Reduction in the electrospray source

Discussions about GC-MS, LC-MS, LC-FTIR, and other "coupled" analytical techniques.

8 posts Page 1 of 1
Hi, I have a question about the electrochemical reactions in the electrospray source. In ES+ reduction reactions take place on the counter electrode (sample cone?), right? Can anyone explain why the positive analytes of interest aren´t reduced or attracted to the surface of the negative cathode but instead pass through the sample cone into the MS?
/shai

Shai,

The reason is that there is a voltage gradient from the edge of your electrospray tip down to the detector of your MS. You may start applying from 2000-5000 volts in your electrospray tip, your sample cone may be grounded but then the rest of ion optics in the MS start having increasingly negative values, attracting the positive ions in the MS.

There are differences depending on what MS you are using but this is the general trend.

Kostas

Thank you for your answer, but I still wonder about the reduction reactions; why don´t all positive ions follow the gradient through the sample cone? The ones that are reduced, is it just the ones that are "sprayed far away" from the orifice?

Maybe it´s my lack of electrochemical knowledge but it´s a bit unclear to me... I´ve read that the reduction reaction "maintains the charge balance to allow continuous production of charged droplets" (Cole, R.B.). Couldn´t this production be maintained when the positive ions leave the source for the mass analyser?

Sofia,

Only a small percentage of the ions generated during the electrospray process will reach the MS detector (with the help of your vaccum). The best eficiency that has been achieved up to now is only 1% (with nanospray etc.). The rest of the ions will end up in the counter-electrode and what will happen in simple words is that they will descharged as the counter electrode is connected to the ground.

There are several reasons that your analyte ions will not make it to the MS (surface tension... big droplets the ions won't be able to get dissolved etc...).

The charged balance is maintened as all ions reaching the counter-electrode or the detector will be descharged (both the counter-electrode and the MS detector are grounded).

Does this makes more sense to you?

Thank you very much! It makes sense!

I am not sure I understood your (initial) question but the reduction/oxidation takes place as far as I am aware in the capillary tip (were the spray is produced). I would say that all ions go through this process one way or another. The "ion selection" takes place first through polarity (+ or -) then mass.

JNF,

the following links take you to articles about the ESI-source as an electrochemical cell. If I understand them correct reduction reactions take place at the counter electrode (sample cone). Please let me know if I´ve misunderstood and I would be interested in a more detailed description about what you mean about ox/red in the capillary tip.

http://www.spectroscopynow.com/Spy/pdfs/JMS35804.pdf
http://www.spectroscopynow.com/Spy/pdfs/JMS35763.pdf

/Sofia

Oxidations occur in the tip of the capillary when you trying to do positive ion MS and reductions when interested in negative ion MS. Thus the process that happens in the tip is what matters. Off course once the ions are ejected they "fly" towards the counter electrode and ideally the counter electrode is your detector (for the correct ions). Off course most ions will hit the orifice plate and the circuit will thus be closed. What happens there is irrelevant to your analysis (roughly speaking).
Regards
JNF
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