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Topic: Aqueous Electrolysis  (Read 676 times)

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Offline NamiraAzeed

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Aqueous Electrolysis
« on: January 20, 2021, 05:36:35 PM »
I'm having trouble in general with electrolysis of aqueous solutions but I was given this question:

Deduce the products formed during the electrolysis of an aqueous solution of sodium fluoride. Write an equation for the reaction nat the positive electrode (the anode) and give your reasoning.

The answer says that oxygen gas and hydrogen gas are the products formed - how can these be the ONLY ones formed? Basically, what are the full reactions formed at the anode and cathode?

And what is the deal with water being oxidized (instead of the halide)?
Answer given was: since EoOX for F- is very negative/Eored for F2 is very high. I have no idea why the difference in electrode potentials result in one being oxidised or reduced over the other as I thought it was which ever value was more positive unless a higher concentration was indicated.

I know I'm having trouble with the concepts and it would be great if someone could explain this to me or give me some links to places that explain this.

« Last Edit: January 20, 2021, 08:26:29 PM by NamiraAzeed »

Offline chenbeier

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Re: Aqueous Electrolysis
« Reply #1 on: January 21, 2021, 02:09:31 AM »
First write down which ions are present in an aquaeous sodium flouride solution.
Then write down which reactions happen or might happen at anode and cathode.
Compare the products and the stability in this envirement.
Additionaly for all reactions look up the redox potential.

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