Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 2 - 8.5 Oxidation and reduction-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 8.5 Oxidation and reduction- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 8.5 Oxidation and reduction- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

Key Concepts:

8.5 understand oxidation and reduction in terms of electron transfer and changes in oxidation number, and the application of these ideas to reactions of s-block and p-block elements

Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

8.5 Oxidation and Reduction 

Oxidation and reduction (redox) reactions involve electron transfer and corresponding changes in oxidation number. These concepts are essential for understanding reactions of s-block and p-block elements.

Definitions (Electron Transfer)

  • Oxidation: loss of electrons
  • Reduction: gain of electrons

Remember: OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Definitions (Oxidation Number)

  • Oxidation: increase in oxidation number
  • Reduction: decrease in oxidation number

  • Oxidation and reduction always occur together.
  • Electrons lost = electrons gained.

Example 1: Reaction of Sodium with Chlorine (s-block)

\( \mathrm{2Na + Cl_2 \rightarrow 2NaCl} \)

  • Na: \( \mathrm{0 \rightarrow +1} \) → oxidation (loses 1 e⁻)
  • Cl: \( \mathrm{0 \rightarrow -1} \) → reduction (gains 1 e⁻)

Half-equations

  • \( \mathrm{Na \rightarrow Na^+ + e^-} \)
  • \( \mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-} \)

Therefore, sodium is oxidised and chlorine is reduced.

Example 2: Reaction of Magnesium with Oxygen (s-block)

\( \mathrm{2Mg + O_2 \rightarrow 2MgO} \)

 

  • Mg: \( \mathrm{0 \rightarrow +2} \) → oxidation
  • O: \( \mathrm{0 \rightarrow -2} \) → reduction

Example 3: Reaction of Chlorine with Water (p-block)

\( \mathrm{Cl_2 + H_2O \rightarrow HCl + HClO} \)

  • Cl in \( \mathrm{Cl_2} \) = 0
  • Cl in \( \mathrm{HCl} \) = −1 (reduction)
  • Cl in \( \mathrm{HClO} \) = +1 (oxidation)

This is a disproportionation reaction (same element oxidised and reduced).

Application to s-block Elements

 

  • s-block metals (Group 1 and 2) are easily oxidised.
  • They lose electrons to form positive ions.
  • Therefore, they are strong reducing agents.

Example: \( \mathrm{Na \rightarrow Na^+ + e^-} \)

Application to p-block Elements

  • p-block elements show variable oxidation states.
  • Can act as oxidising or reducing agents.
  • Can undergo disproportionation reactions.

Key Points

  • Oxidation = loss of electrons / increase in oxidation number.
  • Reduction = gain of electrons / decrease in oxidation number.
  • Redox reactions involve both processes simultaneously.
  • s-block elements are easily oxidised.
  • p-block elements show more varied redox behaviour.

Therefore, redox concepts explain the reactivity and behaviour of elements across the periodic table.

Example 1 :

Identify the oxidising and reducing agents in the reaction:

\( \mathrm{Mg + Cl_2 \rightarrow MgCl_2} \)

▶️ Answer/Explanation

Mg is oxidised (0 → +2) → reducing agent.

Cl₂ is reduced (0 → −1) → oxidising agent.

Example 2 :

Explain why chlorine can act as both an oxidising agent and a reducing agent.

▶️ Answer/Explanation

Chlorine has multiple oxidation states.

It can gain electrons (reduction) or lose electrons (oxidation).

Example: disproportionation in water.

Therefore, it can act as both oxidising and reducing agent.

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