# Would this experiment be electrolysis experiment be viable?

I plan to do a 'research project' or an experiment involving the electrolysis of dilute sulfuric acid.

The idea or 'question' is this:

Would there be a variation in the $$\mathrm{pH}$$ of the electrolyte at different points at different displacements from each electrode.

So I was wondering if $$\mathrm{pH}$$ in specific areas, such as around the cathode would differ from the $$\mathrm{pH}$$ at the exact center between the two electrodes.

One's initial assumption would be that the charged ions would be would be equally dispersed in the circuit path, but there is a reduction/oxidation of ions near the electrodes, in this case hydrogen and oxygen ions, which may mean that there would be some variation of their concentration in the cell.

So is my experiment worth doing, even if it might end up in failure, because I will need to buy a $$\mathrm{pH}$$ meter which my school doesn't have.

• Litmus paper will be quite enough. – Ivan Neretin Jul 3 '19 at 14:38
• Thank you for your comment. Using litmus paper may be fine for preliminary results, but for this project, I plan to 'mathematically' interpret the results, to calculate how far a displacement from a cathode would give me x.x pH . So a pH meter might be useful in that sense. – TheTenthBox Jul 3 '19 at 14:43
• You are aiming to measure something that can't be measured reliably. Liquids mix all the time. The displacement of such-and-such pH from cathode will fluctuate wildly, depending on the weather, any slightest vibration, and how long ago the solution was prepared. No, this is not viable. – Ivan Neretin Jul 3 '19 at 14:53
• Mathematical treatment in electrochemistry may be far more complicated than we can imagine. You would need diffusion equations, Fick's laws, and I believe what not. It is not a trivial mathematical problem. – M. Farooq Jul 3 '19 at 14:57

Diffusion of ions, propelled by the electric field (i.e., current) will occur at a given, calculable rate. Deposition of those ions (and conversion to $$\ce{O2 + H2}$$) will occur and the $$\mathrm{pH}$$ will rise at the cathode (negative electrode) and fall at the anode (positive).
Then back diffusion will occur as the $$\ce{OH-}$$ and $$\ce{H+}$$ ions produced at the electrodes even out. The $$\mathrm{pH}$$ of the starting sulfuric acid solution should be ~3 (i.e., $$\pu{1.0e-3 M}$$), and the current should be ramped up high by using a higher than usual voltage (maybe 4-5 volts).
This will allow the cathode $$\mathrm{pH}$$ to shift by a whole $$\mathrm{pH}$$ or so, and the anode perhaps more (because it is closer to neutrality). The anode could even go alkaline! The center of the solution should have a fairly stable $$\mathrm{pH}$$.
The evolution of gas will cause considerable mixing at both electrodes, so it might help to surge the current for a few seconds, then shut it off to measure the $$\mathrm{pH}$$.
The problem with using an indicator is that you need to avoid having the whole solution being darker than one or the other electrodes, so you can't differentiate the color change. However, if you select an indicator which is colorless at the $$\mathrm{pH}$$ of the original solution, you might find one which turns color at the anode or cathode shift.