In a message dated 6/23/01 10:59:18 AM EST, [email protected] writes: << Subj: RE: CS>Theorizing About the Species Produced at Electrodes During LVDC CS Pro... Date: 6/23/01 10:59:18 AM EST From: [email protected] (Ivan Anderson) Reply-to: [email protected] (*Silver-List*) To: [email protected] A salt bridge is normally a gel containing an electrolyte which will not effect the cell reactions under investigation. The idea is not to let the two half cells interact but still enable a current to flow. The reactions as found in this arrangement are not necessarilly those that would be found if there were two electrodes in the same cell. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: Yes. I agree with what you have there. I think there's more to it than that, as I mentioned in my last post, but I think we have a good working definition for what we are discussing. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > Possible Anodic Side Reactions: > (1) 2[OH-] = H2O + 1/2O2(g) + 2e Eo = -0.40 > (2) 1/2H2O = 1/4O2 + [H+] + 2e Eo = -.70 > > Reaction (2) is the more commonly written reaction because it does not > contemplate the presence of hydroxyl ion. Thus, the E potential, > as you point > out, is about -1.22 volts which is beyond the probable half cell > voltage of > -0.8 dictated by the continuing oxidation of silver to silver ion. > > However, if ones takes into account the (typically) close > proximity of the > cathode and anode, migration of OH- to the anode will occur due to the > following half cell cathodic reaction, OH- may migrate to the anode, if it is not otherwise occupied in surrounding Ag+ ions, but even when it gets there there is an overvoltage required in producing oxygen gas. > Probable Cathode Side Reaction: > (3) 2H2O + 2e- = H2(g) + 2OH- xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: Measuring the pH vs time near the anode would be a good way to determine to what extent side reaction (1) plays a role, don't you think? >xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > In addition, the close proximity of the anode allows Ag+ to > migrate to the > cathode to maintain electrical neutrality so that half cell > reaction (3) can > proceed. If any OH- ions are oxidised at the cathode, then a similar no. of Ag+ are not. If Hydrogen gas is evolved at the cathode then a similar amount of Ag+ is not reduced. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: REDUCTION occurs at the cathode. In any case, I'm not getting your point. We ARE talking about competing reactions aren't we? The extent that one reaction has preference over another is controlled by an assortment of variables, is it not? xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > In summary, just as the migration of Ag+ to the cathode allows > the production The migration of Ag+ to the cathode prevents the formation of OH- surely, as it takes the place of the reduction of water. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Ivan: I would put it differently...The migration of Ag+ to the cathode is a side reaction that COMPETES with the reduction of water to form OH-. The rest of my comments have been included in a separate post since the format here is getting too messy. Roger xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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