Writing half-equations
In order to identify what is occurring in a redox reaction, it is often important to divide the reaction into two half-equations which separate the reduction and the oxidation processes.
For example, alcohol (CH3CH2OH) can be oxidised by oxygen to form acetic acid (CH3COOH). This reaction can cause wine to develop a vinegary taste if exposed to air. The reaction can be shown by the following equation.
CH3CH2OH + O2 → CH3COOH + H2O
This redox reaction can be divided into the two half-equations shown below.
Reduction: O2 + 4 H+ + 4 e-→ 2 H2O
Oxidation: CH3CH2OH + H2O → CH3COOH + 4 H+ + 4 e-
The following shows how these half-equations have been constructed.
Constructing the reduction half-equation
There are rules to follow to help you write half-equations.
Click on the steps to reveal the method.
Write the formula of the reactant and product.
O2 → H2O
Balance the atoms of the element being oxidised or reduced.
O2 → 2 H2O
Balance the oxygen atoms by adding H2O to the side of the equation short of oxygen.
O2 → 2 H2O (no change required – oxygen atoms already balanced)
Balance the hydrogen atoms by adding H+ to the side of the equation short of hydrogen.
O2 + 4 H+ → 2 H2O
Calculate the total amount of charge on each side of the equation.
O2 + 4 H+ (charge = +4) → 2 H2O (charge = 0)
Add enough electrons (e-) to the side of the equation with the most positive charge in order to balance the electrical charge.
O2 + 4 H+ + 4 e-→ 2 H2O
Constructing the oxidation half-equation
Click on the steps to reveal the method.
Write the formula of the reactant and product.
CH3CH2OH → CH3COOH
Balance the atoms of the element being oxidised or reduced.
CH3CH2OH → CH3COOH
Balance the oxygen atoms by adding H2O to the side of the equation short of oxygen.
CH3CH2OH + H2O → CH3COOH
Balance the hydrogen atoms by adding H+ to the side of the equation short of hydrogen.
CH3CH2OH + H2O → CH3COOH + 4 H+
Calculate the total amount of charge on each side of the equation.
CH3CH2OH + H2O (charge = 0) → CH3COOH + 4 H+ (charge = +4)
Add enough electrons (e-) to the side of the equation with the most positive charge in order to balance the electrical charge.
CH3CH2OH + H2O → CH3COOH + 4 H+ + 4 e-
A printable version of the rules for balancing half-equations list is available here 
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Practise you skills in constructing half-equations by completing these consolidation problems .
The answers can be accessed here .
