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4. (a) A table of standard oxidation potentials shows the roltage as metal: give up electrons to form cations If you had two metals_ Ml & M2, with different sta...

Question

4. (a) A table of standard oxidation potentials shows the roltage as metal: give up electrons to form cations If you had two metals_ Ml & M2, with different standard oxidation potentials; VI & V2, explain which one will be oxidized with justification. (b) Consider the dissociation of AgO into metallic silver and oxygen gas. Ifthe free energy of formation at 300K for Ag-O is -11.21 kJ mol. What would be the equilibrium partial pressure of oxygen for this dissociation at 300K?

4. (a) A table of standard oxidation potentials shows the roltage as metal: give up electrons to form cations If you had two metals_ Ml & M2, with different standard oxidation potentials; VI & V2, explain which one will be oxidized with justification. (b) Consider the dissociation of AgO into metallic silver and oxygen gas. Ifthe free energy of formation at 300K for Ag-O is -11.21 kJ mol. What would be the equilibrium partial pressure of oxygen for this dissociation at 300K?



Answers

Using the standard reduction potentials listed in Appendix $\mathrm{E}_{2}$ calculate the equilibrium constant for each of the following reactions at $298 \mathrm{~K}$ : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

Are in this question, were asked to use the standard reduction potentials found in appendix C to calculate the equilibrium constant for each of the following reactions at 298. Kelvin. So there's several equations were going to use for this problem. The 1st 1 that will need is gonna put the equations over to the side here. Um, the first equation that we're gonna be using is my from my pen is that the East cell not is equal to the e the potential for the reduction minus the potential for the oxidation that's gonna give us East cell. Um, then we need to find Delta G. So Delta G not is equal to negative and f That's rate is constant times e cel and then from Delta G, we can get Kay equilibrium constant, which is going to be equal to e to the negative Delta G not over, Artie. So for each one of these, we are going to be using all three of those equations. So for the 1st 1 we can see iron is going from Effie to every two. Plus it's losing two electrons. Therefore, um, that's this is a two electron system. Therefore, it is being oxidized. Nickel is gaining two electrons, again confirming the two electron system, and it's being reduced. So when we find the E cell for this, it will be the reduction. So the nickel, which is negative 0.28 volts minus the oxidation, which is negative 0.44 volts. It's given East Cell of positive 0.16 volts. We then sell for Delta G, again using equation to the side. So that's negative. And then we have. This is a two cyst to electro system. So two molds of electrons times Faraday's constant, which is 96,000 485. And that's Jules per volt times, moles times Um, the East Cell, which we just calculated, which is the point ones 16 volts. And we're going to get the Delta G equals negative. 30 875 undercover about six figs right now. Because this number is gonna go on to be used to find Kay. And finally, K is gonna be equal to e to the negative. Delta G. Um, so that's to the negative. 30875 jewels divided by and we have our which is the 8.314 Jules promote Kelvin times the temperature. If to 98 Kelvin we plug all that into the calculator and we find that the equilibrium constant for this 1st 1 um, is going to be. And where they're rounding it to 16 Figure the answer key. It's gonna be three times 10 to the fifth for our next one. We've got cobalt here and it's going from cobalt cobalt two plus. So again, that is a two electrons system. And if the Coble is being oxidized, hydrogen is going from positive oxidation number plus 1 to 0, therefore is gaining electrons. Therefore, it is being reduced. And again there's two H to elect hydrogen is here, each gaining one electron. So that is another two electron system. So it looks like we have two electrons in this case. So again, I'm going to start by finding e cel. So the cell will be the potential for the reduction, which is the hydrogen so that that's our are zero. Remember the reduction potential for hydrogen since it's what everything else is compared to a zero minus that of, uh, the cobalt, which is negative 0.28 votes. So we get that this is gonna be positive. 0.28 volts. So now we can find Delta, Jeannot and Delta Jeannot is gonna be equal to negative and then the number of moles of electrons. Which again, in this case, was two moles of electrons, Um, times Faraday's constant, which is the 96,000 485 jewels per volt moles times the voltage, which was 0.28 volts. And I get my Delta G in this case equal to negative. That's 5 54,032 Jules. So I put that into my equation for Kay, so K equals E to the negative Negative. 5432 Jules divided by 8.314 Jewels Permal Calvin Times a Tu 98 Kelvin And that gives me a K equal to three times 10 to the ninth. Finally, for part C Um Okay, so we have bro mean going from B R minus, so as a negative one. Oxidation number two b R, which has a zero oxidation number. So it's losing electrons, so it's being oxidized And there's 10 bro. Mean each one losing one electron. So this looks to me to be a 10 electron system. Let's double check with our reduction. So our reduction here, um is the permanganate miss changing oxidation numbers? Um, from if you solved for the oxidation number here, you would find its positive seven. It's when a positive, too. That's five electrons that it's gaining. But there's two of them. So again, that is a 10 electron system. So we start by finding our e cel herself Potential is going to be the reduction, which is tthe e um, the permanganate. So that's 1.51 volts minus the bro Mean, which is 1.7 volts to give up 0.44 volts. Within cell for Delta Jeannot, which is equal to negative in this case, is a 10 miles of electrons times Faraday's constant of 96,485 jewels per mole kelvin times the cell potential of 0.44 volts and I get the Delta G in this case is 4 to 4 534 jewels. So 424,534 jewels. So I'm gonna plug that into my equation for Kay. So K is equal to e to the negative of that sees me. That was a negative here. So e to the negative Negative. 424,534. Jules, her 8.314 Jules promote Kelvin times the temperature of 2 98 Kelvin and put it all into the calculator and you get e are you get K is equal to three times 10 to the 74th. Very big K.

So now we'll work on problem 52 from chapter 20 And this problem, we're asked for each of the following reactions. Right. A balanced equation. Calculate the standard IMF calculate Delta G at 2 98 Calvin and calculate the equilibrium constant at 2 98 Calvin. So our first reaction is obvious. I died ion oxidized. I too by Mercury two plus So we have this eye minus our quiz ion, which is going to be oxidized by Mercury two plus to produce I too. And then the mercury will be reduced. Teo neutral artery and there should be a two here is well to make to be behind on molecule so we can look in appendix d e excuse me and get the reduction potentials for the these 2/2 reactions. And so we'll put the reduction reaction first of the reduction of mercury and we get zero point 79 minus zero point 0.5 for and that's equal to 0.25 boats. So that's our EMF. So then the calculate don't g we need to look no are, uh, equation is negative. End very days, constant times this imf. So this end here represents the number of electrons that are transferred. And since we see Mercury to going to neutral mercury, we should say that there's two electrons transfer, so we have negative, too. And we have Faraday's constant, which is 99 1,000 six. 96,485 units for Faraday's Constant are, uh, Jules. So then we need to multiply by the IMF, which we just generated, which is 0.25 So here we multiply through on we get negative, too. Uh, 0.34 times 10 to the fifth Jules Purple, which is also equal to 234 kill Jules. And the way is correct, but we'll need to use Jules promotes to calculate the K value. So now our equation for K involves Delta Gee being equal to negative r t a line of K. So if we put this around to solve for Ellen of K, we get lnk is equal to Delta G over our tea. Uh, negative. So here we can put. So I actually wrote down the wrong number here. Okay, this here is the for the next one, and I got them mixed up So for the Delta Ji, for part A is negative. 4.82 times 10 to the fourth. And this is negative. 48.0. If, uh, yes. So Delta. Gee, here we have this negative 4.8 times. 10 to the fourth Jules Permal, divided by Artie. So the value of our that we're going to use is 8.314 because it's in units of jewels per children more. And then we're going to multiply that by 298 Kelvin. And that gives value of 19 point for seven. So then the Ellen, uh, this is equal to Elena K. So then, K, if we take the inverse of the Ellen or E, we get 2.86 times 10 to the eh? So now we can move on to Part E where the reaction which we're looking at is, uh, acidic solution in acidic solution copper one ion is oxidized a cover to buy nitrate. So we have cover one, see you plus plus nitrate and then we're in acidic solution. So we had protons present and it's copper. One is being oxidized to cover two plus Plus, you know and water. So now we can balance his equation. We know that cover this nitrate tonight. You know, from the half reaction there, Appendix E will be a transfer of three electrons over this copper. Ah, elect copper One to cover, too, is a transfer of won elections. So we'll need to balance this batting three here and then we also need for protons for this nitrate reaction and that will produce two waters. So now first, calculate the IMF, we first put the reduction. So nature it to an old measures oxide. And we get 0.96 from the appendix. And then for the oxidation, we get 0.15 So we have a value of 0.81 votes. So now we can get you a Delta G, which is equal to negative and f e. In this case, we're transferring three electrons. Saturday's concert is 96 45 and R E, which we just calculated a 0.81 So we go ahead and calculate 2.3 four times 10 to the Yes, Jules, her mall. And then for Kay, we can solve the same way we did before Alan K. Is equal to Delta G over negative, Artie. So this is equal to negative. 2.34 times 10 to the fifth Trooper over in a native, 8.314 Jules Kelvin Moore and to 98 couples, which is the temperature were given in the problem. This gives the value of 94.6. And when we solve for Kay, we get 1.25 times 10 to the 41. Very merry large value. So now, in part, see, yeah, we're given the reaction of again an acidic and basic solution from him hydroxide. It's oxidants crummy by Cielo minus. So for this reaction, here we have crewmate criminal hydroxide uh, we're in. It's going to be oxidized by seal Oh, minus and we're in basic solution. So this produces will leave the sub scripts for the second part. So we have Ah, crummy cr 04 two minus. We produce water and Lorraine chloride. So the chromium hardrock said reaction the oxidation of chromium Jack said that the crew mate involves a transfer of two electrons. The reduction of er co write involves the transfer of three elections. So we're going to need to multiply the chromium by two and the hydroxide in the chlorine by three. So if we had and 1/2 reactions up, some of the hydroxide will cancel from both sides, so we get 400 oxides in five water. So the IMF, the reduction value here is 0.89 for the reduction of the correct and then the reduction or the reduction half reaction. The chromium is zero from 13 and that gives the value of 1.2 bolts for Delta G calculate. In the same way we again have a transfer of six electrons. Faraday's constant in our mess get a value of negative 1st 0.90 times 10 to the fit. Well, Jules promote. Excuse me. So then we calculate lnk. We get negative 5.90 times 10 to the fifth over, our gas constant and our temperature, and we get a value of 238. When will these take the Ellen? We get kay this week. The interest Ellen or the E 2.25 times 10 to the 1 0/3 So very, very large.

Problem 52 party according to question. We died first. Happy question. So two iniciativa course it will give I too solid plus two electron and standard production potential. Or this reaction is zero point I 36 world. No mercury at the two plus two echoes again. Two electrons and it will reform two mg liquid. Each tender reduction potential For this equation is 0.7 89 old. No, we need to add this to reactions for a complete cell reaction. So we can write two are negative a course plus at G two less too a post. It will give I too solid plus to access the liquid and standard Production potential for this cell is equals two. The reduction of get hard minus introduction up and or now protecting the value production up. Sell Calls to a zero 78 nine. This is active cattle and these actors and not minus minus mhm 0.536. That is because 20 253 world. So this is a standard M. For this reaction. No, we need to calculate Gibbs. Free energy for this reaction gives free energy. Your standard give give free And the internet gives free energy calls to delta Z. Not that is equals two minus N. F. He not now. We clearly see here two more electrons are transferred. So minus two more electron multiply value of Faraday in Faraday is 96.5 kilo two. Part of the world. More electron multiply way. Standard standard salim. There it is calculated 0.253 World. After calculating we get standard Gibbs. Free energy for this reaction is equals two -48 eight. Kill a june. So this is also answer for party. Now we need to calculate value of King. What this reaction? So mm We know that Delta jeannot equals two minus Artie Ellen key. Okay so we can write alan Kay calls to delta psI Not divided by RT. Now put the value value greater. Do not already calculated. So that is minus -48.8 8-9. We need to convert in a jewel so can return to the power for jewel divided the value of our he's eight point 314 june caramel Calvin. Multiplying their temperature 2 98 Calvin. Now after calculating we can write alan Kay calls to 17.9 so we'll take anti log on both sides so we can write as Kay calls to Italy power 17.9. That is it close to 3.6 into 10 to the power. This is a value of including constantly at 2 98 Calvin. So this is also answered. Part A. Now in part B. According to question posed to be right. Oh hops L. Jackson, C R. O. H hole tries solid combined with five hydroxy echoes and it will form C. R. O. Poor minus two It cause plus poor as to liquid Plus three electron and each tender ducks. Sorry, sorry, this is something mistake. Okay, in part to be cooper plus echoes. Gibbs call put pledged to echoes plus one electron and each tender reduction potential power. This reaction is equals two 0.153 Bulls. No Under reaction is an 0. 3 minus echoes plus poor at plus echoes last three electrons and it will give and no gas plus adds to liquid and extender reduction potential, Which it calls to 0.96 world. Okay. No, We need to balance the questions. 1st 2 questions. We multiply by three so you can write three copper plus echoes plus nitrate iron echoes bless four at plus a ghost. It will give three c. U. Plus to a cost yes and no gas plus too much too liquid and not sell because to be simply huge. Introduction it doesnt potential of catheter minus it doesn't potential of energy. We simply put the value. So this is the get hard and this is the energy. So 0.96 old minus 0.153 World. That is equals two 0.81 old. This is standard sally mm. So this is all this answer for part B. Now we need to calculate gives extended gibbs free energy. So standard gifts planers E. Yeah there is a genetic goes to minus and f. He not now simply put the value That is clearly see a three multiple opponent transport. So -3. multiply value of Friday is 96.5 jewel Permal per Calvin. Sorry, 90 96.5 kg. Will permit world Permal electron multiply by 0.81 world. After calculating we get -2.3 45 and do 10 to the power two kg. So this value of IMF sorry this value of gibbs free energy for this reaction nine. We need to calculate value. Okay who directly? Right here Lnk calls to We simply use this formula. Okay. Alan Kay calls to minus -2.3 45 into 10 to the power five jewels. This divided by eight point 314 jules. Partner mall Calvin multiplied by 2 98 Calvin. That is equals two 95 approximately goes to So we'll take anti law on both sides so we can write Kay calls to italy power 95 30 equals two 10 to the power approximately. Go support even so this is will you okay for part B? No in part C again we right. Absolutely excellent. That is chromium who actually tries solid plus five hydroxy in echoes medium it will form CR 04 two minus echoes plus Poor is too liquid Last three Electrons. Standard reduction potential for this Equals and equals 2 -0.31 sorry 0. 1 three world. And again, second absolute reaction is cl oh negative across plus adds to liquid Combined with two electron and it will form Cl negative across Plus two hydroxy in and cause an external reduction potential for this reaction as it calls to 0.89 old. So we need to balance the equations. So first to christian multiplied by two and no second question multiplied by three. So we can write fully question that is those C r O h whole try solid last three. Cielo negative act cost blessed four. Hydroxy and Echoes. It will lead to To. c. Two minus echoes Plus three cl negative echoes plus highway two liquid. No. Mhm. We can write standard for this reaction. Was simply used Alexa potential for cattle minus the reduction potential for and or so this is a Catherine and this is a not so simply put the value 0.8 nine minus -0.13. That is equals to 1.0 to hold. So this is a standard mm for this reaction. So for this cell no we need to calculate delta. Not again we use minus N. F. Not simply put the value -66 left and six more electronic. Transfer the call to multiply by 3 6. Multiplied by 96 five. Multiplied by the value of P M. F. One point here to africa liberating we get -900 .58 kg joules. So this is value of gibbs. Free energy for this this part part C. Now we need to calculate value. Okay so again we use this formula so we can directly right. Ellen key calls to minus minus pi 0.9 058 into 10. to the power Hi jules leave hide it by eight point 8.31. Poor jule caramel Calvin multiply their temperature to 98 Calvin. Pre calculating we get 2 38. So now we can take anti log on both sides. So we can write he goes to three point three into 10. to the power 103 That is approximately close to 10 to the power 103. So this is final answer. The value of care temperature to 98 Calvin for Part C.

This problem gives us a disproportion. Ation reaction. Um, and asked us to answer a few questions. Parts A through C based on the information contained in this equation So we can start with part A. What are the half reactions that are contained within this overall reaction, we can assign oxidation states. I think that would be a good place to start to make this pretty easy if you two plus as a plus two oxidation state solid iron has a zero and three plus as a plus three. So let's use this plus two oxidation state, um, and think of it in terms off to different reactions. Okay, um, we're going from a plus two to a zero. So we have a reduction, and then we're going from a plus to from a plus to a plus three. Excuse me. So that's an oxidation reaction. So and just explaining those guys, we've determined which half reactions make up this overall reaction. So we know that one reaction must be f e three plus in a career solution. Plus, an electron will yield F B two plus iniquitous solution. So that's our first half reaction in our second half reaction is gonna be ft two plus an angry A solution, plus an electron or two electrons. Excuse me. There's a solid iron. That's all you need to do for part a. Okay. For part B were asked to find standard reduction potentials to determine whether the reaction is reactant or product favored. Okay, so the values that you're gonna have in your book are probably going to be a little bit different than the ones that I'm using. I just pulled them from the Internet because I don't have access to your textbook, but I did find these values for the reduction potentials of these reactions. Okay, One thing that we need to go back to because we need to go back to this original equation where we label their oxidation states and remember which one is getting reduced. Which ones getting oxidized? Because we're gonna be using our you sell people t e plus minus e minus equation to determine if the reaction is reactant or product favored. E plus is going to be the half reaction that, uh, is a reduction in the overall reaction. So, um, you Plus, in this case is going to be negative 0.44 volts because we're starting out with iron two plus and we're getting a reduction. Two solid iron that's gonna be e plus. So he so it's equal to negative 0.44 volts minus. And then our, um, e minus is going to be 0.77 volts because that's the oxidation reaction. So you sell. If you do, the math is going to be equal to negative 1.21 volts. And because this value is negative, um, that means that the equilibrium will shift to the reactant. Okay. And now, for part C were asked to find the equilibrium constant get if we look back at this equation, we see, um, that we have to forms of iron, that Aaron Equis solution, and one that is a solid. We do not ever include solids in reaction quotients or equilibrium, constants and then equilibrium. Constants are always going to be our products over our reactant. So we see here that we have two moles of iron, three plus and three moles of iron two plus. It's okay. It's just going to be equal to 2/3 because we are generating two moles of our product here from three moles of are reacting here


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