How do you work out whether reduction would occur based on reduction potentials
First, using observations, since the E value for fluorine's reduction is higher, fluorine more preferentially undergoes reduction compared to chlorine. Furthermore, if you know your periodic table, fluorine being higher in the group more readily gains a valence electron (with fewer electron shells and hence a stronger force of attraction) and would be reduced preferentially and oxidise chloride ions.

Now using the numbers

We know the following formula: $$E^o_{cell}=E^o_{cathode}-E^o_{anode}$$
And also that oxidation is at the anode and reduction at the cathode.

Now, if and only if $E^o_{cell}$ is positive would we have a spontaneous reaction.
All that is left is to consider the two cases and see which would work.

Case 1: $\ce{2 F- +  Cl2 -&gt; 2Cl- + F2 }$

Reduction Equation (Chlorine): $\ce{Cl2 + 2e- -&gt; 2Cl-  }$,  $E^o_{cathode} = 1.36\,V$

Oxidation Equation (Fluorine): $\ce{2F- -&gt; F2 +2e-  }$,  $E^o_{anode} = 2.87\,V$

Then $$E^o_{cell}=1.36\,V-2.87\,V=-1.51\,V\lt0\,V$$ 

Hence this reaction would  not be spontaneous and not take place in this direction.

Case 2: $\ce{2 Cl- +  F2 -&gt; 2F- + Cl2 }$

Reduction Equation (Fluorine): $\ce{F2 + 2e- -&gt; 2Cl-  }$,  $E^o_{cathode} = 2.87\,V$

Oxidation Equation (Chlorine): $\ce{2Cl- -&gt; Cl2 +2e-  }$,  $E^o_{anode} = 1.36\,V$

Then $$E^o_{cell}=2.87\,V-1.36\,V=1.51\,V\gt0\,V$$

Hence this reaction would be spontaneous and take place in this direction.