Determining the charge on an ion through mole calculation in electrolyis During electrolysis $\pu{0.02 mol}$ of metal $\ce{X}$ is deposited on cathode when $\pu{0.12 mol}$ of electrons are passed through molten chloride of $\ce{X}.$ What is the formula of the chloride?


I have tried solving it and my answer is $\ce{XCl6}$ which is incorrect. The correct answer given is $\ce{XCl3}.$

In the explanation the book says that $\pu{0.06 mol}$ of electrons have flown to cathode and goes on to divide the amount of electrons at cathode by the amount of metal $\ce{X}$ $(0.06/0.02),$ so it's $3$ which means $1$ atom of $\ce{X}$ gains $3$ electrons, giving it $+3$ charge.

What I do not get is what about the other $\pu{0.06 mol}$ of electrons. Isn't metal $\ce{X}$ supposed to gain all of the $\pu{0.12 mole}$ of electrons? Shouldn't cathode have the total $\pu{0.12 mol}$ of electrons?
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.