This question has been argued to the death. I will attempt to give some closure, but this will most probably be the last comment I make on this topic. m@tty have been giving some dead-accurate spot-on responses, his understanding of this is almost the same as mine.
Firstly, to debunk option D. No, 'recharging'
does not occur. For reasons being:
1. You will have to modify all the pumps/extractors in the fuel cell (in a secondary cell, you simply apply the opposite voltage. In this case, you have to do that, AND modify everything)
2. You will have to be pumping in the products (in a secondary cell, you don't need to 'add' products. In this, you will have to add CO2 and H2O continuously into the fuel cell)
3. You have no control over what is being produced (the recharging reaction involves what I can only describe as the most energetically-unfavourable polymerization reaction where you have no control over how long the final product will be, as far as recharging is concerned this doesn't give you the original reactants back)
4. You are not even sure if the reaction will proceed (not only do you have to overcome the fact that the reaction is kinetically unfavourable, and any products produced will almost spontaneously re-combust because of the temperature/pressure required for such a reaction to happen in the first place, you run into huge problems with equilibria [as equilibrium of a combustion reaction is all the way to the right])
So my argument here says fuel cells in general do not 'recharge' in the sense that by definition it is no longer recharging (at least not in a fashion at all similar to secondary cells), and even if you do it will still be extremely difficult, if not impossible. The only case where this might work is the hydrogen fuel cell. Every other type of fuel cells (methanol/ethanol/hydrocarbons) will not work. So by saying that "fuel cells are rechargeable", you are saying "only a particular type of fuel cell might be rechargeable", and even so it won't work well, and is not what the industry adopts. That, in my opinion, constitutes an incorrect answer.
As for option C. I would first have to point out that
- Just because it's not in the textbook doesn't make it 'wrong'
- Just because it's not in the textbook doesn't make it non-assessable (VCAA doesn't examine by the textbook definitions, they examine by the study design)
- Just because it's not taught doesn't mean it's bad science
- Just because you have always been taught something doesn't mean it's right on every level
The last point, especially last point, is true for almost all scientific knowledge. If something doesn't work with a definition, it is often that the definition is incorrectly understood/defined. If you go on to do science at university, you will find that a lot of first year (and second year) is spent on "everything you learnt from year 12 is wrong and over-simplified. This is the proper version:..."
Now, as for the actual process, I would like to do this in four parts:
1. reductant is an aqueous species, anode is a non-reactive electrode
In this case, you should most definitely agree that electrons flow from the reductant to the anode (reductant-->anode-->cathode). Because the Pt or graphite electrode do not give off electrons by their own accord. The reductant has to donate electrons to the anode first.
2. Reductant is a metal, coated on the bottom half of a non-reactive electrode, i.e. a layer of Fe(s) coated on the bottom half of a stick of Pt(s)
In this case, you should also agree that electrons go from Fe(s) --> Fe2+(aq) + e(adsorbed onto Pt), because for electrons to flow from anode-->cathode, it has to go through the electrode. Again, the same reasoning applies, the reductant has to donate electrons to the electrode first before any electron flow can happen.
3. Reductant is a metal, and is also the electrode. The wires connecting to the anode is connected to the center (inside the electrode) and is insulated so no charge can be passed between the wire and the electrode, thus the only electron flow will have to be through the center.
In this case, you should also agree that electrons go from the surface to the core, because oxidation can only take place on the surface, and electrons have to go through the core.
4. Reductant is a metal, the wire is a normal alligator clip that connects to the electrode.
In this case, I can argue that the metal on the base of the electrode will react and give off electrons, this electron will thus have to travel through the center-core of the electrode to reach the wire. In actuality, due to coulombic repulsion and entropy, electrons don't travel in a simple fashion, and a large number of electrons end up slowly drifting up the electrode in an almost uniform fashion (
here is an introductory page). Electron transfer is not limited to the surface, it is present in the core as well. For a transfer of electrons in the core to occur, the electrons have to come from, you guessed it, the surface.
So here, are at least two cases where electrons definitely flow from reductant to anode (cases 1 and 2). In all other cases, there is always a flow of electrons through the core, and on an atomic level electrons will have to go from surface of metal to embedded inside the metal.
At this point, we look at what a reductant is: a reductant undergoes oxidation. In a Fe(s) electrode, what is the reductant?
It is only the surface Fe(s) atoms, because they are the only ones that undergo oxidation. Fe(s) in the core is not a reductant, as it does not undergo oxidation. We then define what an anode is: the electrode where oxidation is taking place (from IUPAC gold book), but what is an electrode? An electrode is a conductive material that is in contact with a non-conductive material, and in this case, the entire piece of Fe(s), except for the atoms that are reacting/have reacted, because at that point they stop being conductive.
To paint a picture:
Initially:
Fe-Fe-Fe-Fe-Fe-Fe-Fe
Oxidation:
Fe-Fe-Fe-Fe-Fe-Fe-2e Fe2+
Transfer
Fe-Fe-Fe-Fe-Fe-2e-Fe Fe2+
What has happened? The reductant (Fe on the surface) gave electrons to the electrode (the rest of the Fe(s)). What is this electrode? Anode. What has happened? Electron flowed from inside the reductant (inside the atom/molecule) to the anode.
Another query would be 'electrons are delocalized, they are not bound to the surface Fe'. Whilst this is one way of looking at it, if we have to remove a single Fe atom, we will see that the electron count of the metal decrease by 2. Thus when we define the reductant as Fe(s) [note that it is neutral], we have included two electrons as part of the 'package', and these electrons leave the package and is donated to the electrode.
So what I have shown here is that given the correct definitions:
-Reductant is a compound/atom that undergoes oxidation, i.e. only applies to the surface layer at any given point in time
-Anode is a conductive material at which oxidation occurs
We will always see the reductant giving electrons to the anode, and there will always be a flow of electrons from reductant to the anode.
And back to the original question, does something which work in all cases (option C) seem like a 'more correct' option than something which may or may not work for a particular case (option D)? I think so.
If your reply is 'We were never taught like that' or 'my definition has always been blah blah' or 'it's not in the textbook', I can only say that kind of rebuttal is more a complaint/whinge than a logical rebuttal.
That is all. /longpost.
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