[DavidCM]

Hi. I'm trying to solve this problem but I am not able to find what can I do.

How many grams of Zn metal and chlorine are released at the electrodes of an electrolytic cell containing an aqueous solution of ZnCl₂, if 1.80 faradays are passed through it?

First of all, the semi-reactions:

Zn²⁺ + 2e Zn(s)
2Cl^- Cl₂(g) + 2e-

1 Faraday = 1 C/mol e, so we have 1.80 · (96485) C/mol e.

But now... I can't find what should I do.

[Borek]

1 Faraday = 1 C/mol e, so we have 1.80 · (96485) C/mol e.

Something is wrong here, even if you are on somewhat right track that moles, Coulombs and faradays are related.

1 Faraday is just a mole of electrons. You need two electrons per Zn²⁺, you can easily calculate how many moles of Zn were deposited (this is actually Faraday's law of electrolysis).

[DavidCM]

Thank you so much!! I got it.