Lecture 7

Oxidation-Reduction Reactions

 

In the Acid-Base reactions, there was a transfer of a proton (H+) between chemical species. In Oxidation-Reduction reactions, there is a transfer of electrons (e-) between species. The reaction of solid zinc with acid is a good example. The figure, below, shows a strip of solid zinc in a beaker of acid. Bubbles of gas are being given off during the reaction. What is happening?

 

 

Zn (s) + H+ (aq) products

 

The gas being given off is hydrogen, H2, one of the products. The second product, which is not readily visible, is Zn2+ (aq).

 

Zn (s) + H+ (aq) H2 (g) + Zn2+ (aq)

 

This needs to be balanced:

 

Zn (s) + 2 H+ (aq) H2 (g) + Zn2+ (aq)

 

The hydrogen in this reaction is going from H+ (aq) to H2 (g). It has a +1 charge in the reactants and is neutral (no charge) in the products. So, as a reactant, H+ possesses no electrons. As a neutral product H2 must possess 2 protons and 2 electrons. Each hydrogen atom has gained one electron. Where did the electrons come from? Zn (s) is neutral, so each zinc atom in the reactants must possess 30 protons and 30 electrons. During this reaction, the Zn(s) is consumed, and Zn2+ (aq) ions are formed. In Zn2+, each zinc atom still possesses 30 protons, but only 28 electrons. Each Zn atom has given up two electrons. Electrons have been transferred from zinc to hydrogen.

 

Zinc lost electrons. It was oxidized.

Hydrogen gained electrons. It was reduced.

 

Oval Callout: Ger

 

                                                                                                Lose Gain

Electrons Electrons

Oxidation Reduction

 

It is possible to keep track of the electrons in these reactions by using oxidation numbers. These are the number of charges that an atom in a molecule would have if the electrons that it is sharing were transferred completely toward or away from it. This sounds more complicated than it is.

 

Oxidation Numbers

 

There is a set of rules, but only two of them are really necessary, most of the time.

 

1. The oxidation number of an element in its elemental form is zero. Examples of this are N2 (g), O2 (g), Na (s), Cl2 (g), etc. The elemental atomic species, like Na, obviously have no charges, so their oxidation numbers are zero. In diatomic elements, like O2, there is an even sharing of the electrons, since both of the bonded atoms are the same element and have exactly the same tendency to lose or gain electrons.

 

2. The oxidation number of a monatomic ion is exactly the same as its charge. So, Group IA ions will all have an oxidation number of +1, since they all lose one electron. Group IIA ions will all have an oxidation number of +2. Aluminum ions only exist as Al3+ and will have an oxidation number of +3.

 

3. The oxidation state of oxygen, in most compounds is -2, i.e., it tends to pull 2 shared electrons toward itself. The exceptions are H2O2, hydrogen peroxide and O22-, peroxide, when it is -1. In O2 it is zero (see rule 1.).

 

4. The oxidation state of hydrogen is almost always +1. The exceptions are H2 (oxidation number = zero , rule 1.) and when hydrogen is bonded to metals in binary compounds, like LiH, when it is -1.

 

5. Fluorine is always -1. The other halogens are -1, except when bonded to oxygen (rule 3. gives oxygen an oxidation number of -2, making halogens bonded to oxygen positive).

 

6. Oxidation number of the atoms in a compound must add up to the total charge on that molecule or ion.

 

Let's work a few examples:

 

Assign oxidation numbers to all of the elements in the following compounds:

 

a) Lithium oxide Li2O O = -2. The molecule is neutral, so the lithium atoms must have a total oxidation number of +2.

Since there are two of them, each one must be +1.

Li = +1

b) Nitric acid HNO3 O = -2. There are three oxygen atoms, so they contribute -6.

The molecule is neutral, so the H and N must contribute +6

H = +1 hydrogen + nitrogen = +6. H = +1, so N = +5.

N = +5

 

What is the oxidation number of sulfur in each of the following compounds?

 

a) hydrogen sulfide H2S H = +1. Overall, the molecule is neutral. The hydrogen atoms contribute +2.

S = -2

 

b) sulfate ion SO42- O = -2. There are four oxygen atoms, contributing -8.

The overall charge is -2.

-8 + S = -2.

S = +6

In the sulfur examples, you can see how the oxidation state of an atom in a molecule depends very strongly on what it is bonded to.

 

What is the oxidation number of the elements in the following compounds?

 

a) sodium hydride NaH In this compound, hydrogen is bonded to a metal in a binary compound (rule 4.) In this case, the hydrogen atom contributes -1.

H = -1 The compound is neutral, so the sodium atom must contribute +1

Na = +1

 

b) barium peroxide BaO2 In this compound, we have a O-O bond (rule 3.) In this case, each oxygen atom contributes -1

O = -1 The compound is neutral, so, the barium atom must contribute +2

Ba = +2

 

Let's look at another oxidation-reduction reaction, and assign oxidation numbers to all of the atoms involved. This is the reaction involved in Ni-Cd batteries.

 

Cd(s) + NiO2 (s) + 2H2O (l) Cd(OH)2 (s) + Ni(OH)2 (s)

 

Cd(s) = 0 This is elemental cadmium (rule 1).

 

 

NiO2 The oxygen atoms each contribute -2 (this is nickel dioxide, not nickel peroxide)

O = -2 The compound is neutral, and each oxygen atom contributes 2-. So, the nickel atom must contribute +4

Ni - +4

 

H2O The oxygen atom contributes -2.

O = -2 The molecule is neutral, so the two hydrogen atoms must contribute +2

H = +1

 

Cd(OH)2 The oxygens contribute -2

O = -2 The hydrogens contribute +1

H = +1 The overall compound is neutral, so (2 x -2) + (2 x +1) + Cd = 0

Cd = +2

 

Ni(OH)2 The oxygens contribute -2

O = -2 The hydrogens contribute +1

H = +1 The overall compound is neutral, so (2 x -2) + (2 x +1) + Ni = 0

Ni = +2

 

We can now assign these values to the elements in the reaction:

 

Cd(s) + NiO2 (s) + 2H2O (l) Cd(OH)2 (s) + Ni(OH)2 (s)

0 +4 -2 +1 -2 +2 -2 +1 +2 -2 +1

 

None of the oxygen (-2) or hydrogen atoms (+1) change oxidation number.

Cadmium goes from 0 to +2. Its oxidation number increases. It is oxidized.

Nickel goes from +4 to +2. Its oxidation number decreases. It is reduced.