In contrast to acid-base reactions, there is distinct class of reaction, known as reduction-oxidation (or redox) reactions, that obey a different pattern. In a redox reaction, polar products are generated from non-polar reactants. You may have run into such reactions already (even if you did not know what they are called!) When iron is left in contact with oxygen (in air) and water, it rusts. The iron is transformed from a hard, non-polar metallic substance (Fe solid) into a powdery substance Fe₂O₃.nH₂O(s). Rusting is mechanistically similar to the reactions that occurs when copper turns green, silver tarnishes (turns black), and perhaps the favorite reaction of chemists everywhere, the explosive reaction that occurs when sodium metal is added to water.

7.1 Reactions
7.2 Types (Acid-Base)
7.3 Lewis Acid-Base
7.4 Nucleo- & Electrophiles
7.5 Oxidation-Reduction
7.6 Energetics

All of these reactions start off with a metal in its elemental form. Pure metals have no charge or permanent unequal distribution of charge (that makes them different, for example, from salts, like NaCl). As the reaction proceeds (here we will use the sodium-water reaction as the focus of our discussion), the metal atoms becomes cations. They lose electrons to another molecule and becomes positively charged. In our example, sodium metal Na(s) reacts with water to form Na+(aq) – a completely different (and much less reactive) species. That is, of course, only half the story. The “lost” electrons from sodium must go somewhere (since, in chemical reactions, we don’t need to consider matter-energy inter-conversion). The question is where?

The overall reaction of sodium and water is: 2Na(s) + 2H_wO(l) ↔ 2Na+(aq) + 2–OH(aq) + H₂(g)
It turns out that this reaction is actually somewhat complicated (although we like it because it is explosive), so we will come back to it shortly.

An easier reaction to analyze is that between sodium and chlorine to form sodium chloride.
2Na(s) + Cl₂(g) ↔ 2NaCl(s)

We have already looked at the structure of ionic compounds (in chapter 4) and have seen that they are best modeled (that is, represented and thought about) by considering NaCl as a three dimensional lattice of alternating positive (Na+) and negative (Cl–) ions. Here is the answer to where the electrons from the sodium atoms go, they are transferred to another reactant in the mixture, namely chlorine atoms. This transfer leads to the breakup of a Cl2 molecule into two Cl– ions. We define a process in which a species loses electrons an oxidation, and the corresponding gain of electrons is called a reduction (even though that seems somewhat counterintuitive).

At the simplest level we could write:
Na → Na+ +e– (an oxidation reaction) and Cl + e– ↔ Cl – ( a reduction reaction).

It turns out that all reactions in which elements react with each other to form compounds are redox reactions. For example: the reaction of molecular hydrogen and molecular oxygen is also a redox reaction:
2H₂(g) + O₂(g) ↔ 2H₂O(l)

The problem here is that there is no obvious transfer of electrons. Neither is there an obvious reason why these two elements should react in the first place, since neither of them has any charge polarity that might lead to an initial interaction. That being said, there is no doubt that H₂ and O₂ react, in fact they (like sodium) and water) react explosively. When we look a little more closely at the reaction, we can see that there is a shift in electron density on individual atoms as they move from being reactants to products. The reactants contain only pure covalent (H–H and O–O) bonds, while in the product (H₂O) the bonds are polarized: Hδ+ and Oδ– (recall that oxygen is a highly electronegative atom because of its high effective nuclear charge). There has been a shift in overall electron density towards the oxygen. While this is a bit more subtle than the NaCl case, the oxygen has gained some “extra” electron density, and so it has been reduced, but not by a whole electron, just a part of one. Similarly, the hydrogen has been oxidized, but again, not by a whole electron, but by part of one. This can be confusing, since electrons are elementary particles, they do not have parts - we are really talking about where the electron spends most of its time, that is the electron density. In order to keep this straight, chemists have developed a system to keep track of the losses and gains in electron density, they use oxidation numbers.

Oxidation: states and numbers

Now we may seem to be moving into the area of 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.

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, in water the oxygen is the more electronegative atom. We therefore imagine that the bonding electrons are on oxygen, this means that the hydrogen atoms have no electrons (and so has a +1 charge). The oxidation number of H (in water) is +1, while oxygen, while two (imagined) extra electrons as a –2 charge – its oxidation number is –2.
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 just 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 to +1 in the products, and the oxygen goes from zero to –2. When oxidation numbers change during a reaction, the reaction is a redox reaction.
Now back to sodium and water reaction. Let us see if we can spot what which reduced (we know the sodium gets oxidized, because it loses an electron).

Obviously it must be the oxygen or hydrogen (or could it be both?)
2Na(s) + 2H₂O(l) → 2Na+(aq) + 2–OH(aq) + H₂(g)

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 meaning that it started at +1 and ends at zero: it is these two hydrogen atoms that have been reduced!

The historical origins of the term oxidation involve reaction with oxygen, for example in combustion reactions, the simplest of which can be written as: CH₄(g)+ O₂(g) → CO₂(g) + H₂O(g)

Oxidation reactions like this provide major sources of energy, not only by burning fuel (natural gas, gasoline, coal etc), but also in biological systems where carbon containing molecules, such as sugars and lipids, react with molecular oxygen, to form compounds with very stable bonds (CO₂ and H₂O) and releasing energy that can be used to break bonds and rearrange molecules.

7.1 Reactions
7.2 Types (Acid-Base)
7.3 Lewis Acid-Base
7.4 Nucleo- & Electrophiles
7.5 Oxidation-Reduction
7.6 Energetics

Question to answer:

• For the reaction CH₄(g)+ O₂(g) → CO₂(g) + H₂O(g), which atoms are oxidized and which 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)
  H+(aq) + –OH(aq) → H₂O(l)
  

Questions to ponder:

• What would happen if you could separate out the oxidation reaction (where electrons are lost) and the reduction reaction (where electrons are gained). Would that be possible?
• What if you separate the two reactions but join them by an electrical connection - what do you think would happen? (See chapter 10 for the answer)