Carboxyllic acid Test 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?)
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!