The Wright Stuff. Chemistry Tutorial. REDOX REACTIONS by Dr John G Wright. Press PgDn or left click for the next slide. Redox Reactions. Oxidation and Reduction Oxidation Numbers (this set of 18 slides) Disproportionation Balancing Half Equations Electrochemical Cells
The tutorials are divided into five main sections, each covering a separate topic concerning redox reactions, plus a problems section.
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.)
First, the definition.
But where do you start? We find that some atoms always have the same oxidation number in their compounds or ions. As stated earlier, the first rules overrule the later rules.
K and O are in our table of known oxidation numbers, and KMnO4 is a compound.
K is +1 (Group I metal) and oxygen is almost always -2.
K + Mn + (4 x O) = 0 (that’s a zero)
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.)
Easy? It’s just simple arithmatic, not chemistry.
Let’s look at these rules in action in a simple problem.
Oxygen comes before chlorine in our known 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.
Oxygen is almost always -2, chlorine is usually -1 and fluorine is always ALWAYS -1. As P is the unknown, we can reason that although Cl can have other values, then -1, the commonest value, is the one to use (otherwise the problem is unsolvable).
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.
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
K and O are in the table of fixed values.
2K + 2Cr + 7O = 0
+ 2 + 2Cr - 14 = 0
2Cr - 12 = 0, so 2Cr = +12, and Cr = +6
KMnO4 potassium manganate(VII)
NaClO3 sodium chlorate(V)
POCl2F phosphorus(V) oxydichlorofluoride
NaH2PO3 sodium dihydrogenphosphate(III)
K2Cr2O7 potassium dichromate(VI)
NaNO2 sodium nitrate(III), NaBrO4 sodium bromate(VII)
POCl3 phosphorus(V) oxychloride
CrCl3 chromium(III) chloride, TiO titanium(II) oxide
TiO2 titanium(IV) oxide
Did you get them all right?
An important use of oxidation numbers is in the recognition ofoxidation and reduction. They let us quickly see when anelement has gained or lost electrons, and hence been oxidisedor reduced. If the oxidation number becomes more positive, oxidation has occurred, becoming more negative means that reduction has occurred.
Calculating the oxidation numbers reveals all.
Consider a reaction where the following change occurs (the other reagents are not important at this stage).
Here are a few more examples showing the use of oxidationnumbers to discover whether oxidation or reduction occurs.