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,

$$\ce{M^{n+} + n e^- -> M}$$

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.

• Ah, that makes perfect sense. Thank you! – Tetsu Dau Sep 4 '19 at 5:52
• Often, $z$ is used for the stoichiometric factor, and $n$ for the amount of substance, so it becomes $n = \frac{Q}{z F}$, see e.g. bipm.org/utils/en/pdf/SIApp2_mol_en.pdf @rv7 – Karsten Theis Sep 4 '19 at 9:59