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4.(17 pts) Consider the following cell notation: Zn(s) Zn?t(aq) || 0z(g) , Ht(aq) Pt(s) a) (4 pts) Make a sketch of this voltaic cell. Label the anode and the catho...

Question

4.(17 pts) Consider the following cell notation: Zn(s) Zn?t(aq) || 0z(g) , Ht(aq) Pt(s) a) (4 pts) Make a sketch of this voltaic cell. Label the anode and the cathode and indicate the charge on each electrode_ b) (3 pts) Write the anode and cathode half-reactions and the overall cell reaction if 02 gas is reduced to HzO() c) (3 pts) Calculate the emf 'of the cell under standard conditions using the standard reduction potentials from Appendix E_ d) (4 pts) Calculate the values of AG" an

4.(17 pts) Consider the following cell notation: Zn(s) Zn?t(aq) || 0z(g) , Ht(aq) Pt(s) a) (4 pts) Make a sketch of this voltaic cell. Label the anode and the cathode and indicate the charge on each electrode_ b) (3 pts) Write the anode and cathode half-reactions and the overall cell reaction if 02 gas is reduced to HzO() c) (3 pts) Calculate the emf 'of the cell under standard conditions using the standard reduction potentials from Appendix E_ d) (4 pts) Calculate the values of AG" and Kp for the redox reaction at 25"C (298.15 K). e) (3 pts) If the concentrations of the Zn?+ and Ht ions are standard (1.00 M each) , what pressure of 02 gas should be used to create the cell potential of +1.80 V?



Answers

A voltaic cell utilizes the following reaction:
$$
4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)
$$
(a) What is the emf of this cell under standard conditions?
(b) What is the emf of this cell when $\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=$ $0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=0.50 \mathrm{atm}$ , and the $\mathrm{pH}$ of the solution in the cathode half-cell is 3.50$?$

Problem 67 party first we need to balance this salary action. So first we right hub salary action that is. I've been blessed to echoes give And people are three ecos plus one electron. And for two guests Plus four plus echoes plus poor electron. It will give two molars edge to liquid. No external production potential for this reaction is he's not production equals two 0.771 old and extended direct reduction potential for this reaction. Is it close to 1.23 World now we need to balance the equation so we might reply this first equation. The sorry reaction by food after we need to add this to reaction. After adding we can right poor every blessed to echoes Plus or two gas plus pulled edge plus echoes. It will give Poor iron plus three a course Plus two more ads to liquid and is and mm pop sell at extender dot com potential we need to use mm not obsolete goes to in our production of cattle minus in or production of energy. So in order ducts not scattered is cattle actors. This capsule reaction and not act as this capsule reaction. So we simply subs tract 1.1 point 23 Hold minus 0.77 world. So PMF of cell. At a standard condition is it calls to 0.4 59. That is approximately equals to 0.46. So this is answer for part E. No. In part B. We need to calculate mhm. We need to calculate E. M. F. Of this cell at specific iron concentration iron plus two concentration and iron plus three concentration. No, we can write us mm. Pop cell equals two. You know what minus 0.059. two. Divided by an into log concentration of F plus three to the power poor. Divided by concerned prison of iron Plus 2 to the powerful multiply by concentration of hydrogen nine to the powerful into partial pressure of or two. Now here you have u. of n equals two to whereas value of hydrogen iron concentration Equals to 10 to the power minus 3.50. That is equals two 3.2 in 2. 10. to the power minus pulled em. Now we need to pull the value equals standard mm. Upset. We already calculated here. So That is zero four 59 minus 0.0592, divided by the value of N. Is we clearly see here four electrons are transferred. So divided by poor in two log concentration of every plus three is given question that is 0.010 to the power poor divided the concentration of iron plus two is one point 3 to the power four into 3.2 into 10. to the power minus food to the power pool Into partial pressure of co two Is also given question that is zero 50 Now we need to calculate we can right Mm sell it goes to 0.4 59 After calculating mhm. This quantity we can simplify and write 0.0592 divided by food long seven in log seven into 10. to the power hype. Okay. No we can write equals two zero point 459 minus 0.5 92, divided by four. and value of log seven into 20. The power -5 with five point five point 84 pipe. That is it close to 0.4 59 minus 0.0 865. After calculating we get the value of Yemen capsule that is 0.37 old. So this is final answer for part B.

Among the two reactions. One is the higher reduction of hydrogen to gain two electrons and from hydrogen gays. So the the reduction of hydrogen is zero world. On the other hand, in another reaction were yellow mini. Um hello, Mini. Um reduces Can in three electron to farm L A mini, um e reduction off aluminum is minus 1.66 Word looking to the reduction potential of board have civil reaction so we can say the reduction potential up Hydrogen is greater than aluminum. Therefore, hydrogen will reduces while aluminum will oxidize so we can write that to hell. Amini Um toe l. Amenia will oxidize toe form toe a three plus and release a total of six electron. So the oxidation will be no plus 1.66 were changing. The sign is we have it in this reaction and reverse direction. On the other hand, six hydrogen ions again six electrons perform Ah, hydrogen guests, The e the reduction is zero work heading there too, or half cell reactions. We can like that Hello medium. React with the hydrogen on forming a low mini um three pless irons and hydrogen case. He can cross electrons on either side sell standard self potential. Not off the city is the sum of the standards. Potential have potential approximation and reduction, which is 1.66 world. This is the standards and potential for artists no to re termine the non standard self potential A, which is equal toe in art, minus 0.592 divided by and and log off que. Where is equal toe in art is 1.66 world minus 0.592 and it's equal to the number of electron last and again, which is six. Malls up electron and log off. Kim is equal toe the concentration on the of the substances on the product hand side. So on the product hand side, we has a low mini um three. Plus a race to the Power X number of morals, which is equal to two hydrogen gas rest to the power three divided by under react hair inside his only concentration of hydrogen, which is six more. So that's the power six is in the equation in the question we have given that the concentration up aluminum is 0.1 Moeller similarly the concentration off hydrogen. I'll is also equal to 0.1 Moeller while we have given the pressure up hydrogen So pressure up hydrogen is 10. 80 m using the ideal guests equation PV is equal toe end rt r p is equal toe end where we are t we're in where we is. Concentrations off p is equal to see Artie are the concentration is equal toe be divided by rt So ps 10 atmosphere we're are 0.8 to 1 letter Permal letter atmosphere Permal Park alien and temperature is 25 C, which is equal to 98 Kelvin calculating the concentration off hydrogen which is for model So the concentration of hydrogen cases. You're a point for mother now putting the values in the equation. We get there e is equal toe 1.66 minus 0.592 divided by six is equal toe 9.87 and tow tenderest power minus three. Log up 0.1 square multiplied by 0.4 cube divide by 0.1 rest to the power six. So e is equal toe 1.66 minus 9.87 and pretend this par minus three and log up. All these values is equal to 2.8, so is equal to 1.66 minus. The multiple of these two values is zero point 02 77 So subjecting 0.2 to 7 from 1.66 we get dust self potential 1.6 32 world. This is the required solution to this part. Off the question. We have determined the now standard self potential and the standard self attention now to calculate the Delta Jeannot for the reaction which is equal to minus and, if in art. So putting the value of minus and his six malls off electron. If it's fair artist Constant, which is 96,500 Coolum par mola electrons and in art is equal toe 1.66 world now minus six and 96,000 500 multiplied by 1.66 old and stayed upward. We can write travel Park column is we know one world is equal to one job per column. Multiplying these value. We get the value of TelDta jean art, which is equal toe minus nine, six 1140 jowls and we revived by 1000. So we will get a Delta G not minus nine six one 0.41 Killer Joe, this is the required answer for the Delta. Jeannot off this part off the question now to determine the equilibrium. Question que We know that in art is equal toe 0.592 Do you want to buy and log up? Okay, are rearranging the equation and taking the anti log We can write that k is equal toe empty Log up e in art multiplied by and do advise 0.592 So k is equal toe anti log up in art is equal toe 1.66 old and is the number of electron last and again which is six miles of electron divided by 0.592 So k is equal toe empty Log up 1. 68 point to r K is equal toe taking the anti log up 1. 68.2, which is equal to one and 2 10 Rest power 1 68.2. So this is the answer to the equilibrium constant off this reaction now to determine toe calculate to calculate the mosque off. Hello medium where the current I is then impaired. Time T is 25 minute which is equal to 60 seconds for per minute. This is the Time t to determine the mask. We will use the formula that mass up aluminum is equal toe molecular weight Are Moeller weight off the divide by N and F multiplied by I and to de so the mass up aluminum is equal toe molecular weight up our model Massa Aluminum is 27 in this three because hello medium form a three plus and release a total of three electrons. If it's Ferrari constant, which is 96,500 Cool, um, are more electrons reporting the values 27 Grandpa Roman Multiply by 10 m pair multiplied by 25 2 62nd time Divide by six moles up electrons and divided by 96,500. Cool, um, are mullah electrons so mass is equal. Toe 27 multiplied by 10 and stood up. Empire one m pair is equal toe cool. Um, part second, So cool them per 2nd and 25 2 60 seconds. Divided by six and two. 96,500. Cool. Um, So calculating the mask. We will get one 139 creams off. Hello, Mini. Um will deposit it. This is the required solution. Toe the party of this question. Sorry. There is three moles of electron. We have written six miles up electron. So correcting the value. This is three moles up electrons. The answer is correct. Which is 1.39 g. It is the calculation for three months of electrons.

The question we have to unbalanced redox reactions for which we need to balance and draw the electrochemical cell identifying the cathode and the an ode and the direction of electron flow. Do. This will draw a generic electrochemical cell and then identify the 2/2 reactions that comprised the entire reaction that it can be helpful to simply look at the appendix on. In Appendix 5-5 you'll see a table that contains these 2/2 reactions and their corresponding standard, uh, reduction potentials. Because the standard reduction potential is greater for chlorine than it is going to serve as the cathode as it serves as the cathode than to get the cell potential. It'll be Catholic potential minus Deanna with potential to get the cell potential. So to finish up the electrochemical cell, we have the chlorine and the cool ride with platinum electrode to carry the electron serving as the cathode, where reduction occurs with a higher reduction potential and then, uh, chromium three plus being oxidized to die Chrome eight. Serving as the node where oxidation occurs, electrons always slow from the an ode to the cathode for the balance chemical reaction. If we then multiply this chemical reaction by three so that the electrons cancel and recognize that the direction of this half reaction needs to be switched in order to get, uh, in order for oxidation to occur and then some the reactions. Together, the electrons will cancel, and this will be our balanced overall Redox reaction. We then look at the 2nd 1 Also draw generic electrochemical cell and look up the 2/2 reactions. We have copper two plus going to copper solid with the reduction potential. Appoint 34 bolts and magnesium magnesium two plus going to magnesium solid with a reduction potential of negative 2.37 volts. We see that the copper two plus wants to be reduced more than the magnesium, so it will be reduced. And if we look at the balance chemical reaction, it is being reduced, going from to plus to oxidation state of zero for just copper by itself. So where reduction occurs, that is the cathode. So cover two plus is being reduced to copper solid in the cathode compartment, and then magnesium solid is being oxidized to magnesium two plus in the an old compartment, the electron thin travel from the an ode to the cathode. Always to get to sell potential, we simply take the Catholic potential 0.34 minus the hallowed potential negative 2.37 And we get to sell potential the overall balance. Chemical reaction is already given to us. If you look at it closely, you'll see that the chemical reaction given has Adams balanced, and it also has charges balanced.

Okay, So first part, we have an equation. Chromium three, Positive a course. Plus Ian, too. Gas shoes, gifts. CR 272 negative. Occurs to see a negative. I was Okay, so first we'll draw, um to reaction containers on in this one. Let's see. Like this. Okay, let's see. This one is C R. Three positive on this one is seal negative. So this is where the gas would be evolved. Right? So the electrons will floor like this to write the situation. We will first write the oxidation. Right? So oxidation is C R. Three positive ACLU's close water perform cr two or seven negative plus hedge positive plus six electrons. Okay. And just to balance it, it will be to see our positive seven inch tool. This would be 14 on six electrons. Similarly, reduction will be three c l two ashes plus six electrons giving out six c and negative act was so overall equation movie 203 positive actors plus seven inch too. Ik word. Just three siento gashes. This would get canceled. We'll give cr two or seven are close plus 14 edge positive. A course plus 60 and negative equals. Okay, so clearly, this is and node. And this is get old because this is where the gas is released. So this is Katherine. And this is and what way can also calculate the overall, um, en arc overall will be equal to okay. Production minus e oxidation. So reduction is one point 36 minus 1.33 So you get 0.3 volts as your value moving on to the next part. Um, we are given this equation. Okay, So if you like this equation, Okay, this is reduction, right? Andi, this is oxidation. So overall, yeah, you get what is given this would get canceled. And also, you don't need to, um, balance it. It is balanced. Um, for them. Better understanding. We can always draw it. Laters. So this is magnesium on. This is called to positive. So the electrons flew in this direction. This is a nod on. This is get food


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