2015 AP Chemistry free response 1a

Metal air cells are a relatively new type of portable energy source consisting of a metal anode, an alkaline electrolyte paste that contains water, and a porous cathode membrane that lets in oxygen from the air. A schematic of the cell is shown above, and so we see the different parts. Right over here, we see the metal anode. It's an anode, so this would be the negative terminal of our power source—it'll be the source of electrons.

We have an alkaline electrolyte paste that contains water. Let me underline that—there's a lot of interesting words there! Alkaline electrolyte paste that contains water—that's this right over here in the middle. When they say it's alkaline, that means that it's going to be basic versus acidic. It's going to have a pH higher than 7. An electrolyte is something that, if you dissolve it in a polar substance, a polar substance like water, well then the solution is going to be good at conducting electricity.

So, what they tell us is this electrolyte paste is going to be something that's good at conducting electricity, and they tell us it's alkaline, so it's going to be basic; it's going to have a pH higher than 7. And there's a porous cathode membrane that lets in oxygen from the air. So, this is the cathode—this is going to be the positive terminal. This is where the electrons are going to be attracted to.

It's a porous cathode membrane that allows oxygen from the air to seep in because it is porous. Reduction potentials for the cathode and three possible metal anodes are given in the table below. So, remember reduction potentials—you can view these as the gaining of electrons. This is the potential to gain electrons, which is one way to think about a reaction that somehow involves electrons being incorporated in some way.

Right over here, you have molecular oxygen reacting with water. For every molecule of molecular oxygen, you have two molecules of water, four electrons, and then they react to form four hydroxide anions. So once again, this is a reduction reaction because those electrons are being incorporated into the molecule, and so you're left with the hydroxide right over here.

If you look at our metal air cell up here, where is that occurring? Well, you need oxygen, you need water, and you need electrons. If you look at the cathode right over here, you've got your electrons coming in, you've got your oxygen coming in, and water we can assume is available in this area. The electrolyte paste has water in it, and the cathode is porous—it's a porous membrane-type substance.

So, they are all available over here, and you can assume that that reaction could happen right over here in the cathode. Let me just write it: O2 in as a gas plus two molecules of H2O in liquid form, and then you've got your electrons coming in on the wire, I guess you could say. That is going to yield four hydroxide anions.

Four hydroxide anions are in a solution. They're part of that electrolyte paste that's in water—they're an aqueous solution, a water-based solution. So that's going to happen right over there. The hydroxide anions—and I could even say four hydroxide anions—are going to be produced every time this reaction is happening.

They tell us the reduction potential has a positive voltage, and notice they say the reduction potential at pH 11. So that's consistent with alkaline conditions because this paste might seep in through here a little bit, and obviously you have all this hydroxide being produced. So, you're going to have a higher pH and more basic pH.

The fact that this is a positive voltage—plus 0.34 volts—means that the potential is going in this direction. The electric potential favors this reaction to happen going from left to right. Now, they also said they gave us the reduction potentials for the cathode, which we just talked about. This is the reduction potential for the cathode, and three possible metal anodes are given in the table below.

Here, there are three possible metal anodes. So, we could have zinc, we could have sodium, we could have calcium. Let's just go with zinc since it's the first one listed here. So if we assumed that the metal here was zinc, what's going to be going on? We're going to have the zinc reacting with these hydroxides that are being produced over in the cathode.

Then you're going to use that to produce zinc oxide, water, and electrons. So you're actually going to have the reverse of this reaction. Let me write the reverse: you're going to have zinc in the solid state. This whole anode is made out of metal zinc, and then for every molecule of that, you're going to have two hydroxide anions that are dissolved in water—that's what makes the electrolyte paste alkaline.

You are going to then go to—you're going to react and you're going to form zinc oxide, zinc oxide in the solid state, plus water in the liquid state, plus two electrons. So this reaction that I just described is going to be happening right over here. You can think of it as the hydroxide anions getting formed at the cathode, and then they move their way over to the left to the anode where they react with the zinc and form zinc oxide.

You have more and more zinc oxide being formed, water which can then seep its way back into the electrolyte paste, and then it can eventually react again at the cathode, and then you have these two electrons. This gives rise to an electron source right over here. The electrons would migrate, and then they can go to the positive side, the cathode, to react again.

You can start to see how this will be an energy source that you can tap into—this current that'll form to do some useful work. That's why you have an energy source. Alright, so let's read the questions now that we have a decent understanding of what's going on.

Early forms of metal air cells used zinc as the anode. As the example we just thought about, zinc oxide is produced as the cell operates according to the overall equation below. For every two molecules of zinc and one molecule of molecular oxygen, you produce two molecules of zinc oxide.

Use the data in the table above to calculate the cell potential for the zinc air cell. Let's think about—let's break down this reaction. Actually, we can just break it down into these two steps here because notice this—you can't see them both at the same time, but the things that are reacting are zinc. Let me underline the things that are reacting: you have zinc, you have oxygen.

So if you return this, if you return the second reaction around like we did before, let me rewrite it. Let me rewrite both of them, actually, a little bit lower right over here. So you have this top reaction: O2 in gaseous state plus two H2O in liquid plus four electrons yields four hydroxide anions in an aqueous solution.

Then this one, we said we were going to go in the other direction: zinc in the solid state plus two hydroxide anions yields zinc oxide in the solid state plus liquid water plus two electrons. When you see this, you see that we are reacting—we have the oxygen, we have the zinc, which are the two things we want to react here.

If you want to make them equivalent in terms of the number of molecules, we could say, alright, we have one molecule of molecular oxygen and we need two zincs. Let me multiply this whole reaction by two since two zincs react with four hydroxides to produce two zinc oxides, two molecules of water, and four electrons. I just said, oh, if this reaction happened twice, you'd have just twice as many of the reactants and the products here.

And then notice now you have a molecule of molecular oxygen and two molecules of zinc. If all of this were to happen, you are going to produce two molecules of zinc oxide. What about everything else that I put that were in these reactions? Well, the water is here, but then water also gets—two molecules of water are here.

I guess if you look at the total reaction, we're neutral there; we haven't produced or lost any water. You have four electrons, you have four electrons, so that's why they didn't write it in this total reaction here. And these four hydroxide ions—well, they're going to be used up in this second part of the reaction, so we're neutral there. They get produced, but then they get used.

If you look at the total reaction, you're using the molecular oxygen and the zinc to produce zinc oxide. They say use the data in the table above to calculate the cell potential for the zinc air cell. The potential for this top reaction is positive 0.34 volts.

Now what's the reaction—what's the potential for this bottom reaction? Well, it's the reverse of this reaction, so this reaction has this reaction has a negative potential. But since we reverse it, it's going to be a positive potential, so it's going to be positive 1.31 volts.

Now some of you might be saying, well, not only did we reverse it, we multiplied it by two. We said two of those reactions when we multiply the voltage by two. Well, if we were just thinking about the energy released from a reaction, well, yeah. Absolutely, if you're going to have the reaction twice or twice as much of the reaction, you're going to have twice as much of the energy.

But voltage, you have to remember that you can view it as potential energy per unit charge. Since we're not—this isn't kind of your absolute energy; this is your energy per unit charge is one way to think about it. When you're doing more of it, it doesn't change the actual voltage, so that's an important thing to think about.

If this was talking about total energy or enthalpy or something like that, then if you were to multiply both sides by two, you would multiply the energy released or taken by two. But since we're talking about voltage, voltage isn't a quantity that depends on the number of charges or the number of molecules you're doing. It's a per—what is the potential per unit charge, is one way to think about it.

For this whole reaction, we would just add the combined voltage of both of these and then you would get 1.65 volts. Now, let's do part two. The electrolyte paste contains hydroxide and hydroxide ions. On the diagram of the cell above, draw an arrow to indicate the direction of migration of hydroxide ions through the electrolyte as the cell operates.

Well, we've already thought about that. The hydroxide gets produced at the cathode with this reaction and then it gets used up in the anode. So the hydroxide is going to be moving in that direction.