Oxidising and Reducing Agents

Oxidising Agents and Reduction Agents

  1. In a redox reaction, a compound that is reduced is the oxidizing agent. An oxidising agent is a substance which oxidises something else.
  2. Inversely, a compound that is oxidised is the reducing agent. A reducing agent reduces something else.

Example:

  1. In this reaction, iron(III) oxide is reduced. Therefore it is the oxidising agent. It has oxidised carbon monoxide to become carbon dioxide.
  2. The carbon monoxide is oxidised. Therefore it acts as the reducing agent. It has reduced iron(III) oxide to become iron metal.

Example:

In this reaction:

  1. copper(II) oxide is reduced, hence it is the oxidising agent.
  2. carbon is oxidised, hence it is the reducing agent.
Commonly Used Oxidising and Reducing Agent and Their Half Equation

Below are the oxidising and reducing agent that normaly in use

Oxidising Agent

Acided Potassium Manganate (VII)

MnO4 + 8H+  +  5e   →      Mn2+ +      4H2O

Observation:

Acided Potassium Dicromate (VI)

Cr2O72- +   14H+ +  6e   →      2Cr3+ +     7H2O

Observation:

Hydrogen Peroxide

H2O2 +     2H+ +   2e  →     2H2O

Concentrated Nitric Acid

NO3 +    4H+ +   3e  →      NO +   2H2O

Reducing Agent

Sulphur Dioxide

SO2 +     2H2O →      SO42- +   4H+ +  2e

Hydrogen Sulphide

H2S →     2H+ +  S +  2e

Sodium Sulphite Aqueous

SO32- +  H2O →     SO42- + 2H+ +  2e

Tin(II) Chloride Aqueous

Sn2+ →     Sn4+ +  2e
Forming Ionic Equation from the Half Equation
  1. Ionic equation of a redox reaction can be formed from the half equations of the reaction.
  2. When writing the ionic equation, make sure that the number of electrons in both the oxidation reaction and reduction reaction is balanced.

Example:
Redox reaction between potassium iodide and potassium manganate (VII)

Half equations
2I → I2 + 2e ——-(1)

MnO4 +  8H+ + 5e → Mn2+ +  4H2O ——-(2)

To make the number of electrons in both chemical equation equal
(1) x 5

10I → 5I2 + 10e

(2) x 2

2MnO4 +  16H+ + 10e → 2Mn2+ +  8H2O

Ionic Equations
Add the 2 equations together. Exclude the electrons.

10I + 2MnO4 +  16H+ → 5I2 2Mn2+ +  8H2O

Example:
Redox reaction between iron(II) sulphate and potassium dicromate(VI)

Half equations

Fe2+ → Fe3+ + e ——-(1)

Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O ——-(2)

To make the number of electrons in both chemical equation equal
(1) x 5

6Fe2+ → 6Fe3+ + 6e

(2)

Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O

Ionic Equations
Add the 2 equations together. Exclude the electrons.

6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

Example:
Redox reaction between iron(III) nitrate and sulphur dioxide gas
Half equations

Fe3+ → Fe2+ + e ——-(1)

SO2 + 2H2O → SO42- + 4H+ + 2e ——-(2)

To make the number of electrons in both chemical equation equal
(1) x 2

2Fe3+ → 2Fe2+ + 2e

(2)

SO2 + 2H2O → SO42- + 4H+ + 2e

Ionic Equations
Add the 2 equations together. Exclude the electrons.

2Fe3+ + SO2 + 2H2→ 2Fe2+ SO42- + 4H+

Example:
Redox reaction between iron(III) chloride and hydrogen sulphide gas
Half equations

Fe3+ → Fe2+ + e ——-(1)

H2S → 2H+ + S + 2e ——-(2)

To make the number of electrons in both chemical equation equal
(1) x 2

2Fe3+ → 2Fe2+ + 2e
(2)

H2S → 2H+ + S + 2e

Ionic Equations

Add the 2 equations together. Exclude the electrons.

2Fe3+ + H2S → 2Fe2+ + 2H+ + S

Leave a Comment