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To explain why the combustion of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]) is a redox reaction, we need to identify changes in oxidation numbers of the elements involved. Here's the balanced chemical equation for the reaction:
[tex]\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
1. Assign oxidation numbers to all elements in the reactants and products:
- In [tex]\(\text{C}_2\text{H}_2\)[/tex] (acetylene):
- Hydrogen (H) always has an oxidation number of +1.
- Carbon ([tex]\(\text{C}\)[/tex]): Let's assign it an oxidation number [tex]\(x\)[/tex].
[tex]\[ 2x + 2(+1) = 0 \Rightarrow 2x + 2 = 0 \Rightarrow 2x = -2 \Rightarrow x = -1 \][/tex]
- Hence, the oxidation numbers are:
[tex]\[ C: -1, \; H: +1 \][/tex]
- In [tex]\( \text{O}_2 \)[/tex] (dioxygen):
- Oxygen ([tex]\(\text{O}\)[/tex]) in its elemental form has an oxidation number of 0.
[tex]\[ O: 0 \][/tex]
- In [tex]\(\text{CO}_2\)[/tex] (carbon dioxide):
- Oxygen is generally -2 in compounds.
- Carbon ([tex]\(\text{C}\)[/tex]): Let's assign it an oxidation number [tex]\(y\)[/tex].
[tex]\[ y + 2(-2) = 0 \Rightarrow y - 4 = 0 \Rightarrow y = +4 \][/tex]
- Hence, the oxidation numbers are:
[tex]\[ C: +4, \; O: -2 \][/tex]
- In [tex]\(\text{H}_2\text{O}\)[/tex] (water):
- Hydrogen (H) is +1.
- Oxygen (O) is -2.
[tex]\[ H: +1, \; O: -2 \][/tex]
2. Identify the changes in oxidation numbers to determine what is oxidized and what is reduced:
- Carbon in [tex]\(\text{C}_2\text{H}_2\)[/tex] changes from -1 to +4:
[tex]\(-1 \rightarrow +4\)[/tex], an increase in oxidation number. This indicates that carbon is oxidized.
- Oxygen in [tex]\( \text{O}_2 \)[/tex] changes from 0 to -2:
[tex]\(0 \rightarrow -2\)[/tex], a decrease in oxidation number. This indicates that oxygen is reduced.
3. Define the terms oxidation and reduction:
- Oxidation involves the loss of electrons.
- Reduction involves the gain of electrons.
4. Determine the transfer of electrons:
- Each carbon atom in [tex]\(\text{C}_2\text{H}_2\)[/tex] relinquishes 5 electrons (since going from -1 to +4 means each carbon loses 5 electrons). For two carbon atoms, this amounts to a total of [tex]\(2 \times 5 = 10\)[/tex] electrons lost (oxidation).
- Each oxygen molecule ([tex]\(\text{O}_2\)[/tex]) gains 2 electrons (since going from 0 to -2 means each oxygen atom gains 2 electrons). For five molecules of [tex]\(\text{O}_2\)[/tex], this amounts to a total of [tex]\(5 \times 2 = 10\)[/tex] electrons gained (reduction).
Conclusion:
The combustion reaction of acetylene is a redox reaction because there is a transfer of electrons between the reactants. Carbon in [tex]\(\text{C}_2\text{H}_2\)[/tex] is oxidized (loses electrons), and oxygen in [tex]\( \text{O}_2 \)[/tex] is reduced (gains electrons).
[tex]\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
1. Assign oxidation numbers to all elements in the reactants and products:
- In [tex]\(\text{C}_2\text{H}_2\)[/tex] (acetylene):
- Hydrogen (H) always has an oxidation number of +1.
- Carbon ([tex]\(\text{C}\)[/tex]): Let's assign it an oxidation number [tex]\(x\)[/tex].
[tex]\[ 2x + 2(+1) = 0 \Rightarrow 2x + 2 = 0 \Rightarrow 2x = -2 \Rightarrow x = -1 \][/tex]
- Hence, the oxidation numbers are:
[tex]\[ C: -1, \; H: +1 \][/tex]
- In [tex]\( \text{O}_2 \)[/tex] (dioxygen):
- Oxygen ([tex]\(\text{O}\)[/tex]) in its elemental form has an oxidation number of 0.
[tex]\[ O: 0 \][/tex]
- In [tex]\(\text{CO}_2\)[/tex] (carbon dioxide):
- Oxygen is generally -2 in compounds.
- Carbon ([tex]\(\text{C}\)[/tex]): Let's assign it an oxidation number [tex]\(y\)[/tex].
[tex]\[ y + 2(-2) = 0 \Rightarrow y - 4 = 0 \Rightarrow y = +4 \][/tex]
- Hence, the oxidation numbers are:
[tex]\[ C: +4, \; O: -2 \][/tex]
- In [tex]\(\text{H}_2\text{O}\)[/tex] (water):
- Hydrogen (H) is +1.
- Oxygen (O) is -2.
[tex]\[ H: +1, \; O: -2 \][/tex]
2. Identify the changes in oxidation numbers to determine what is oxidized and what is reduced:
- Carbon in [tex]\(\text{C}_2\text{H}_2\)[/tex] changes from -1 to +4:
[tex]\(-1 \rightarrow +4\)[/tex], an increase in oxidation number. This indicates that carbon is oxidized.
- Oxygen in [tex]\( \text{O}_2 \)[/tex] changes from 0 to -2:
[tex]\(0 \rightarrow -2\)[/tex], a decrease in oxidation number. This indicates that oxygen is reduced.
3. Define the terms oxidation and reduction:
- Oxidation involves the loss of electrons.
- Reduction involves the gain of electrons.
4. Determine the transfer of electrons:
- Each carbon atom in [tex]\(\text{C}_2\text{H}_2\)[/tex] relinquishes 5 electrons (since going from -1 to +4 means each carbon loses 5 electrons). For two carbon atoms, this amounts to a total of [tex]\(2 \times 5 = 10\)[/tex] electrons lost (oxidation).
- Each oxygen molecule ([tex]\(\text{O}_2\)[/tex]) gains 2 electrons (since going from 0 to -2 means each oxygen atom gains 2 electrons). For five molecules of [tex]\(\text{O}_2\)[/tex], this amounts to a total of [tex]\(5 \times 2 = 10\)[/tex] electrons gained (reduction).
Conclusion:
The combustion reaction of acetylene is a redox reaction because there is a transfer of electrons between the reactants. Carbon in [tex]\(\text{C}_2\text{H}_2\)[/tex] is oxidized (loses electrons), and oxygen in [tex]\( \text{O}_2 \)[/tex] is reduced (gains electrons).
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