Redox (OCR A Level Chemistry)

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 state of a single atom

Oxidation Numbers

• The oxidation state 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

Oxidation Rules Table

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

• From its position in the periodic table and/or
• The other element(s) present in the formula
• The oxidation states of all other atoms in their compounds can vary
• By following the oxidation number rules, the oxidation state of any atom in a compound or ion can be deduced
• The position of an element in the periodic table can act as a guide to the oxidation number

Oxidation Numbers & the Periodic Table

• Test your understanding on the following examples:

Answers

Are oxidation numbers always whole numbers?

• The answer is yes and no
• When you try and work out the oxidation numbers of sulfur in the tetrathionate ion S₄O₆^2- you get an interesting result!

The oxidation number of sulfur in S₄O₆^2- is a fraction

• The fact that the oxidation number comes out to +2.5 does not mean it is possible to get half an oxidation number
• This is only a mathematical consequence of four sulfur atoms sharing +10 oxidation number
• Single atoms can only have an integer oxidation number, because you cannot have half an electron!

Roman numerals

• Roman numerals are used to show the oxidation states of transition metals which can have more than one oxidation state
• Iron can be both +2 and +3 so Roman numerals are used to distinguish between them
  • Fe²⁺ in FeO is written as iron(II) oxide
  • Fe³⁺ in Fe₂O₃ is written as iron(III) oxide

• Metals can react with acid to form a salt and hydrogen

Metal + acid → salt + hydrogen

•  During this reaction, there are changes in oxidation number 
  • This means that the reaction can be classified as a redox reaction

Answer 1

Step 1: Write the balanced symbol equation

• • Zn + 2HCl → ZnCl₂ + H₂ 

Step 2: Deduce the changes in oxidation number

• • Zinc - starts at 0, changes to +2
  • Hydrogen - starts at +1, changes to 0
  • Chlorine - remains at -1 throughout

Step 3: Explain which species is reduced / oxidised

• • Zinc is oxidised as its oxidation number increases from 0 to +2
  • Hydrogen is reduced as its oxidation number decreases from +1 to 0

Answer 2

Step 1: Write the balanced symbol equation

• • Mg + H₂SO₄ → MgSO₄ + H₂ 

Step 2: Deduce the changes in oxidation number

• • Magnesium - starts at 0, changes to +2
  • Hydrogen - starts at +1, changes to 0
  • Sulfate ion - remains at -2 throughout

Step 3: Explain which species is reduced / oxidised

• • Magnesium is oxidised as its oxidation number increases from 0 to +2
  • Hydrogen is reduced as its oxidation number decreases from +1 to 0

We can identify the oxidation and reducing agents in a reaction by using the oxidation state.

• For example

• If we look at zinc, Zn, in the reaction above we can see that it increases from 0 to +2 in zinc sulfate, ZnSO₄  
• An increase in oxidation number indicates oxidation has occured
• Therefore zinc is the reducing agent
• If we look at sulfuric acid, H₂SO₄, the oxidation state of hydrogen has decreased from +1 to 0 in H₂
• A decrease in oxidation number indicates reduction has occurred 
• Therefore sulfuric acid is the oxidising agent 

Answer
Step 1: Deduce the oxidation numbers of nitrogen and chlorine in the equation (hydrogen = +1, oxygen = -2, sodium = +1                                   
   N in NH₃ is -3
   Cl in NaClO is +1
   N in N₂H₄ is -2
   Cl in NaCl is -1                

Step 2

Identify which species has been oxidised and which has been reduced by looking at the oxidation numbers
   Nitrogen is increasing in oxidation number, therefore has been oxidised
   Chlorine is decreasing in oxidation number, therefore has been reduced

Step 3

Identify the oxidising and reducing agent 
   NH₃ is the reducing agent (it has been oxidised itself)
   NaClO is the oxidising agent (it has been reduced itself) 

Remember, the whole species is the reducing agent, not just the element (e.g. NH₃ is the reducing agent, not N on its own)