A friend and I recently produced some aluminum hydroxide, $\ce{Al(OH)3}$ for a project, and we wondered how it functions as an electrolyte. We know it has a low conductivity and solubility, but we wanted to know for knowledge's sake. We put some aluminum hydroxide in a bowl and used a power supply running at 31.9 volts (The resistance was high enough for the current to be immeasurable). We also used aluminum electrodes (aluminum foil).

The electrodes were bubbling and a white precipitate was forming on the top, and it seems we have less hydroxide than we started with...

So what exactly did we do? $%edit$


1 Answer 1


When you are applying 31.9V across I am assuming cm-ish distances, you are definitely causing a lot of electrochemistry to happen in the solution.

You are oxidizing one of your aluminum electrodes to $\ce {Al^3+}$, therefore precipitating $\ce{Al(OH)_3}$ and probably reducing water on the other electrode to evolve $\ce{H_2}$.Along with these, you are probably doing a lot of other reactions with any and every impurity around, but these two should dominate.

Measuring ionic conductivity properly requires doing an AC measurement since you cannot have a sustained DC current without doing electrochemistry of some sort. The electrochemistry on the electrode surface ends up the main factor influencing your measurement in any DC setup.


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