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?
2 answers
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.
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.
1 answer