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