9.1 - Introduction to oxidation and reduction
9.1.1: Define oxidation and reduction in terms of electron loss and gain.
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)
This is the name given to removal of electrons from a species - the reagent causing the loss of electrons is called the oxidising agent
Mg(s) Mg2+ + 2e
In this (half) equation the magnesium atom loses electrons and becomes an ion.
This is the gain of electrons - the species donating the electrons is called the reducing agent
Fe3+ + 3e Fe(s)
In this (half) equation the iron III ion gains three electrons to become an atom.
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
3Mg(s) + 2Fe3+ 2Fe(s)+ 3Mg2+
The electrons from the magnesium are transferred to the iron III ions
Loss of electrons = Oxidation
Gain of electrons = Reduction
Mnemonic (memory aid)
Oxidation Is Loss
9.1.2: Deduce 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.
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 H2SO4
Hydrogen = +1 oxidation number
Oxygen = -2 oxidation number
(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(NO3)2
Calcium is in group 2 = +2 oxidation number
Oxygen = -2 oxidation number
(+2) + [(2 x N) + (6 x -2)] = 0
+2 + 2N -12 = 0
2N = 10
N = +5
9.1.3: State the names of compound using oxidation numbers. 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
9.1.4: Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
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
redox reaction animation