Determining the charge on an ion through mole calculation in electrolyis
The suppose total and half reaction are:

$$\begin{align}
\ce{2 MCl_n &amp;-&gt; 2 M(s) + $n$ Cl2} \\
\ce{M^n+ + $n$ e- &amp;-&gt; M(s)} \\
\ce{2 Cl- &amp;-&gt; Cl2 + 2 e-}\\
\end{align}$$

The question is rather tricky.

Note that I am addressing the supposed question in hand, not the electrochemical evaluation.

In some sense, if we consider large scale motion, no electrons passed through the electrolyte, they just got on one side in and on the other side out.

In usual sense, there is 3:1 electron:halogenide molar ratio for $\ce{MCl3},$ therefore result would we $\pu{0.06 mol}.$

But in yet other sense, if we consider nanoscale motion, the $\pu{0.06 mol}$ of electrons passed through the electrolyte at the anode, and the same amount through the electrolyte at the cathode, therefore in total $\pu{0.12 mol}.$

By other words, $\pu{0.12 mol}$ is total molar amount of electrons involved in net electrochemical reaction, passing through the electrolyte between electroactive substances and electrodes.

I would say, the question is either intentionally tricky either badly formulated.