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kmav83
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I am a PhD student and I am trying to decipher what is going on in my system, which consists of a platinum electrode, an electrolyte (varying pH, buffer solution with KCl salt and TRIS buffer), thin SiO2 and Si when low bias is applied to the Pt electrode (0 to 0.5 V) with respect to the Si. I'm sure a double layer will form at the electrolyte-SiO2 interface...and since charge cannot pass between the Si and the SiO2, reactions that contribute to ion current in the electrolyte will only be occurring at the Pt electrode. But I am quite confused because as you can see, this is not a typical electrochemical cell and it becomes hard to imagine operation of such a system.
First, at the Pt electrode, an oxidation reaction involving dissolution of Pt cannot occur because the standard potential of Pt oxidation is -1.18V and therefore the reaction is non-spontaneous. And the applied bias is less than +1.18V, so this reaction cannot even be driven by applied bias. So, the only other reaction I can see happening is 2 H+(aq) + 2e− → H2(g), for which the standard potential is 0V? OR 2H2O(l) + 2e- → H2(g) + OH-(aq)...but the standard potential for this is -0.83V and the applied potential is in the opposite direction...so this reaction becomes impossible. Are there any other reactions that can occur? XXXX This is wrong...See EDIT! XXXX
So, I guess my first question is, if there are no ions in solution that can partake in a redox reaction involving the metal of the electrode and if applied bias is not sufficient to drive that reaction (dissolution of Pt in my case), then the only other possibility is electrolysis of water? Also, which electrolysis half reaction should occur...only hydrogen gas evolution (since applied bias is low and positive)?EDIT: Actually according to E-pH diagrams, water is stable (at neutral pH) between 0 and 1 V applied bias. So...there can be no electrolysis reaction at the metal electrode. So, this makes things even more confusing. At low positive voltages, even electrolysis cannot occur at the Pt electrode surface. But electrons are building up at the Pt surface...causing charging of the double layer. But these electrons cannot build up for ever...what happens when no redox reaction can happen and electrons keep building up? Am I making any sense?
First, at the Pt electrode, an oxidation reaction involving dissolution of Pt cannot occur because the standard potential of Pt oxidation is -1.18V and therefore the reaction is non-spontaneous. And the applied bias is less than +1.18V, so this reaction cannot even be driven by applied bias. So, the only other reaction I can see happening is 2 H+(aq) + 2e− → H2(g), for which the standard potential is 0V? OR 2H2O(l) + 2e- → H2(g) + OH-(aq)...but the standard potential for this is -0.83V and the applied potential is in the opposite direction...so this reaction becomes impossible. Are there any other reactions that can occur? XXXX This is wrong...See EDIT! XXXX
So, I guess my first question is, if there are no ions in solution that can partake in a redox reaction involving the metal of the electrode and if applied bias is not sufficient to drive that reaction (dissolution of Pt in my case), then the only other possibility is electrolysis of water? Also, which electrolysis half reaction should occur...only hydrogen gas evolution (since applied bias is low and positive)?EDIT: Actually according to E-pH diagrams, water is stable (at neutral pH) between 0 and 1 V applied bias. So...there can be no electrolysis reaction at the metal electrode. So, this makes things even more confusing. At low positive voltages, even electrolysis cannot occur at the Pt electrode surface. But electrons are building up at the Pt surface...causing charging of the double layer. But these electrons cannot build up for ever...what happens when no redox reaction can happen and electrons keep building up? Am I making any sense?
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