Oxidation and reduction
Define oxidation and reduction in terms of electron loss and gain. Introduce the concept of the halfequation.
Oxidation and reduction reactions
These are reactions where electrons are transferred from one species (atom, molecule or ion) to another.
We can write 'half' equations to show only what happens to the species losing electrons or a different
'half' equation to show the species gaining electrons.
The whole equation is put together by making sure that the numbers of electrons are balanced in each
half equation and adding them together (when the electrons will cancel out)
Oxidation
This is the name given to removal of electrons from a species - the reagent causing the loss of electrons
is called the oxidising agent
Example:
Mg(s)
Mg2+ + 2e
In this (half) equation the magnesium atom loses electrons and becomes an ion.
Reduction
This is the gain of electrons - the species donating the electrons is called the reducing agent
Example
Fe3+ + 3e
Fe(s)
In this (half) equation the iron III ion gains three electrons to become an atom.
Redox reactions
Obviously the electrons leave one species and go to another. Consequently reduction has to be
accompanied by oxidation and vice versa. For this reason reactions involving transfer of electrons are
called reduction and oxidation or redox for short
Example
3Mg(s) + 2Fe3+
2Fe(s)+ 3Mg2+
The electrons from the magnesium are transferred to the iron III ions
Summary
Loss of electrons = Oxidation
Gain of electrons = Reduction
Mnemonic (memory aid)
OIL-RIG
Oxidation Is Loss
Reduction Is Gain
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.
Oxidation number
This is the apparent valency of an atom within a compound. It is usually considered as if the element
were bonded ionically to allow the apparent number of electrons gained or lost to be assessed.
The sum of all the oxidation numbers in a compound must add up to 0. By convention, the oxidation
number is written as a Roman numeral in the name, eg. iron II sulphate, sulphur VI oxide.
The oxidation number of an uncombined element is always zero (0)
Calculating the oxidation number
There are some elements that virtually always have the same oxidation number and these can be used to
calculate the oxidation numbers of the atoms in question.
Hydrogen, for example always has an oxidation number of -1 when bonded to a metal (more
electropositive element) and +1 when bonded to a more electronegative element (non-metal). Oxygen is
always -2 (except when in the form of the peroxide ion when it has an O-O bond giving it an oxidation
number of -1). Group 1 and 2 metals usually have an oxidation number of 1+ and 2+ respectively.
Example
Calculate the oxidation number of sulphur in sulphuric acid H 2SO4
Hydrogen = +1 oxidation number
Oxygen = -2 oxidation number
Therefore:
(2 x H) + S + (4 x O) = 0
2 + S -8 = 0
S=6
Example
Calculate the oxidation number of nitrogen in calcium nitrate Ca(NO 3)2
Calcium is in group 2 = +2 oxidation number
Oxygen = -2 oxidation number
Therefore:
(+2) + [(2 x N) + (6 x -2)] = 0
+2 + 2N -12 = 0
2N = 10
N = +5
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.
Names of compounds
Where there is any doubt about the oxidation state of an element within a compound it is stated using
Roman numerals immediately after the ambiguous element. For example Iron compounds may be iron in
the oxidation state +2 or +3 - it must therefore be stated as iron II or iron III in the compound name.
In the examples above the full systematic name for sulphuric acid is sulphuric VI acid and calcium nitrate
is calcium nitrate V
Example
Name the following compound - FeSO4
Oxidation state of the oxygen = -2 Oxidation state of the sulphur = +6
Therefore oxidation state of the iron = - (+6 - 8) = +2
The name of the compound FeSO4 is iron II sulphate
Example
Name the following compound - TiCl4
Oxidation state of the chloride = -1
Therefore oxidation state of the titanium = - (- 4) = +4
The name of the compound TiCl4 is titanium IV chloride
Identify whether an element is oxidised 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 sulphur and
oxyacids, halogens and halide ions.
Oxidation and reduction
As stated above, for the purposes of oxidation and reduction the oxidation number can be thought of as
the apparent ionic charge of an atom within a compound. For example, in sulphuric acid the sulphur is in
the VI (6+) oxidation state. For the purposes of redox we can consider that it has an ionic charge of +6
(even though it is clearly covalently bonded). This makes it easier to follow any transfer of electrons.
If the sulphur changes to an oxidation state of IV during a chemical reaction then it has gone from an
apparent ionic charge of +6 to a charge of +4, i.e. it has gained two electrons (negative charges). It has
therefore been reduced (gain of electrons) in the process.
Examples
2FeCl2 + Cl2
2FeCl3
The iron changes from 2+ to 3+ and is therefore oxidised (removal of electrons)
The chlorine gains an electron to go from 0 to -1 and is therefore reduced (addition of electrons)
Zn + CuSO4
Cu + ZnSO4
The zinc changes from oxidation state 0 to +2 (removal of electrons) it is oxidised (animation)
The copper changes from 2+ to 0 and is oxidised and is therefore reduced (addition of electrons)
Cr2O72- + 3SO2 + 2H+
2Cr3+ + 3SO42- + H2O
The chromium changes from +6 to +3 and is therefore reduced (gain of electrons)
The sulphur changes from +4 to +6 and therefore loses electrons = oxidation (loss of electrons)
2KI + Br2
2KBr + I2
The iodide ions (oxidation number = -1) change to iodine (oxidation number = 0) : oxidation
Bromine (element, oxidation number = 0) changes to bromide ions (oxidation number = -1) : reduction
5Fe2+ + MnO4- + 8H+
5Fe3+ + Mn2+ + 4H2O
The iron changes from 2+ to 3+ and is oxidised (removal of electrons)
The manganese atom changes from +7 to +2 and is therefore reduced (addition of electrons)
Define the terms oxidising agent and reducing agent.
Oxidising agents
These are the chemicals that cause the oxidation in a redox reaction. We call the reacting compounds in
a reaction the reagents (short form of the words reacting agents).
We consider that the removal of electrons from a species is oxidation and these electrons have to be
taken away by another compound or species. This species that attracts the electrons is said to be the
oxidising agent i.e. the reagent that causes the oxidation.
Reducing agents
Similarly the reagent that causes reduction in a redox reaction is said to be the reducing agent.
The oxidising agent takes the electron and is itself reduced, the reducing agent loses the electrons and is
itself oxidised.
2KI
+ Br2
Iodide ions get oxidised
Bromine gets reduced
Iodide - reducing agent
Bromine - oxidising agent
Cr2O72-
+ 3SO2 + 2H+
Chromium VI gets reduced
Sulphur IV gets oxidised
Chromium VI oxidising agent
Sulphur IV reducing agent
2KBr
+ I2
2Cr3+
+ 3SO42- + H2O