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For each of the following cell reactions, write each half-cell reaction, and calculate E and Eofor the cell. a) Hg2Cl2 (s) → 2Hg (l) + Cl2 (0.80atm) b) 6Fe2+...

Question

For each of the following cell reactions, write each half-cell reaction, and calculate E and Eofor the cell. a) Hg2Cl2 (s) → 2Hg (l) + Cl2 (0.80atm) b) 6Fe2+ (1.0M) + Cr2O72-(0.50M) + 14H+(3.0M) → 2 Cr3+ (0.71M ) + 6Fe3+ (2.0M) + 7H2O (l)

For each of the following cell reactions, write each half-cell reaction, and calculate E and Eo for the cell. a) Hg2Cl2 (s) → 2Hg (l) + Cl2 (0.80atm) b) 6Fe2+ (1.0M) + Cr2O7 2- (0.50M) + 14H+ (3.0M) → 2 Cr3+ (0.71M ) + 6Fe3+ (2.0M) + 7H2O (l)



Answers

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate $\mathscr{C}_{\text {cell. }}$ b. Calculate $\Delta G^{\circ}$ and $K$ for the cell reaction at $25^{\circ} \mathrm{C}$ c. Calculate $\mathscr{E}_{\text {cell }}$ at $25^{\circ} \mathrm{C}$ when $\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}$ and $\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}$

In this problem were given these 2/2 reactions that occur within a galvanic cell in part A. We want to write out what the overall cell reaction is as well as the cell potential. We know that for galvanic cells, the cell potential has to be positive. So therefore, the F reaction with the larger standard reduction potential will be the half reaction that undergoes reduction while the other one will be the one undergoes oxidation. So based on the differences in the standard reduction potentials, we know that the second half reaction will have to be reversed, since it is smaller and therefore will undergo oxidation. So the top half reaction will undergo reduction. So that's a U three plus plus three electrons goes to solid gold. The standard reduction potential for that was 1.50 volts, and now the second half reaction will be flipped to undergo oxidation. So we have TL going to TL Plus was one electron, and now we flip the sign of the standard reduction potential. Since this half reaction has been reversed to undergo oxidation. So instead of negative 0.34 it's positive 0.34 so Now we combine these two together to get the overall reaction as well as the standard cell potential. We have to multiply the second reaction by three so that we have three electrons on each side to cancel out. So the overall cell reaction we're left with is a you three plus a qui ists waas three solid Valium goes to au solid plus three pl waas quickest in the standard cell potential has done just equal to 1.50 plus 0.34 and that comes out to 1.84 volts. Now in Part B, we want to calculate the value of the change in Gibbs Free energy at standard conditions Delta G, which is equal to negative and f e. And so that's equal to negative. And it's a number of moles of electrons. We see that we had three electrons cancelling out, so that's three moles of electrons times Faraday's constant 96,000 400 85 cool ums her mole of electrons times the standard self potential, which we just calculated to be 1.84 boats or jewels per Coolum. And I only calculate out Delta G. We're left with energy units of jewels. And if we divide by 1000 we can get Delta Jean units of killer jewels come out to about negative 500 33 killer JAL's. And now we also want to find the value of K, the equilibrium constant. We know that when a galvanic cell is an equilibrium, cell potential is equal to zero. Surfing the nursed equation. We can say that zero is equal to be self potential. It's standard conditions, which was again 1.84 boats, minus 0.591 divided by N, which again is three electrons, three moles of electrons. And then we do the log of Q. But since the sell potential zero, this corresponds to the equilibrium constant K so we can rearrange that equation now to Sulphur K. And when we do, we should get K is equal to about 2.5 times 10 to the power of 93 and on part C. You're given nonstandard concentrations for both T l plus and a U three plus, and we want to determine the cell potential. So we again use an Ernst equation, and we need to calculate que We know that from the reaction this is equal to t o plus ions cubed, divided by the AU. Plus I owns concentration A from this reaction a u three plus ions and were given the molar concentrations. Reach one of those imports, see? And so that is 1.0 times 10 Do the negative fourth moller cubed, divided by 1.0 times 10 duty negative to Moller for a U three plus and Sue. When we plug in that value of Q into the nurse equation, we have the nonstandard to sell potential equals the standard cell potential. Again, 1.884 volts, minus 0.591 divided by N Moles of electrons again is three. And then we multiply that by the log of cute again we calculated que based on this ratio. And so when we saw that all out for the sell potential, we should get a final answer of about 2.4 volts

And increase in oxidation state goes to oxidation on the decrease in oxidation. State goes to reduction. Let me represent the expression for self potential. We hav e not sell as e off. Get towed minus e an old. The expression between cell potential and free energy can be represented by Delta. Gee gives free energy equal negative. And if and e not off cell, the condition for spontaneous and no one's pointing its reaction. I represented us, Del Apache. You're not still less than greater than zero will be equal. Do spontaneous. On the second part off the artist gun Tucci on e knocks ill Greater than lesser than zero will be non spontaneous. Considering the given reaction, we write the oxidation and reduction half off the cell so oxidation it will be toe. See you gives me too. See you positive plus two electron and reduction half will be M and two positive plus two electron. Give me Mm. We calculate the e not sell using the standard half till reaction so not sell. We use this formula as represented earlier. Equal negative 1.1 made war minus zero point 52 bulls in order of the cell turn out to be negative 1.70 board. So the E North is negative, which indicates that the reaction is known spontaneous. We have to for no exact same steps for Option B and Ocean Sea, too. As shown for option B. The reaction is spontaneous because the value off E note is positive for auction. See, the reaction is known spontaneous because e note is negative.

Alright. So from this half reactions and the standard reduction potentials, we are supposed to uh find the they sell potential. Alright, this self potential um give free energy and Casey, the equilibrium constant. Alright, so, first of all, we need to find the cell potential. Now there's some potential. If we look at the reduction potentials, we can see that the first reaction has the lowest reduction potential. So it is that one which should be oxidized. So we are going to reverse that one. So at the end node, that's where oxidation takes place. We will have C. L. Two plus six miles of water, produce two months of cl 32. So, the three miners. Alright, Last 12. So more stuff hydrogen. Mhm. Thank you. Mm. And because of that, the because we're interested, we would change the sign of the Reduction potential which remind us 1.4 seven vaults then. And the cattle does the reduction takes place. Mhm. They don't get to spend it. Mhm. Yeah. Good. All right. The cathode we have and so we retain that one does reduction. So we have 2708, Sorry. So it's tremendous. Plus two electrons will form yeah. Form to a war 2-. And the standard reduction potential will be 2.01 votes. So if we add those two equations now, before we do that we will multiply this by five. Do you have five moles of electrons? 10 miles of lectures? So when we we do that, we have now have C L two plus six moles of H 20 plus five. Asked to the eight to remind us for to c L three minus. Mhm plus 12 hydrogen areas less five times to 10. As for Yeah, 2-. And the sub potential stand at some potential will be the star style potential. B zero point 54 vaults. Right, So that's the standard cell potential then that's a then be we want uh give free energy delta G. And those are G is giving us -3F. Thanks Tyne as potential. And its number of most of electrons f is for a discuss stint. Right? So this occurred to minus we have 10 most of electrons. So at 10 most of electors times Friday's customs, which is 96 485 cool. And spare more of electrons times this potential 0.5 or vaults which is jews cooler. All right. So if we do that, we were going to get mhm minus five points 2. 1 time's temper. Five Jews, which is in killer jersey to B -5.21 times 10 power to kill a juice. All right. So that is the jew. The sea. Yeah, We want the equilibrium constant. Now the formula or the equation we are going to use is some potential because 0.0 591 over in look, kid, right? It's where K is the equipment custom? Right. So, yeah, means that zero points by four. Because 0.05 my one We were in. Which is 10. Okay. Yeah. So if we do this, we'll get okay. Yeah. Okay. Because 91.37. So, okay. We got to the anti lock of listen, 91 point 37. Okay. Which is 2.35 times then four 9-1. Right? Yeah. So that is K. Mhm equilibrium constant. And that is delta G. And there's a cell potential, right? Mhm mm hmm.

Okay, So for this problem, we're going to calculate the sander's self potential, right? The overall balance cell reaction. So the first step is to go to table 20.1 in your textbook and look up the values for these reduction potentials for the half reactions. So you'll see for the reaction with platinum. Yeah. Have you? Value is one point one age and the reaction of 10. You have negative 0.1 three 75 volts. Eso the half reaction with lower reduction potential will proceed in the opposite direction. It will be the oxidation reaction. So, as you see, negative 0.1 375 is less than 1.18 So this one, the 10 reaction is going to proceed in the opposite direction. So she draw the to write the overall reaction. First thing we need to do is, um, reverse the tin reactions that we're gonna have. Yeah, solid tune on this side on, then we can add that to the platinum reaction is gonna proceed as written. Okay. Yeah. Okay. So the electrons, we're gonna cancel, and we can rewrite this as the final answer. Mhm. No. All right. Mhm. Yeah, Yeah. Yeah. Okay, So this is the overall cell reaction. Now, to write this using sell notation, you find the the reaction that took place of the node goes on the left, which is the accid ation reaction. She start with tin solid. Yeah, you dropped one parallel line to represent the phase boundary the difference between the two phases and then mhm two parallel lines. Then we write the reaction that occurs at the CAFO, which is the reduction. Mhm another wine. And then you write the last part right here. Mhm. Now, to calculate the standard self potential. Yeah, just, um, what you're gonna do is this is e equal. Yeah. Uh huh. To the reduction minus No. Um, so you need the standard self if you look at the values we got from the chart, so we would use the value that goes with the reduction reaction, which was 1.18 and then you're going to subtract it from the value that we saw for the oxidation zero. Thanks. Mhm. Now, to my choices are going to cancel each other so you end up adding them together. And you she at one point 316 volts


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