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Course: Chemistry library > Unit 19
Lesson 1: 2015 AP Chemistry free response questions- 2015 AP Chemistry free response 1a
- 2015 AP Chemistry free response 1b and c
- 2015 AP Chemistry free response 1d
- 2015 AP Chemistry free response 1e
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- 2015 AP Chemistry free response 2c
- 2015 AP Chemistry free response 2d and e
- 2015 AP Chemistry free response 2f
- 2015 AP Chemistry free response 3a
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- 2015 AP Chemistry free response 4
- 2015 AP Chemistry free response 5
- 2015 AP Chemistry free response 5a: Finding order of reaction
- 2015 AP Chemistry free response 6
- 2015 AP Chemistry free response 7
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2015 AP Chemistry free response 1a
Analyzing reactions and cell potential of a metal-air cell. From 2015 AP Chemistry free response 1a. .
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- potassium sulphate + copper sulphate (delta) = what is the answer after balancing?(3 votes)
- The reaction wouldnt occur because the deltaH needed to start the decomposition of potassium sulfate would be the same energy release to form the bonds in the product potassium sulfate(3 votes)
- how do we know we should use the second half-reaction? 4:30(3 votes)
- To find a total cell potential we should take into account both oxidation and reduction reactions. It's not enough to know how much an element "wants" to be reduced, we also should check how strongly the other element wants to be oxidized, as electrons lost by one element are the electrons gained by another.(3 votes)
- Why do you reverse the reaction for the Zinc Oxide equation?(3 votes)
- The reaction in the table is a reduction reaction (where zinc is being reduced), but in this cell, zinc is being oxidized, so the reaction given in the table is happening in reverse, making it an oxidation reaction.(3 votes)
- Since the equation is Ecell=Ecath-Eanode, why does he add the two reduction potentials together?
Thanks!(2 votes)- Sal did this in two steps.
First he reversed the reaction for the anode and rewrote the potential (for what is now an oxidation reaction) as the negative of the reduction potential, which gives us–Eanode
.
Second he added the reductionEcathode
and the oxidation–Eanode
potentials to getEcathode + –Eanode
.
So this works out to be the same as your equation.
Note that theE
values here are all for reductions ...(2 votes)
- The first equation in the table is really confusing. Why would molecular oxygen grab hydrogen from the water? Even more so, it seems it only happens in basic solution, where we already have an excess of hydroxide ions! I can imagine that since there's not enough protons, oxygen reacts with water instead, but oxygen is solved in regular water all the time, and we don't see a spontaneous increase in basicity.(2 votes)
- actually that happens in the cathode and the equation is already given in the question paper so no worries!!
Just learn how to use it rather trying to derive it might confuse you......
anyways, hopefully my suggestion helps....... : )(1 vote)
- For the first part of A so (i) when you flip the half reaction and multiply it by two why don't you do the the same for the potential.(1 vote)
- The electric potential is an intrinsic property that is not affected by the amount of material present.(3 votes)
- what is an alkene and how do i tell if it is so?(1 vote)
- An alkene is an unsaturated hyrdrocarbon that contains at least one carbon-carbon double bond. You can identify an alkene by its' double bond.(3 votes)
- Atwhy do you add the cell potentials? I thought the formula for E cell was 11:29
Ecell= E cathode - E anode... So shouldn't you subtract?(2 votes)- Sal did this in two steps.
First he reversed the reaction for the anode and rewrote the potential (for what is now an oxidation reaction) as the negative of the reduction potential, which gives us–Eanode
.
Second he added the reductionEcathode
and the oxidation–Eanode
potentials to getEcathode + –Eanode
.
So this works out to be the same as your equation.
Note that theE
values here are all for reductions ...(2 votes)
- What is the difference between chemistry and AP chemistry? I've noticed that the videos in both topics are the same. I know that AP chemistry gives you credits for college but does the actual content change?(1 vote)
- Why is anode considered the negative terminal and cathode the positive terminal when thinking about galvanic cells?
Thanks(1 vote)
Video transcript
- [Voiceover] 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 seven. Electrolyte paste. 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,
this electrolyte paste is gonna 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 seven. And a porous cathode
membrane that let's 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, and it's a porous cathode
membrane that lets in oxygen from the air. And so you can see that
oxygen is available in this membrane. It somehow is allowed 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. Reduction is the gaining of electrons. So this is the potential
to gain electrons, is one way to think about it, or a reaction that
somehow involves electrons being incorporated in some way. And right over here you
have molecular oxygen reacting with water. For every molecule of molecular oxygen, you get 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. And 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. Well, if you look at the
cathode right over here, you got your electrons coming in, you got your oxygen coming
in, and water, we can assume, is available in this area. The electrolyte paste has water in it. The cathode is porous, it's a porous membrane-type substance. So they're all available over here. So you can assume that, that reaction could happen right over
here in the cathode. So let me just write it, 02 in as a gas. So for every molecule of O2,
you're gonna have two molecules of H2O in a liquid form. And then you got your electrons
coming in on the wire, I guess you could say,
and that is going to yield four hydroxide anions. Four hydroxide anions. And they're in a solution. They're part of that electrolyte
paste that's in water. They're an aqueous solution,
a water-based solution. And so that's going to
happen right over there so the hydroxide anions, I can even say four hydroxide anions are going to be produced
every time this reaction is happening. And they tell us the reduction potential. This has a positive 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 gonna have a
higher pH, a more basic pH. And so the fact that this
is a positive voltage, so 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 give
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. So here there are three
possible metal anodes. So we could have zinc,
we could have sodium, we could have calcium, and so 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? Well, you're going to
have the zinc reacting with these hydroxides
that are being produced over in the cathode, and
then you're going to use that to produce zinc oxide water and electrons. So you're actually gonna have
the reverse of this reaction. You're going to have,
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 gonna have two hydroxide
anions that are dissolved in water. That's what makes the
electrolyte paste alkaline. And you are going to then go to, they'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, plus two electrons. And so this reaction
that I just described, this is going to be
happening right over here. So you could think of it
as the hydroxide anions get formed at the cathode,
and then they move their way over to the left to the anode
where they react with the zinc and they turn, or the zinc
reacts and forms zinc oxide, so 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. So these electrons, this
turns into an electron source right over here, and then
the electrons would migrate and then they can go to the
positive side, the cathode, to react again. And so 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. All right, so let's
read that questions now that we have a decent
understanding of what's going on. Early forms of metal air
cells use zinc as the anode. Well, that's 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, calculate the cell potential
for the zinc air cell. So 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 is, we can't see them both at the same time but this, the things that are reacting, you have zinc. Let me underlie the
things that are reacting. You have zinc, you have zinc. You have oxygen, you have oxygen. And so if you return this, if you return the second reaction around like we did before, and let me rewrite it, let me rewrite both of them
actually a little bit lower right over here. So you have this top reaction. So you have O2 gaseous state plus two H2O liquid, plus four electrons. Whoops, four electrons, yields four hydroxide anions in aqueous solution. And then this one, we said we're gonna go in the other direction. So you're going to have zinc, you're going to have zinc in the solid state, plus two hydroxide anions. And then that's going to yield zinc, zinc oxide in the solid state, plus liquid water plus two electrons. So when you see this, you
see that we are reacting, we have the oxygen, we have the zinc, which are the two things
(mumbles) interact here. And if you want to make them equivalent in terms of the number of molecules, we could say all right, we have one molecule of molecular oxygen and we need two zincs. So let me multiply this
whole reaction by two. So two zincs react with four hydroxides, react to produce two zinc oxides, two molecules of water, two molecules of water, and then four electrons. I just said, well, if this
reaction happened twice, you'd would have just twice
as many of the reactants and the products here. And then notice, now you have a molecule of molecular oxygen, two molecules of zinc, molecule of molecular oxygen,
two molecules of zinc. If all of this were to happen, you are going to produce
two molecules of zinc oxide. And well, what about
everything else that I put, that were in these reactions? Well, the water is here, but then water also gets
two modules of water, two molecules of water. So 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
molecules, anions, well, they're going to be
used up in this second part of the reactions. So we're neutral there. They get produced, but then they get used. So if you look at the total reaction, you're using the molecular
oxygen and the zinc to produce zinc oxide. And so they say use the
data in the table above, calculate the cell potential
for the zinc-air cell. Well, the potential for
this top reaction is zero, is positive .34 volts. And 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 a negative potential, but since we reversed 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, wouldn't we multiply the voltage by two? Well, if we were just
talking about the energy released from a reaction,
well, yeah, absolutely. If you're gonna have the
reaction twice or twice as much of the reaction, you're
gonna have twice as much of the energy. But voltage, you have to remember, you can view it as potential
energy per unit charge. So since we're not, this isn't (mumbles) 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 entropy 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 we're doing to, it's a per. It's what is the potential per unit charge is one way to think about it. And so for this whole
reaction, we would just add the combined voltage of both of these, and then you would get 1.65, 1.65 volts. Now let's do part two. The electrolyte paste
contains 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 a 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, the hydroxide is going to be moving in that direction.