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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: DavidCM on December 21, 2023, 11:47:06 AM

Title: Electrolysis and Faradays
Post by: DavidCM on December 21, 2023, 11:47:06 AM

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 :rarrow: Zn(s)
2Cl^- :rarrow: 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.

Title: Re: Electrolysis and Faradays
Post by: Borek on December 21, 2023, 12:09:00 PM

Quote from: DavidCM on December 21, 2023, 11:47:06 AM

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).

Title: Re: Electrolysis and Faradays
Post by: DavidCM on December 21, 2023, 12:26:23 PM

Thank you so much!! I got it.