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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: faust on April 28, 2006, 02:26:38 PM

Title: Cyclic voltammetry of a solution of LiClO4
Post by: faust on April 28, 2006, 02:26:38 PM

Hello , I have a small question :

First I use european conventions, which means : reduction is a negative current, oxydation is a positive current

So I have a solution 0.1 M of LiClO4 in CH3CN.

When I sweep the potential from 0 to -3V (vs SHE) I see the solvant/Electrolyte wall (may be a reduction of ClO4^- to Cl-?)

Then when a go back from -3V to 0V I see a peak at about -2.5V , what is this peak? (Oxidation of Cl- to Cl2?! at such low potential?)

If I do 0--> -2V (I don't reach the solvant wall), then -2V--> 0 I don't see this peak...

So , what could be this peak?

Thank you
 

Title: Re: Cyclic voltammetry of a solution of LiClO4
Post by: Borek on April 28, 2006, 02:43:49 PM

Have you checked what's going on without LiClO₄? Or if the peak height is LiClO₄ concentration dependent?

Title: Re: Cyclic voltammetry of a solution of LiClO4
Post by: faust on April 29, 2006, 03:55:26 AM

yes of course. I have done the same experiment with Bu4NBF4 and I don't have this peak...

Title: Re: Cyclic voltammetry of a solution of LiClO4
Post by: Borek on April 29, 2006, 08:30:23 AM

So at least you know it is either Li⁺ or ClO₄^- :)

Does it look like something reversible?

Title: Re: Cyclic voltammetry of a solution of LiClO4
Post by: faust on April 29, 2006, 10:30:53 AM

well it's difficult to say if it's reversible or not.. because the reduction wave is in the solvant wall.

But after reading something about electrolysis of LiClO4 in acetonitrile with platinium electrodes, I think it's the reduction of Li+ than its reoxydation!

Title: Re: Cyclic voltammetry of a solution of LiClO4
Post by: Borek on April 29, 2006, 07:10:30 PM

That was my first idea too - Li should react at about -3.0 V. But my voltammetry is so rusty as if it was not used in the last 15 years... Wait! It WAS not used in the last 15 years ;)