Chemistry Discussion #9, Chapter 16 Student name TA name Section. Things you should know when you leave Discussion today:
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1 Chemistry Discussion #9, Chapter 16 Student name TA name Section Things you should know when you leave Discussion today: 1. Potential E cell at standard conditions.(all concentrations are at 1M) 2. Cell potential E [J/C] is a measure of electrical potential difference. a. E cell = E red (cathode) - E red (anode) b. E = logk (at 25 C); 3. Magnitude of the cell potential is a measure of the available energy from the reaction. RT E= E 2.3logQ n F E= E - e Q logq = log at 25 C only K a. If Q=1 then E= E indicating we are at standard state conditions. (All concentrations are 1M and all pressures are 1atm) b. If Q=K then E= 0 indicating we are at equilibrium i. E>0 forward reaction proceeds ii. E<0 reverse reaction proceeds iii. E=0 no electron flow (battery is dead) 1. Is the following expression at at 25 C E= E - logq 2. Is the following expression at 25 C E = logk 3. Is the following expression E=0 4. Is the following expression E= E 5. If E = V and E = 1.00 V circle everything that must be true: (Hint: remember what does it mean for Q to be equal to1?) Q=1 Q>1 Q<1 Q<K eq Q=K eq Q>K eq a. (at home) Assuming the temperature is 25ºC and = 2.00 mol calculate the values of K eq and Q to 1 sig. fig. 1
2 6. For a reaction A(aq) + B(aq) <==> 2C(aq) at 298K a. E = 2.00 V and E = -1.00V and = 2.00mol circle everything that must be true: Q=1 Q>1 Q<1 Q<K eq Q=K eq Q>K eq b. If the concentration of C is doubled what is the Q new /Q old? c. If the concentration of C is doubled what is the new value of E. d. If concentration of the C is tripled how will the magnitude of E increase or decrease? e. (At home calculate the new value of E)(Answer:1.98) 7. You have alectrochemical cell consisting of two separate solutions. Coming out of the first solution is a lead electrode and a platinum electrode is coming out of the second solution. From last week handout: Pt(s) H 2 (g) H 3 O + (aq) SO 2 4 (aq) PbSO 4 (s), Pb(s) Anode (oxidation takes place): H 2 (g) 2H + (aq) +2e- E red = 0V 2- Cathode (reduction takes place): PbSO 4 (s) +2e- Pb(s) + SO 4 (aq) E red = V E cell = E red (cathode) E red (anode)= V-0V= V Net reaction: 2H 2 O +H 2 (g) + PbSO 4 (s) 2H 3 O + (aq) + Pb(s) + SO 4 2- (aq) =2 logq Q = 2 2 H 3O SO E = E 2( ) 4 H g Check-off one of the following: Change in the cell Increase in E Decrease in E No effect on E 1) increase in ph of the solution [H 3 O + ] decreases 2) dissolving Na 2 SO 4 (s) in the solution K sp >>1 3) increase in size of the Pb(s) electrode 4) decrease in H 2 gas pressure 5) increase H 2 gas pressure 6) increase in the amount of PbSO 4 (s) K sp <<1 7) adding HCl in the solution 8) addition of water to the solution ( hint: how does concentration vary with volume?) 2
3 Net reaction: 2H 2 O +H 2 (g) + PbSO 4 (s) 2H 3 O + (aq) + Pb(s) + SO 4 2- (aq) =2 2 2 logq Q = H 3O SO V E = E a. At Q=1 was the reaction a spontaneous proces? H 2( g) For the following questions [SO 4 2 (aq)] and p(h 2 (g)) are kept at standard states. You can only change concentration of [H 3 O + ]. b. What ph is needed for E cell = E cell? c. Calculate equilibrium constant K= d. What ph is needed for E cell = 0? e. Give axample of the ph that will make this reaction spontaneous. (Explain why): f. If the cell ph=10, what is E cell? Was this process spontaneous at ph=10? 8. Answer the questions for the following redox reaction at 25 o C. (E o red(zn 2+ Zn) = 0.76 V; E o red(ag + Ag) = 0.80 V) a. Cathodic RXN: Zn(s) Zn 2+ (aq, 0.01M) Ag + (aq, 1M) Ag(s) b. Anodic RXN: c. Net RXN: d. E o cell = e. Q= f. E cell = 3
4 9. Answer the questions for the following redox reaction: Cu(s) Cu 2+ (aq, 0.001M) Cu 2+ (aq, 1M) Cu(s) E o red(cu 2+ Cu) = 0.34V a. Cathode RXN: b. Anode RXN: c. Net RXN: d. E o cell = e. Q= f. E cell = 10. Answer the questions for the following redox reaction at 25 o C. Ag(s) Ag + (aq, 0.001M) Ag + (aq, 1M) Ag(s) a. Cathodic RXN: b. Anodic RXN: c. Net RXN: d. E o cell = e. Q= f. E cell = 11. Calculate the voltage (E) of a concentration cell constructed with the Cl concentration difference between sea water and river water at 25 o C. Assume that the Cl concentration (due to dissolved NaCl) of sea water is 35 g/l and than that of river water is 1.0 mg/l. a. Cathode RXN: b. Anode RXN: c. E o cell = d. Q= e. E cell = 4
5 12. The standard cell potential for the process of A(aq) + B(aq) <==> 2C(aq) at 300 K in which three moles of electrons are transferred is E o cell = 3.00 V. Alectrochemical cell for this process is constructed and the measured voltage is E cell = 5.00 V. Circle all the correct statements. a. Q<1 Q = 1 Q > 1 Q = K Q > K Q < K b. If concentration of the C is tripled will the new E o cell Increase Decrease Stay the same c. If concentration of the C is tripled how will the magnitude of E change? Increase Decrease Stay the same d. What is a new value of E? e. If the concentration of C is doubled (assuming the temperature is 298K), calculate the new value of E. f. If concentration of the C is tripled what is the new Q new /Q old = 13. ADP is converted to ATP in mitochondria of human cells. The energy required for this process is provided in part by the concentration of H 3 O + being high outside than inside the inner membrane of the mitochondria. The concentration difference results in alectrochemical potential E = V across the membrane. Calculate the ratio [H 3 O + (outside)]/[h 3 O + (inside)] that accounts for this electrochemical potential. Assume E = (0.060/ ) V log(q/k). [H 3 O + (outside)]/[h 3 O + (inside)] = 14. The voltage of alectrochemical cell for the reaction 2 C(s) + D 2+ (aq) 2 C + (aq) + D(s) is E = 0.80 volts when Q = 0.10 at 25 o C. Calculate the voltage at 25 o C after the solution in the cathode is diluted so that the ion concentrations in the cathode are reduced to exactly half their starting concentrations. 5
6 E (volts) 15. Below is a plot of the measured voltage at 298 K of alectrochemical cell at different values of log(q) log (Q) a) Determine Eº cell from the plot. b) How many moles of electrons are transferred per mole of reaction? Hand out Answers: 2*10 33, ,4,-1, 1.982, 0, 10-12, 6, 0.24,1.56, 1.62, 0, 0.001, 0.09,0.18, 0.27, 3.00, 9, 3*10 2 ; 0.791, 0.12, 3 Exam 2 Answers: a) 0.082%; b) M M M M a M b M c. Stays the same a. Acidic b. Neutral c. Basic d. Basic e. Acidic K 9. 3 sp y K M sp 11. 6H 2 O(l) + 4 SbO + (aq) + 3Mo(s) 4Sb(s) + 3MoO 2 (s)+4h 3 O + (aq) Mo(s) reducing agent 6
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