What kind of rate of oxygen production could be achieved with a simple 9V electrolysis setup using water?
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$\begingroup$ Adding to what @Poutnik said, a 9 V battery is literally a battery of 6 AAAA cells. If you dissect a 9 V battery, you will see this. Better to use a surplus DC “wall wart” from some dead (or gone) electrical gizmo. And 3 or 4 volts is usually fine, if the electrolyte concentration is not too low. $\endgroup$– Ed VCommented Apr 24, 2020 at 18:56
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$\begingroup$ By the way, this is a trivially simple experiment to try at home. Put water in a drinking glass, dissolve Epsom salt (magnesium sulfate heptahydrate, from the grocery store) in the water and completely submerge a new 9V battery in, standing upright. You can collect either gas using an inverted test tube or substitute. You can figure out which battery terminal has oxygen bubbles rising from it. The battery will have to be tossed later. Anyway, don’t expect anything but a slow gas evolution rate. $\endgroup$– Ed VCommented Apr 24, 2020 at 19:10
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2 Answers
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If you use pure water, you will not get any H2 or O2 at all, because no electrolysis can be done in pure water, whatever the voltage. You have to add some solute, like sulfuric acid ou sodium sulfate. And the yield in H2 and O2 will be proportional to the concentration of this added substance. Anyway, the yield in H2 and O2 is proportional to the surface of the electrode and inversely proportional to the distance between them.
So there are three parameters that should be known to be able to answer your question : the concentration of the solute, the surface of the electrode, and the distance between them. The voltage is important, but the three mentioned parameters are much more important.
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There is too many unknown variables, like composition and conductivity of electrolyte plus internal resistance of battery.
Generally, these batteries do not provide high current and most of the too high voltage is wasted on heat production.
