Now, we may seem to be deploying more arcane terms designed to confuse the non- chemist, but in fact, oxidation numbers (or oxidation states) can be relatively easy to grasp as long as you remember a few basic principles:145
For an ion, the charge is the oxidation number. The oxidation number of Na+ is +1, the oxidation number of the oxide ion (O2–) is –2.
For elements that are covalently bonded to a different element, we imagine that all the electrons in the bond are moved to the most electronegative atom to make it charged. As an example, the oxygen in water is the more electronegative atom. Therefore, we imagine that the bonding electrons are on oxygen and that the hydrogen atoms have no electrons (rather, they have a +1 charge). The oxidation number of H (in water) is +1, whereas in oxygen it is -2, because of the -2 charge of the two imagined extra electrons that came from the bond.
- Elements always have an oxidation number of zero (because all of the atoms in a pure element are the same, so none of the bonds are polar).
Remember this is just a way to keep track of the electrons. Oxidation numbers are not real; they are simply a helpful device. It is also important to remember that the oxidation number (or state) of an atom is dependent upon its molecular context. The trick to spotting a redox reaction is to see if the oxidation number of an atom changes from reactants to products. In the reaction:
2H2(g) + O2(g) ⇄ 2H2O(l)
H changes from zero in the reactants (H2) to +1 in the products (H2O), and the oxygen goes from zero (O2) to –2 (H2O). When oxidation numbers change during a reaction, the reaction is a redox reaction.
Now lets look at the reaction sodium and water, which is a bit more complicated to see if we can spot what is oxidized and what is reduced.
2Na(s) + 2H2O(l) ⇄ 2Na+(aq) + 2–OH(aq) + H2(g)
It is relatively easy to see that the sodium gets oxidized, because it loses an electron, going from Na to Na+. But which species gets reduced? Is it the oxygen or the hydrogen? Or could it be both? If we check for changes in oxidation state, the oxygen in water starts at –2 and in hydroxide (–OH) it is still –2 (it has not been reduced or oxidized). If we check the hydrogens, we see two distinct fates. One of the hydrogen atoms stays bonded to the oxygen atom (in hydroxide); it starts at +1 and stays there. However, the other type ends up bonded to another hydrogen atom; it starts at +1 and ends at zero. It is these latter two hydrogen atoms that have been reduced!
Historically, the term oxidation has denoted a reaction with oxygen. For example, in simple combustion reactions:
CH4(g)+ O2(g) ⇄ CO2(g) + H2O(g)
Oxidation reactions like this provide major sources of energy, in the burning of fuel (natural gas, gasoline, coal, etc.) and also in biological systems. In the latter, carbons containing molecules such as sugars and lipids react with molecular oxygen to form compounds with very stable bonds (CO2 and H2O), releasing energy that can be used to break bonds and rearrange molecules. In a similar vein the original meaning of reduction was reaction with hydrogen, for example acetic acid can be reduced to ethanol by reacting with hydrogen:
CH3CO2H + H2(g) ⇄ CH3CH2OH
What is important to note is that, there cannot be an oxidation without a reduction – and vice-versa Just like there can be no acid without a base.
Questions to Answer
- For the reaction CH4(g)+ O2(g) ⇄ CO2(g) + H2O(g), which atoms are oxidized and which are reduced?
- For the reaction CH3CO2H + H2(g) ⇄ CH3CH2OH which atoms are oxidized and which are reduced?
- Write an explanation to a friend who has no chemistry background to explain the difference between these two reactions that give the same product:
- 2H2(g) + O2(g) ⇄ 2H2O(l) and H+(aq) + –OH(aq) ⇄ H2O(l)
Questions for Later
- Is it possible to separate out the oxidation reaction (where electrons are lost) and the reduction reaction (where electrons are gained)? What would happen?
- What if you separate the two reactions but join them by an electrical connection? What do you think would happen?