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Oxidation & Reduction in Terms of Electrons (AQA GCSE Chemistry)
Revision Note
Oxidation & reduction in terms of electrons
Higher tier only
- As well as understanding oxidation and reduction in terms of oxygen, you need to understand these reactions in terms of electrons
- When a substance loses electrons it is oxidised
- When a substance gains electrons it is reduced
- If this occurs in the same reaction, the reaction is a redox reaction
- For example, when iron reacts with a compound of copper such as copper sulfate a displacement reaction occurs
iron + copper sulfate → iron(II) sulfate + copper
Fe + CuSO4 → FeSO4 + Cu
- We can write this as an ionic equation
Fe + Cu2+ + SO42– → Fe2+ + SO42– + Cu
- The sulfate ions, SO42-, appear on both sides of the equation unchanged
- This means that they are spectator ions and do not participate in the chemistry of the reaction
- So, they can be removed from the equation
Fe + Cu2+→ Fe2+ + Cu
- This balanced ionic equation can be further split into two half equations illustrating oxidation and reduction individually
Fe → Fe2+ + 2e–
Cu2+ + 2e–→ Cu
- The iron has lost electrons to become a positive ion, so has been oxidised
- The positive copper ion has gained electrons to become an atom, so have been reduced
The redox reaction between Fe and Cu2+
The Fe atom is oxidised (loses electrons) and the Cu2+ ion is reduced (gains electrons)
Electrolysis
- Oxidation and reduction take place during the process of electrolysis at the anode (positive electrode) and the cathode (negative electrode)
- Positive ions are attracted towards the cathode
- Reduction (gain of electrons) takes place here
- E.g. Pb2+ + 2e– → Pb
- Reduction (gain of electrons) takes place here
- Negative ions are attracted towards the anode
- Oxidation (loss of electrons) takes place here
- E.g. 2Br– → Br2 + 2e–
- Oxidation (loss of electrons) takes place here
Examiner Tip
Remember: OIL RIG - Oxidation Is Loss, Reduction Is Gain of electrons
Worked example
Which change in the following equation is oxidation?
V3+ + Fe3+ → V4+ + Fe2+
Answer:
- Step 1 - Identify the changes for each species
-
- V3+ to V4+
- V3+ has lost 1 electron
- Fe3+ to Fe2+
- Fe3+ has gained 1 electron
- V3+ to V4+
- Step 2 - Identify each change as either oxidation and reduction
- V3+ to V4+ is oxidation
- Fe3+ to Fe2+ is reduction
- Therefore, V3+ has been oxidised
Identifying oxidised & reduced species
Higher tier only
- 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)
Examiner 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.
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