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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

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Chemistry tutorial

The Wright Stuff

Chemistry Tutorial



Dr John G Wright

Press PgDn or left click for the next slide

Redox reactions
Redox Reactions

  • Oxidation and Reduction

  • Oxidation Numbers (this set of 18 slides)

  • Disproportionation

  • Balancing Half Equations

  • Electrochemical Cells

  • ProblemsThe individual topics can be viewed as separate slide shows. If a slide show is missing from the web page, it means it is being updated and improved. (Or perhaps I just haven’t finished writing it yet!)

The tutorials are divided into five main sections, each covering a separate topic concerning redox reactions, plus a problems section.

Oxidation and reduction
Oxidation and Reduction

  • It is assumed that you have read the first tutorial on Oxidation and Reduction and are familiar with the meaning of these terms, especially the electron-based definitions.

  • Students on introductory level courses may only need to read the previous section on Oxidation and Reduction, but those on advanced courses, such as A-level and above, will need to study both tutorials, with the emphasis being placed on the modern electron-based definition of oxidation and reduction.

Oxidation numbers
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 numbers1
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.

First, the definition.

Oxidation numbers2
Oxidation Numbers

  • The rules are given in order of importance. If you reach a rule which gives you data to work from and a later rule appears to contradict the earlier result, just ignore it, you use the earlier result. I.e. later rules do not overrule earlier decisions that you have already made.

Oxidation numbers3
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 numbers4
Oxidation Numbers

  • Fluorine is always ALWAYS -1 in its compounds.

  • Group I metals are always 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.)

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.

Oxidation numbers5
Oxidation 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 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.

Oxidation numbers6
Oxidation Numbers

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

    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.

Oxidation numbers7
Oxidation Numbers

  • Calculate the oxidation number of phosphorus in POCl2F

    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.

Oxidation numbers8
Oxidation 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 ?

    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

Oxidation numbers9
Oxidation Numbers

  • To avoid any confusion when an element can have several oxidation numbers, the oxidation number is usually mentioned in the compound’s name, written as Roman numerals in brackets after the element to which it refers. In names like “elementate(X)”, the number refers to “element” and not the associated oxygens.

  • 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 numbers10
Oxidation numbers

  • Check for yourself that the numbers in the following compounds’ names are the same as the oxidation number of the element with which they are associated

  • LiNO3 lithium nitrate(V), PtCl4 platinum(IV) chloride

    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?

Oxidation numbers11
Oxidation 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.

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.

Oxidation numbers12
Oxidation 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 as I(I) ® I(V) + 6e-

Consider a reaction where the following change occurs (the other reagents are not important at this stage).

Oxidation numbers13
Oxidation 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 = +4i.e. Fe has lost electrons and been oxidised, while the oxygen has gained electrons and been reduced. The carbon hasn’t changed. Check these oxidation numbers for yourself.

Here are a few more examples showing the use of oxidationnumbers to discover whether oxidation or reduction occurs.

The end
The End

  • I hope you have enjoyed this tutorial. It is the second in a series of tutorials on oxidation and reduction. The tutorials become progressively more advance, but unless it says so on the frist couple of pages, are all intended for A-level students.

  • I also hope you have increased your understanding of chemistry, learned something useful about chemistry and that it will increase your marks in examinations.Bye for now, Dr John G Wright, The Wright Stuff