Hi Tanya, you are right, balancing organic oxidation-reduction reactions is less clear than doing inorganic reactions in aqueous solution, where real ions and charges are present. Organic reactions such as the combustion of propane involve covalent, molecular reactants and products, and not being in aqueous solution means we can't use the familiar method of balancing the half-reactions using H+ or OH- as we do with inorganic reactions in aqueous solution. Nevertheless, we can still use that standard method of balancing half-reactions here also, with a little "gimmicking" to make it work.
For review, here are the steps we normally take to balance a redox reaction in aqueous solution, using the method of "half-reactions" (which I hope you are familiar with! - if not, let me know):
1. Determine which reactant is oxidized and which is reduced by checking the changes in oxidation numbers in the reactants and products.
2. Write the unbalanced oxidation and reduction half-reactions.
3. Balance the elements other than H and O in each half-reaction by inspection.
4. Balance the oxygens by adding H2Os to the appropriate side of each half-reaction.
5. Balance the hydrogens by adding H+ ion to the appropriate side of each half-reaction.
6. Balance the charge by adding electrons (e-) to the appropriate side of each half-reaction. The electrons will go on the right (product) side of the oxidation half-reaction and on the left (reactant) side of the reduction half-reaction.
7. Multiply one or both half-reactions through by an integer so that the the number of electrons are the same in each half-reaction.
8. Now add the two half-reactions together. The electrons lost and gained are the same and "cancel", leaving the balanced reaction in acidic aqueous solution. (To convert to basic solution, we can "neutralize" the H+ ions by adding the appropriate number of OH- ions (forming H2O) to both sides of the equation (add the same number of OH- ions to each side now, to keep the reaction balanced)).
OK! We can use a similar approach for organic reactions. There are different versions favored by different authors, but here is the general idea, using the propane combustion as an example.
C3H8 + O2 --> CO2 + H2O
1. In this reaction, propane is oxidized and O2 is reduced, but we need to be more specific about what is oxidized. The elements that actually get oxidized are the carbons. In propane, the oxidation states of the hydrogens are the usual +1. If each H is +1 in the formula C3H8, the average oxidation state of the carbons is -8/3, but more accurately, the oxidation states of the end carbons are -3 each and the oxidation state of the central carbon is -2.
The carbon-containg product is CO2 with the carbon in an oxidation state of +4, so we know the carbons are becoming oxidized. The hydrogens do not change their oxidation state in this reaction.
Oxygen goes from an oxidation state of zero in O2 to -2 in CO2 and H2O, so the oxygens are getting reduced by gaining 2 electrons each.
2. Therefore we have the unbalanced half-reactions:
Oxidation half-rxn: C3H8 --> CO2
Reduction half-rxn: O2 --> H2O
3. Balance the elements other than H and O:
Oxidation half-rxn: C3H8 --> 3CO2
Reduction half-rxn: O2 --> H2O
4 & 5. We will "balance" the hydrogens and oxygens by adding hydrogen and oxygen atoms ("H" and "O") to the appropriate sides. These are more like temporary "filler" labels used in order to balance the reaction; they will not appear in the overall reaction. Realistically however, hydrogen and oxygen atoms are likely to be intermediates, so perhaps this idea is not so outrageous. Note for the reduction reaction, it is simpler to simply balance the oxygens by putting a coefficient of 2 in from of the H2O, instead of adding an "O". There are various modifications of this approach out there.
Oxidation half-rxn: C3H8 + 6O --> 3CO2 + 8H
Reduction half-rxn: O2 + 4H --> 2H2O
6. Balancing the charge. Here, we will only look at the "charges", which are not really charges but oxidation numbers, of the atoms actually being oxidized and reduced in each half-reaction, the carbons and oxygens, respectively.
In the oxidation half-reaction, we have 3 carbons with an average oxidation state of -8/3 for a total "charge" of the carbons of -8 on the left side. On the right side, we have 3 carbons with oxidation states of +4 each, for a total of +12. Therefore, we need to add 20 electrons to the right-hand side of the oxidation half-reaction to have a total "charge" of -8 on each side, with respect to the oxidations states of the carbons.
Oxidation half-rxn: C3H8 + 6O --> 3CO2 + 8H + 20e-
In the reduction half-reaction, we have O2 with each oxygen in a zero oxidation state on the left, and two oxygens at -2 each in the water molecules on the right side, for a total of -4. Therefore, we need to add 4 electrons on the left side to get a -4 sum of the "charge" on each side, with respect to the oxidation states of the oxygens.
Reduction half-rxn: O2 + 4H + 4e- --> 2H2O
7. Balancing electron lost and gained. We need to multiply the reduction half-reaction through by 5 to have 20 electrons lost and gained:
Oxidation half-rxn: C3H8 + 6O --> 3CO2 + 8H + 20e-
Reduction half-rxn: 5O2 + 20H + 20e- -> 10H2O
8. Now (finally!) we can add our two half-reactions together:
C3H8 + 5O2 + 20H + 6O + 20e- --> 3CO2 + 10H2O + 8H + 20e-
which simplifies to
C3H8 + 5O2 + 12H + 6O --> 3CO2 + 10H2O
Combining 12H and 6O to make 6H2Os, we have
C3H8 + 5O2 + 6H2O --> 3CO2 + 10H2O
Which finally simplifies to
C3H8 + 5O2 --> 3CO2 + 4H2O
Which, in this case, is just what you get by trial and error! This overall process can be "steamlined" somewhat, and as I mentioned, there are different modifications out there depending on the author's preference, but the above illustrates the general principles involved. Hope it made sense!
Steve
P.S. I forgot that you were combusting 1-propanol, not propane, sorry about that! The average oxidation state of carbon in C3H8O will be -2 this time, if the hydrogens are their usual +1 and oxygen its usual -2. Here are the unbalanced half-reactions to start with:
C3H8O + O2 --> CO2 + H2O
Oxidation half-rxn: C3H8O --> 3CO2
Reduction half-rxn: O2 --> 2H2O
Balancing the oxygens and hydrogens with H and O as needed gives
Oxidation half-rxn: C3H8O + 5O --> 3CO2 + 8H
Reduction half-rxn: O2 + 4H --> 2H2O
Adding the electrons:
Oxidation half-rxn: C3H8O + 5O --> 3CO2 + 8H + 18e-
Reduction half-rxn: O2 + 4H + 4e- --> 2H2O
Multiplying the oxidation half-reaction through by 2 and the reduction half-reaction through by 9 gives
Oxidation half-rxn: 2C3H8O + 10 O --> 6CO2 + 16H + 36e-
Reduction half-rxn: 9O2 + 36H + 36e- --> 18H2O
Adding the two reactions together gives
2C3H8O + 9O2 + 20H + 10 O + 36e- --> 6CO2 + 18H2O + 36e-
and further combining the Hs and Os to more waters gives and cancelling the 36 electrons lost and gained;
2C3H8O + 9O2 + 10H2O --> 6CO2 + 18H2O
gives finally,
2C3H8O + 9O2 --> 6CO2 + 8H2O
Whew!! I didn't look for any yet, but there are probably some Internet sites out there that give a more simplified procedure for balancing organic reactions like these. The method may vary, but the principle will be the same.
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It still requires that we first figure out what the oxidation and reduction half-reactions are. In the examples accompanying the procedure, the oxidation half-reaction is the "organic" half-reaction, and the reduction half-reaction is the "inorganic" half-reaction, because, as is commonly the case, an orgnic molecule is oxidized with inorganic oxidants such as MnO4-, Cr2O72-, or IO4-. However, it looks like with this procedure, you really can write the two half-reactions based merely on one being "organic" and the other "inorganic", as the author has some reduction examples also using BH4-, Zn, and Fe as the inorganic reductant.
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