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Aqueous NaCl and overvoltage

In lab we constructed an electrolytic cell using a U tube, an adapter, aligator clips, and graphite electrodes (pencil lead).

we added 3g NaCl to 150mL of deionized water and used this to fill the U tube.  At the cathode there was immediate evolution of gas (H2), while nothing happened at the anode for some time...eventually larger bubbles that were much slower to form appeared on the electrode.  The lab instructor told us that the reaction at the anode was the oxidation of H20.

The lab manual mentioned that overvoltage is something to be considered in the electrolysis of H20.  And from the documents I found it seemed that Cl2(g) not O2(g) should have been forming at the anode.  I emailed my professor to ask about what I had found and she asserted that it was in fact O2(g). So now the professor, the lab instructor, and the tutor in the science center all claim it was the oxidation of H2O that formed O2(g).  But no one can tell me why they have ruled Cl2(g) out as a possible product of the reaction.  We only tested pH with litmus paper and while the cathode side showed plenty of activity clearly OH- ions were present the Anode solution did not have much of an effect on the litmus paper.  

So ....How quickly might NaOH build up on the cathode side?  Is a U tube effective in keeping the OH- Ions from migrating over to the anode side?  Is it possible that both gasses were being released?  Does Cl- have a significant overvoltage with graphite electrodes? Does its overvoltage increase with increasing current like H2O's does?  Does concentration have a significant effect?

Thank you in advance for any help offered!

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New Contributor

Re: Aqueous NaCl and overvoltage

No Chlorine did not form. The reason is that in solution you have two species which are competing at the electrode Cl- and your water molecule. We simply have to inspect the Standard reduction potential table to see that Cl2 will not form. 2H2O (l)  -----> O2 (g)  + 4H+ (aq)  + 4e ;   E0 = -1.23 V

and        2 Cl- (aq)   ------>   Cl2(g)  + 2e                  E0 = -1.36 V

The reaction  with the lower voltage will prevail and so  oxygen will form rather than  chlorine.

If you were to replace the sodium chloride with sodium Iodide  you will see a different result ( I2 will form).

I have not gone into the exact concentrations which will mean using the Nernst equation, but the result will be the same.  Hope this helps.

New Contributor

Re: Aqueous NaCl and overvoltage

I have read from a few university websites as well as a patent and an electrochemistry text that

H2O has a significant overvoltage under nonstandard conditions.  One even suggested a range of .4v-.6v per electrode so at the anode with E(naught) = +.82  an added .4-.6v brings that to +1.22 to +1.42

+1.42 > +1.36  which leads me to wonder if chlorine has a significant overvoltage as well (with a graphite electrode) and also what about H+ concentration if some H2O is oxidized at the electrode it must be leaving H30+ around if that concentration increases around the electrode wouldn't it be increasingly favorable for the formation of Cl2(g)?

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New Contributor II

Re: Aqueous NaCl and overvoltage

I agree with Adolf King and your instructor that you are seeing oxygen being formed at the anode, not chlorine.   The important part of this demonstration of the electrolysis of water is that the volume of hydrogen produced is twice that of the oxygen.  This is why you are seeing more bubbles evolving from the cathode than the anode.

The best way to do this experiment is with the Hoffman apparatus using platinum electrodes.  I do this with my students in middle school where I volunteer as a science partner to a physical science teacher.  With this apparatus, I use a 10% solution of sodium sulfate as the electrolyte, collect the hydrogen and oxygen gases, transfer them to test tubes and show how the hydrogen burns and the oxygen supports combustion.  If the Hoffman apparatus is not available, you can use a simple set-up with two pencils sharpened at both ends, a 9- volt battery, two wire connectors with alligator clips, and a beaker containing a 10% solution of sodium sulfate. You cannot collect the gases, but you can clearly see that twice as many bubbles emanate for the cathode than the anode.  Both of these procedures are available to the public on our website: Click on the tab entitled, "Laboratory Procedures," scroll down to "The Electrolysis of Water" and click on the highlighted PDF documents, "Water Electrolysis" and "Water Electrolysis Simple."

I hope this is helpful.