox reactions, even if elemental oxygen is not one of the reactants.
    ♦ An example is: $\mathbf\small{\rm{3\,\overset{0}{Mg}\,(s)+\overset{0}{N}_2\,(g)\rightarrow \overset{+2}{Mg}_3\,\overset{-3}{N}_2\,(g)}}$ 
• Here, N gets reduced. It is the oxidant.

B. Decomposition reactions
This can be explained in 5 steps:
1. A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances.
2. The general form of a combination reaction is:
C → A + B
• For this decomposition to take place, energy input (in the form of heat, light or electricity) is required.     
3. In order that, a decomposition reaction is a redox reaction, at least one of the products must be in the elemental form. That means, both A and B cannot be compounds.
4. Let us see some examples:
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow 2\,\overset{0}{H}_2\,(g)+\overset{0}{O}_2\,(g)}}$
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{Na}\,\overset{-1}{H}\,(l)\rightarrow 2\,\overset{0}{Na}\,(g)+\overset{0}{H}_2\,(g)}}$
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{K}\,\overset{+5}{Cl}\,\overset{-2}{O}_3\,(s)\rightarrow 2\,\overset{+1}{K}\,\overset{-1}{Cl}\,(s)+3\,\overset{0}{O}_2\,(g)}}$
5. Let us see a decomposition reaction in which both the products are compounds:
    ♦ $\mathbf\small{\rm{\overset{+2}{Ca}\,\overset{+4}{C}\,\overset{-2}{O}_3\,(s)\rightarrow \overset{+2}{Ca}\,\overset{-2}{O}\,(s)+\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
• This is not a redox reaction because, oxidation numbers remain the same.

C. Displacement reactions
This can be explained in 5 steps:
• A displacement reaction is a reaction in which one element replaces a similar element in a compound.
• The general form of a displacement reaction is:
X + YZ → XZ + Y
   ♦ Here YZ is a compound. The element Y in that compound is replaced by X.
• Note that, only similar elements can do the replacement. That is.,
   ♦ If Y is a metal, X must also be a metal.
   ♦ If Y is a non metal, X must also be a non metal.
• There are three types of displacement reactions:
I. Metal displacement reaction
II. Hydrogen displacement reaction
III. Non metal displacement reaction 

I. Metal displacement reaction
• This can be explained in 4 steps:
1. Consider the general form: X + YZ → XZ + Y
   ♦ In a metal displacement reaction, both X and Y will be metals.
2. We have seen some examples in section 8.1
Example 1:
(i) Consider fig.8.1 in section 8.1
• When metallic zinc is placed in the CuSO₄ solution, Zn replaces Cu.
• We can write the reaction as:
Zn + CuSO₄  → ZnSO₄ + Cu
(ii) Comparing with the general form in (1), we get:
   ♦ X is Zn
   ♦ YZ is CuSO₄
   ♦ XZ is ZnSO₄
   ♦ Y is Cu
• So this reaction is a displacement reaction.
(iii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+\overset{+2}{Cu}\,\overset{6}{S}\,\overset{-2}{O}_4\,(aq)\rightarrow\overset{+2}{Zn}\,\overset{6}{S}\,\overset{-2}{O}_4\,(aq)+\overset{0}{Cu}\,(s)}}$
• We see that:
   ♦ Cu is reduced.
   ♦ Zn is oxidized.

Example 2:
(i) Consider fig.8.2 in section 8.1
• When metallic copper is placed in the AgNO₃ solution, Cu replaces Ag.
• We can write the reaction as:
Cu + 2AgNO₃  → Cu(NO₃)₂ + 2Ag
(ii) Comparing with the general form in (1), we get:
   ♦ X is Cu
   ♦ YZ is AgNO₃
   ♦ XZ is Cu(NO₃)₂
   ♦ Y is Ag
• So this reaction is a displacement reaction.
(iii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+2\,\overset{+1}{Ag}\,\overset{5}{N}\,\overset{-2}{O}_3\,(aq)\rightarrow \overset{+2}{Cu}\,(\overset{5}{N}\,\overset{-2}{O}_3)_2\,(aq)+2\,\overset{0}{Ag}\,(s)}}$
• We see that:
   ♦ Ag is reduced.
   ♦ Cu is oxidized.

3. A few more examples are given below:
Example 3:
(i) 5Ca (s) + V₂O₅ (s)  → 5CaO (s) + 2V (s)
• Comparing with the general form in (1), we get:
   ♦ X is Ca
   ♦ YZ is V₂O₅
   ♦ XZ is CaO
   ♦ Y is V
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{5\,\overset{0}{Ca}\,(s)+\overset{+5}{V}_2\,\overset{-2}{O}_5\,(s)\rightarrow 5\,\overset{+2}{Ca}\,\overset{-2}{O}\,(s)+2\,\overset{0}{V}\,(s)}}$
• We see that:
   ♦ V is reduced.
   ♦ Ca is oxidized.

Example 4:
(i) 2Mg (s) + TiCl₄ (l)  → 2MgCl₂ (s) + Ti (s)
• Comparing with the general form in (1), we get:
   ♦ X is Mg
   ♦ YZ is TiCl₄
   ♦ XZ is MgCl₂
   ♦ Y is Ti
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{2\,\overset{0}{Mg}\,(s)+\overset{+4}{Ti}\,\overset{-1}{Cl}_4\,(l)\rightarrow 2\,\overset{+2}{Mg}\,\overset{-1}{Cl}_2\,(s)+\overset{0}{Ti}\,(s)}}$
• We see that:
   ♦ Ti is reduced.
   ♦ Mg is oxidized.

Example 5:
(i) 2Al (s) + Cr₂O₃ (s)  → Al₂O₃ (s) + 2Cr (s)
• Comparing with the general form in (1), we get:
   ♦ X is Al
   ♦ YZ is Cr₂O₃
   ♦ XZ is Al₂O₃
   ♦ Y is Cr
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{2\,\overset{0}{Al}\,(s)+\overset{+3}{Cr}_2\,\overset{-2}{O}_3\,(s)\rightarrow \overset{+3}{Al}_2\,\overset{-2}{O}_3\,(s)+2\,\overset{0}{Cr}\,(s)}}$
• We see that:
   ♦ Cr is reduced.
   ♦ Al is oxidized.
4. We see that, metals replace other metals in the compounds. This property is very useful in metallurgy. It can be explained in 3 steps:
(i) Consider the general form that we wrote in (1): X + YZ → XZ + Y
   ♦ YZ is an ore of the metal Y.
         ✰ Y is the useful metal.
         ✰ Z is the impurity which is chemically combined with Y.
• If we can make YZ to react with X, we will get Y in free form
(ii) Recall the series that we wrote in section 8.1: Zn > Cu > Ag
• We saw that, any metal in the series can displace any other metal on it’s right side in the series.
(iii) Scientists have added many more metals to the above series.
• So in metallurgy, to get the metal Y in free form, we select X in such a way that, 'X lies to the left of Y' in the series.

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In the next section, we will see hydrogen displacement reactions.


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