Here is some extra information that you should know:
1)When you do half reactions one equation should have electrons on the left side and the other should have electrons on the right side. If you end up with them on the same side then you did something wrong.
2)When you add the 2 half reactions together you need to have the same number of electrons on both sides so they cancel
Example given these half reactions:
Cu--->Cu(2+) + 2e- and 6e- + 6H(+) + 2HNO3---> 2NO + 4H2O
If you look at these you have 2e- on the right side of the first equation but 6e- on the left side of the other. To make sure that the electrons would cancel when you add them together you have to multiply the first half reaction by 3 to get:
3Cu--->3Cu(2+) + 6e- This way the electrons cancel when you add the 2 half reactions together getting a final equation of:
3Cu + 6H(+) + 2HNO3---> 3Cu(2+) + 2NO + 4H2O
This is why you have 5's in your Fe equation, because the original equation would have just been
Fe(2+)---->Fe(3+) +1e- so it was multiplied by 5 to cancel with the 5e- in the other half reaction.
I hope this helped to clarify some things. I'm not really sure how deep you went into redox reactions so I hope I didn't confuse you with some of the extra stuff I gave you.