Why is Q/nF equivalent to the amount of substance in electrolysis?
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.