How can I work out what reaction will happen?
There are no 'free electrons' that I can see in ethene


That's correct, but...  In ethene there are two bonds between the two carbons that hold the two carbons together.  A chemist calls this a "double" bond.  The important thing about a double bond (that's not obvious from the drawing) is that the two bonds are not the same.  One is a strong sigma bond, just like in ethane; the other is a weaker pi bond.  While the electrons in the pi bond are not free, they are not held as tightly 



as the electrons in the sigma bond and are, therefore, more available for reaction.  This is why ethene is so much more reactive than ethane.


  Am I supposed to apply some logic... Or, am I just overthinking it?


Yes, there is logic and that body of logic is what organic chemistry is all about.  In the case of bromine reacting with ethene, the reaction belongs to a class of reactions known as electrophilic additions.  Most reactions in organic chemistry involve a molecule that has electrons available to share (a nucleophile) and a molecule that would like more electrons (an electrophile).  These two molecules react (form a chemical bond) by the one donating\sharing its electrons and the other molecule accepting\sharing those electrons.  Here, the pi bond in ethene has electrons that it is willing to share (they're not free, but they're not tightly held either), while the bromine molecule is highly polarizable, much like $\ce{Br^{\delta +}-Br^{\delta -}}$, with the positively polarized bromine atom wanting electrons.  The reaction proceeds as pictured below, to yield 1,2-dibromoethane.




  A side problem to this is why, if we place an alkene in bromine water,
  we don't get the same thing happening with ethane.


I bet you can answer that one now.  Ethene has a pi bond with electrons loosely held and available for reaction.  Ethane, on the other hand, only has a sigma bond where the electrons are more tightly held and less available for reaction.