What does iron (III) get reduced to in a galvanic cell: solid iron or iron (II)?
$$\begin{align}
\ce {Fe^{2+} {(aq)} + 2e^{-} &amp;&lt;=&gt; Fe {(s)}} &amp;E^⦵= \pu{-0.44V}\\
\ce {Fe^{3+} {(aq)} + e^{-} &amp;&lt;=&gt; Fe^{2+} {(aq)}} &amp;E^⦵= \pu{+0.77V}\\
\ce {Mg^{2+} {(aq)} + 2e^{-} &amp;&lt;=&gt; Mg {(s)}} &amp;E^⦵= \pu{-2.37V}\\
\end{align}$$

Thus 

$$
\begin{align}
\ce {Mg {(s)} + 2Fe^{3+} {(aq)} &amp;&lt;=&gt; Mg^{2+} {(aq)} +2Fe^{2+} {(aq)}}&amp;&amp;E^⦵_{\text{cell}}= 0.77--2.37=\pu{+3.14V}\\
\ce {Mg {(s)} + Fe^{2+} {(aq)} &amp;&lt;=&gt; Mg^{2+} {(aq)} +Fe {(s)}}&amp;&amp;E^⦵_{\text{cell}}= -0.44--2.37=\pu{+1.93V}\\
\end{align}$$

Both reactions are feasible, and I'd imagine the first would proceed until the concentration of $\ce {Fe^{3+} {(aq)}}$ is very low, and then the second would start at a noticeable rate.
This assumes, of course, the iron solid is able to deposit on the carbon cathode, and I don't see why not.

Hope this was helpful.

(Thank you Dr. J for pointing out my typo)