Preview

The lectures in this unit cover balancing & classifying chemical reactions: including redox, acid- base, and precipitation reactions. This lecture covers oxidation-reduction reactions.

Oxidation-Reduction

All chemical reactions occur when the surface of one atom or ion interacts with the surface of another atom or ion. In some cases ions swap partners but they maintain the same charge. This is what happens in acid-base and precipitation reactions. In other reactions atoms gain or lose electrons by trading or sharing in different ways. We call these types of reactions Oxidation- Reduction reactions, or redox reactions.

I. Oxidation Numbers

In order to correctly identify and characterize redox reactions we need a way to track electron ownership and transfer. To do this we calculate the oxidation number of each atom. This is very much like ionic charge except that we use it for covalent compounds too. In this respect it is more of a bookkeeping method than a representation of actual electron ownership, but it works for our purposes. To assign oxidation numbers, first determine if the species is a neutral (elemental), an ion or ionic compound, or a covalent compound. Then use the following rules to calculate the oxidation number for each atom present:

For any neutral element in its standard state (sodium metal (Na), oxygen gas (O2), elemental carbon (C), the oxidation number is zero. For any ion or ions in an ionic compound, the oxidation number is the charge of the ion. The sum of the oxidation numbers in any molecule or polyatomic ion must equal the total overall charge of that molecule or polyatomic ion. For covalent compounds, use the following assignments in exactly this order (the first assignment applies first, so if two of these elements are bound to each other, use the number of the first to calculate the number of the second.) H = +1, F = -1, O = -2, Cl, Br, I = -1.

For example, let’s consider the molecule CaCl2. This molecule is an ionic compound. We need only determine the charge on each ion and that will be the oxidation number. The charge on Ca is +2 because it is a main group element in group 2. The charge on Cl is -1 because it is a main group element in group 7. We know we assigned the charges correctly because the 2+ charges cancel out to make the neutral ionic formula of CaCl2: meaning one Ca ion +2 -1 for every 2 Cl- ions. Therefore the oxidation number for Ca is +2 and the oxidation CaCl number for Cl is -1.

2- If we instead consider the polyatomic ion CO3 we must use the rules for covalent compounds. This is because the C and O inside of the polyatomic ion are bound together with covalent bonds. The rules state that H should be assigned first, then F, then O. Since this formula does not have

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an H or an F we can go ahead and assign the oxygen atom an oxidation number of -2. Now we use the third rule (sum of oxidation numbers must equal overall charge) to determine the oxidation number on carbon. Because each oxygen atom has an oxidation number of -2, and there are three of them, there is a total contribution of -6 from the oxygen atoms. Since the charge on the polyatomic ion is -2 we can see the carbon +4 -2 atom must have an oxidation number of +4 to cancel out 4 of the -6 contributed 2- CO3 from oxygen. That leaves a 2- charge on the overall ion.

II. Oxidation and Reduction within reactions

Once we have assigned individual oxidation numbers we can determine which element was oxidized and which was reduced within a chemical reaction.

Oxidation occurs when something loses electrons Reduction occurs when something gains electrons (the charge is reduced; hence the name)

If the oxidation number became more positive the element must have lost electrons. This is oxidation. If it became more negative the element must have gained electrons. This is reduction.

One way to remember this is to memorize the mnemonic device OiL RiG: where Oxidation is Losing and Reduction is Gaining. Just remember you are Losing and Gaining electrons.

Consider the following equation:

Na(s) + Cl2(g) → NaCl

On the left hand side we have elemental Na and elemental Cl2. They have no charges and thus their oxidation numbers are 0. On the right hand side, the NaCl is an ionic compound and each ion has a charge of +1 and -1, respectively. This is the same value as their oxidation numbers.

Oxidation numbers: 0 0 +1 -1 Na(s) + Cl2(g) → NaCl

This chemical reaction includes electron transfer because the oxidation numbers for the elements on the left are different from the oxidation numbers for the same elements on the right. Since sodium was 0 on the left hand side and is now +1 on the right hand side, we see it lost an electron. Since each chlorine atom was 0 on the left hand side and it is now -1 on the right hand side, we see that each chlorine atom gained an electron.

So for the example of Na and Cl2 combining to form NaCl: which element was oxidized and which was reduced?

Na goes from 0 to +1: it lost an electron and was oxidized Cl goes from 0 to -1: it gained an electron and was reduced

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Oxidation and Reduction always occur together. This is because when you lose an electron it must go somewhere and when you gain an electron it must come from somewhere. The two processes always occur at the same time. When one species is oxidized, another must be reduced and vice versa. For this reason sometimes the species being oxidized is referred to as the reducing agent. By losing electrons (being oxidized) it helps something else be reduced. In the same manner the species being reduced is sometimes referred to as the oxidizing agent. By gaining electrons it allows something else to lose them, or be oxidized. If it helps you to remember this: A real-estate agent does not usually go around buying real estate herself. Instead she helps other people buy real estate. So when you see the word agent remember it is helping something else do that thing.

Let’s practice assigning oxidation numbers and determining which species were oxidized and which were reduced in a few more problems. For the following reactions, tell what the oxidation number is for each atom on both sides of the equation. Then identify which element was oxidized and which was reduced!

Ca + Cl2 → CaCl2 KOH → K2O + H2O SO2 + O2 → SO3

Be sure to try these by yourself before looking at the answers! For the reaction Ca + Cl2 → CaCl2

On the left hand side: calcium is by itself with no charge. This means it is elemental and the oxidation number is 0 (same as the charge). The same goes for Cl2, which has a charge of 0. Each of the two chlorine atoms must also have a charge of 0. It is in its elemental state.

On the right hand side: CaCl2 is an ionic compound where the charge on Ca is +2 and the charge on Cl is -1. That means the oxidation number for Ca is +2 and the oxidation number for Cl is -1.

Oxidation numbers: 0 0 +2 -1

Ca + Cl2 → CaCl2

Ca went from 0 to +2: it lost electrons and was oxidized. Cl went from 0 to -1: it gained electrons and was reduced.

For the reaction KOH → K2O + H2O

On the left hand side: KOH is an ionic compound consisting of potassium (K+) ion and the hydroxide (OH-) ion. The potassium ion has an oxidation number of +1 (same as the charge). Since hydroxide ion is a covalent polyatomic ion we must use the rules for covalent compounds. The hydrogen in the hydroxide ion must have an oxidation number of +1. Since the polyatomic ion as a whole has a charge of -1, the oxygen must then have an oxidation number of -2 (because the sum must equal the charge: -2 + 1 = -1).

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On the right hand side: K2O is an ionic compound where potassium has a charge (and therefore oxidation number) of +1. The oxygen atom in K2O has a charge (and therefore oxidation number) of -2. For H2O, the oxidation number on hydrogen must be +1 and since the molecule is neutral, the oxidation number on oxygen must be -2.

Oxidation numbers: +1 -2 +1 +1 -2 +1 -2 KOH → K2O + H2O

Potassium was +1 on the left and +1 on the right: it did not change. Oxygen was -2 on the left and -2 on the right: it did not change. Hydrogen was +1 on the left and +1 on the right: it did not change.

Because the oxidation numbers did not change, this is not an oxidation-reduction reaction! But we had to compare oxidation numbers to see that clearly.

For the reaction SO2 + O2 → SO3

On the left hand side we have two different compounds: SO2 and O2. For the SO2 we do not know the oxidation number on sulfur, but we follow the rules for assigning oxidation numbers to covalent compounds and find that oxygen has an oxidation number of -2. That means since the molecule is neutral (and the sum of oxidation numbers must equal the charge) sulfur must have an oxidation number of +4. For the molecule O2, oxygen is in its elemental form and the oxidation number is 0.

On the right hand side, we have SO3. We cannot make any assumptions about the oxidation number on sulfur but we can deduce it from oxygen which again has an oxidation number of -2. That means for the neutral molecule SO3, sulfur must have an oxidation number of +6.

Oxidation numbers: +4 -2 0 +6 -2

SO2 + O2 → SO3

Sulfur went from +4 to +6: it was oxidized. The oxygen atoms in SO2 started out and ended up at -2: they did not change The oxygen atoms from elemental oxygen started out at 0 and ended up at -2: they were reduced.

------Take a minute and complete section III of your Chemical Equations worksheet. This has you assign oxidation numbers to several chemical reactions. Can you identify which species are oxidized and which are reduced?

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