What happens to the cell potential when there is less oxygen in a zinc-air cell. From 2015 AP Chemistry free response 1b and 1c.
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- At6:40it is claimed that the cell potential will be lower because there are fewer oxygen molecules available. Is that really correct? Even if there would be only one oxygen molecule available to oxidize (2) zinc molecules, the potential (being an intensive property) should still be the same as under STP. The difference to STP is that the electric current (number of transferred electrons over a unit time frame) would be very small.(14 votes)
- Pressure isn't the same however. So it wouldn't be standard STP. Think of cell potential in relation to Q. Q would be larger due to lower concentration of reactants. This in turn would lead to lower instantaneous cell potential. E = E* - .592/n * log Q. Log Q increases as Q increases which decreases E.(12 votes)
- Why is the partial pressure of O2 lower?(2 votes)
- The atmosphere gets less dense as you go higher (i.e. the pressure gets lower). This means that there is a lower partial pressure for each component of the atmosphere.
This is why people carry oxygen when they climb mountains such as Everest and why jets have those little masks that supposedly drop down if the cabin springs a leak at high altitude.(2 votes)
- To calculate Q , we transform the partial pression to molar concentration ?
PV= nRT <-----> n/V = P/RT(1 vote)
- Acc. to Le chatelier's principle shouldn't there be an increase in the rate of reaction, therefore increasing potential of the cell? Because they haven't specified the decrease of moles of Oxygen at higher elevation(1 vote)
- Wait, I'm confused. The scoring guidelines for this question explained it in terms of Q and K and equilibrium shifting. Would explaining it without terms of Q changing closer to K still earn me a point? Also, why does Q changing closer to K make the cell have less potential and why does Q change? How do we know that Q is closer to K now (and not further)?(1 vote)
- Can you also justify c)ii using La Chatlier's Principle? Lower pressure would favor reactants due to more molecules.(0 votes)
- [Voiceover] All right, part b. A fresh zinc air cell is weighed on an analytical balance before being placed in a hearing aid for use. As the cell operates, does the mass of the cell increase, decrease, or remain the same? Justify your answer to part b one, or b i, b one in terms of the equation for the overall cell reaction. So let's just think about it for a little bit. Does the mass of the cell increase, decrease, or remain the same? Well, let's go to our original reaction over here. This is our total reaction. The zinc is part of the metal air cell, so that would contribute to its weight beforehand, and then it reacts with the oxygen. Well, where is this oxygen coming from? Well, this is oxygen is coming from the air and essentially gets incorporated into the zinc. So, before, the oxygen is in the air, but then after the reaction, after the total reaction, it gets incorporated with the zinc, I guess I should say, not into the zinc, of course. It gets incorporated with the zinc as zinc oxide, and so now, the oxygen is part of the cell. Part, or contributes to weight of the cell. So, contributes, contributes to the weight, the weight of cell. It doesn't contribute before the reaction, but it contributes after the reaction. So our, the mass, and I guess you could say, the weight, is going to increase. So I'll just write increase. And part two, we need to justify our answer in terms of the equation for the overall cell reaction. So let me just write the overall cell reaction. So, two zinc molecules react with one molecular oxygen, one O_two, to yield two zinc oxide, two zinc oxide molecules. Well, what we could say is, is that this was not part of cell before reaction, not part of cell before reaction. It's coming from the air. Before reaction. And then we could say, part of cell after reaction. Part of cell after reaction. And we could write it out in words, if we like. After reaction. And so I could say, oxygen, oxygen comes from air, from, from air, air to be incorporated into the zinc oxide, to be incorporated into zinc oxide which then contributes to weight of cell, or to the mass of cell. Which then contributes to mass of cell, to mass of cell. All right, I think I feel pretty good about that. Now let's do part c. The zinc-air cell is taken to the top of a mountain where the air pressure is lower. Will the cell potential be higher, lower, or the same as the cell potential at the lower elevation? Justify your answer to the first part based on the equation for the overall cell reaction and the information above. Well, let me just write the overall cell reaction again. So let me, the overall cell reaction. I'll rewrite it again. So we have two zinc molecules. Let me write that a little bit neater. Two, and I wrote it messier. Two zinc molecules. One molecular oxygen. We're going to have this ingrained in our brain by the time this is done, I'm writing it so frequently. We get two zinc oxides. Two zinc oxides. And so what's going to happen if we go to the top of a mountain where the air pressure is lower, we have just fewer air molecules in general. That means we're also going to have lower oxygen molecules. So we could say that the partial pressure of oxygen is going to be lower. So we could say, partial pressure of reactant will be lower. Partial pressure. There's just going to be fewer oxygen molecules from the air being able to bump and react with the zinc molecules. Well, actually, it's not reacting with the zinc molecules, as we saw in the original reactions, when we look at the half reactions. Fewer reactions from the air just bumping in the right way with the water molecules and actually the electrons that are coming in through the circuit, you're just going to have fewer of them bouncing around, being able to do this reaction. So you could say partial pressure, partial pressure of reactant. Or maybe I'll say partial pressure, or you could also do the concentration of the reaction, there's going to be just, reactant. There's going to be less of it around. Partial pressure or we can say concentration, concentration lower, lower, so lower, so lower cell potential. Lower cell, cell potential. So actually, I just did it in reverse. I answered part two first. And so part one is lower. So we're gonna have a lower cell potential. Lower cell potential. It's really important to get the conceptual understanding of why it's happening. Remember, when we look at the half reaction, what's happening, what is happening at our cathode, we have oxygen coming in from the air, and it gets into the kind of pores of this porous substance, and when the oxygen, the O_2 molecules bump in just the right way with the water and the electrons coming in, they just kind of go together, and react, you get your hydroxide, for they react to form the hydroxide. Now if you have less, you know, at a lower elevation, you're gonna have a lot of oxygen. You're gonna have a high partial pressure of the molecular oxygen. So there are a lot of them (mumbles) moment in time are going to be available to bounce around and react. But if you go to a higher elevation, you're just gonna have fewer of these characters. So you're gonna have fewer of these characters. So you're going to have a lower partial pressure of oxygen, or you could think of it as you have a lower concentration of oxygen inside the pores, so it's gonna be harder for this reaction to move forward. And you can think about the extreme situation. What if you were going to a super high elevation. If you were eventually going into space, where you have very little oxygen or no oxygen at all, well then, then the reaction's going to stop altogether. So hopefully, hopefully that, that makes, let me just go back to part c here. I'm having trouble with my, oh, this, then I go back to part c. And there you go.