Calculate the standard oxidation potential from given potentials
You can write down the three half reactions:

\begin{aligned}
 \ce{Fe^{2+} + 2e- &amp;-&gt; Fe}      &amp; E_1&amp;= x\,\mathrm{V}\\
 \ce{Fe^{3+} + e-  &amp;-&gt; Fe^{2+}} &amp; E_2&amp;= y\,\mathrm{V}\\
 \ce{Fe^{3+} + 3e- &amp;-&gt; Fe}      &amp; E_3&amp;= z\,\mathrm{V}
\end{aligned}

The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry 


$x$ joules per coulomb, or
$96500x$ joules per mole of electrons, or
$2\times 96500x$ joules per mole of reaction.


In the same fashion, the second half reaction has the electrons carrying


$y$ joules per coulomb,
$96500y$ joules per mole of electrons,
the same $96500y$ joules per mole of reaction.


The third half reaction is the sum of the first two, so its electrons carry


$2\times 96500x + 96500y$ joules per mole of reaction, which is
$3\times 96500 z$ joules per mole of reaction, or
$96500z$ joules per mole of electrons, or
$z$ joules per coulomb.


Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.