I will list the questions directly, if you want to know details please read below.

As in the figure, we have water+baking soda in a plastic cup, there are 2 versions in figure that affects LED's brightness and current that flows in circuit.

  1. In 1st version of figure, LED is bright and more current flows (measured 9.76 mA with multimeter)
  2. In 2nd version of figure, LED is dull and less current flows (measured 0.22 mA with multimeter)

So my question is, why changing aluminium's place from negative to positive affects current flow of circuit? is it related with aluminium's chemical properties?

Details that explain how i got to this question:

My main aim was to make a home made diode by following instructions in this wonderful site, i did one and it worked.To understand the working mechanism better, i decided to setup things as in the figure upper and faced with this question that looks like main working logic of this home-made diode.In site there was an explanation but i couldn't understand it.

  • 1
    $\begingroup$ Hopefully, what I wrote as a comment, since converted to an answer, will not turn out to be duplicative of an existing answer: I wrote it just before calling it the end of a long day finishing my basement echelle spectrograph construction (thanks, @M.Farooq !). $\endgroup$ – Ed V May 16 at 10:58

When aluminum is the anode (connected to the positive battery terminal), a thin layer of insulating aluminum oxide is produced via the “anodization”. It serves as a dielectric, so basically you have a leaky capacitor: current is low. But when the aluminum is connected to the negative terminal, it is the cathode and then you get reduction of hydrogen ions to hydrogen gas. So considerably more current can flow. This is essentially how tantalum and aluminum electrolytic capacitors have rectifying properties: their oxide films are the cause. Fun link, by the way.


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