In the previous section, we completed a discussion on Equilibrium. In this chapter, we will see redox reactions.
1. In early days, scientists used the word oxidation to denote those simple reactions in which, an element combines with oxygen.
• Some examples are:
2Mg (s) + O2 (g) → 2MgO (s)
S (s) + O2 (g) → SO2 (s)
2. As time passed, scientists became aware of a bit more complex reactions. For example:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
♦ In this reaction, oxygen reacts with the compound CH4.
♦ As a result, hydrogen in that compound is replaced by oxygen.
• In the reactions in (1), oxygen reacts with elements.
♦ But here, oxygen reacts with a compound.
• Such reactions involving replacement of hydrogen, were also put under the category of oxidation reactions.
3. With the passage of time, even more complex reactions came to be known. Some examples are:
Mg (s) + F2 (g) → MgF2 (s)
Mg (s) + Cl2 (g) → MgCl2 (s)
Mg (s) + S (g) → MgS/ (s)
• In these reactions, oxygen is not present. But the behaviour of fluorine, chlorine and sulfur are similar to that of oxygen.
♦ Both fluorine and oxygen attracts electrons possessed by Mg.
♦ Chlorine also attracts electrons possessed by Mg.
♦ Sulfur also attracts electrons possessed by Mg.
(Later in our discussions, we will see how and why oxygen, fluorine, chlorine, sulfur etc., attract electrons from other elements)
• Due to the ‘oxygen-like’ behaviour, such reactions were also put under the category of oxidation reactions.
4. So we can prepare a list containing the above three items.
• Types of reactions which early scientists considered as oxidation reactions:
(a) Simple reactions in which oxygen combines with other elements.
(b) More complex reactions in which oxygen replaces hydrogen from other compounds.
(c) Even more complex reactions in which elements other than oxygen behave like oxygen.
Modern day scientists explain oxidation reactions on the basis of electron transfer. Such an explanation can be written in 11 steps:
1. Consider the reaction: 2Na (s) + Cl2 (g) → 2NaCl (s)
• We see that, on the product side, there are two NaCl molecules.
• Each of those NaCl molecules are independently stable. They need no external help to stay stable.
• So we will consider one molecule of NaCl.
2. How sodium attain octet:
• Sodium is mono atomic. So in a sample of sodium, there will be individual Na atoms. Each of those Na atoms has to donate one electron to attain octet.
(Recall that, elements situated towards the left of the periodic table, are metallic in nature and they tend to donate electrons. The number of electrons which they donate can be determined from their electronic configurations)
3. How chlorine attain octet:
• Chlorine is di atomic. So in a sample of chlorine gas, there will be individual Cl2 molecules. Each Cl atom has to accept one electron to attain octet.
(Recall that, elements situated towards the right of the periodic table,
are non-metallic in nature and they tend to accept electrons. The number of
electrons which they accept can be determined from their electronic
configurations)
• We cannot send a sample of individual Cl atoms to react with Na atoms. We will be sending a sample of individual Cl2 molecules.
• So we can write: Each Cl2 molecule can accept two electrons.
◼ That means, two Na atoms will react with each Cl2 molecule.
4. The electron donation by Na atoms can be represented as:
2Na (s) → Na+ (s) + 1e- + Na+ (s) + 1e-
⇒ 2Na (s) → 2Na+ (s) + 2e-
5. The electron acceptance by Cl2 molecule can be represented as:
Cl2 (g) + 2e- → Cl- (g) + Cl- (g)
⇒ Cl2 (g) + 2e- → 2Cl- (g)
6. If we add the reactions in (4) and (5), we get:
2Na (s) + Cl2 (g) + 2e- → 2Na+ (s) + 2e- + 2Cl- (s)
• The electrons on both sides will cancel each other. So we get:
2Na (s) + Cl2 (g) → 2Na+ (s) + 2Cl- (s)
• The Na+ and Cl- ions on the product side will stick together because of the electrostatic force of attraction. It is an ionic bond.
• One Na+ ion will combine with one Cl- ion to form one Na+Cl- molecule.
• The remaining Na+ ion will combine with the other Cl- ion to form another Na+Cl- molecule.
• Note that, the charges in Na+Cl- balance perfectly. Thus we get stable molecules.
• Two molecules of NaCl will be formed. So the final added can be modified as:
2Na (s) + Cl2 (g) → 2Na+Cl- (s)
• This is the same balanced equation that we wrote in (1). So the reactions that we wrote in (4) and (5) are correct.
7. Modern day scientists consider:
♦ Loss of electrons as oxidation.
♦ Gain of electrons as reduction.
• So in the present case, we can write:
♦ Na is oxidized.
♦ Cl is reduced.
8. We arrived at the final equation in (6) by adding the two equations in (4) and (5)
• Each of those two reactions which were added, is called a half reaction.
• So the half reactions in our present case are:
♦ 2Na (s) → 2Na+ (s) + 2e-
♦ Cl2 (g) + 2e- → 2Cl- (g)
9. Now we can write the definition of oxidation and reduction:
♦ The half reaction which involve loss of electron(s) is called oxidation reaction.
♦ The half reaction which involve gain of electron(s) is called reduction reaction.
10. Next we can write about oxidizing agent and reducing agent:
• We see that, the ‘electron donating character’ of Na leads to the reduction of Cl.
♦ So Na is called the reducing agent.
◼ In general, the atom which donates electron(s) is called reducing agent.
• We see that, the ‘electron accepting character’ of Cl leads to the oxidation of Na.
♦ So Cl is called the oxidizing agent.
◼ In general, the atom which accepts electron(s) is called oxidizing agent.
11. We can write a summary as follows:
Oxidation: Loss of electron(s) by any species.
♦ The species which loses electron(s) is said to be oxidized.
Reduction: Gain of electron(s) by any species.
♦ The species which gains electron(s) is said to be reduced.
Oxidizing agent: The species which accepts electron(s).
Reducing agent: The species which donates electron(s).
Let us see another example. It can be written in 10 steps:
1. Consider the reaction: 4Na (s) + O2 (g) → 2Na2O (s)
• We see that, on the product side, there are two Na2O molecules.
• Each of those Na2O molecules are independently stable. They need no external help to stay stable.
• So we will consider one molecule of Na2O.
2. How sodium attain octet:
•
Sodium is mono atomic. So in a sample of sodium, there will be
individual Na atoms. Each of those Na atoms has to donate one electron
to attain octet.
3. How oxygen attain octet:
• Oxygen is di atomic. So in a sample of oxygen gas, there will be individual O2 molecules. Each O atom has to accept two electrons to attain octet.
• We cannot send a sample of individual O atoms to react with Na atoms. We will be sending a sample of individual O2 molecules.
• So we can write: Each O2 molecule can accept four electrons.
◼ That means, four Na atoms will react with each O2 molecule.
4. The electron donation by Na atoms can be represented as:
4Na (s) → 4Na+ (s) + 4e-
5. The electron acceptance by O2 molecule can be represented as:
O2 (g) + 4e- → O2- (s) + O2- (g)
⇒ O2 (g) + 4e- → 2O2- (g)
6. If we add the reactions in (4) and (5), we get:
4Na (s) + O2 (g) + 4e- → 4Na+ (s) + 4e- + 2O2- (s)
• The electrons on both sides will cancel each other. So we get:
4Na (s) + O2 (g) → 4Na+ (s) + 2O2- (s)
7. On the product side we have:
• Four Na+ ions.
♦ These ions are individually stable.
• Two O2- ions.
♦ These ions are individually stable.
8. Formation of product molecule:
• Two Na+ ions will combine with one O2- ion to form one (Na+)2O2- molecule.
• The remaining two Na+ ions will combine with the other O2- ion to form another (Na+)2O2- molecule.
• Note that, the charges in (Na+)2O2- balance perfectly. Thus we get stable molecules. It is an ionic bond.
9. So the final equation in (6) can be modified as:
4Na (s) + O2 (g) → 2(Na+)2O2- (s)
• This is the same balanced equation that we wrote in (1). So the reactions that we wrote in (4) and (5) are correct.
10. Now we can write the summary:
Oxidation: Loss of electron(s) by any species.
♦ The species which loses electron(s) is said to be oxidized.
• The half reaction in (4) is: 4Na (s) → 4Na+ (s) + 4e-
♦ This reaction indicates loss of electrons. So it is the oxidation reaction.
♦ Na is oxidized.
Reduction: Gain of electron(s) by any species.
♦ The species which gains electron(s) is said to be reduced.
• The half reaction in (5) is: O2 (g) + 4e- → 2O2- (g)
♦ This reaction indicates gain of electrons. So it is the reduction reaction.
♦ O2 is reduced.
Oxidizing agent: The species which accepts electron(s).
♦ Clearly, O2 is the oxidizing agent.
Reducing agent: The species which donates electron(s).
♦ Clearly, Na is the reducing agent.
Let us see one more example. It can be written in 10 steps:
1. Consider the reaction: 2Na (s) + S (s) → Na2S (s)
• Consider one molecule of Na2S.
2. How sodium attain octet:
•
Sodium is mono atomic. So in a sample of sodium, there will be
individual Na atoms. Each of those Na atoms has to donate one electron
to attain octet.
3. How sulfur attain octet:
• Sulfur is poly atomic. But from a sample of solid sulfur, we need to consider only one atom of sulfur to be reacting with Na. Each S atom has to accept two electrons to attain octet.
◼ That means, two Na atoms will react with each S atom.
4. The electron donation by Na atoms can be represented as:
2Na (s) → 2Na+ (s) + 2e-
5. The electron acceptance by S atom can be represented as:
S (s) + 2e- → S2- (s)
6. If we add the reactions in (4) and (5), we get:
2Na (s) + S (s) + 2e- → 2Na+ (s) + 2e- + S2-
• The electrons on both sides will cancel each other. So we get:
2Na (s) + S (s) → 2Na+ (s) + S2-
7. On the product side we have:
• Two Na+ ions.
♦ These ions are individually stable.
• One S2- ions.
♦ These ions are individually stable.
8. Formation of product molecule:
• The two Na+ ions will combine with the one S2- ion to form one (Na+)2S2- molecule.
• Note that, the charges in (Na+)2O2- balance perfectly. Thus we get stable molecule. It is an ionic bond.
9. So the final equation in (6) can be modified as:
2Na (s) + S (s) → (Na+)2S2- (s)
• This is the same balanced equation that we wrote in (1). So the reactions that we wrote in (4) and (5) are correct.
10. Now we can write the summary:
Oxidation: Loss of electron(s) by any species.
♦ The species which loses electron(s) is said to be oxidized.
• The half reaction in (4) is: 2Na (s) → 2Na+ (s) + 2e-
♦ This reaction indicates loss of electrons. So it is the oxidation reaction.
♦ Na is oxidized.
Reduction: Gain of electron(s) by any species.
♦ The species which gains electron(s) is said to be reduced.
• The half reaction in (5) is: S (s) + 2e- → S2- (s)
♦ This reaction indicates gain of electrons. So it is the reduction reaction.
♦ S is reduced.
Oxidizing agent: The species which accepts electron(s).
♦ Clearly, S is the oxidizing agent.
Reducing agent: The species which donates electron(s).
♦ Clearly, Na is the reducing agent.
◼ A reaction in which one reactant undergoes oxidation whereas the other gets reduced during the course of reaction are termed as oxidation-reduction reactions or redox reactions.
Let us see a solved example:
Solved example 8.1
Justify that the reaction: 2Na (s) + H2 (s) → 2NaH (s) is a redox change.
Solution:
1. We see that, on the product side, there are two NaH molecules.
• Each of those NaH molecules are independently stable. They need no external help to stay stable.
• So we will consider one molecule of NaH.
2. How sodium attain octet:
•
Sodium is mono atomic. So in a sample of sodium, there will be
individual Na atoms. Each of those Na atoms has to donate one electron
to attain octet.
3. How hydrogen attain octet:
• Hydrogen is di atomic. So in a sample of hydrogen gas, there will be individual H2 molecules. Each H atom has to accept one electron to attain octet.
• We cannot send a sample of individual H atoms to react with Na atoms. We will be sending a sample of individual H2 molecules.
• So we can write: Each H2 molecule can accept two electrons.
◼ That means, two Na atoms will react with each H2 molecule.
4. The electron donation by Na atoms can be represented as:
2Na (s) → 2Na+ (s) + 2e-
5. The electron acceptance by H2 molecule can be represented as:
H2 (g) + 2e- → H- (g) + H- (g)
⇒ H2 (g) + 2e- → 2H- (g)
6. If we add the reactions in (4) and (5), we get:
2Na (s) + H2 (g) + 2e- → 2Na+ (s) + 2e- + 2H- (s)
• The electrons on both sides will cancel each other. So we get:
2Na (s) + H2 (g) → 2Na+ (s) + 2H- (s)
7. On the product side we have:
• Two Na+ ions.
♦ These ions are individually stable.
• Two H- ions.
♦ These ions are individually stable.
8. Formation of product molecule:
• One Na+ ion will combine with one H- ion to form one Na+H- molecule.
• The remaining Na+ ion will combine with the other H- ion to form another Na+H- molecule.
• Note that, the charges in Na+H- balance perfectly. Thus we get stable molecules. It is an ionic bond.
9. So the final equation in (6) can be modified as:
2Na (s) + H2 (g) → 2Na+H- (s)
• This is the same balanced equation that we wrote in (1). So the reactions that we wrote in (4) and (5) are correct.
10. Now we can write the summary:
Oxidation: Loss of electron(s) by any species.
♦ The species which loses electron(s) is said to be oxidized.
• The half reaction in (4) is: 2Na (s) → 2Na+ (s) + 2e-
♦ This reaction indicates loss of electrons. So it is the oxidation reaction.
♦ Na is oxidized.
Reduction: Gain of electron(s) by any species.
♦ The species which gains electron(s) is said to be reduced.
• The half reaction in (5) is: H2 (g) + 2e- → 2H- (g)
♦ This reaction indicates gain of electrons. So it is the reduction reaction.
♦ H2 is reduced.
Oxidizing agent: The species which accepts electron(s).
♦ Clearly, H2 is the oxidizing agent.
Reducing agent: The species which donates electron(s).
♦ Clearly, Na is the reducing agent.
◼ We can write:
One reactant is oxidized and the other is reduced. So it is a redox reaction.
•
In the next section, we will see competitive electron transfer reactions.
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