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How do you do a balanced redox equation of methane and oxygen gas?
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+ Chemical reactions
+ Natural gas
+ Methane
+ Oxygen
+ Chemistry
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Matthew Althouse
How do you do a balanced redox equation of methane and oxygen gas?
To clarify, Oxidation is the loss of electrons, not the gain of oxygen. Reduction is the gain of electrons, not the loss of oxygen. When something is oxidized, that means it is losing electrons. That usually happens in a reaction with some element X and oxygen in which X loses electrons and oxygen gains those electrons. Thus X is being oxidized. If you don’t understand this, I can clarify in the comment section.
When an element is oxidized the oxidation number increases. When it is reduced, the oxidation number decreases (decreases, reduces, get it?).
In this case, 2Cu + O2 -> 2CuO.
Elements in elemental form (not ions and not in compounds) have an oxidation number of 0. Elements in Diatomic form (N2, O2, F2, Cl2, etc.) have oxidation numbers of 0. Thus the oxidation number of Cu = 0 and oxidation number for O in O2 = 0.
So our starting oxidation numbers are 0 for both Cu and O.
Now, let’s calculate the oxidation numbers in CuO. O in most compounds has an oxidation number of -2. The exception to this are peroxides. H2O2, Na2O2, etc. Where there’s two oxygen and two of some other element. In peroxides, O has an oxidation number of -1. But CuO is not a peroxide, thus O has an oxidation number of -2. And since the overall compound is neutral (the compound is not a polyatomic ion), All the oxidation numbers must add up to 0. So 0 = X + -2. X is 2. The Cu has an oxidation number of 2.
So our ending oxidation numbers are 2 for Cu and -2 for oxygen.
Our oxidation number went from 0 -> 2 for Cu.
Our oxidation number went from 0 -> -2 for O.
Thus Cu is being oxidized (the oxidation number increased) and O is being reduced (the oxidation number decreased). Since something is being oxidized and something is being reduced, it IS a redox reaction.
To clarify, Oxidation is the loss of electrons, not the gain of oxygen. Reduction is the gain of electrons, not the loss of oxygen. When something is oxidized, that means it is losing electrons. That usually happens in a reaction with some element X and oxygen in which X loses electrons and oxygen gains those electrons. Thus X is being oxidized. If you don’t understand this, I can clarify in the comment section.
When an element is oxidized the oxidation number increases. When it is reduced, the oxidation number decreases (decreases, reduces, get it?).
In this case, 2Cu + O2 -> 2CuO.
Elements in elemental form (not ions and not in compounds) have an oxidation number of 0. Elements in Diatomic form (N2, O2, F2, Cl2, etc.) have oxidation numbers of 0. Thus the oxidation number of Cu = 0 and oxidation number for O in O2 = 0.
So our starting oxidation numbers are 0 for both Cu and O.
Now, let’s calculate the oxidation numbers in CuO. O in most compounds has an oxidation number of -2. The exception to this are peroxides. H2O2, Na2O2, etc. Where there’s two oxygen and two of some other element. In peroxides, O has an oxidation number of -1. But CuO is not a peroxide, thus O has an oxidation number of -2. And since the overall compound is neutral (the compound is not a polyatomic ion), All the oxidation numbers must add up to 0. So 0 = X + -2. X is 2. The Cu has an oxidation number of 2.
So our ending oxidation numbers are 2 for Cu and -2 for oxygen.
Our oxidation number went from 0 -> 2 for Cu.
Our oxidation number went from 0 -> -2 for O.
Thus Cu is being oxidized (the oxidation number increased) and O is being reduced (the oxidation number decreased). Since something is being oxidized and something is being reduced, it IS a redox reaction.
The complete combustion of methane. Despite combustion being a redox reaction, the equation is most easily balanced by inspection.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
You can go through the half-reaction method as you might for other redox reaction, but there is really not point in using that to balance this equation. Yet, if you did you would pretend that it is an acidic reaction and use H+ and H2O.
The complete combustion of methane. Despite combustion being a redox reaction, the equation is most easily balanced by inspection.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
You can go through the half-reaction method as you might for other redox reaction, but there is really not point in using that to balance this equation. Yet, if you did you would pretend that it is an acidic reaction and use H+ and H2O.
To clarify, Oxidation is the loss of electrons, not the gain of oxygen. Reduction is the gain of electrons, not the loss of oxygen. When something is oxidized, that means it is losing electrons. That usually happens in a reaction with some element X and oxygen in which X loses electrons and oxygen gains those electrons. Thus X is being oxidized. If you don’t understand this, I can clarify in the comment section.
When an element is oxidized the oxidation number increases. When it is reduced, the oxidation number decreases (decreases, reduces, get it?).
In this case, 2Cu + O2 -> 2CuO.
Elements in elemental form (not ions and not in compounds) have an oxidation number of 0. Elements in Diatomic form (N2, O2, F2, Cl2, etc.) have oxidation numbers of 0. Thus the oxidation number of Cu = 0 and oxidation number for O in O2 = 0.
So our starting oxidation numbers are 0 for both Cu and O.
Now, let’s calculate the oxidation numbers in CuO. O in most compounds has an oxidation number of -2. The exception to this are peroxides. H2O2, Na2O2, etc. Where there’s two oxygen and two of some other element. In peroxides, O has an oxidation number of -1. But CuO is not a peroxide, thus O has an oxidation number of -2. And since the overall compound is neutral (the compound is not a polyatomic ion), All the oxidation numbers must add up to 0. So 0 = X + -2. X is 2. The Cu has an oxidation number of 2.
So our ending oxidation numbers are 2 for Cu and -2 for oxygen.
Our oxidation number went from 0 -> 2 for Cu.
Our oxidation number went from 0 -> -2 for O.
Thus Cu is being oxidized (the oxidation number increased) and O is being reduced (the oxidation number decreased). Since something is being oxidized and something is being reduced, it IS a redox reaction.
To clarify, Oxidation is the loss of electrons, not the gain of oxygen. Reduction is the gain of electrons, not the loss of oxygen. When something is oxidized, that means it is losing electrons. That usually happens in a reaction with some element X and oxygen in which X loses electrons and oxygen gains those electrons. Thus X is being oxidized. If you don’t understand this, I can clarify in the comment section.
When an element is oxidized the oxidation number increases. When it is reduced, the oxidation number decreases (decreases, reduces, get it?).
In this case, 2Cu + O2 -> 2CuO.
Elements in elemental form (not ions and not in compounds) have an oxidation number of 0. Elements in Diatomic form (N2, O2, F2, Cl2, etc.) have oxidation numbers of 0. Thus the oxidation number of Cu = 0 and oxidation number for O in O2 = 0.
So our starting oxidation numbers are 0 for both Cu and O.
Now, let’s calculate the oxidation numbers in CuO. O in most compounds has an oxidation number of -2. The exception to this are peroxides. H2O2, Na2O2, etc. Where there’s two oxygen and two of some other element. In peroxides, O has an oxidation number of -1. But CuO is not a peroxide, thus O has an oxidation number of -2. And since the overall compound is neutral (the compound is not a polyatomic ion), All the oxidation numbers must add up to 0. So 0 = X + -2. X is 2. The Cu has an oxidation number of 2.
So our ending oxidation numbers are 2 for Cu and -2 for oxygen.
Our oxidation number went from 0 -> 2 for Cu.
Our oxidation number went from 0 -> -2 for O.
Thus Cu is being oxidized (the oxidation number increased) and O is being reduced (the oxidation number decreased). Since something is being oxidized and something is being reduced, it IS a redox reaction.
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The complete combustion of methane. Despite combustion being a redox reaction, the equation is most easily balanced by inspection.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
You can go through the half-reaction method as you might for other redox reaction, but there is really not point in using that to balance this equation. Yet, if you did you would pretend that it is an acidic reaction and use H+ and H2O.
CH4 + 2H2O → CO2 + 8H+ + 8e- …….. oxidation half-reaction
2(O2 + 4H+ + 4e- → 2H2O) ……………… reduction half-reaction
————————————————-
CH4 + 2O2 + 2H2O + 8H+ → CO2 + 8H+ + 4H2O
simplify
CH4 + 2O2 → CO2 + 2H2O
The complete combustion of methane. Despite combustion being a redox reaction, the equation is most easily balanced by inspection.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
You can go through the half-reaction method as you might for other redox reaction, but there is really not point in using that to balance this equation. Yet, if you did you would pretend that it is an acidic reaction and use H+ and H2O.
CH4 + 2H2O → CO2 + 8H+ + 8e- …….. oxidation half-reaction
2(O2 + 4H+ + 4e- → 2H2O) ……………… reduction half-reaction
————————————————-
CH4 + 2O2 + 2H2O + 8H+ → CO2 + 8H+ + 4H2O
simplify
CH4 + 2O2 → CO2 + 2H2O
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