I am toying with the idea of a koh solution using aluminum electrodes that have been nickel plated before hand. I would clean the nickel oxide off of the anode periodically using a dilute acid solution then plate more nickel on the aluminum electrodes. The aluminum would be aluminum foil, and i would hope that the nickel would block the koh from attacking the aluminum. But i am curious what would happen if the nickel layer is breached and the koh is allowed to react with the aluminum. If I am correct, potassium aluminate and hydrogen will be formed. I am unsure if the addition of potassium aluminate will affect the electrolysis or if it will just act as an electrolyte. I would have multiple anodes and cathodes so i am curious is the loss of one electrode would destroy my solution. or if i can just replace it and not worry.
First, I am a little concerned with the expression "toying" as the compounds mentioned are quite hazardous. If you are unsure of all of the potential electrochemical reactions that could occur with this setup, I would recommend that you should not simply jump in and try it just yet. The MSDS for Potassium Aluminate is at the following link. However you derive it, that is what will end up in your solution. Solutions of KOH are also nothing to handle without extreme care and protective measures in place.
We encourage amateur science inquiries and experiments. However, we must emphasize SAFETY above all else. This is particularly true if you are simply reproducing some experiment or process that may already have been determined at some considerable risk or injury.
Whether aluminum will react electrolytically with K+ or OH- ions depends on the applied potential. Aluminum will chemically react with aluminum to form Aluminum Hydroxide and eventually more potassium aluminate. You will see the bubbles of hydrogen, and a white precipitate for a short time. After that, the precipitate will gradually dissolve in the solution. Some possible pathways are presented below.
KOH dissociates into K+ and OH-. Then, the aluminum reacts with the hydroxide ion to form Al(OH)3. Then the OH is replaced in the reaction by the K+ ions breaking down water molecules. This leads to the evolution of Hydrogen gas which will be given off. (From Equation help for reaction of aluminum foil in KOH solution | Yeah Chemistry )
The explanation for the dissolving of Al in a strong basic solution is:
Al (s) + KOH(aq) --> K+(aq) + Al(OH)4-(aq)
This equation is strongly exothermic and is the reaction that takes place when crystal Drano (which is primarily a metal hydroxide with bits of Al in it) is added to a drain.
Stoichiometry involves the ratio of the reactants and products. These ratios are given by the coefficients in a balanced equation. I did not balance the above equation because the original post said that it was not necessary to do so. However, the balanced equation for the above reaction would be:
Al (s) + 4 KOH(aq) --> 4 K+(aq) + Al(OH)4-(aq)
This indicates that for every Al atom that reacts (or for every mole of Al that reacts) 4 formula units of KOH are required (or four moles of KOH are required).
Given this equation you can then determine the amounts of reactants and products that are required if I know the amount of just one of the reactants or products.
For instance, 2 moles of KOH will react with .5 moles of Al (4:1 ratio)
or 8 moles of Al will react with 32 moles of KOH (1:4 ratio)
Al + 3KOH --> 3K+ + Al(OH)3
2Al + 6H2O ==> 2Al(OH)3 (precipitate) + 3H2 (bubbles)
Al(OH)3 + KOH ==> KAlO2 (potassium aluminate) + 2H2O
AGAIN - with this type of reaction system, if you are UNSURE, DO NOT attempt it! The system or procedure can be modified or replaced. Physical injury to yourself may be permanent and debilitating.