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Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10 PDF Download

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  • 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:
    Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10
  • 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:
    Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10
  • This equation is an example of a balanced ionic equation which can be further split into two half equations illustrating oxidation and reduction individually:
    Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10
  • 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

Solved Example

Zinc displaces copper from a solution of copper(II)sulfate. Using ionic equations, determine which species undergoes oxidation and which species undergoes reduction.

  • Write the full equation
    • Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
  • Write the ionic equation
    • Zn (s) + Cu2+ (aq) + Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10(aq) → Zn2+ (aq) + Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10 (aq) + Cu (s)
  • Use the ionic equation to rule out / ignore spectator ions that are present as reactants and products
    • Oxidation & Reduction in Terms of Electrons | Chemistry for Grade 10is 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)

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.

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