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My experimental setup looks at using different pH solutions (of $\ce{H2SO4}$ and $\ce{NaOH}$) for electrolysis of water with a Hoffman setup and then measuring the rate of production of gas (note this is proportional to the current) at a fixed voltage.

Here are the results from my testing:

pH vs Rate of Electrolysis

To make it clearer, here is a graph with logarithmic scale:

pH vs Rate of Electrolysis (Logarithmic)

It is slightly misleading to look at the trend of oxygen as it redissolves into the solution much more than hydrogen, which is why its rate of production is slightly less than half of that of hydrogen.

My question is, what graph fit should I expect to see, I can't find anything on what it should theoretically be.

Should the best fit line be something like $y=e^{\lvert x \rvert}$?

If we look at the logarithmic graph, should it be in the shape of a quadratic ($e^{(x^2)}$) that goes through the points, should it be two straight lines ($e^{\lvert x \rvert}$)?

I'm not sure on what the theory says it should be. All help appreciated, thanks!

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    $\begingroup$ Is not better to be first aware of what the theory predicts and THEN making an experiement? // Hints: check how conductivity and pH depend on concentration of NaOH and H2SO4. $\endgroup$
    – Poutnik
    Commented May 31 at 18:48
  • $\begingroup$ Hi there, sorry if i come off as not having looked but I've tried to find stuff and I have't been able to. If you come across a specific article / link that would be very helpful $\endgroup$
    – Eshwar Kolli
    Commented May 31 at 18:48
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    $\begingroup$ It is rather textbook knowledge of electrochemistry. Try targeted keyword searching with search modifier like site:stackexchange.com OR libretexts.org OR wikipedia.org . // Specifically the site libretexts.org is focused on education community driven textbook-quality articles. // Gas gaining rate is roughly proportional to conductivity, which is roughly proportional to molar concentration for low/medium concentration, and - approximately - pH = - log (molar concentration) $\endgroup$
    – Poutnik
    Commented May 31 at 18:52
  • $\begingroup$ You may find helpful the following article from the Journal of Chemical Education: Water Electrolysis Accompanied by Side Reactions $\endgroup$
    – PAEP
    Commented Oct 29 at 10:19
  • $\begingroup$ I would expect that once you fix $\Delta \mathrm{V}$ in your experiment, the production rate (i.e. reaction rate) would mainly depend on the concentration of $\ce{H^+}$ or $\ce{OH^-}$, depending on the pH. $\endgroup$
    – PAEP
    Commented Oct 29 at 10:23

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The pH is not the main factor in electrolysis. The yield in $\ce{H2}$ and $\ce{O2}$ dépends above all on the voltage and on the inner resistance of the solution. You don't mention these parameters. The yield of gases does not depend primarily on the pH. Electrolyze of a neutral solution of $\ce{Na2SO4}$ can produce both gases even if the pH is nearly $7$.

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    $\begingroup$ pH is a really important factor for water electrolysis. He applied a constant voltage over a cell, when near pH 7 the amount of OH- or H+ is ridiculously low. This means the hydrogen evolution or oxygen evolution reaction will use water as the reagent for a significant amount of current. This changes the required cell potential from 1.23V to around 2V. You basically lose 800 mV to concentration polarization. $\endgroup$
    – Noah
    Commented May 31 at 13:48
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    $\begingroup$ Of course at pH new 7, the amount of $\ce{H+}$ and $\ce{OH-}$ is extremely low. But it is not important for electrolysis if the solution contains high amounts of ions having no effect on pH. For example, a solution of $\ce{Na2SO4}$ may be neutral and be electrolyzed without difficulty. $\endgroup$
    – Maurice
    Commented May 31 at 13:54
  • $\begingroup$ I think you're missing the point I was making. I am not talking about the ionic resistance if no ions are present. I am talking about "concentration polarization" since there is basically no reagent present to do hydrogen evolution from protons or oxygen evolution from hydroxide you will need to do these from water. If we look at the potential required for either reaction from water we find a cell potential beyond the thermodynamic potential (1.23V) but a cell potential of around 2V. So you lose 800 mV just because of the pH (concentration of reagents), not the ionic conductivity. $\endgroup$
    – Noah
    Commented May 31 at 14:01
  • $\begingroup$ The voltage is held constant, the resistance of the solution changes with pH - that is essentially what im trying to find as resistance is inversely proportional to both the current and rate of gas production $\endgroup$
    – Eshwar Kolli
    Commented May 31 at 18:47
  • $\begingroup$ I think you are talking about faradaic resistance. The faradaic resistantance is related to the velocity of the the redox reactions on the electrodes. Since you are working with a constant potential, this is related to the concentration of reactants ($\ce{H^+, OH^-, ...}$). Check the Tafel equation and kinetic of redox electrode reactions. $\endgroup$
    – PAEP
    Commented Oct 29 at 20:54

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