How do you work out whether reduction would occur based on reduction potentials I'm having a bit of trouble with standard reduction potentials. Specifically, when you're given a table of half reactions and reduction potentials...is the potential of the reverse reaction (the oxidation potential) just the negative of the reduction potential? So, say you were given the two reactions :

$\ce{Cl2 + 2e- -&gt; 2Cl- }$   $     E = 1.36$V

$\ce{F2 + 2e- -&gt; 2F-  } $   $  E = 2.87$V

and the question asked whether F- would reduce Cl2 to Cl-, so

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

...how would you work this out?
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.