Why is Q/nF equivalent to the amount of substance in electrolysis? I am told that

$$Q = nF$$

where $Q$ is the quantity of current, $n$ is the amount of electrons and $F$ is Faraday's constant. Since Faraday's constant is the charge carried by 1 mol of electrons, multiplying it by the amount of electrons to give the total charge makes perfect sense. But why does dividing $Q$ by $nF$ gives the amount of substance? Won't the answer just be 1 for all possible values?
For a general electrolytic reaction,

$$\ce{M^{n+} + $n$ e^- -&gt; M}$$

where $n$ moles of electrons are consumed in depositing 1 mole of substance ($n$ is also known as $n$-factor). So, for $x$ moles of substance, $nx$ moles of electrons would be consumed. Therefore,

$$Q = nxF,$$

dividing by $nF$ on both sides, we have

$$\frac{Q}{nF} = x,$$

which is the amount of substance.