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7.1    Oxidation and Reduction

7.1    Oxidation and Reduction

Redox reactions

  • Oxidation is:
    1. gain of oxygen
    2. loss of hydrogen
    3. loss of electrons
    4. increase in oxidation number

 

  • Reduction is:
    1. gain of hydrogen
    2. loss of oxygen
    3. gain of electrons
    4. decrease in oxidation number

 

  • Oxidation and reduction always occur simultaneously in a chemical reaction. Such reaction is called a redox reaction. In a redox reaction, one substance must be oxidised and another must be reduced

  • An oxidising agent is a substance which oxidises another substance and itself is reduced.
  1. A reducing agent is a substance which reduces another substance and itself is oxidised.

 

  • In short, an oxidising agent undergoes reduction while a reducing agent undergoes oxidation

 

  • Disproportionation is a redox reaction in which both oxidation and reduction occurs on the same The atom is simultaneously oxidised and reduced.

 

Oxidation state (oxidation number) of a substance

  • Oxidation state shows that total number of electrons which have been removed from or added to an element to get to its present state
  • When electrons have been removed, the oxidation number increases.(positive)
  • When electrons have been added, the oxidation number decreases.(negative)
  • Since removing electrons is an oxidation process, therefore oxidation is the increase in oxidation number
  • Since adding electrons is a reduction process, therefore reduction is the decrease in oxidation number.
  • For example, from V to V²⁺, two electrons have been removed, therefore the oxidation state of is +2. From V to V³⁺, three electrons have been removed, therefore the oxidation state is +3. Removing another electron gives:

V³⁺ + H2O → VO²⁺ + 2H⁺ + e⁻

 

Four electrons have been removed starting from V, therefore the oxidation state is +4. In all cases, V has been oxidised.

 

  • Another example, from S to S²⁻, two electrons have been added, therefore the oxidation state is -2. S is said to have been reduced

 

Rules to determine oxidation numbers of a substance

  • All atoms in an atom, molecule or ion can be given an oxidation number
  • The rules to determine the oxidation number of a substance:
  1. All free atoms in elements have an oxidation number of zero. e.g.
  2. For simple ions, the oxidation number is the same as the charge on the ion. e.g.
  3. For polyatomic ion, the sum of all the oxidation numbers of the atoms in the ion is equal to the charge of the ion e.g. 
  4. For a neutral covalent molecule, the sum of all the oxidation numbers of the atoms in the molecule is equal to zero. e.g. 
  5. For Group I and Group II elements, their oxidation number are always +1 and +2 For aluminium, it is always +3.
  6. For hydrogen, its oxidation number is always +1 except in metal For example, NaH, where its oxidation number is -1.
  7. For oxygen, its oxidation number is always -2 except in peroxides and fluorine For example, BaO2, where its oxidation number is -1.
  8. For fluorine, its oxidation number is always -1, with no exceptions
  • A summary:

  • To work out the oxidation number of a particular atom in a molecule/ ion, find the sum of all the oxidation number of the atoms present and equate it to zero/ charge of the ion. An example:

 

Balancing redox equations

1) There are two methods to balance complicated redox equations:

  1. Using electron half-equations.
  2. Using changes in oxidation number

2)  Using electron half-equations:

  1. In this method, the redox equation is divided into two half-equations. One for oxidation and another for reduction
  2. Steps(in acidic condition):
    • Divide the equation or information given into two half-equations.
    • Balance all other elements other than oxygen and hydrogen
    • Balance the oxygen by adding H2O to the appropriate side of the equation
    • Balance the hydrogen by adding H⁺ to the appropriate side of the equation
    • Balance the charge by adding electrons to the appropriate side of the equation.
    • Combine two half-equations such that the electrons cancel out each other
  3. An example:
  4. Steps(in alkaline condition):
  • Balance the equation as if it happens in an acidic condition first
  • Add OH⁻ to both sides of the equation to react with all the H⁺ to form H2O.
  • Cancel the excess H2O on either side of the equation

Note: If it is obvious enough that OH⁻ must be added in order to balance the equation, then add OH⁻ instead.

3) Using changes in oxidation number:

  1. This method utilises the fact that an increase in certain amount of oxidation number in a substance must be accompanied by a decrease in same amount of oxidation number in another substance
  2. An example: