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For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.(a) $\operatorname{Mn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Ni}(s)$(b) $3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Cu}(s)$(c) $\mathrm{Na}(s)+\mathrm{LiNO}_{3}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{Li}(s)$(d) $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ba}(s) \longrightarrow \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ca}(s)$

a. $E(c e l l)=0.928 V$ This reaction is spontaneous.b. $E(c c l l)=2.00 V$ This reaction is spontaneous.c. $E(c e l l)=-0.33 V$ This reaction is not spontaneous.d. $E(c c l)=0.084 V$ This reaction is spontaneous.

Chemistry 102

Chapter 17

Electrochemistry

Carleton College

Drexel University

University of Maryland - University College

University of Kentucky

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03:07

A liquid is a nearly incom…

04:38

A liquid is a state of mat…

01:52

For each reaction listed, …

01:22

Determine the overall reac…

17:43

Calculate $\mathscr{E}^{\c…

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Use the data in Appendix D…

02:29

02:01

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Calculate Ecell for each …

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08:10

03:29

02:04

00:51

Calculate the cell potenti…

05:12

Use the following half-rea…

07:31

Use tabulated electrode po…

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Starting with standard pot…

03:52

Balance each skeleton reac…

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given a full reaction. How do we tell it? The standard cell potential is of this reaction. And given that sell potential, how do we know the reaction of spontaneous? Well, the first step is dividing the reaction into the half reactions. I've already done this and balance this. And if you have any questions about that reference a previous video I have several videos about how to balance Redox reactions and sweating on behalf reactions of the first steps that is outlined quite clearly, I'm gonna drop the states and most of these problems. But remember that you should be adding those, um, for your homework and tests. And okay, so we have the oxidation reaction in green and the reduction reaction in blue. And now, basically, we should refer to ascend a reduction potential table in order to see what the potentials are for each of these, um, half reactions. Just be aware that it is. It is a standard production potential tables. That means you're not going to get the Senate oxidation potential directly. You have to flip the sign on it because it's gonna show this reactions. Oxidation, reacts, oxidation, reaction backwards. I can not talk today. Okay, so I looked those up in the book and we're gonna see the value I saw for the reduction version of this reaction was negative. 1.185 waltz. So I changed its eye on that. And then the reduction I just took out right on the table. Negative 0.257 bowls. Now, to get the potential of the entire cell, we're going to add these values together, which is, you know, really subtraction. So we add and we get a cell potential, a 0.9 to beat wolves. So now to tell us that his spontaneous or not, it's really just the cell potential is greater than zero of spontaneous is we have been zero. So if it's positive and it's not spontaneous, if it's a negative value so we can see that 0.928 is indeed positive. So we can say that Yes, it is spontaneous. Okay, so this is spontaneous. What's another example? So again, we're gonna be given a reaction this time. We have copper being reduced in aluminum being oxidized, and we could see that in these balance, half reactions. And then we're the look of the cell potentials for each one. And there's no even though, in this balance form we have coefficients and same with appear in this final reaction. There's coefficients does not change the cell potential, so you don't want to multiply the cell potential by anything. In either case, you just take out right on the table for ale. Three plus tale of the reduction so flipped the sign on the oxidation in the table of said negative 0.34 But we put a sign on it. Same took the exact value for the luxury of copper to plus the reduction of copper two plus two copper solid. And we have We add these together, hold on at them together and we get tools to bolt is greater than zero. So this reaction is spontaneous. Cool. Another example. This time we have sodium being oxidize and lithium being reduced. You see the reduction for any plus to an A has the potential of negative 2.71 The sign on that and then we see that lithium lithium plus one toe lithium solid has negative 3.4 volts in the table. We add these together and we get negative. 0.33 evils. Right, Because the reduction value, the negative value is larger than the positive value. So we're gonna have a negative number. This is less than zero, Tono wise not letting me type less than zero. So it is not spontaneous. All right. Last example. Same thing. Okay, so you are going to split it up into the half reactions. It's important to notice here that the nitrate is not being oxidized, a reduced in the adams in the country. I on the things happening to them there inert. So they're not gonna be a part of her half reactions. They won't be looking at what is being oxidized and reduced. So we have Jerry. I'm going to bury him two plus that it's worth being oxidized. Then we have calcium to plus going to calcium that's being reduced again. Any confusions with making the half reactions right, Friends? One of my other videos on balancing Madox reactions to clarify that. So we see a value of negative too 0.91 for barium two plus two barium. So sign we copied the exact value here. Negative point. Negative. 2.868 volts. We Abbas together so functionally were subtracted and we get 0.44 bolts. This is even though very close to zero. It is greater than zero. So this reaction is spontaneous. That's how you find the cell potential of, ah, reaction, given the balance reaction, and it's spontaneous or no. All right.

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