How do you figure out the oxidation half-reaction and reduction half-reaction of this chemical equation

5Fe+2 + MnO4- + 8H+ -> 5Fe+3 +Mn+2 + 4H2O

I don't understand half reactions at all and I need to understand them for my test tomorrow. Please explain how to do this if you can. Thanks in advance!

Ok here is some basic information on oxidation reactions.

Oxidation: This is a loss of electrons( or an icnrease in the oxidation number of an atom)
Example 2+ to 3+

Reduction: This is the gain of electrons( or a decrease in the oxidation number of an atom)
Example 2+ to 1+

*Hint: I usually just look at whether the number goes up or down. If it goes up then it is oxidized. If it goes down it is reduced.

By looking at this formula Fe is being oxidized and Mn is being reduced and this is how you tell.

Looking at the charges Fe is the easiest one to recognize because it is a lone atom. The oxidation number goes up from 2+ on the left side to 3+ on the right side so it is being oxidized.

The oxidation half reaction is as follows:

5Fe(2+)---->5Fe(3+) To balance this you need to balance out the electrons on both sides.
(5)*(2+)---->(5)*(3+)
So the left side is 10+ and the right side is 15+ to balance this add 5e- to the right side and you get:

5Fe(2+)<---->5Fe(3+) + 5e- and that is the easy half reaction.

This leaves you with the reduction half reaction.
In MnO4(-) the oxidation # of Mn is 7+ because O always has a -2 and it has to equal
Mn + (4*)(-2)= -1 so Mn has to be 7+
Then on the right side Mn has an oxidation # of 2+ so it goes down from 7+ to 2+ meaning it is reduced.

Since this is a final equation and you don't have to do any redox reaction work for it the reduction half reaction is:

MnO4(-) + 8H(+) ----> Mn(+2) + 4H2O and you just need to find the electrons
-1 + (8)*(1+) ----> +2 + 0 leaving you with
+7---->+2 so add 5 electrons on the left side to get the final answer:

5e- + MnO4(-) + 8H(+) ----> Mn(+2) + 4H2O