Chemistry Tutorial 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 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 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: 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 (H 2 O 2, 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 Numbers Calculate the oxidation number of Mn in KMnO 4. K and O are in our table of known oxidation numbers, and KMnO 4 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 Mn - 8 = 0 i.e. Mn - 7 = 0 so Mn = +7 (We say +7 and not 7, because it is a charge.)

Oxidation Numbers Calculate the oxidation number of chlorine in the ion ClO 3 - 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 Numbers Calculate the oxidation number of phosphorus in POCl 2 F 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 = 0 P - 5 = 0 P = +5 But phosphorus can have other values for it’s oxidation number, depending on the compound under investigation

Oxidation numbers What is the oxidation number for P in NaH 2 PO 3 ? Na = +1, H = +1 usually, and O = -2 almost always. Na + 2H + P + 3O = P - 6 = 0 P - 3 = 0, so P = +3 What is charge on the Cr in K 2 Cr 2 O 7 2K + 2Cr + 7O = Cr - 14 = 0 2Cr - 12 = 0, so 2Cr = +12, and Cr = +6

Oxidation 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:- KMnO 4 potassium manganate(VII) NaClO 3 sodium chlorate(V) POCl 2 Fphosphorus(V) oxydichlorofluoride NaH 2 PO 3 sodium dihydrogenphosphate(III) K 2 Cr 2 O 7 potassium dichromate(VI)

Oxidation numbers What are the names and Ox numbers of the following? LiNO 3 PtCl 4 NaNO 2 lithium nitrate(V), platinum(IV) chloride, sodium nitrite NaBrO 4 POCl 3 sodium bromate(VII), Phosphorus(V)oxychloride CrCl 3 TiOTiO 2 chromium(III) chloride, TiO titanium(II) oxide, titanium(IV) oxide

Oxidation Numbers Consider the equation below. 2FeCl 2 + Cl 2  2FeCl 3 The oxidation number of the iron changes thus: Fe 2+  Fe 3+ ( + e - of course) sometimes written as Fe(II)  Fe(III) + e - i.e. Fe 2+ 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 an element has gained or lost electrons, and so been oxidised or reduced. If the oxidation number becomes more positive, oxidation has occurred, becoming more negative means that reduction has occurred.

Oxidation Numbers I -  IO 3 - Is this oxidation or reduction? Calculating the oxidation numbers reveals all… In I -, the iodine is -1, while in IO 3 - 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 -  IO e - which could also be written as I(I)  I(V) + 6e -

Oxidation Numbers 3MnO 2 + KClO 3 + 6KOH  3K 2 MnO 4 + KCl + 3H 2 O Mn = +4  Mn = +6 Cl = +5  Cl = -1 i.e Mn has lost electrons and been oxidised, while Cl has gained electrons and been reduced. 4FeCO 3 + O 2  2Fe 2 O 3 + 4CO 2 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) SnCl 4 b)CrO 3 c)VOCl 3 d) V 2 O 3 e) HNO 3 f) FeSO 4

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