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Quantitative electrolysis

Explore the electrifying world of electrolytic cells! Discover how a non-spontaneous redox reaction is driven by current from an external voltage source. Learn about the formation of solid zinc at the zinc electrode and the oxidation of solid copper. Get charged up with quantitative electrolysis problems, using Faraday's constant and mole ratios to calculate the mass of zinc deposited.
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Created by Jay.

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  • leaf yellow style avatar for user Drew Frase
    Rather than go step by step to determine the amount of grams plated (which I would not advise against until you feel comfortable with the concepts), you can always utilize the following equation to save yourself some work:
    It = nF
    where I is the current expressed in amperes or coulombs/second, n is the moles of electrons and F is Faraday’s constant, which has the value of 96,485 C/mol e–.

    Just something to keep in mind!
    (21 votes)
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  • piceratops ultimate style avatar for user Jeds
    Since you are plating ~10 grams of zinc onto the electrode over this hour, is it reasonable to assume that the concentration of zinc ions in solution changes over time?

    If so, would that just mean that in order to maintain a constant current of 5 amps over the hour long reaction you would need to steadily increase the applied voltage?
    (2 votes)
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  • aqualine ultimate style avatar for user Daniel Li
    Couldn't you use the electrodeposition equation, It/nF?
    (1 vote)
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Video transcript

- Here's the electrolytic cell we talked about in the previous video. Remember, an electrolytic cell uses current to drive a non-spontaneous redox reaction. We need an external voltage source; here's our battery. We know electrons come from the negative terminal of the battery, and electrons are forced onto the zinc electrode. We know there are zinc 2+ ions in solution. So, if zinc 2+ gains two electrons, zinc 2+ is reduced to solid zinc. So we're going to form solid zinc at our zinc electrode. Solid zinc forms here. At our other electrode, the battery pulls electrons away from copper. Solid copper is oxidized; we lose two electrons to form cu 2+. So, the copper electrode loses mass over time. When we look at our problem, this is a quantitative electrolysis problem, because they're telling us what the current is: 5.0 amps. We're applying a current of 5.0 amps; how much zinc, in grams, deposits on the zinc electrode after one hour? We have to figure out how much zinc forms on our zinc electrode. First, we need to think about the definition for current. Current; lemme write this down here. Current is equal to charge over time. In physics, I is equal to current; charge is represented by Q; and time is lower case t. We have already seen charge before, in earlier videos, and we know that's measured in coulombs, and time is in seconds. Coulombs per second gives us an ampere, or an amp, for short. Let's plug in what we know; we know the current is five amps, so we plug that into here. We get 5.0 amps. We don't know what the charge is, but we do know the time. How much time are we talking about here? One hour; we need to convert one hour into seconds. How many seconds is one hour? There are 60 minutes in an hour, and each minute is 60 seconds. 60 times 60 gives us 3600, so there are 3600 seconds in one hour. Now we can solve for Q; we can solve for the charge. Five times 3600 is equal to 18,000. After one hour, we're talking about 18,000 coulombs. From the charge, we can figure out how many moles of electrons we're dealing with here, because of Faraday's constant. Remember, Faraday's constant tells us that one mole of electrons has a charge of 96,500 coulombs. If we have 18,000 coulombs and we're trying to find how many moles of electrons that is, we would need to divide by Faraday's constant. 18,000 divided by 96,500, which is Faraday's constant, the charge of one mole of electrons. If you do it this way, you can see that coulombs would cancel out, and you would get moles of electrons. So this is equal to .19 moles of electrons. .19 moles of electrons were forced through our electrolytic cell because of the battery. Next, we need to relate the moles of electrons to the moles of zinc that are formed. We can do that by remembering our reduction half-reaction: zinc 2+ plus two electrons forms solid zinc. Let's write down our reduction half-reaction here. We know that zinc 2+ ions are being reduced to form solid zinc. Let's think about those mole ratios here. Two moles of electrons are needed to reduce one mole of zinc 2+ ions to form one mole of solid zinc. We now have the relationship. We know the mole ratio of electrons to moles of solid zinc; it's a mole ratio of two to one. One mole of zinc is produced for every two moles of electrons that we have. Let's set up a proportion to figure out how many moles of zinc are produced. We'll put electrons over solid zinc, so we have a mole ratio of two to one. Our mole ratio is two to one. Then, on the right side of our proportion, we know that .19 moles of electrons were forced through our cell. So we can write, down here, .19 moles of electrons, and that would be over x. X represents the moles of solid zinc. Let's solve our proportion: that's 2x is equal to .19. 2x is equal to .19. .19 divided by two is equal to .095. So, x is equal to .095, and x represents the moles; this is the moles of zinc that are produced in our reaction. Now, again, we got this from our mole ratio. One mole of zinc forms for every two moles of electrons. So, if you have .19 moles of electrons, half that is how many moles of zinc that are formed? Finally, we need to go from moles of zinc to grams of zinc. Our problem asked us for grams of zinc that were deposited. Going from moles to grams is pretty easy; you just multiply by the molar mass. So, if you have .095 moles of zinc and we multiply by the molar mass of zinc, which is 65.39, the molar mass of zinc is 65.39 grams per mole, if you multiply, the moles cancel out, and that gives you grams. So we get 6.2 grams of zinc. That's our final answer; that's how much zinc deposits on the zinc electrode. There are other ways to think about these types of problems; I prefer to go through everything step by step and logically come to the final answer. You could also just think about units and do all this math; that's another way to do it. This is just the way that I prefer to do a quantitative electrolysis problem.