Edexcel International AS Chemistry

Revision Notes

2.3.1 Oxidation Numbers - Introduction

Oxidation Numbers - Introduction

  • There are three definitions of oxidation and reduction  used in different branches of chemistry
  • Oxidation and reduction can be used to describe any of the following processes

Definitions and Examples of Oxidation & Reduction

Electrochemistry OIL RIG Diagram, downloadable AS & A Level Chemistry revision notes

Use the acronym "Oil Rig" to help you remember the definitions of oxidation and reduction

Oxidation Number

  • The oxidation number of an atom is the charge that would exist on an individual atom if the bonding were completely ionic
  • It is like the electronic ‘status’ of an element
  • Oxidation numbers are used to
    • Tell if oxidation or reduction has taken place
    • Work out what has been oxidised and/or reduced
    • Construct half equations and balance redox equations

Oxidation Numbers of Simple Ions

Worked example

What are the oxidation numbers of the elements in the following species?

a) C                b)  Fe3+                       c)  Fe2+

d) O2-             e)  He                          f)  Al3+

Answers:

a) 0     b) +3    c) +2   

d) -2    e) 0      f) +3

 

  • So, in simple ions, the oxidation numbers of the atom is the charge on the ion:
    • Na+, K+, H+ all have an oxidation number of +1
    • Mg2+, Ca2+, Pb2+ all have an oxidation number of +2
    • Cl, Br, I all have an oxidation number of -1
    • O2-, S2- all have an oxidation number of -2

Oxidation Number Rules

  • A few simple rules help guide you through the process of determining the oxidation number of any element
  • Remember, you are determining the oxidation number of a single atom
  • The oxidation number (ox.no.) refers to a single atom in a compound

Oxidation Number Rules Table

Electrochemistry Table 1_Oxidation Numbers, downloadable AS & A Level Chemistry revision notes

Roman Numerals 

  • Transition metals are characterised by having variable oxidation numbers.
  • Oxidation numbers can be used in the names of compounds to indicate which oxidation number a particular element in the compound is in
  • Where the element has a variable oxidation number, the number is written afterwards in Roman numerals.
  • This is called the STOCK NOTATION (after the German inorganic chemist Alfred Stock), but is not as widely used for non-metals, so SO2 is often referred to as sulfur dioxide rather than sulfur(IV) oxide
  • For example, iron can be both +2 and +3 so Roman numerals are used to distinguish between them
    • Fe2+ in FeO can be written as iron(II) oxide
    • Fe3+ in Fe2O3 can be written as iron(III) oxide
  • More complicated examples include other atoms / ions as part of the formula 
    • Potassium manganate(VII) implies that the manganese is in a +7 oxidation
    • Potassium manganate(VII) contains the potassium ion K+ and the manganate ion MnO4 
      • Since the oxygen in the manganate ion is in the -2 oxidation state, there is a total of -8 from the oxygen
      • The manganate ion has an overall -1 charge, which means that the manganese ion must be in the +7 oxidation state

Oxidation Numbers - Calculations

Molecules or Compounds

  • In molecules or compounds, the sum of the oxidation numbers on the atoms is zero

Oxidation Number in Molecules or Compounds Table

  • Because CO2 is a neutral molecule, the sum of the oxidation states must be zero
  • For this, one element must have a positive oxidation number and the other must be negative

How do you determine which is the positive one?

  • The more electronegative species will have the negative value
  • Electronegativity increases across a period and decreases down a group
  • O is further to the right than C in the periodic table so it has the negative value

How do you determine the value of an element's oxidation number?

  • From its position in the periodic table and / or the other element(s) present in the formula

Hydrogen in metal hydrides, H-

  • An example of a metal hydride is sodium hydrogen, NaH which is a neutral compound 
  • The hydrogen atom is present as a hydride ion 
    • This has the symbol H- and therefore the oxidation state -1
  • We can also think of this as a sum
    • Since group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0)

Oxygen in peroxides, O-

  • Peroxides include hydrogen peroxide, H2O2, which again is a neutral compound 
  • The sum of the oxidation states of the hydrogen and oxygen must be 0 
  • Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to accommodate the neutral charge of the compound

Ions

  • The oxidation number of elements within an ion can be calculated using the rules shown in the Oxidation Number Rules Table (above)

Worked example

Deduce the oxidation number of the highlighted element in:

  1. The nitrite ion: NO2 
  2. The nitrate ion: NO3 
  3. The thiosulfate ion: S2O32– 
  4. The tetrathionate ion: S4O62– 

Answers

    • In all of the answers, the O is more electronegative than the other element, so the oxidation number of O will be –2

Answer 1:

    • The nitrite ion contains two oxygen atoms
      • 2 x –2 = –4
    • The overall ion has an oxidation state of –1, which means that the oxidation numbers of the elements in the ion must add up to –1
    • Therefore, the oxidation number of N is +3
      • N + (2 x –2) = –1
        N = +3

Answer 2:

    • The nitrate ion contains three oxygen atoms
      • 3 x –2 = –6
    • The overall ion has an oxidation state of –1, which means that the oxidation numbers of the elements in the ion must add up to –1
    • Therefore, the oxidation number of N is +5
      • N + (3 x –2) = –1
        N = +5

Answer 3:

    • The thiosulfate ion contains three oxygen atoms
      • 3 x –2 = –6
    • The overall ion has an oxidation state of –2, which means that the oxidation numbers of the elements in the ion must add up to –2
    • The oxidation number of both S atoms is +4
      • 2S + (3 x –2) = –2
        2S = +4
    • Therefore, the oxidation number of S is +2
      • 2S = +4
      • S = +2

Answer 4:

    • The tetrathionate ion contains six oxygen atoms
      • 6 x –2 = –12
    • The overall ion has an oxidation state of –2, which means that the oxidation numbers of the elements in the ion must add up to –2
    • The oxidation number of all four S atoms is +10
      • 4S + (6 x –2) = –2
        4S = +10
    • Therefore, the oxidation number of S is +2.5
      • 4S = +10
      • S = 2.5
    • This specific question is a rare example showing a fractional oxidation number for the element in question / sulfur
      • This is because the sulfur atoms are in two different environments 
        tetrathionate-ion-s-environments
      • The fractional oxidation number shows the average oxidation number of the sulfur environments 

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