The terms oxidation number and oxidation state are used interchangeably to indicate the apparent charge of an atom within a species, as if all atoms are ionic. The IUPAC does differentiate between the two terms, representing oxidation numbers using Roman numerals and oxidation states with numbers preceded by a sign.
Syllabus reference R3.2.1Reactivity 3.2.1 - Oxidation and reduction can be described in terms of electron transfer, change in oxidation state, oxygen gain/loss or hydrogen loss/gain.
- Deduce oxidation states of an atom in a compound or an ion.
- Identify the oxidized and reduced species and the oxidizing and reducing agents in a chemical reaction.
Guidance
- Include examples to illustrate the variable oxidation states of transition element ions and of most main group non-metals.
- Include the use of oxidation numbers in the naming of compounds
Tools and links
- Structure 3.1 - What are the advantages and limitations of using oxidation states to track redox changes?
- Structure 2.3 - The surface oxidation of metals is often known as corrosion. What are some of the consequences of this process?
Oxidation
Historically, the concept of oxidation meant reaction with oxygen. Many elements react directly with oxygen forming oxides. The metal was said to be "oxidised" in this reaction.
2Cu(s) + O2(g) → 2CuO(s)
It was known that hydrogen could be used to reverse the oxidation reaction and hence "reduce" the metal oxide to pure metal. Reduction was considered to be the opposite of oxidation.
CuO(s) + H2(g) → Cu(s) + H2O(l)
For this reason, the definition of oxidation was extended to include the removal of hydrogen.
We now know that oxidation involves the removal of electrons from a species, while reduction involves the addition of electrons to a species.
Hence, electrons are transferred from one species to another in a redox reaction.
Example:
removal of electrons
Mg(s) → Mg2+ + 2e
addition of electrons
½O2(g) + 2e → 2O2-
Summary of oxidation
- Addition of oxygen
- Removal of hydrogen
- Removal of electrons
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.
Example:Identify the oxidising agent in the following redox reaction:
Cu2+(aq) + 2I-(aq) → Cu+(s) + I2(aq)
The species that loses electrons in this reaction, i.e. the species getting oxidised, is the iodide ions.
2I-(aq) - 2e → I2(aq)
Therefore the copper(II) ions are the oxidising agent.
Reduction
Following on from the section on oxidation above, reduction is the gain of electrons - the species donating the electrons is called the reducing agent
Example
Fe3+ + 3e → Fe(s)
In this equation the iron(III) ion gains three electrons to become an atom.
Summary of reduction
- Removal of oxygen
- Addition of hydrogen
- Addition of electrons
Reducing agents
The reagent that causes reduction in a redox reaction is called the reducing agent.
The oxidising agent takes the electron(s) and is itself reduced, the reducing agent loses the electron(s) and is itself oxidised.
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2KI
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+ Br2
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→
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2KBr
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+ I2
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||||
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Iodide ions get oxidised
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Bromine gets reduced
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|||||||
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Iodide - reducing agent
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Bromine - oxidising agent
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Cr2O72-
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→
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2Cr3+
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||||
| Chromium(VI) gets reduced | sulfur(IV) gets oxidised | |||||||
| Chromium(VI) oxidising agent | sulfur(IV) reducing agent |
Example: Identify the reducing agent in the following redox reaction:
I2(aq) + 2S2O32-(aq) → S4O62-(aq) + 2I-(aq)
The species that gains electrons in this reaction, i.e. the species getting reduced, is the iodine.
I2(aq) + 2e → 2I-(aq)
Therefore the thiosulfate ions are the reducing agent.
Oxidation state
The oxidation state this is the apparent valency of an atom within a compound.
This can be determined by treating all atoms as if they are ionic, to allow the apparent number of electrons gained or lost to be assessed.
The redox rules
- The sum of all the oxidation states in a compound must add up to 0.
- The sum of all of the oxidation states in an ion must add up to the charge on the ion.
- The oxidation state of an uncombined element is always zero (0).
- By convention, when naming compounds the oxidation number is written as a Roman numeral in the compound name immediately after the atom to which it refers.
Examples:
- The oxidation state of iron in iron(II) sulfate is +2
- The oxidation state of phosphorus in phosphorus(V) oxide is +5
- The oxidation state of chromium in potassium dichromate(VI) is +6
Calculating the oxidation state
There are some elements that nearly always have the same oxidation state. These can be used to calculate the oxidation states of the atoms to which they are bonded.
Hydrogen, for example has an oxidation state of -1 when bonded to a metal (more electropositive element) and +1 when bonded to a more electronegative element (non-metal). Oxygen is nearly always -2 (the exception is when it is in the form of the peroxide ion, O-O2-, it has an oxidation state of -1).
Group 1 and 2 metals have an oxidation state of +1 and +2 respectively.
Example: Calculate the oxidation state of sulfur in sulfuric acid H2SO4
- Hydrogen = +1 oxidation state
- Oxygen = -2 oxidation state
(2 x H) + S + (4 x O) = 0
2 + S - 8 = 0
S = +6
Example: Calculate the oxidation state of nitrogen in calcium nitrate Ca(NO3)2
- Calcium is in group 2 = +2 oxidation state
- Oxygen = -2 oxidation state
Therefore:
(+2) + [(2 x N) + (6 x -2)] = 0
+2 + 2N -12 = 0
2N = 10
N = +5
Oxidation state in ions
When you are dealing with an ion, the process for working out the oxidation state of the atoms within the ion is the same, with the important exception that the sum of the oxidation states must add up to the charge on the ion.
Example: Calculate the oxidation state of the chloride atom in the chlorate ion, ClO3-
Each of the oxygen atoms has an oxidation state of -2
Therefore as Cl + (3 x -2) = -1
Then Cl = -1 + 6 = +5
Roman numerals
As we have seen, some chemical elements have a variable oxidation number. The only way to know an element's oxidation number is to work it out from known oxidation numbers. To avoid confusion in the name, elements with variable oxidation numbers have the number included.
The oxidation state or number of any element within a compound is shown by a Roman numeral immediately after the element in question. Roman numerals are used to avoid confusion. The numerals are only used in the names of the compounds, NOT the formulae.
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Number
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Roman Numeral
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| 1 | I |
| 2 | II |
| 3 | III |
| 4 | IV |
| 5 | V |
| 6 | VI |
| 7 | VII |
| 8 | VIII |
Examples
- copper(II) oxide
- copper(I) oxide
- Sodium sulfate(VI)
- Sodium sulfate(IV)
Naming compounds
In each case the valency or oxidation state of the element immediately prior to the Roman numeral is defined.
- copper(II) oxide has the formula CuO - the oxidation state of the copper is 2
- copper(I) oxide has the formula Cu2O - the oxidation state of the copper is 1 in this case.
- Sodium sulfate(IV) has the formula Na2SO3 - the sulfur is the element with a variable oxidation number; in this case 4.
- Sodium sulfate(VI) has the formula Na2SO4 - the sulfur is the element with a variable oxidation number; in this case 6.
In some situations there may even be two Roman numerals required to prevent any kind of ambiguity.
Example: copper(I) sulfate(IV) - Cu2SO3
Examples: Give systematic names for the following compounds from the formula.
K2MnO4 and KMnO4
K2MnO4 contains a transition element with a variable oxidation number. In terms of valencies (oxidation numbers)
(2 x K) + (4 x O) - Mn = 0; therefore Mn = +6
The name of K2MnO4 is potassium manganate(VI)
KMnO4 contains a transition element with a variable oxidation number. In terms of valencies (oxidation numbers)
(1 x K) + (4 x O) - Mn = 0; therefore Mn = +7
The name of KMnO4 is potassium manganate(VII)
Example: Name the following compound - FeSO4
Oxidation state of the oxygen = -2; Oxidation state of the sulfur = +6 (in the sulfate ion)
Therefore oxidation state of the iron = - (+6 - 8) = +2
The name of the compound FeSO4 is iron(II) sulfate
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
Naming ions
Exactly the same convention applies to ions. If there is an element within the ion that has a variable oxidation state then the oxidation state is included in the ion name as a Roman numeral. Radical negative ions are named differently from the elements that are present, so a certain amount of flexibility of thought is needed; they end with -ATE.
Example: Name the following ions.
ClO4- and ClO3-
ClO4- contains an element with a variable oxidation number (chlorine). In terms of valencies (oxidation numbers)
Cl + (4 x O) = -1; therefore Cl = +7
The name of the ClO4- ion is the chlorate(VII) ion
ClO3- contains an element with a variable oxidation number (chlorine). In terms of valencies (oxidation numbers)
Cl + (3 x O) = -1; therefore Cl = +5
The name of the ClO4- ion is the chlorate(V) ion
Complex ions
The names of positive complex ions are fairly straightforward. The ligands (species that are bonded to the central metal ion) are named first followed by the metal ion with its oxidation state.
Example: Name the following complex ion.
[Cu(H2O)6]2+
Copper is an element with a variable oxidation state. We know that the water molecules have already cancelled out the oxidation numbers of oxygen and hydrogen (as the water molecule is neutral)
Therefore copper is in the 2+ oxidation state
The name of the complex ion is the hexaaqua copper(II) ion
If the complex ion is negative (an anion) then the metal changes its name:
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Metal
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Metal in negative ion
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|---|---|
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Titanium
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Titanate
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Chromium
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Chromate
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Manganese
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Manganate
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Iron
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Ferrate
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Copper
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Cuprate
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Cobalt
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Cobaltate
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Zinc
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Zincate
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Aluminium
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Aluminate
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Lead
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Plumbate
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Example:Name the following complex ion.
[NiCl4]2-
Nickel is an element with a variable oxidation state. We know that the each chloride ion has a -1 oxidation state.
Therefore Nickel is in the 2+ oxidation state
The name of the complex ion is the tetrachloro nickelate(II) ion
Reduction and oxidation 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. Logically when a species loses or gains electrons, it must change its oxidation number or state.
The whole oxidation and reduction equation (redox equation) is put together by balancing the number of electrons on both sides in each half-equation and adding them together (when the electrons cancel out).
Redox reactions
In a redox reaction one or more electrons leave one species and go to another. Consequently, reduction of one species has to be accompanied by oxidation of the other (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+
Two electrons from each magnesium atom are transferred to the iron(III) ions
Finding the oxidised and reduced species in a redox reaction
This is a case of checking the oxidation state or number of an atom on the left hand side and comparing it with the same atom on the right hand side. If the oxidation number increases then the element has been oxidised, if it decreases then it has been reduced. The process can be done mechanically, which may be rather laborious, although after a little practise you start to recognise the atoms that are most likely to be involved.
Example: Which element is oxidised in the following redox equation?
H2O2(aq) + 2KI(aq) → 2KOH(aq) + I2(aq)
Oxidation causes an atm to increase its oxidation number. Hydrogen is rarely involved unless the element appears on one side or the other. In this equation we can get a clue that iodine is part of the redox pair as there is elemental iodine on one side, but not the other. The oxidation state of an uncombined element is always zero.
In potassium iodide, the iodine has an oxidation number of -1 (potassium is always +1). On the right hand side iodine has an oxidation number of zero. It has, therefore, increased its oxdation number from -1 to zero, it has been oxidised.