I'm currently in high school doing a project in chemistry. So I had to make a project where concepts from physical chemistry will be used alongside organic chemistry.

After going through my book I decided to create a test for carboxylic acids using electrochemistry. The basis of the test is:

Aldehydes, ketones and alcohols are non ionic in aqueous solution while carboxylic acids are ionic. Therefore carboxylic acid can be used as an electrolyte.

So I finalized the following:

Lead, copper (my school doesn't have much chemicals and reagents and mostly available one was lead and copper so I stuck with it). The thing is, boiling the given organic compound with elemental lead, hydrogen peroxide and taking the precipitate obtained and measure the weight and dissolve it in test tube with water this acts as an electrolyte for lead half cell and make a copper half cell and connect the copper electrode and lead electrode with wires and put a suitable salt bridge. If there is reading in ammeter connected to the wires then the given organic compound contains carboxylic acid.

Is my method okay? Edit:

1)since someone pointed out a lot of flaws , can I do like this: Taking the aqueos solution of the organic compound as electrolyte and putting Zn,Cu electrodes in it and connecting the electrodes using a wire and connecting the wires to an ammeter ,(will this work like a lemon battery?)

  • $\begingroup$ Try to be clear. Which organic compound do you boil with lead and hydrogen peroxide ? What is the nature of the precipitate obtained ? Why do you weigh it ? Why and how is this precipitate suddenly soluble into water ? Another fundamental question : Why do you choose a carboxylate salt ? Of course carboxylates can be used as electrolyte. But carboxylates are decomposed at the anode through the Kolbe reaction, which gives CO2 and a hydrocarbon. Why use lead ? It is toxic ! Why not use Cu in CuSO4 at both electrodes ? And you weigh both electrodes before and after electrolysis. $\endgroup$
    – Maurice
    Commented Nov 5, 2020 at 9:59
  • 2
    $\begingroup$ while carboxylic acids are ionic. Therefore carboxylic acid can be used as an electrolyte. This is not correct. Carboxylic acids CAN dissociate to ions, but there is a equilibrium and are called for the reason weak electrolytes. See Study.com - lessons - weak-electrolyte-definition-examples E.g just about 1% of acetic acid is dissociated at pH 2.75. 6% vinegar has pH about 2.38, so just a fraction of % is present as ions. . $\endgroup$
    – Poutnik
    Commented Nov 5, 2020 at 10:09
  • $\begingroup$ if the compound I took contains COOH then it boils with H2O2 and lead to form lead acetate, this precipitate is soluble in water, and it is white in colour. I didn't choose a carboxylate salt but I'm testing for it. If in the first step itself I'm not getting a precipitate then can I say compound doesn't contain COOH? Please clarify me , I'm not clear that's why asked for help. And I'm weighing them because to record what molar aqueous solution I'm taking. $\endgroup$
    – Jagan
    Commented Nov 5, 2020 at 10:40
  • 1
    $\begingroup$ @Jagan. It is the first time you are speaking of acetic acid ! I could not guess it. Anyway electrolyzing lead acetate is not a good idea. Specially with a membrane. Lead is toxic. Acetate ions are decomposed by electrolysis. The membrane increases seriously the resistance of the bath. $\endgroup$
    – Maurice
    Commented Nov 5, 2020 at 10:44
  • $\begingroup$ I'm sorry but what exactly is a membrane here xD, I'm still at school and in my textbooks they haven't mentioned about membranes in electrochemistry $\endgroup$
    – Jagan
    Commented Nov 5, 2020 at 10:46

1 Answer 1


Let's be positive. The edited proposal has merit. You can pass a current thru a solution of a carboxylic acid and measure it with an ammeter (a more sensitive milliammeter would be better).

How about this: check for conductivity. Commercial conductivity meters use AC current and an AC ammeter. Use two inert, identical electrodes for this, and use DC if you can't arrange an AC apparatus. Do a conductivity measurement (i.e., measure the ohms resistance of the solution with the two electrodes taped to a certain distance separation (e.g., two copper wires (or stainless steel) 1 cm apart, 1-2 cm immersion, taped to a plastic ruler) so you can do the identical measurement on another solution.

Then, for a second test, examine the conductivity of that same solution diluted by a factor of 10. Now here's where you are on your own: the conductivity will be less, but not exactly 10 times less, because as the carboxylic acid is diluted, it will ionize to a greater percentage, and therefore carry more current at the same voltage than its concentration reduction predicts.

I say you are on your own, because the experimental results will not be as clear as you might wish (lots of opportunity for scatter), but this will enable you to explore the physical chemistry idea of the ionization of carboxylic acids and dilution and the question of voltage vs current, DC vs AC, electrode composition (reactivity); also bring up pH.

Above all, keep it simple! Consider two variables and plot one against another. You would be surprised how difficult that will be and how there are many more variables just waiting to hop into your experiment!

But you will have fun learning about all the variables and writing up the results and impressing your teacher with your comprehension. Good luck!


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