How do I find out if ions are oxidised/ reduced based on half cell potentials?
Look at this table of half-reactions: http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm

I'll explain how it contradicts your answer to the first question; you can then rework your answers and update your question with justifications.


  Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?


One entry of the table says $$ \ce{Cr^{3+}(aq) + e- -&gt; Cr^{2+} (aq)} \quad –0.424.$$    

The $-0.424$ is what's called the standard potential, expressed in volts.  It is a measure of how favorable the half-reaction is (compared with a standard half-reaction).  The higher the number, the more favored the products will be.

Now you have to find a half-reaction involving $\ce{Fe^{2+}}$.  Since you want Fe to reduce something, it will itself be oxidized.  The table shows no half-reaction where $\ce{Fe^{2+}}$ is oxidized, but that's because by convention only reduction half-reactions are shown; to get the opposite, oxidation, half-reactions, just swap the terms and negate the potential.  That is, take this line from the table:
$$
\ce{Fe^{3+}(aq) + e^{–}  -&gt; Fe^{2+}(aq) } \quad 0.771
$$
and turn it into 
$$
 \ce{Fe^{2+}(aq) -&gt; Fe^{3+}(aq) + e^{–}} \quad {-}0.771
$$

Now add this, term by term, to the Cr half-reaction. The result is
$$
\ce{Fe^{2+} + Cr^{3+} -&gt; Fe^{3+} + Cr^{2+}} \quad {-}1.195
$$
If the potential in the final reaction is negative, the reaction is impossible (more precisely, equilibrium lies to the left --- in the case very far to the left, becase 1 volt goes a long way).

That's the basics.  There is a subtlety involved when the number of electrons exchanged is not the same in the two half-reactions; you'll find it in your textbook.  You could try answering the other 5 questions now and if you get stuck, tell us precisely where.