Redox (OCR A Level Chemistry A): Revision Note
Oxidation Number
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
Molecules or Compounds In molecules or compounds, the sum of the oxidation number on the atoms is zero
Oxidation Number in Molecules or Compounds
Because CO2 is a neutral molecule, the sum of the oxidation number 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
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:
Worked Example
Deducing oxidation numbers Give the oxidation number of the elements in bold in these compounds or ions:
a. P2O5
b. SO42-
c. H2S
d. Al2Cl6
e. NH3
f. ClO2-
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 S4O62- you get an interesting result!
The oxidation number of sulfur in S4O62- 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
Redox Reactions & Equations
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
Worked Example
Explain why each of the following reactions is a redox reaction:
Zinc + hydrochloric acid → zinc chloride + hydrogen
Magnesium + sulfuric acid → magnesium sulfate + hydrogen
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
Examiner Tips and Tricks
Remember that oxidation number increases in oxidation reactions and decreases in reductions reactions
If you are asked to explain why a reaction is a redox reaction, you should always talk about one of the following:
Gain / loss of oxygen
Gain / loss of hydrogen
Gain / loss of electrons
Changes in oxidation numbers
Simply saying that a reaction is a redox reaction because "reduction and oxidation happen at the same time" is describing, not explaining
Interpreting Redox
We can identify the oxidation and reducing agents in a reaction by using the oxidation state.
For example
Zn (s) + H2SO4 (aq) → ZnSO4 (aq) + H2 (g)
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
Worked Example
Identify the oxidising agent and reducing agent in the following reaction:
2NH3 + NaClO → N2H4 + NaCl + H2O
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)
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