Halfequations show only what happens to either the oxidised or the reduced species in a redox reaction. 
Halfequations
The oxidised species in a reaction loses electrons. This may be written down as a process that doesn't show where the electrons go to, or come from. Such an equation is termed a halfequation, as it only tells half of the story.
For the reaction:
CuSO_{4}(aq) + Zn(s)
ZnSO_{4}(aq) + Cu(s)

Electrons are transferred from the zinc metal to the copper(II) ions in solution. The process of reduction (addition of electrons) that happens to the copper (II) ions can be represented by the following halfequation:
Cu^{2+}(aq) + 2e
Cu(s)

Notice that there is no mention of the zinc in this halfequation. However, a similar halfequation could be written to represent the oxidation of Zinc in the reaction:
Zn(s)  2e
Zn^{2+}(aq)

This could also be written as:
Zn(s)
Zn^{2+}(aq) + 2e

Notice also that the sulfate ions do not appear in either halfequation. This is because they are not involved in the original reaction. The sulfate ions are merely spectators to the whole process. At the start of the reaction they are ions in solution and they remain unchanged at the end of the reaction. For this reason they are sometimes called 'spectator ions'.
Example: Write the halfequation for the reduction that occurs in the following reaction:
Reduction is gain of electrons. The hydrogen ions turn to hydrogen gas by gaining electrons:

Constructing the ionic equation
The full equation for a reaction can be constructed from the two halfequations by adding them together. However, before two halfequations can be added together the number of electrons in each halfequation must be made the same.
It is important to note here that only reduction halfequations can be added to oxidation halfequations and vice versa. It would be impossible to add two reduction halfequations together as we would end up with electrons that have no home to go to.
For the following two halfequations we can construct the full (ionic) equation by simple addition. We add each left hand side together, then we add each right hand side together, as if it were a simple maths sum:
Cu^{2+}(aq) + 2e
Cu(s) Cu^{2+}(aq) + Zn(s)
Zn^{2+}(aq) + Cu(s)

The electrons that appear on either side of the reaction arrow cancel out, leaving only the ions that react.
The above reaction is uncomplicated, as the number of electrons on each side of the reaction cancel out. The following example shows a situation in which the electrons do not initially cancel out, in which one halfequation has to be manipulated to equalise the electrons.
halfequation 1: Cu(s)
Cu^{2+}(aq)
+ 2e

halfequation 2: Ag^{+}(aq)
+ 1e Ag(s)

It should be seen that if we simply add these equations together there will be one extra electron remaining on the right hand side. Before the equations can be added, the second equation must be multiplied through by 2 to give:
halfequation 2: 2Ag^{+}(aq)
+ 2e 2Ag(s)

The halfequations can now be added to give:
2Ag^{+}(aq) + Cu(s)
2Ag(s) + Cu^{2+}(aq)

This is now the balanced ionic redox reaction.
Example: Given the two following halfequations, construct the full ionic equation for the redox reaction:
We can see that the number of electrons is the same on both sides so the halfequations can be added together directly:

Example: Given the two following halfequations, construct the full ionic equation for the redox reaction:
The number of electrons is NOT the same on both sides so the halfequations can NOT be added together directly. First the silver halfequation must be doubled to make the electrons the same in both equations:
Now they can be added together and the electrons cancelled out:
