Q:

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?

A:

For a general electrolytic reaction,
$$\ce{M^{n+} + $n$ e^- -> M}$$
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.

