Chemistry 102 2016 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 = -2.00 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
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 = -0.360V E cell = E red (cathode) E red (anode)= -0.360V-0V= -0.360V 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
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 4 0.06V 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
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
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 = 0.150 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
E (volts) 15. Below is a plot of the measured voltage at 298 K of alectrochemical cell at different values of log(q). 0.12 0.08 0.04 0-0.04-0.08-0.12 0 1 2 3 4 5 6 7 8 9 10 11 12 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, 10 100,4,-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: 1. 8.8 10-7 2. a) 0.082%; b) 2.0 10-8 3. 5.0 10-12 M 4. 1.0 10-4 M 5. 2.4 10-5 M 6. 7. 8. 7 10-13 M a. 1.4 10-5 M b. 1.5 10-5 M c. Stays the same a. Acidic b. Neutral c. Basic d. Basic e. Acidic K 9. 3 sp y 0. 975 3 K 1.08 10. 1.32 10-16 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