Topic 10: Oxidation and Reduction
10.1 Oxidation and Reduction
10.1.1. Define oxidation and reduction in terms of electron loss and gain.
Introduce the concept of the half-equation.
Oxidation-reduction (redox) reactions involve a transfer of electrons from the reducing
agent to the oxidizing agent. So, oxidation involves a loss of electrons (an increase in the
electron number) and reduction involves a gain of electrons (a decrease in oxidation
number.) The best and most basic example of a redox reaction is the formation of table
salt:
2Na(s) + Cl2(g) 2NaCl(s)
In this reaction, sodium is literally giving an electron to chlorine, so sodium is being
oxidized and chlorine is being reduced. Ionic reactions are literal obvious redox
reactions; however covalent bonds can also be considered redox reactions. Often times,
when working with redox reactions, we split them up into two different reactions, the
reduction reaction and the oxidation reaction. So, let’s use for example the reaction:
Ce4+(aq) + Sn2+(aq)  Ce3+(aq) + Sn4+(aq)
To work with this reaction (usually in an effort to balance it) we often times split it up.
The oxidation reaction is:
Sn2+(aq)  Sn4+(aq)
And the reduction reaction is:
Ce4+(aq)  Ce3+(aq)
To learn how to balance these kinds of reactions, see 19.1.
10.1.2. Calculate the oxidation number of an element in a compound.
Oxidation numbers should be shown by a sign (+ or -) and a number, eg +7 for Mn in
KMnO4.
The concept of oxidation states (also called oxidation numbers) provides a way to keep
track of electrons in oxidation-reduction reactions, particularly redox reactions involving
covalent bonds. We define the oxidation states (or oxidation numbers) of the atoms in a
covalent compound as the imaginary charges the atoms would have if the shared
electrons were divided equally between identical atoms bonded to each other or, for
different atoms, were all assigned to the atom in each bond that has the greater attraction
for electrons. Of course, for ionic compounds containing monatomic ions, the oxidation
states of the ions are equal to the ion charges. These considerations lead to a series of
rules for assigning oxidation states:
1. The oxidation state of an atom in an element is 0. For example, the oxidation
state of each atom in the substances Na(s), O2(g), O3(g), and Hg(l) is 0.
2. The oxidation state of a monatomic ion is the same as it charge. For example, the
oxidation state of the Na+ ion is +1 and of the Cl- ion is -1.
3. Oxygen is assigned an oxidation state of 02 in its covalent compounds, such as
CO, CO2, SO2, and SO3. An exception to this rule occurs in peroxides
(compounds containing the O22- group), where each oxygen is assigned an
oxidation state of -1. The best-known example of a peroxide is hydrogen
peroxide (H2O2).
4. In its covalent compounds with nonmetals, hydrogen is assigned an oxidation
state of +1. For example, in the compounds HCl, NH3, H2O and CH4, hydrogen is
assigned an oxidation state of +1.
5. In its compounds, fluorine is always assigned an oxidation state of -1.
6. The sum of the oxidation states must be zero for an electrically neutral compound.
For an ion, the sum of the oxidation states must equal the charge of the ion. For
example, the sum of the oxidation states for the hydrogen and oxygen atoms in
water is 0; the sum of the oxidation states for the carbon and oxygen atoms in
CO32- is -2; and the sum of oxidation states for the nitrogen and hydrogen atoms
in NH4+ is +1.
These rules do not cover every possible case, but they should be sufficient for anything
IB throws at you. It is worthwhile to note at this point that the convention is to write
actual charges on ions as n+ or n-, the number being written before the plus or minus
sign. On the other hand, oxidation states (not actual charges) are written +n or –n, the
number being written after the plus or minus sign. Who knows if IB follows this
convention, but you should.
10.1.3. State and explain the relationship between oxidation numbers and the names of
compounds.
Oxidation numbers in names of compounds are represented by Roman numerals, eg iron
(II) oxide, iron (III) oxide.
Remember (if you’ve learned yet, if not learn) that d-block elements can have multiple
oxidation states. The way you specify which oxidation state you want when writing the
name is by putting in a roman number which signifies the oxidation state of the element.
So for example, if you want to write the name for FeO, the oxidation state of Fe in that
compound is +2, so the name is iron (II) oxide, pronounced “iron two oxide”. This is a
different compound then iron (III) oxide (pronounced iron three oxide), which would be
Fe2O3. This is because the iron now has an oxidation state of +3, and since the net charge
has to be zero, there must be three oxygen (total of -6 charge) and two iron (total of +6
charge).
10.1.4. Identify whether an element is oxidized or reduced in simple redox reactions,
using oxidation numbers.
Appropriate reactions to illustrate this can be found in topics 3 and 11. Possible
examples include: iron (II) and (III), manganese (II) and (VII), chromium (III) and (VI),
copper (I) and (II), oxides of sulfur and oxyacids, halogens and halide ions.
To do this, you find the oxidation states of all the elements on both sides. Then compare,
whichever element’s oxidation state got smaller (more negative) was reduced, while
whichever element’s oxidation state got bigger was oxidized. We will use the following
reaction for an example:
2Al(s) + 3I2(s)  2AlI3(s)
In the reactants side, the oxidation state of all of the substances are 0 because they are in
element reform (refer to rules under 10.1.2. if you don’t understand this). In the products,
Al has an oxidation state of +3, so that means iodine must have an oxidation state of -1,
because three of them neutralize the +3. (See how these were figured out, basically the
charge the elements would receive if they became ions). So, aluminum went from an
oxidation state of 0 to +3, so it is being oxidized, while Iodine when from an oxidation
state of 0 to -1, so it is being reduced.
10.1.5. Define the terms oxidizing agent and reducing agent.
An oxidizing agent is an electron acceptor and a reducing agent is an electron donor. For
the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
We can say the following (figure out the oxidation numbers for yourself and see why):
 Carbon is oxidized because there has been an increase in ist oxidation state
(carbon has formally lost electrons).
 Oxygen is reduced because there has been a decrease in its oxidation state
(oxygen has formally gained electrons.)
 CH4 is the reducing agent.
 O2 is the oxidizing agent.
Note that when the oxidizing or reducing agent is named, the whole compound is
specified, not just the element that undergoes the change in oxidation state.
10.2 Reactivity
10.2.1. Deduce a reactivity series based upon the chemical behavior of a group of
oxidizing and reducing agents.
Displacement reactions of metals and halogens (see 3.3.1.) provide a good experimental
illustration of reactivity. Standard electrode potentials or reduction potentials are not
required.
10.2.2. Deduce the feasibility of a redox reaction given reactivity series.
10.2.3. Describe and explain how a redox reaction is used to produce electricity in a
voltaic cell.
Students should be able to draw a diagram of a simple half-cell, and show how two halfcells can be connected by a salt bridge to form a whole cell. Suitable examples of halfcells are Mg, Zn, Fe and Cu in solutions of their ions.
A voltaic cell (also called a galvanic cell) is a device in which chemical energy is
changed to electrical energy. When Zn and Cu2+ are in solution together, the electrons
naturally flow from Zn to Cu2+ to create Zn2+ and Cu. Under these conditions, no useful
work is obtained from the chemical energy involved in the reaction, which instead is
released as heat. In order to harness this energy, we separate the oxidizing agent from the
reducing agent, thus requiring the electron transfer to occur through a wire. The current
produced in the wire by the electron flow can then be directed through a device, such as
an electric motor, to provide useful work. However, in order to keep this running, we
have to connect the two solutions with a salt bridge, which allows ions to flow from one
to cell to another in order to stabilize the charges (otherwise charges would build up in
the two cells and electron flow would stop. So, a voltaic cell can be drawn as this:
10.3 Electrolysis
10.3.1. Draw a diagram showing the essential components of an electrolytic cell.
An electrolytic cell converts electrical energy to chemical energy. The diagram should
include the source of electric current and conductors, positive and negative electrodes
and the electrolyte.
An electrolytic cell uses electrical energy to produce chemical change. The process of
electrolysis involves forcing a current through a cell to produce a chemical change for
which the cell potential is negative; that is, electrical work causes an otherwise
nonspontaneous chemical reaction to occur. Electrolysis has great practical importance;
for example, charging a battery, producing aluminum metal, and chrome plating an object
are all done electrolytically.
10.3.2. Describe how current is conducted in an electrolytic cell.
Current is conducted in an electrolytic cell in the opposite direction to that in a voltaic
cell. So the electrons come from the new anode (what was the cathode in the voltaic
cell), flow through the wire by the power of the external energy source, and into the
cathode (what was the anode in a voltaic cell.) Also, ion flow is opposite through the salt
bridge then in the voltaic cell.
10.3.3. Deduce the products for the electrolysis of a molten salt.
Equations showing the formation of products at each electrode should be given.
We will study the electrolysis of an imaginary molten salt X2Y. At the cathode positively
charged ions gain electrons, i.e. X+ + e-  X. At the anode negatively charged ions lose
electrons, i.e. Y2-  Y + 2e-. The circuit is completed by the two oxidation and reduction
half equations creating an effectively complete circuit. In this example, X will be
produced at the cathode and Y at the anode. Since only two species are present in a
molten salt, there is no possibility for any other chemicals to be produced (i.e. water
cannot be involved, etc.)
10.3.4. Distinguish between the use of a spontaneous redox reaction to produce
electricity in a voltaic cell and the use of electricity to carry out a non-spontaneous
redox reaction in an electrolytic cell.
Some teachers may wish to describe reactions at the electrodes in a cell in terms of
reduction at the cathode and oxidation at the anode, but this is not required.
In a voltaic cell, the reaction that is taking place is a spontaneous redox reaction. It
requires no external power to take place, it would happen of its own accord if left by
itself with both species present. We just use such a reaction to gain power by using the
energy electrons create while they flow. Adversely, the in a electrolytic reaction, we use
electricity to carry out a non-spontaneous redox reaction in an electrolytic cell. This
means that the reaction, left to itself would never occur. It only really happens because
we apply energy to make it happen, this absorbs energy but it sometimes has good
products as you will see below.
10.3.5. Describe and explain the use of electrolysis in electroplating.
Restrict this to copper plating.
Plating means depositing the neutral metal on the electrode by reducing the metal ions in
solution. For our examples above, the Cu2+ ion requires two electrons to become an
atom of copper metal: Cu2+(aq) + 2e-  Cu(s). So, if we want to plate something with
copper, we make that something the electrode in the anode of a electrolytic cell, and as
the copper is reduced, it will plate onto the electrode and cover the electrode in copper,
thus we now have a copper plated electrode.