C3H8 + 5O2 --> 3CO2 + 4H2O what are the oxidation #'s of each? What is being oxidized? reduced? Please explain
To determine the oxidation numbers of each element in the given chemical equation, we first need to understand the rules for assigning oxidation numbers.
1. The oxidation number of an atom in its elemental form is zero. Therefore, the oxidation number of O₂ and H₂O is zero.
2. The sum of the oxidation numbers in a neutral compound is zero. In this equation, CO₂ is a neutral compound; therefore, the sum of the oxidation numbers of C and O in CO₂ is zero.
3. Group 1 elements, such as H, have an oxidation number of +1 when combined with nonmetals, and -1 when combined with metals. In this case, H₂O is a neutral compound, so H has an oxidation number of +1.
Now, let's assign oxidation numbers to each element:
C₃H₈:
Since C₃H₈ is a hydrocarbon, hydrogen always has an oxidation number of +1.
Hence, H in C₃H₈ has an oxidation number of +1.
To find the oxidation number of carbon, we know that hydrogen normally has an oxidation number of +1. Since there are 8 hydrogen atoms in C₃H₈, their total oxidation number is (+1) × 8 = +8.
The sum of all oxidation numbers in a compound should be equivalent to the compound's overall charge. Since C₃H₈ is a neutral molecule, the total oxidation number should be zero.
So, the oxidation number of carbon in C₃H₈ is -8.
2. Now let's consider the products of the reaction:
CO₂:
The sum of the oxidation numbers in CO₂ should be zero since it is a neutral compound.
We know that O normally has an oxidation number of -2. Carbon should have an oxidation number to balance out the total:
Let the oxidation number of carbon in CO₂ be x.
Therefore, the equation would be: x + (2 × -2) = 0.
Solving for x, we find that the oxidation number of carbon in CO₂ is +4.
H₂O:
As mentioned earlier, hydrogen in H₂O has an oxidation number of +1.
Since H₂O is a neutral compound, the sum of the oxidation numbers in it should be zero.
Considering that oxygen normally has an oxidation number of -2, let's assign an oxidation number of x to hydrogen.
The equation becomes: 2x + (-2) = 0.
By solving for x, we find the oxidation number of hydrogen in H₂O is -1.
Now let's determine what is being oxidized and reduced:
In the given equation: C₃H₈ + 5O₂ → 3CO₂ +4H₂O
Carbon (C) in C₃H₈ is going from an oxidation number of -8 to an oxidation number of +4 in CO₂. Hence, carbon is being oxidized.
Oxygen (O) in O₂ is going from an oxidation number of 0 to an oxidation number of -2 in CO₂ and H₂O. Thus, oxygen is being reduced.
To recap:
Oxidation numbers:
- Carbon (C) in C₃H₈: -8 → +4
- Hydrogen (H) in C₃H₈: +1
- Carbon (C) in CO₂: +4
- Oxygen (O) in CO₂: -2
- Hydrogen (H) in H₂O: -1
- Oxygen (O) in H₂O: -2
Oxidized: Carbon (C) in C₃H₈
Reduced: Oxygen (O) in O₂
To determine the oxidation numbers, we need to assign an oxidation number to each element in the equation. The oxidation number is a number that indicates the electron distribution in a compound or molecule.
In the given balanced equation: C3H8 + 5O2 → 3CO2 + 4H2O
Let's determine the oxidation numbers for each element.
Starting with C3H8 (propane):
Since hydrogen (H) is in compound form, its oxidation number is +1.
Since the overall molecule is neutral, the sum of the oxidation numbers of the carbon atoms must be zero. Therefore, each carbon will have an oxidation number of -2.
Moving on to CO2 (carbon dioxide):
Since oxygen (O) is in compound form, its oxidation number is -2. The carbon atom will have an oxidation number of +4 because it has two oxygen atoms bonded to it.
Finally, H2O (water):
Since oxygen (O) has an oxidation number of -2, the hydrogen (H) atoms will have an oxidation number of +1 because water is a neutral molecule.
Now let's identify which element is being oxidized and which is being reduced.
In this reaction, the carbon in C3H8 is being oxidized from an oxidation number of -2 to +4 in CO2. So, carbon is being oxidized. (Oxidation involves an increase in oxidation number).
On the other hand, in C3H8, hydrogen is being reduced from an oxidation number of +1 to 0 in H2O. So, hydrogen is being reduced. (Reduction involves a decrease in oxidation number).
To summarize:
- The oxidation number of carbon is being increased from -2 to +4, so carbon is being oxidized.
- The oxidation number of hydrogen is being decreased from +1 to 0, so hydrogen is being reduced.
You need to know how to do this; therefore, please tell me what trouble you're having. I expect you know more about this than you think.
Here is a site to show you how to do oxidation states.
http://www.chemteam.info/Redox/Redox-Rules.html