oxidation number

OXIDATION STATES (OXIDATION NUMBERS)

This page explains what oxidation states (oxidation numbers) are and how to calculate them and make use of them.
Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they are in any quick way.

Explaining what oxidation states (oxidation numbers) are
Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about:
• oxidation and reduction in terms of electron transfer
• electron-half-equations

We are going to look at some examples from vanadium chemistry. If you don't know anything about vanadium, it doesn't matter in the slightest.
Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons:

The vanadium is now said to be in an oxidation state of +2.
Removal of another electron gives the V3+ ion:

The vanadium now has an oxidation state of +3.
Removal of another electron gives a more unusual looking ion, VO2+.

The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one).
The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element.
It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5.

Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1.
Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero.
What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur.

The sulphur has an oxidation state of -2.











Oxidation Numbers (Oxidation States)
Key Concepts

An oxidation number (oxidation state) is the charge an atom would carry if the molecule or ion were completely ionic.
Rules for Assigning Oxidation Numbers
a. Fluorine is assigned an oxidation number of -1 in compounds
HF oxidation number of fluorine = -1
b. Oxygen is assigned an oxidation number of -2 in compounds
H2O oxidation number of oxygen is -2
Except
o Peroxides: oxidation number of oxygen is -1
H2O2 oxidation number of oxygen is -1
o Superoxides: oxidation number of oxygen is -½
KO2 oxidation number of oxygen is ½
o Oxygen fluorides: OF2 oxidation number of oxygen is +2
O2F2 oxidation number of oxygen is +1
c. Hydrogen is assigned an oxidation of +1 in compounds
HCl oxidation number of hydrogen is +1
d. Group I elements (Alkali Metals) are assigned an oxidation number of +1 in compounds
NaNO3 oxidation number of sodium is +1
e. Group II elements (Alkaline-earth metals) are assigned an oxidation number of +2 in compounds
MgBr2 oxidation number of magnesium is +2
f. An atom of any element in the free state has an oxidation number of 0
S8 oxidation number of each sulfur atom is 0
g. Any monatomic ion has an oxidation number equal to its charge
H- oxidation number of hydrogen is -1
h. The sum of the oxidation numbers of all the atoms in a formula equals the electrical charge shown with the formula
o The sum of the oxidation numbers of all the atoms shown in a formula of a compound is 0
CO2: let x be the unknown oxidation number of carbon
0 = x + (2 x -2)
So x (oxidation number of C) = 0 + 4 = +4

o The sum of the oxidation numbers of all the atoms shown in the formula for a polyatomic ion or complex ion equals the electrical charge on the ion
Cr2O72-: let y = unknown oxidation number of chromium
-2 = 2y + (7 x -2)
2y = -2 + 14
2y = +12
y (oxidation number of Cr) = 12 ÷ 2 = +6
Oxidation
An increase in oxidation number which corresponds to a loss of electrons
(or to an addition in oxygen or loss of hydrogen)
Example

Zn(s) -----> Zn2+ + 2e
Zinc has been oxidised since there has been a loss of electrons to form Zn2+.
The oxidation number of zinc has increased from 0 to 2+
Reduction
A reduction in oxidation number which corresponds to a gain of electrons
(or to an addition of hydrogen or loss of oxygen)
Examples

Ag+ + e -----> Ag(s)
The silver ion has been reduced since it has gained an electron to form Ag(s)
The oxidation number of silver has been reduced from +1 to 0
MnO4- + 8H+ + 5e -----> Mn2+ + 4H2O

Oxidation number of manganese (x) in MnO4- is:
-1 = x + (4 x -2)
x = -1 + 8 = +7

oxidation number of manganese in MnO4- = +7
manganese has been reduced since there is a reduction in oxidation number from +7 to +2





Summary
Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers.

Working out oxidation states
You don't work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums!
• The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.
• Some elements almost always have the same oxidation states in their compounds:
element usual oxidation state exceptions
Group 1 metals always +1
Group 2 metals always +2
Oxygen usually -2 except in peroxides and F2O (see below)
Hydrogen usually +1 except in metal hydrides where it is -1 (see below)
Fluorine always -1
Chlorine usually -1 except in compounds with O or F (see below)


The reasons for the exceptions
Hydrogen in the metal hydrides
Metal hydrides include compounds like sodium hydride, NaH. In this, the hydrogen is present as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion - in this case, -1.
Alternatively, you can think of it that the sum of the oxidation states in a neutral compound is zero. Since Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides
Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound and so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O
The problem here is that oxygen isn't the most electronegative element. The fluorine is more electronegative and has an oxidation state of -1. In this case, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen
There are so many different oxidation states that chlorine can have in these, that it is safer to simply remember that the chlorine doesn't have an oxidation state of -1 in them, and work out its actual oxidation state when you need it. You will find an example of this below.

Warning!
Don't get too bogged down in these exceptions. In most of the cases you will come across, they don't apply!



Examples of working out oxidation states
What is the oxidation state of chromium in Cr2+?
That's easy! For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n:
n + 3(-1) = 0
n = +3 (Again, don't forget the + sign!)
What is the oxidation state of chromium in Cr (H2O)63+?
This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.
The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.
What is the oxidation state of chromium in the dichromate ion, Cr2O72-?
The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.
2n + 7(-2) = -2
n = +6

What is the oxidation state of copper in CuSO4?
Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulphur) whose oxidation states can both change.
The only way around this is to know some simple chemistry! There are two ways you might approach it. (There might be others as well, but I can't think of them at the moment!)
• You might recognise this as an ionic compound containing copper ions and sulphate ions, SO42-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2.
• You might recognise the formula as being copper (II) sulphate. The "(II)" in the name tells you that the oxidation state is 2 (see below).
You will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion.
Using oxidation states
In naming compounds
You will have come across names like iron (II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.
This can also be extended to the negative ion. Iron (II) sulphate is FeSO4. There is also a compound FeSO3 with the old name of iron (II) sulphite. The modern names reflect the oxidation states of the sulphur in the two compounds.
The sulphate ion is SO42-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate (VI) ion.
The sulphite ion is SO32-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion.
So FeSO4 is properly called iron (II) sulphate (VI), and FeSO3 is iron (II) sulphate (IV). In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.


Using oxidation states to identify what's been oxidised and what's been reduced

This is easily the most common use of oxidation states.
Remember:
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced.
Example 1:
This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:

Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:.

The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced.
Example 2:
The reaction between sodium hydroxide and hydrochloric acid is:

Checking all the oxidation states:

Nothing has changed. This isn't a redox reaction.
Example 3:
This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is:

Obviously the chlorine has changed oxidation state because it has ended up in compounds starting from the original element. Checking all the oxidation states shows:

The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? Yes! Both! One atom has been reduced because its oxidation state has fallen. The other has been oxidised.
This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced.







Using oxidation states to work out reacting proportions
This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation.
Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons.
Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else.
This example is based on information in an old AQA A' level question.
Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions?
The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate.
But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion.
The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.