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- 2015 AP Chemistry free response 1d
2015 AP Chemistry free response 1d
Calculating which metal will transfer more electrons in a metal-air battery. From 2015 AP Chemistry free response 1d.
Want to join the conversation?
- I'm confused as to what the voltage is and why it doesn't come into play in the problem. What is voltage, and what does it mean in the context of these reactions?
Thank you.(7 votes)
- My understanding is that voltage is the willingness of the reaction to occur and is used to determine in which direction a reaction would occur. The question is asking for the number of electrons that result from 1 g of metal in the reaction, so the voltage is thus irrelevant.(6 votes)
- Why isn't the mol/g multiplied by how many electrons are present in the reaction? I got the same answer as Sal, but I structured my stoichiometry different.... Is it incorrect to do so?(1 vote)
- He did account for that in his equations, see the "mol e-" term.
There's often multiple ways to arrive at the correct answer for these questions.(1 vote)
- why Na + OH- gives Na2O but not NaOH+e-(1 vote)
- When it comes to chemical reactions, you don't need to balance individual elements or ions of reactions. See a bigger picture.
2Na+ + 2OH- = Na2O + H2O
# of Na: 2(left) = 2(right)
# of O : 2(left) = 2(right)
# of H : 2(left) = 2(right)
Balancing a chemical equation isn't different from balancing a mathematical equation.
(2x+3y)+(3x+2y) = (4x+1y)+(1x+4y)
What matters is the total number of a variable on both sides than some part of them(1 vote)
- Is this right for me to write down 2H2SO4 instead of either 2H+ + 2HSO4- / 4H+ + 2SO4^2- ?(0 votes)
- They mean the same thing.
It all depends on whether you are expected to write an ionic or a molecular equation.(1 vote)
- The Cl oxidation number from NaOCl on the AP Chem FRQ was a +1(0 votes)
- [Voiceover] Metal-air cells need to be lightweight for many applications. In order to transfer more electrons with a smaller mass, sodium and calcium are investigated as potential anodes. A 1.0 gram anode of anode of which of, (laughs) A 1.0 grm anode of which of these metals would transfer more electrons, assuming that the anode is totally consumed during the lifetime of a cell? Justify your answer with calculations. So let's go to the top where they showed the reactions involving sodium and calcium. And actually, let me just copy and paste this so it's easier for me to reference near the bottom, where we're actually doing the question. So let me just copy this. And then let's paste it down here, that I can reference it. Okay, I'll even make it a little bit smaller just so I don't take up too much much real estate. And so the reactions that we care about are actually going to go in the reverse direction of these half-reactions. Remember, when we looked at our power source right over here, this first half-reaction is what produces the hydroxide ions. The hydroxide ions migrate their way to the left-hand side. And then they perform, then they are involved in oxidizing, in oxidizing whatever metal you have here. If this was zinc, the hydroxide combines with the zinc. They're going to oxidize the zinc. And those extra electrons that are released in that reaction, those are the ones that are going to travel across the wire and provide the current. But the hydroxide, it could react with sodium, if this metal is sodium. It could react with calcium, if this metal is calcium. So the reactions we care about are going from this side, the right side, to the left side. And so, let me just rewrite that again. And actually, just the act of rewriting it will help, will, my pen has been having trouble, is having trouble, actually scrolling down. But just the act of rewriting it will help with digesting what's going on. So the sodium reaction. For every two molecules of sodium, it's solid, it's actually forming our anode, plus two molecules of hydroxide, dissolved in the water, aqueous solution, we are going to yield sodium oxide, a solid version of that, plus water in the liquid state, and plus two electrons. And this is really important 'cause we wanna think about one gram of each of these metals. How many electrons would they yield? How many electrons are they going to be able to transfer through, potentially through, that wire? So that's the first relationship up there. And then, actually, let me just leave the sodium there. And then let me do the calcium over here. And remember, we wanna go from the right to the left, so let me rewrite it that way. Calcium, a solid state, plus two hydroxides, dissolved in water, is going to yield calcium oxide, solid, plus liquid water, liquid, plus two electrons. So, once again, I just rewrote each of these in the opposite direction. And, if I wanna write the voltage, I would flip the sign now, 'cause I have written them in the other direction. But they're not asking us about the voltage. They want us to know a one gram anode of which of these metals would transfer more electrons? So if we think about sodium, the case with sodium, we have 1.0 grams, 1.0 grams of sodium. Well, let's think about how many moles of sodium that is. Well, that's going to be, well let's see. We're going to have, we could look up sodium on our Periodic Table of Elements. And this is actually the one that they give you when you take the AP Test. And you see sodium right over here. We could read its atomic weight, sometimes now called relative atomic mass. And that's the weighted average of the atomic masses of the isotopes that you would typically find of sodium on Earth, if you just took a random sample of sodium, and you were, average their atomic masses, it would be 22.99. So that tells us that a mole of sodium is going to weigh, is going to have a mass of 22.99 grams. Or you could say, so a mole is going to, a mole is going to have a mass of 22.99 grams. So I could write one mole, one mole of sodium is going to have a mass of 22.99 grams. And of course, this is still talking about sodium. And so I made sure that I have it in terms of moles per gram, because when you do our dimensional analysis, grams divided by grams, or grams of sodium divided by grams of sodium, they're gonna cancel out, and I'm gonna be left with moles of sodium. So this part of the expression, this is gonna give me how many moles of sodium we are dealing with if I have one gram of sodium. But we need to think about how many electrons, how many electrons are we going to be able to transfer? And that's where we actually have to look at this reaction. So for every two, for every two moles of sodium, we are going to transfer two moles of electrons. Or another way you can think about it is for every mole that, here you have a two to two relationship, or you could say for every mole of sodium, we are going to transfer one mole of electrons. And so let's do that. We're gonna transfer one mole, one mole of, let me write it this way. One mole of electrons per mole of sodium. If you have two moles of sodium, you're going to get two moles of electrons. If you have one mole of sodium, you're gonna have two. (laughs) If you have one mole of sodium, you're gonna have one mole of electrons. And so, what is this going to give us? Well, if you just multiply out. Actually, let's look at the units first. Grams of sodium cancel out with grams of sodium. Moles of sodium cancel out with moles of sodium. And so you're left with just moles of electrons, moles of, (scoffs) why am I making an apostrophe? Moles of electrons. And it's going to be 1.0 times one, divided by 22.9, times one. So it's 1.0 divided by 22.9. And let's see. I'm gonna have two significant digits right over here, two significant figures. So let me get the calculator out. And it is, let's see, so it's just one, I could just say one divided by 22.9 is equal to, so if we have two significant figures, it's approximately .044, .044 So, I'll say approximately 0.044 moles of electron. Remember, moles. All we've got is numbers. This is still a lot of electrons. A mole is 6.022 times 10 to the, you know, or roughly that, times 10 to the 23rd. So it's a very, very large number of electrons that we are going to potentially transfer over here, if we have one gram of sodium. Now let's do the same thing with the calcium. So if we have calcium, it's going to be the same exercise here. We have one gram of calcium times one mole of calcium, has what mass? One mole of calcium has a mass of how many grams? So let's go back to the Periodic Table. And there you have calcium right over here. The atomic weight is 40.08, or oftentimes, or now the more modern name is relative atomic mass, the weighted average of the typical, of the isotopes. 40.08. So I could write 40.08 right over here. And then, in our reaction, for every mole of calcium, how many moles of electrons are we going to generate? Well, we have in this reaction, one, for every one calcium, we have two electrons. So for every mole of calcium, we're gonna have two moles of electrons. So we could say times two moles of electrons for every mole of calcium that we have as a reactant over here. And so this is going to give us, this is going to be approximately, see, our units, grams of calcium cancels with grams of calcium. Moles of calcium cancel with moles of calcium. And so we're gonna have one times one, divided by 40.08, times two. So it's going to be two divided by 40.08, which is, so, two divided by 40.08. Now we're gonna have two significant figures right over here. 'Cause this two and those ones are kind of pure numbers, while these are from measurements. So, and this is the fewest number of significant figures, so we have two. So two divided by that. Let's see, so it's gonna be 0.050. So 0.050 moles of electrons. So you can see, right from this calculation, that one gram of calcium can transfer more electrons than one gram of sodium, if sodium was the anode. So we see very clearly that calcium, so one gram of calcium can transfer more electrons, can transfer more electrons, more electrons than, 1.0 gram of sodium. And you can see it right over here. We have done the calculation.