4
$\begingroup$

I created an electrolysis setup to study how different electrolytes, their concentrations, and the electric current affect the "speed" of the electrolysis process. With that in mind, I built a simple apparatus:

  1. A beaker
  2. two test tubes
  3. PVC-covered copper wires for electrodes.

For each electrolyte and concentration, I then apply different voltages: $\pu{1.5 V, 3 V, 4.5 V, 6 V}$ and $\pu{9 V}$

I think I am missing some detail or not controlling some variable because some of my results were not expected.

For instance, in the electrolysis of aqueous $\ce{NaCl}$ (of all cases!).
I was expecting to have $\ce{H2}$ created at the cathode and $\ce{Cl2}$ at the anode. However, while I experienced $\ce{H2}$ evolution in the cathode (which was collected in the upside-down test tube over), I did not notice any gas evolution at the anode.

I reviewed Jan's response to a similar question in this forum who explained that, because to elevation of the pH due to the generated $\ce{NaOH}$, the $\ce{Cl2}$ would react further resulting in the production of $\ce{NaOCl}$, and in no gas being produced at the anode.

However, Jan previously indicated that in the beginning

"...chlorine gas at the anode should bubble after some time".

In my case, I never noticed any bubbling at the anode during the experiment. I varied the concentration of the aqueous $\ce{NaCl}$ solution, from very weak to saturated; tried different voltages, changed the temperature of the solution. Nevertheless, the behavior was the same in each scenario. I also tested the voltage at different points with a voltmeter (everything checks), made sure I was using pure salt (without Iodine or additives).

One interesting "clue" is that I noticed some "blue powder" mostly at the bottom of the beaker (some suspended too) but none inside of the test tube over the anode. I also noticed some white build up on the copper wire at the anode, which I think to be $\ce{NaOH}$.

Right now, I am at a loss for ideas and would appreciate any suggestion or explanation.

$\endgroup$
4
$\begingroup$

As is often the case with chemistry, you can't ignore the fact that your instruments are also made of atoms and have a chemistry of their own.

Naturally, in this setup you just dissolve copper from the wires: $$\ce{Cu -> Cu^2+ + 2e-}$$

You may check the table of electrode potentials and verify this with numbers. Being more energetically favorable than the other options, this reaction occurs first. Chlorine never had any chance to appear.

That's pretty much the size of it. Voltage is irrelevant (as long as it is high enough for the reaction to occur, which is true even for 1.5V). Salt concentration is irrelevant. Additives are irrelevant. Temperature is irrelevant, as long as the solution is liquid. Time is irrelevant; wait longer, and you'll get all copper eaten up, which will break the circuit and thus force the process to stop.

Due to the lack of acidity, much of the dissolved copper hydrolyzes and finally ends up in the form of hydroxide down at the bottom, which explains the mentioned blue powder.

With electrodes of unobtanium or other inert substance (graphite being a moderately good approximation) you'd get your chlorine all right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.