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REDOX REACTIONS

Oxidation Numbers

This section is about calculating the oxidation state or oxidation number of an element in a compound. It uses a set of rules which require little or no previous knowledge of the chemical being examined. (The oxidation number is similar to the valency of the element but has a + or - sign.)

Oxidation NumbersOxidation Numbers

• The oxidation number of an element is the charge which the atom would have if the element was acting as an ion in the species being studied.

• This can seem a bit strange at times, especially when you know that the compound you are examining is covalent. But it enables us to easily work out whether an element is oxidised or reduced in a reaction. We just compare the oxidation number before and after the reaction and work out whether it has gained or lost electrons.There are a series of simple rules we use to calculate the oxidation number of an element.

Oxidation NumbersOxidation Numbers

• The oxidation number of an uncombined element is zero. I.e. the element in the element itself is zero.

• The algebraic sum of the oxidation numbers of all the atoms in a compound equals zero.

• The algebraic sum of the oxidation numbers of all the atoms in an ion equals the charge on the ion.

• Algebraic sum means take account of the sign of the charge. Learn these rules.

Oxidation Numbers: RulzOxidation Numbers: Rulz

• Fluorine is always -1 in its compounds.

• Group I metals are always +1 in compounds.

• Group II metals are always +2 in their compounds.

• Oxygen is almost always -2 in compounds, except in peroxides (H2O2, ROOR, etc.) where it is -1. (Or when overruled by one of the above rules.)

• Halides are often -1, but other numbers are often possible.

• Hydrogen can be +1 or -1. (If it is the first element given in a formula, it is usually +1, if given second or third, it may be -1. If in doubt, the most electronegative element is +ve and the other element will be -ve.)

Oxidation NumbersOxidation Numbers

Calculate the oxidation number of Mn in KMnO4.

K and O are in our table of known oxidation numbers, and KMnO4 is a compound.

K = +1 (Group I metal) and oxygen is almost always -2.

K + Mn + (4 x O) = 0

So +1 + Mn + (4 x -2) = 0

i.e. +1 + Mn - 8 = 0

i.e. Mn - 7 = 0

so Mn = +7 (We say +7 and not 7, because it is a charge.)

Oxidation NumbersOxidation Numbers

• Calculate the oxidation number of chlorine in the ion ClO3-

Oxygen comes before chlorine in our table, so it’s rule takes preference, and it has a charge of -2.

Cl + (3 x O) = -1 ( the -1 is the charge on this ion)

Cl + (3 x -2) = -1

Cl - 6 = -1

Cl = +5 (Again, it’s +5 and not 5 as it’s the charge on the “ion”.)

Here’s another example, this time involving an ion.

Oxidation NumbersOxidation Numbers

• Calculate the oxidation number of phosphorus in POCl2F

• Oxygen is usu. -2, chlorine is usu. -1, fluorine is always -1.

As P is the unknown, we can reason that although Cl can have other values, then -1

P + O + (2 x Cl) + F = 0

P - 2 - 2 -1 = 0

P - 5 = 0

P = +5

But phosphorus can have other values for it’s oxidation number, depending on the compound under investigation

Oxidation numbersOxidation numbers

• What is the oxidation number for P in NaH2PO3?

• Na = +1, H = +1 usually, and O = -2 almost always.

Na + 2H + P + 3O = 0

+ 1 + 2 + P - 6 = 0

P - 3 = 0, so P = +3

• What is charge on the Cr in K2Cr2O7

2K + 2Cr + 7O = 0

+ 2 + 2Cr - 14 = 0

2Cr - 12 = 0, so 2Cr = +12, and Cr = +6

Oxidation NumbersOxidation Numbers

• As an element can have several oxidation numbers, the oxidation number is usually given in the compound’s name, written as Roman numerals in brackets after the element name e.g. “elementate(X)”, the number refers to “element” and not the associated oxygen's.

• So if we look at the examples we’ve just done, we get the following names:-

KMnO4 potassium manganate(VII)

NaClO3 sodium chlorate(V)

POCl2F phosphorus(V) oxydichlorofluoride

NaH2PO3 sodium dihydrogenphosphate(III)

K2Cr2O7 potassium dichromate(VI)

Oxidation numbersOxidation numbers

• What are the names and Ox numbers of the following?

• LiNO3 PtCl4 NaNO2

• lithium nitrate(V), platinum(IV) chloride, sodium nitrite

• NaBrO4 POCl3

• sodium bromate(VII), Phosphorus(V)oxychloride

• CrCl3 TiO TiO2

• chromium(III) chloride, TiO titanium(II) oxide, titanium(IV) oxide

Oxidation NumbersOxidation Numbers

• Consider the equation below.2FeCl2 + Cl2 2FeCl3

• The oxidation number of the iron changes thus:Fe2+ Fe3+ ( + e- of course)

sometimes written as Fe(II) Fe(III) + e-

i.e. Fe2+ has been oxidised. This example is easy to spot as an example of oxidation, but sometimes it’s a bit harder.

Oxidation numbers let us quickly see when anelement has gained or lost electrons, and so been oxidisedor reduced. If the oxidation number becomes more positive, oxidation has occurred, becoming more negative means that reduction has occurred.

Oxidation NumbersOxidation Numbers

• I- IO3-

Is this oxidation or reduction?

Calculating the oxidation numbers reveals all…

• In I-, the iodine is -1, while in IO3- it is +5. (Don’t take my word

for it, calculate the oxidation numbers yourself.) So the iodide ion has lost electrons, and been oxidised to +5.

• I- IO3- + 6e-

which could also be written asI(I) I(V) + 6e-

Oxidation NumbersOxidation Numbers

• 3MnO2 + KClO3 + 6KOH 3K2MnO4 + KCl + 3H2O Mn = +4 Mn = +6 Cl = +5 Cl = -1i.e Mn has lost electrons and been oxidised, whileCl has gained electrons and been reduced.

• 4FeCO3 + O2 2Fe2O3 + 4CO2 • Fe = +2 Fe = +3• O = 0 O = -2• C = +4 C = +4• i.e. Fe has lost electrons and been oxidised, while the

oxygen has gained electrons and been reduced. The carbon hasn’t changed.

Problem 4.11 (page 128)

Assign an oxidation number to each atom in the following compounds:

a) SnCl4 b)CrO3 c)VOCl3

d) V2O3 e) HNO3 f) FeSO4

FinFin