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To determine why the combustion of acetylene ([tex]\(\mathrm{C_2H_2}\)[/tex]) is a redox reaction, we need to assign oxidation numbers to each element in the reaction and analyze how these numbers change. A redox reaction involves the transfer of electrons, resulting in changes in oxidation numbers where some elements are oxidized (lose electrons) and others are reduced (gain electrons).
Let’s break down the reaction:
[tex]\[ 2 \mathrm{C_2H_2} + 5 \mathrm{O_2} \rightarrow 4 \mathrm{CO_2} + 2 \mathrm{H_2O} \][/tex]
### Assign Oxidation Numbers
1. Acetylene ([tex]\(\mathrm{C_2H_2}\)[/tex]):
- Hydrogen ([tex]\(\mathrm{H}\)[/tex]) has an oxidation number of [tex]\(+1\)[/tex].
- To balance this, each Carbon ([tex]\(\mathrm{C}\)[/tex]) must have an oxidation number of [tex]\(\frac{-1}{2}\)[/tex].
This is because the molecule is neutral and each hydrogen brings a +1 charge, so in [tex]\(\mathrm{C_2H_2}\)[/tex] we have:
[tex]\[ 2 \cdot x + 2 \cdot (+1) = 0 \Rightarrow 2 \cdot x + 2 = 0 \Rightarrow 2 \cdot x = -2 \Rightarrow x = -1 \][/tex]
where [tex]\(x\)[/tex] is the oxidation number of each Carbon. Thus, each Carbon in [tex]\(\mathrm{C_2H_2}\)[/tex] has an oxidation number of [tex]\(-1\)[/tex].
2. Oxygen ([tex]\(\mathrm{O_2}\)[/tex]):
- Oxygen in its elemental form ([tex]\(\mathrm{O_2}\)[/tex]) has an oxidation number of [tex]\(0\)[/tex].
3. Carbon Dioxide ([tex]\(\mathrm{CO_2}\)[/tex]):
- Oxygen in compounds has an oxidation number of [tex]\(-2\)[/tex].
- To balance this in [tex]\(\mathrm{CO_2}\)[/tex], each Carbon must have an oxidation number of [tex]\(+4\)[/tex] because:
[tex]\[ x + 2\cdot(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4 \][/tex]
4. Water ([tex]\(\mathrm{H_2O}\)[/tex]):
- Hydrogen has an oxidation number of [tex]\(+1\)[/tex].
- Oxygen has an oxidation number of [tex]\(-2\)[/tex].
### Identify Changes in Oxidation Numbers
- Carbon:
- In [tex]\(\mathrm{C_2H_2}\)[/tex], each Carbon is [tex]\( -1 \)[/tex].
- In [tex]\(\mathrm{CO_2}\)[/tex], each Carbon is [tex]\( +4 \)[/tex].
- Change for each Carbon atom: [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex] (increase, hence oxidation).
- Oxygen:
- In [tex]\(\mathrm{O_2}\)[/tex], each Oxygen is [tex]\( 0 \)[/tex].
- In [tex]\(\mathrm{CO_2}\)[/tex], each Oxygen is [tex]\(-2\)[/tex].
- In [tex]\(\mathrm{H_2O}\)[/tex], each Oxygen is [tex]\(-2\)[/tex].
- Change: [tex]\(0\)[/tex] to [tex]\(-2\)[/tex] (decrease, hence reduction).
### What is Oxidized and What is Reduced?
- Oxidized (Loses electrons):
- Carbon in [tex]\(\mathrm{C_2H_2}\)[/tex] is oxidized. Its oxidation number increases from [tex]\(-1\)[/tex] to [tex]\( +4\)[/tex].
- Reduced (Gains electrons):
- Oxygen in [tex]\(\mathrm{O_2}\)[/tex] is reduced. Its oxidation number decreases from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex].
### Electron Transfer
- Oxidation (Carbon):
- Each Carbon atom goes from [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex], indicating it loses [tex]\(5\)[/tex] electrons per atom ([tex]\(5\)[/tex] electrons per 2 Carbons lead to a transfer of [tex]\(10\)[/tex] electrons).
- Reduction (Oxygen):
- Each Oxygen atom goes from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex], indicating it gains [tex]\(2\)[/tex] electrons per atom. For [tex]\(5 \cdot 2 = 10\)[/tex] electrons are gained in total to balance the [tex]\(2 \times 5 = 10\)[/tex] total loss from 2 Carbons.
### Conclusion
The combustion of acetylene is a redox reaction because there is a transfer of electrons. Carbon gets oxidized (oxidation number increases from [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex]) and oxygen gets reduced (oxidation number decreases from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex]). Therefore, it is verified that the reaction involves both oxidation and reduction processes.
Let’s break down the reaction:
[tex]\[ 2 \mathrm{C_2H_2} + 5 \mathrm{O_2} \rightarrow 4 \mathrm{CO_2} + 2 \mathrm{H_2O} \][/tex]
### Assign Oxidation Numbers
1. Acetylene ([tex]\(\mathrm{C_2H_2}\)[/tex]):
- Hydrogen ([tex]\(\mathrm{H}\)[/tex]) has an oxidation number of [tex]\(+1\)[/tex].
- To balance this, each Carbon ([tex]\(\mathrm{C}\)[/tex]) must have an oxidation number of [tex]\(\frac{-1}{2}\)[/tex].
This is because the molecule is neutral and each hydrogen brings a +1 charge, so in [tex]\(\mathrm{C_2H_2}\)[/tex] we have:
[tex]\[ 2 \cdot x + 2 \cdot (+1) = 0 \Rightarrow 2 \cdot x + 2 = 0 \Rightarrow 2 \cdot x = -2 \Rightarrow x = -1 \][/tex]
where [tex]\(x\)[/tex] is the oxidation number of each Carbon. Thus, each Carbon in [tex]\(\mathrm{C_2H_2}\)[/tex] has an oxidation number of [tex]\(-1\)[/tex].
2. Oxygen ([tex]\(\mathrm{O_2}\)[/tex]):
- Oxygen in its elemental form ([tex]\(\mathrm{O_2}\)[/tex]) has an oxidation number of [tex]\(0\)[/tex].
3. Carbon Dioxide ([tex]\(\mathrm{CO_2}\)[/tex]):
- Oxygen in compounds has an oxidation number of [tex]\(-2\)[/tex].
- To balance this in [tex]\(\mathrm{CO_2}\)[/tex], each Carbon must have an oxidation number of [tex]\(+4\)[/tex] because:
[tex]\[ x + 2\cdot(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4 \][/tex]
4. Water ([tex]\(\mathrm{H_2O}\)[/tex]):
- Hydrogen has an oxidation number of [tex]\(+1\)[/tex].
- Oxygen has an oxidation number of [tex]\(-2\)[/tex].
### Identify Changes in Oxidation Numbers
- Carbon:
- In [tex]\(\mathrm{C_2H_2}\)[/tex], each Carbon is [tex]\( -1 \)[/tex].
- In [tex]\(\mathrm{CO_2}\)[/tex], each Carbon is [tex]\( +4 \)[/tex].
- Change for each Carbon atom: [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex] (increase, hence oxidation).
- Oxygen:
- In [tex]\(\mathrm{O_2}\)[/tex], each Oxygen is [tex]\( 0 \)[/tex].
- In [tex]\(\mathrm{CO_2}\)[/tex], each Oxygen is [tex]\(-2\)[/tex].
- In [tex]\(\mathrm{H_2O}\)[/tex], each Oxygen is [tex]\(-2\)[/tex].
- Change: [tex]\(0\)[/tex] to [tex]\(-2\)[/tex] (decrease, hence reduction).
### What is Oxidized and What is Reduced?
- Oxidized (Loses electrons):
- Carbon in [tex]\(\mathrm{C_2H_2}\)[/tex] is oxidized. Its oxidation number increases from [tex]\(-1\)[/tex] to [tex]\( +4\)[/tex].
- Reduced (Gains electrons):
- Oxygen in [tex]\(\mathrm{O_2}\)[/tex] is reduced. Its oxidation number decreases from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex].
### Electron Transfer
- Oxidation (Carbon):
- Each Carbon atom goes from [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex], indicating it loses [tex]\(5\)[/tex] electrons per atom ([tex]\(5\)[/tex] electrons per 2 Carbons lead to a transfer of [tex]\(10\)[/tex] electrons).
- Reduction (Oxygen):
- Each Oxygen atom goes from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex], indicating it gains [tex]\(2\)[/tex] electrons per atom. For [tex]\(5 \cdot 2 = 10\)[/tex] electrons are gained in total to balance the [tex]\(2 \times 5 = 10\)[/tex] total loss from 2 Carbons.
### Conclusion
The combustion of acetylene is a redox reaction because there is a transfer of electrons. Carbon gets oxidized (oxidation number increases from [tex]\(-1\)[/tex] to [tex]\(+4\)[/tex]) and oxygen gets reduced (oxidation number decreases from [tex]\(0\)[/tex] to [tex]\(-2\)[/tex]). Therefore, it is verified that the reaction involves both oxidation and reduction processes.
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