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Sorry for the barrage of questions, but each of them is relevant.
I am surprised that the color shows at the cathode rather than the anode. If it were forming at the anode, it would just mean that the metal was being oxidized preferentially over the hydroxide, which would be expected for both iron and copper (though stainless steel and nickel do survive this because of passivated oxide surface formation). Given that the reaction is observed at the cathode, it means you are probably reducing something other than the potassium (or that the potassium formed is immediately reacting with your electrode or with the air and the electrode). KOH pulls CO2 out of the air fairly rapidly (though probably not as much of an issue at elevated temperatures), but I could see CO2 reacting with copper and hydroxide to form something similar to the mineral azurite (Cu3(CO3)2(OH)2), which is a very pretty blue (azure). I think it is unlikely that azurite is the product, but it is a possibility. You could also be reducing the CO2 to formate HCO2– and forming copper formate (which is blue), but I'm also not so happy with that answer, as it requires copper2+, which should be unstable at the cathode with a 12V separation from the anode...12V is also probably overkill for this reaction. I suspect 4V to 6V with a higher current is more likely to give good potassium production.e– + K+ → K E0 = –2.93 V2 OH– → O2 + 2H+ + 2 e– E0 = –0.40 VSo thermodynamically, 3.33 V is the minimum potential required, but you can probably expect a couple hundred mV overpotential for each reaction to go, so I estimate 4 V as the likely onset of the desired reaction. (with Pt or Ni as the anode, the overpotential on the oxygen production side will drop slightly, but still probably require at least 3.9 V total)