Oxidation number

Oxidation number

Lessons

In this lesson, we will learn:

  • To recall the oxidation states of common elements.
  • How to find the oxidation number of atoms in full equations.
  • How to use oxidation numbers to show what has been reduced or oxidized in reactions.
  • To identify disproportionation reactions, using oxidation numbers to do so.

Notes:

  • Redox reactions will always have a chemical that is being oxidized and a chemical that is being reduced. You need to find the oxidation number of the atoms in the reaction to see what is being oxidized or reduced.
    The oxidation number is an assigned number that shows the number of electrons gained or lost by an atom compared to its original outer shell configuration.
    There are some key rules of oxidation numbers that help in most situations:
    • ANY ATOM in elemental form always has an oxidation state of zero.
    • Oxygen atoms in a compound have an oxidation state of -2.
    • The oxidation state of ions or atoms in a molecule will add up to the overall charge.
    • Halogen atoms in a compound have an oxidation state of -1.
    • Hydrogen atoms have an oxidation state of +1 unless they are bonded to a metal atom, where they will be -1. Compounds with the formula MHn, where M is a metal, are called hydrides.
    • These oxidation numbers are often written using Roman numerals, especially when writing outside of a chemical equation. Make sure you know Roman numerals for the numbers 1 through 8!

  • Because oxidation number is related to electrons being lost or gained, you can make general predictions on oxidation numbers based on if an element is a metal or non-metal.
    • Non-metals are more electronegative than metals, so in general they will form negative ions in reactions through gaining electron(s) which decreases their oxidation number.
    • Metals are less electronegative than non-metals, so they generally form positive ions in reactions by losing electron(s) which will increase their oxidation number.

  • Just like charge of molecules/ions, the overall change in oxidation state during a reaction will be zero. Because the changes are caused by electrons being lost or gained, any positive change in oxidation state will be mirrored by a negative change in another reactant.

  • WORKED EXAMPLE: Using the rules of oxidation numbers, it is possible to predict the oxidation state of other less predictable elements in a redox reaction. For example:

    Fe2O3 (s) + 3 CO (g) \, \, 2 Fe (s) + 3 CO2 (g)

    Using the rules above, the oxidation numbers of iron and carbon in the reactants and products can be found.
    • Oxygen atoms have an oxidation number of -2 so three of them in Fe2O3 is a total of -6, and the compound is neutral, so the two iron atoms must, combined, have an oxidation number of +6. There are two of them, so Fe in Fe2O3 has an oxidation number of +3.
    • Oxygen has an oxidation number of -2 and there is only one of them in CO, which is neutral, so the C atom in CO must have an oxidation number of +2 to counteract this to bring it back to zero.
    • Fe in the products is elemental iron so it has oxidation number 0. Atoms in any elemental sample will have oxidation number 0.
    • Oxygen has an oxidation number of -2 and there are two O atoms in CO2, which is -4 in total due to the oxygen. The molecule is neutral, so the one C atom in CO2 must have a +4 oxidation number to balance the molecule out.

  • Working out the oxidation numbers in the products and reactants lets you work out the change in oxidation state of the atoms in the reaction – remember, the overall change will be zero.
    • Iron has changed from +3 oxidation state in Fe2O3 to 0 in its elemental form. Therefore iron has been reduced (\triangleOS = -3) in this reaction. The reaction with carbon monoxide is what caused this change, so carbon monoxide is the reducing agent in this process.
    • Carbon has changed from +2 oxidation state in the CO reactant to +4 in the CO2 products. Therefore carbon has been oxidized (\triangleOS = +2) in this reaction. The reaction with iron oxide caused this, so iron is the oxidizing agent, which was itself reduced in the reaction.

    There were two iron atoms in the reaction which is a total change in OS of -6. There were three carbon atoms in the reaction which is a total change in OS of +6.
    These changes in oxidation state balance to zero.

  • Disproportionation reactions are a rare but important type of redox reaction where identical atoms in a chemical compound undergo both reduction and oxidation. One of the most common disproportionation reactions is the decomposition of hydrogen peroxide:

    2H2O2\enspace \enspace 2H2O2 + O2
    Oxidation states: \quad \qquad -1 \qquad \quad \enspace -2 \quad 0 \qquad \qquad \qquad \qquad \enspace \enspace

    In the reactants, hydrogen peroxide has its oxygen atoms in an oxidation state of -1. This is because the oxygen-oxygen bond does not ‘assign’ electrons to either atom. Only the oxygen-hydrogen bonds do that, where one on each oxygen atom gives them an oxidation state of -1 each.
    In the products:
    • One oxygen atom went from a peroxide molecule to elemental oxygen. This takes it from an oxidation state of -1 to 0. This oxygen atom was oxidized.
    • The other oxygen atom went from peroxide to a water molecule. This oxygen atom went from an oxidation state of -1 to -2 and was therefore reduced.
    Disproportionation reactions require an atom to be able to take three different oxidation states: 
    • The reactants will have an atom in a middle oxidation state, where both the reduced and oxidized atoms start. For oxygen this is -1 when it is bonded to another oxygen atom.
    • One higher oxidation state for the oxidized atom. For oxygen, this is 0 in elemental oxygen, O2 (g).
    • One lower oxidation state for the reduced atom. For oxygen, this is -2 such as in H2O.
  • Introduction
    What is the oxidation number?
    a)
    Identifying reduction and oxidation

    b)
    Rules for oxidation numbers.

    c)
    Assigning oxidation numbers in reactions.

    d)
    What has been reduced and oxidized in the reaction?

    e)
    Redox and disproportionation reactions


  • 1.
    Find the oxidation numbers of both elements in the following compounds.
    Find the oxidation numbers of each atom in the following chemical compounds.
    a)
    H2O2

    b)
    MgH2

    c)
    Na2O

    d)
    NCl3


  • 2.
    Recognise and distinguish regular redox reactions from disproportionation reactions.
    a)
    H2 + O2 \enspace \enspace 2H2O
    2H2O2 \enspace \enspace 2H2O + O2
    H2 + Cl2 \enspace \enspace 2HCl