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Why does the open circuit voltage of a battery decrease with increase in $\mathrm{pH}$ of electrolyte, but the voltage across a load increase?

I tested a cell which had aluminum sulfate as its anode electrolyte ($\pu{12 g}$ aluminum in $\pu{50 mL}$ water and a $\mathrm{pH}$ of 3.2), and potassium hydroxide as the cathode electrolyte ($\pu{5 g}$ of $\ce{KOH}$ in $\pu{200 g}$ of water, $\mathrm{pH}$ of 12.3). The anode electrode was aluminum foil and the cathode was platinum. The open circuit voltage (OCV) of the cell was $\pu{0.9 V}$, and when connected to a $\pu{1 k\Omega}$ resistor, the voltage was $\pu{0.324 V}$.

When I increased the concentration of $\ce{KOH}$ from $\pu{5 g}$ to $\pu{10 g}$, and all other material concentrations were kept constant, the OCV decreased from $\pu{0.9 V}$ to $\pu{0.7 V}$, but when I attached the $\pu{1 k\Omega}$ resistor, the voltage increased to $\pu{0.42V}$.

I do not understand why increasing $\mathrm{pH}$ is causing the OCV to decrease, but increase the voltage under a load to increase.

I am guessing for the voltage under the load that maybe the higher concentration of $\ce{KOH}$ leads to more ions in solution thus increasing the current flow. But if this assumption/hypothesis were true, shouldn't the OCV also increase?

Can someone please help?

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  • $\begingroup$ Did you clean/change the aluminum foil in the mean time? It is possible that the aluminum surface is changing/passivating. $\endgroup$ Mar 27 '17 at 13:53
  • $\begingroup$ i cleaned the aluminum, yet the observation was the same $\endgroup$
    – user510
    Mar 27 '17 at 22:04
  • $\begingroup$ A solution containing $5 g$ KOH in $200 mL$ water has a concentration equal to $\frac{5 g}{57 g/mol·0.2 L} = 0.438 M$. Its pH should be $13.6$, and not $12.3$ as given in your text....How do you explain this difference ? $\endgroup$
    – Maurice
    Jan 30 at 21:29
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Your aluminum half cell has a standard OCV (i.e., at 1 M $H^+$) of 1.676 volts. Ref. 1enter image description here

You don't have standard conditions, but that doesn't matter; your conditions do not change for this half cell.

The other half cell is a reduction of water, in alkaline solution, with the electrons from the aluminum to produce hydroxide ions.

enter image description here

The CRC Handbook confirms -0.8277 volts for this reaction. I had intended to use a picture from Ref 1 showing the OCV, but it was incorrect by a factor of 10! Just goes to show you that you can't trust it just because it's in print, or online!

The point is, this reaction wants to consume hydroxide ions and generate electrons, but you are running it in reverse by pumping aluminum electrons into it. When you increase the pH of the half cell, you increase its voltage by giving the half cell more reactant - but then you are fighting it because it is yielding to the aluminum voltage, so your total OCV out decreases. That's why your cell output is so low (that, and the non-standard conditions). But the good thing is that the extra ionic concentration decreases the cell resistance, so it can supply more current to your 1k resistor even with the lower OCV.

Ref 1. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/The_Cell_Potential

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  • $\begingroup$ Not sure why the downvote. The same conclusion is reached by considering Le Chatelier’s Law. This counterintuitive effect is a function of the internal cell resistance and the external load. The initial test gives 324 microamps current and 420 with added KOH. Corresponding internal cell resistances come out to 1777 ohms initially and 666 ohms for the second test. If the external load resistance is raised to 3k ohms or more, the effect disappears. $\endgroup$ Jan 31 at 19:38
  • $\begingroup$ I have not downvoted. But I must state that the redox potential of Aluminum has no influence on the battery, because Aluminum does not exist in the presence of water. It immediately reacts producing $\ce{Al(OH)3}$ and $\ce{H2}$. Usually this reaction does not occur, because the aluminum metal is entirely covered by a protective waterproof thin and transparent layer of aluminum oxide. But as soon as this layer is destroyed, for example in an electrochemical cell, the metal reacts quickly with water, and it is $\ce{H2}$ that works as anode and not $\ce{Al}$. $\endgroup$
    – Maurice
    Mar 1 at 21:16
  • $\begingroup$ @Maurice: Thanks for your interesting comment. But the question involved the concentration increase of KOH in the cathode compartment, which fights whichever anodic process is occurring, yet reduces the cell resistance so the current can increase. Do you think that part reasonable? $\endgroup$ Mar 1 at 22:42
  • $\begingroup$ Are you sure of your measurements ? Because $\pu{5 g KOH }$ is about $\pu{0.09 mol KOH}$. Dissolved in $0.2$ L water it produces a concentration $\pu{0.45 M}$. The pH of such a solution is $\pu{14 + log 0.45 = 13.65}$. And you say that you measured a $\pu{ pH 12.3}$. This corresponds to a concentration $20$ times lower. $\endgroup$
    – Maurice
    Mar 3 at 10:07

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