Redox & Electron Transfer (CIE IGCSE Chemistry)

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Oxidation & Reduction

• Redox reactions can also be defined in terms of electron transfer
• Oxidation is a reaction in which an element, ion or compound loses electrons
  • The oxidation number of the element is increased
  • This can be shown in a half equation, e.g. when silver reacts with chlorine, silver is oxidised to silver ions:

Ag → Ag⁺ + e^-

• Reduction is a reaction in which an element, ion or compound gains electrons
  • The oxidation number of the element is decreased
  • This can be shown in a half equation, e.g. when oxygen reacts with magnesium, oxygen is reduced to oxide ions:

O₂ + 4e^- → 2O^2-

Example: Identifying Redox Reactions

zinc + copper sulphate → zinc sulphate + copper

Zn + CuSO₄ → ZnSO₄ + Cu

• The ions present (with state symbols) in the equation are:

Zn(s) + Cu²⁺(aq) + SO₄^2-(aq) →Zn²⁺(aq) + SO₄^2-(aq) + Cu(s)

• The spectator ions (those that do not change) are SO₄^2-(aq)
• These  can be removed and the ionic equation written as:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

• By analysing the ionic equation, we can split the reaction into two half equations by adding in the electrons to show how the changes in charge have occurred:

Zn(s) → Zn²⁺(aq) + 2e^-

Cu²⁺(aq) +2e^- → Cu(s)

• It then becomes clear that zinc has been oxidised as it has lost electrons 
• Copper ions have been reduced as they have gained electrons

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Oxidation Number

• The oxidation number (also called oxidation state) is a number assigned to an atom or ion in a compound which indicates the degree of oxidation (or reduction)
• It shows the number of electrons that an atom has lost, gained or shared in forming a compound
• The oxidation number helps you to keep track of the movement of electrons in a redox process
• It is written as a +/- sign followed by a number (not to be confused with charge which is written by a number followed by a +/- sign)
• E.g. aluminium in a compound usually has the oxidation state +3
• A few simple rules help guide you through the process of determining the oxidation number of any element

Table of Rules for Assigning Oxidation Numbers 

• Redox reactions can be identified by the changes in the oxidation number when a reactant goes to a product

 

Answer: 

• • The species that has been oxidised is iodine 
    • The oxidation number of I^- is -1
    • The oxidation number of iodine in I₂ is 0
    • The oxidation number has increased so the iodine has been oxidised (lost electrons)
    • 2I^-(aq) → I₂(s) +2e^-
  • The species that has been reduced is chloride ions
    • The oxidation number of chlorine as Cl₂ is 0.
    • The oxidation number of Cl^- is -1
    • The oxidation number has decreased so the Cl^- has been reduced (gained electrons)
    • Cl₂(g) + 2e^-  → 2Cl^-(aq)

Identifying Redox Reactions by Colour Changes

• The tests for redox reactions involve the observation of a colour change in the solution being analysed
• Two common examples are acidified potassium manganate(VII), and potassium iodide
• Potassium manganate(VII), KMnO₄, is an oxidising agent which is often used to test for the presence of reducing agents
• When acidified potassium manganate(VII) is added to a reducing agent its colour changes from purple to colourless

Diagram to show the colour change when potassium manganate(VII) is added to a reducing agent

• Potassium iodide, KI, is a reducing agent which is often used to test for the presence of oxidising agents
• When added to an acidified solution of an oxidising agent such as aqueous chlorine or hydrogen peroxide (H₂O₂), the solution turns a red-brown colour due to the formation of iodine, I₂:

2KI (aq) + H₂SO₄ (aq) + H₂O₂ (aq) →  I₂ (aq) + K₂SO₄ (aq) + 2H₂0 (l)

• The potassium iodide is oxidised as it loses electrons and hydrogen peroxide is reduced, therefore potassium iodide is acting as a reducing agent as it will itself be oxidised:

2I^- →  I₂ + 2e^-

Diagram to show the colour change when potassium iodide is added to an oxidising agent

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Oxidising agent

• A substance that oxidises another substance, and becomes reduced in the process
• An oxidising agent gains electrons as another substance loses electrons
• Common examples include hydrogen peroxide, fluorine and chlorine

Reducing agent

• A substance that reduces another substance, and becomes oxidised in the process
• A reducing agent loses electrons as another substance gains electrons
• Common examples include carbon and hydrogen
• The process of reduction is very important in the chemical industry as a means of extracting metals from their ores 

Example

CuO + H₂ → Cu + H₂O

• In the above reaction, hydrogen is reducing the CuO and is itself oxidised as it has lost electrons, so the reducing agent is therefore hydrogen:

H₂ → 2H⁺ + 2e^-

• The CuO is reduced to Cu by gaining electrons and has oxidised the hydrogen, so the oxidising agent is therefore copper oxide

Cu²⁺ +2e^- →  Cu

Answer

Step 1 - Write half equations to work out what has gained/lost electrons

Fe → Fe²⁺ + 2e^-

Br₂ + 2e^- → 2Br^-

Fe loses electrons; Br₂ gains electrons

Step 2 - Deduce what has been oxidised/reduced (remember OIL RIG)

Fe has been oxidised as it has lost electrons

Br₂ has been reduced as it has gained electrons

Step 3 - Identify the reducing agent

Fe is the reducing agent as it has been oxidised by losing electrons and caused Br₂ to be reduced as it gained electrons