How to find the potential of a mixed Iron/ Cer solution?
First it helps to have an idea of what will happen. Since 1.72 V is a
lot more than 0.771 V (a difference of 1V in chemistry goes a
long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much
better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of
enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$.
Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$
will be reduced to $\ce{Ce^{3+}}$.

In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$
left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look
at the number of moles of each ion that went into the reaction at the
moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts
with one unit of $\ce{Ce^{4+}}$.



So now you've figured out which is the limiting reagent (the one that gets
consumed almost entirely). Let's suppose for the sake of argument it's
the $\ce{Fe^{2+}}$. Because you know how many
moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know
how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can
find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the
equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$,
because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's
all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.

If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same
logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and
$\ce{Ce^{3+}}$ in the equilibrium mix.

So now you have 3 concentrations out of 4.  Finding the 4th requires
knowing the equilibrium constant, which you can calculate from the
difference in potentials. (Write the combined full reaction based on the half-reactions.)

You'll find that the 4th concentration is
very small, justifying the claim that one of the reagents
was almost entirely consumed.

If you're with me so far, please update the question with these
results, and if you still need help with finding the potential at that
point, we can take it from there...