How many electrons are transferred when C3H7OH burns in oxygen to form carbon dioxide and water?
I have no idea where to start. Do I need to do half reactions? Can I even do half reactions for this?
Yes, you may do half reactions for this but my first answer is no electrons are transferred. Electron transfer takes place in the formation of ionic bonds and both CO2 and H2O are covalent and the bonds in propane are covalent. Unless there is additional information I would go with zero.
Not arguing with DrBob. Just presenting another way of looking at it using change in oxidation numbers as the yardstick for electrons transfer.
The balanced reaction is:
2C3H7OH + 9O2 --> 6CO2 + 8H2O
Each of the 18 oxygen atoms in O2 changes from an oxidation state of 0 to an oxidation state of -2. A gain of 2 electrons per oxygen atom.
(18)(2)= 36 electrons "gained".
The source of those 36 electrons gained by oxygen are the 6 carbon atoms. So, 36 electrons are transferred from 6 carbon atoms to 18 oxygen atoms. That is 2 electrons per oxygen atom. The oxygen atom in C3H7OH is not included because its oxidation state of -2 does not change.
To determine the number of electrons transferred in a reaction, you can use the method of balancing oxidation-reduction (redox) reactions. In this case, the reaction involves the burning of C3H7OH (isopropyl alcohol) in oxygen to form carbon dioxide (CO2) and water (H2O).
To balance the redox reaction, we divide it into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
First, let's write the unbalanced equation for the reaction:
C3H7OH + O2 → CO2 + H2O
The oxidation half-reaction involves the loss of electrons, and the reduction half-reaction involves the gain of electrons.
The oxidation half-reaction for the carbon in C3H7OH can be represented as follows:
C3H7OH → CO2
The carbon in C3H7OH is oxidized from an oxidation state of -3 in C3H7OH to +4 in CO2. Therefore, the carbon loses 7 electrons.
The reduction half-reaction for the oxygen in O2 can be represented as follows:
O2 → H2O
The oxygen in O2 is reduced from an oxidation state of 0 in O2 to -2 in H2O. Therefore, each oxygen atom gains 2 electrons.
Since each O2 molecule contains 2 oxygen atoms, a total of 4 electrons are gained in the reduction half-reaction.
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 4 and the reduction half-reaction by 7:
4(C3H7OH → CO2) + 7(O2 → H2O) → 4CO2 + 7H2O
So, in the balanced equation, 7 electrons are transferred when C3H7OH burns in oxygen to form carbon dioxide and water.
To determine the number of electrons transferred in a chemical reaction, you can use the concept of balancing oxidation-reduction reactions. In this case, you can indeed use half-reactions to analyze the electron transfer process.
Here's an outline of how to determine the number of electrons transferred when C3H7OH burns in oxygen to form carbon dioxide (CO2) and water (H2O):
1. Write the balanced equation for the reaction:
C3H7OH + O2 -> CO2 + H2O
2. Divide the overall equation into two half-reactions: oxidation and reduction. In the oxidation half-reaction, one reactant loses electrons, and in the reduction half-reaction, another reactant gains those electrons.
3. Determine the oxidation states for each element in both the reactants and the products in the overall equation.
4. Identify the atoms that undergo changes in oxidation states. In this case, we observe the following changes:
- Carbon in C3H7OH goes from an oxidation state of +3 to +4 (carbon is oxidized).
- Oxygen in O2 goes from an oxidation state of 0 to -2 (oxygen is reduced).
5. Write the half-reactions for these changes:
Oxidation: C3H7OH -> CO2 (oxidation of carbon)
Reduction: O2 -> H2O (reduction of oxygen)
6. Balance the number of atoms in each half-reaction, ensuring the conservation of mass.
7. Add the necessary electrons to balance the charges in each half-reaction. The electrons added in the oxidation half-reaction correspond to the number of electrons transferred in the overall reaction.
8. Compare the number of electrons in both half-reactions. The number of electrons in the oxidation half-reaction should be equal to the number in the reduction half-reaction.
By following these steps, you will be able to determine the number of electrons transferred when C3H7OH burns in oxygen to form carbon dioxide and water.