Oxidation & Reduction in Terms of Electrons (AQA GCSE Chemistry)
Revision Note
Author
StewartExpertise
Chemistry Lead
Oxidation & Reduction in Terms of Electrons
Higher Tier Only
- Displacement reactions can be analysed in terms of redox reactions by studying the transfer of electrons
- For the example of magnesium and copper sulfate, a balanced equation can be written in terms of the ions involved:
Mg (s) + Cu2+ (aq) + SO42- (aq) → Mg2+ (aq) + SO42- (aq) + Cu (s)
- The sulfate ions, SO42-, appear on both sides of the equation unchanged hence they are spectator ions and do not participate in the chemistry of the reaction so can be omitted:
Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s)
- This equation is an example of a balanced ionic equation which can be further split into two half equations illustrating oxidation and reduction individually:
Mg → Mg2+ + 2e–
Cu2+ + 2e–→ Cu
- The magnesium atoms are thus oxidised as they lose electrons
- The copper ions are thus reduced as they gain electrons
Exam Tip
Remember: OIL RIG - Oxidation Is Loss, Reduction Is Gain of electrons
Identifying Oxidised & Reduced Species
- Using the principles of electron loss and gain it is possible to identify which species undergo oxidation and reduction in redox reactions
Worked example
Zinc displaces copper from a solution of copper(II)sulfate. Using ionic equations, determine which species undergoes oxidation and which species undergoes reduction.
Answer
-
- Write the full equation
- Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
- Write the ionic equation
- Zn (s) + Cu2+ (aq) + (aq) → Zn2+ (aq) + (aq) + Cu (s)
- Use the ionic equation to rule out / ignore spectator ions that are present as reactants and products
- is present as a reactant and a product so it can be ignored
- Use the ionic equation to identify the species that is oxidised (OIL)
- Zn (s) → Zn2+ (aq) + 2e–
- Use the ionic equation to identify the species that is reduced (RIG)
- Cu2+ (aq) + 2e– → Cu (s)
- Write the full equation
Exam Tip
After writing half equations, you can see if they are correct by checking that the number of electrons on either side is the same, which should combine to give 0 charge.
You've read 0 of your 0 free revision notes
Get unlimited access
to absolutely everything:
- Downloadable PDFs
- Unlimited Revision Notes
- Topic Questions
- Past Papers
- Model Answers
- Videos (Maths and Science)
Did this page help you?