# Electrolysis puzzle

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.

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