Based on the chart we saw at the beginning of the previous section, the next type that we have to learn is Disproportionation reactions.
This can be explained in 5 steps:
1. We know that, the most common oxidation state of chlorine is -1.
• But if we write all the possible oxidation states of chlorine, we will get:
-1, 0, +1, +3, +5, +7
2. Similarly, the most common oxidation state of manganese is +2.
• But if we write all the possible oxidation states of manganese, we will get:
+2, +3, +4, +6, +7
3. Thus we see that, some elements can exist in more than one oxidation states.
◼ Disproportionation reactions will contain an element which has a minimum of three possible oxidation states.
• On the reactant side, that element will be possessing a single oxidation state, say ‘x’
• On the product side, that element will be having two oxidation states
♦ One of them will be less than ‘x’
♦ The other will be greater than ‘x’
4. Let us see some examples:
Example 1:
• The decomposition of hydrogen peroxide is:
$\mathbf\small{\rm{2\,\overset{+1}{H}_2\,\overset{-1}{O}_2\,(aq)
\rightarrow 2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)
+\overset{0}{O_2}\,(g)}}$
• Here, oxygen experiences disproportionation. It can be explained in 3 steps:
(i) In the reactant side,
♦ O is in the peroxide.
✰ The oxidation state of O there, is -1
(ii) In the product side,
♦ O is in O2
✰ The oxidation state of O there, is 0
♦ O is in H2O
✰ The oxidation state of O there, is -2
(iii) So the change in oxidation state can be written as:
♦ Original oxidation state of -1 (reactant side)
✰ Increases to 0 (product side)
✰ Decreases to -2 (product side)
Example 2:
• Phosphorus undergoes disproportionation in the following reaction:
$\mathbf\small{\rm{\overset{0}{P}_4\,(s)+3\,OH^-\,(aq)+3\,H_2O\,(l)
\rightarrow \overset{-3}{P}\,H_3\,(g)+3\,H_2\,\overset{+1}{P}\,O_2^-\,(aq)}}$
• It can be explained in 3 steps:
(i) In the reactant side,
♦ P is in the elemental form.
✰ The oxidation state of P there, is 0
(ii) In the product side,
♦ P is in PH3
✰ The oxidation state of P there, is -3
♦ P is in H2PO2-
✰ The oxidation state of P there, is +1
(iii) So the change in oxidation state can be written as:
♦ Original oxidation state of 0 (reactant side)
✰ Increases to +1 (product side)
✰ Decreases to -3 (product side)
Example 3:
• Sulfur undergoes disproportionation in the following reaction:
$\mathbf\small{\rm{\overset{0}{S}_8\,(s)+12\,OH^-\,(aq)\rightarrow 4\,\overset{-2}{S^{2-}}\,(aq)+2\,\overset{+2}{S}_2\,O_3^{2-}\,(aq)+6\,H_2O\,(l)}}$
• It can be explained in 3 steps:
(i) In the reactant side,
♦ S is in the elemental form.
✰ The oxidation state of S there, is 0
(ii) In the product side,
♦ S is in S2-
✰ The oxidation state of S there, is -2
♦ S is in S2O3-
✰ The oxidation state of S there, is +2
(iii) So the change in oxidation state can be written as:
♦ Original oxidation state of 0 (reactant side)
✰ Increases to +2 (product side)
✰ Decreases to -2 (product side)
Example 4:
• Chlorine undergoes disproportionation in the following reaction:
$\mathbf\small{\rm{\overset{0}{Cl}_2\,(g)+2\,OH^-\,(aq)\rightarrow
\overset{-1}{Cl^{-}}\,(aq)+\overset{+1}{Cl}\,O^{-}\,(aq)+H_2O\,(l)}}$
• It can be explained in 3 steps:
(i) In the reactant side,
♦ Cl is in the elemental form.
✰ The oxidation state of Cl there, is 0
(ii) In the product side,
♦ Cl is in Cl-
✰ The oxidation state of Cl there, is -1
♦ Cl is in ClO-
✰ The oxidation state of Cl there, is +1
(iii) So the change in oxidation state can be written as:
♦ Original oxidation state of 0 (reactant side)
✰ Increases to +1 (product side)
✰ Decreases to -1 (product side)
5. In all the above examples, we see that:
♦ One oxidation state on the product side, is greater than the original state.
♦ The other oxidation state on the product side, is lesser than the original state.
The hypochlorite ion
This can be explained in 4 steps:
1. Consider the last example 4 that we saw above.
♦ The ClO- ion is called hypochlorite ion.
2. It is a powerful oxidizing agent. It can oxidize the chemicals which cause coloration in fabrics. When those chemicals are oxidized, the fabric will become white.
3. It can also be used for removing odor from industrial waste water.
4. Thus we see that, ClO- is a very useful substance. From the equation, it is clear that, the ClO- ion can be produced from the reaction between Cl2 and an alkali.
Behavior of Halogens
• This can be explained in 7 steps:
1. We know that, Cl belongs to the halogen family. F, Br and I are the other important members of the halogen family.
2. Consider the last example 4 that we saw above. It shows the reaction between Cl and alkalies.
• Br and I also reacts in the same manner with alkalies.
3. But F reacts in a different manner. Consider the following reaction:
$\mathbf\small{\rm{\overset{0}{F}_2\,(g)+2\,OH^-\,(aq)\rightarrow
2\,\overset{-1}{F^{-}}\,(aq)+\overset{-2}{O}\,\overset{-1}{F_2}\,(aq)+H_2O\,(l)}}$
• This reaction can be explained in 3 steps:
(i) In the reactant side,
♦ F is in the elemental form.
✰ The oxidation state of F there, is 0
(ii) In the product side,
♦ F is in F-
✰ The oxidation state of F there, is -1
♦ F is in OF2
✰ The oxidation state of F there, is also -1
(iii) So the change in oxidation state can be written as:
♦ Original oxidation state of 0 (reactant side)
✰ Decreases to -1 (product side)
4. Recall the condition for disproportionation:
♦ One oxidation state on the product side, is greater than the original state.
♦ The other oxidation state on the product side, is lesser than the original state.
5. But in our present case,
♦ Both the oxidation states on the product side, are lesser than the original state.
6. Our original oxidation state is zero.
• So, if there is to be an oxidation state greater than the original state, that greater state must be greater than zero. That means, that greater oxidation state must be a positive number.
7. But F is the most electronegative element that, no other element can take away electron from it.
• Consequently, F can never attain a positive oxidation state.
◼ So it is clear that, F can never undergo disproportionation.
Solved example 8.6
Which of the following species, do not show disproportionation reaction and
why ?
ClO- , ClO2- , ClO3- and ClO4-
Also write reaction for each of the species that disproportionates.
Solution:
1. Let us write the oxidation states in each species. We get:
♦ $\mathbf\small{\rm{\overset{+1}{Cl}\,\overset{-2}{O^{-}}}}$
♦ $\mathbf\small{\rm{\overset{+3}{Cl}\,\overset{-2}{O_2^{-}}}}$
♦ $\mathbf\small{\rm{\overset{+5}{Cl}\,\overset{-2}{O_3^{-}}}}$
♦ $\mathbf\small{\rm{\overset{+7}{Cl}\,\overset{-2}{O_4^{-}}}}$
2. Consider the first species: $\mathbf\small{\rm{\overset{+1}{Cl}\,\overset{-2}{O^{-}}}}$
• Recall the condition for disproportionation:
♦ One oxidation state on the product side, is greater than the original state.
♦ The other oxidation state on the product side, is lesser than the original state.
• So for this species,
♦ One oxidation state of Cl on the product side, must be greater than +1.
♦ The other oxidation state of Cl on the product side, must be lesser than +1.
• Satisfying the conditions:
♦ Condition 1 can be satisfied because, Cl can have an oxidation states of +3, +5 and +7.
♦ Condition 2 can be satisfied because, Cl can have an oxidation states of -1 and 0
• So this species can undergo disproportionation.
3. Consider the second species: $\mathbf\small{\rm{\overset{+3}{Cl}\,\overset{-2}{O_2^{-}}}}$
• For this species,
♦ One oxidation state of Cl on the product side, must be greater than +3.
♦ The other oxidation state of Cl on the product side, must be lesser than +3.
• Satisfying the conditions:
♦ Condition 1 can be satisfied because, Cl can have an oxidation states of +5 and +7.
♦ Condition 2 can be satisfied because, Cl can have an oxidation states of -1, 0
and +1.
• So this species can undergo disproportionation.
4. Consider the third species: $\mathbf\small{\rm{\overset{+5}{Cl}\,\overset{-2}{O_3^{-}}}}$
• For this species,
♦ One oxidation state of Cl on the product side, must be greater than +5.
♦ The other oxidation state of Cl on the product side, must be lesser than +5.
• Satisfying the conditions:
♦ Condition 1 can be satisfied because, Cl can have an oxidation state of +7.
♦ Condition 2 can be satisfied because, Cl can have an oxidation states of -1, 0, +1 and +3.
• So this species can undergo disproportionation.
5. Consider the fourth species: $\mathbf\small{\rm{\overset{+7}{Cl}\,\overset{-2}{O_4^{-}}}}$
• For this species,
♦ One oxidation state of Cl on the product side, must be greater than +7.
♦ The other oxidation state of Cl on the product side, must be lesser than +7.
• Satisfying the conditions:
♦ Condition 1 cannot be satisfied because, Cl cannot have an oxidation state greater than +7.
♦ Condition 2 can be satisfied because, Cl can have an oxidation states of -1, 0, +1, +3 and +5.
• Since both the conditions cannot be satisfied, this species cannot undergo disproportionation.
6. We see that, the first three species can undergo disproportionation. Their reactions are as follows:
Species 1:
$\mathbf\small{\rm{3\,\overset{+1}{Cl}\,\overset{-2}{O^{-}}\rightarrow
2\,\overset{-1}{Cl^{-}}+\overset{+5}{Cl}\,\overset{-2}{O_3^{-}}}}$
Species 2:
$\mathbf\small{\rm{6\,\overset{+3}{Cl}\,\overset{-2}{O_2^{-}}\rightarrow
2\,\overset{-1}{Cl^{-}}+4\,\overset{+5}{Cl}\,\overset{-2}{O_3^{-}}}}$
Species 3:
$\mathbf\small{\rm{4\,\overset{+5}{Cl}\,\overset{-2}{O_3^{-}}\rightarrow
\overset{-1}{Cl^{-}}+3\,\overset{+7}{Cl}\,\overset{-2}{O_4^{-}}}}$
Solved example 8.7
Suggest a scheme of classification of the following redox reactions.
(a) N2 (g) + O2 (g) → 2 NO (g)
(b) 2Pb(NO3)2 (s) → 2PbO (s) + 4 NO2 (g) + O2 (g)
(c) NaH(s) + H2O (l) → NaOH (aq) + H2 (g)
(d) 2NO2 (g) + 2OH- (aq) → NO2- (aq) + NO3- (aq) + H2O (l)
Solution:
(a) This is a combination reaction.
• Reason can be explained in 2 steps:
(i) The combination reaction is the reaction in which two or more substances react to form a single new substance.
♦ Here, N2 and O2 react to form the new substance NO
(ii) In order that, a combination reaction is a redox reaction, at least one of the reactants must be in the elemental form.
♦ Here, both the reactants are in elemental form.
(b) This is a decomposition reaction.
• Reason can be explained in 2 steps:
(i) A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances.
♦ Here, the compound Pb(NO3)2 breaks down into three simpler substances: PbO, NO2 and O2
(ii) In order that, a decomposition reaction is a redox reaction, at least one of the products must be in the elemental form.
♦ Here, the product O2 is in elemental form.
(c) This is a displacement reaction.
• Reason can be explained as follows:
The H atom in NaH is displaced by OH- ion.
(d) This is a disproportionation reaction.
• Reason can be explained in 2 steps:
(i) Writing the oxidation numbers, we get:
$\mathbf\small{\rm{2\,\overset{4}{N}\,\overset{-2}{O}_2\,(g)+2\,\overset{-2}{O}\,\overset{+1}{H^{-}}\,(aq)\rightarrow \overset{3}{N}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{5}{N}\,\overset{-2}{O_3^{-}}\,(aq)+\overset{+1}{H}_2\,\overset{-2}{O}\,(l)}}$
(ii) Here, nitrogen experiences disproportionation.
• In the reactant side,
♦ N is in NO2.
✰ The oxidation state of N there, is 4
• In the product side,
♦ N is in NO2-
✰ The oxidation state of N there, is 3
♦ N is in NO3-
✰ The oxidation state of O there, is 5
• So the change in oxidation state can be written as:
♦ Original oxidation state of 4 (reactant side)
✰ Increases to 5 (product side)
✰ Decreases to 3 (product side)
In the next section, we will see balancing of redox reactions.
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