Name:____________________ Electrochemical Cells II – Supplemental Worksheet KEY All the electrochemical cells on this worksheet are the same ones on the first Electrochemical Cells worksheet. To make the work on this worksheet easier, refer to the work you did on the previous Electrochemical Cells worksheet. Experimental Observations of Electrochemical Cells 1. Consider the voltaic cell that contains standard Co2+/Co and Au3+/Au electrodes. The following experimental observations were noted: (1) Metallic gold plates out on one electrode and the gold ion concentration decreases around that electrode, and (2) The mass of the cobalt electrode decreases and the cobalt (II) ion concentration increases around that electrode. a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? <Co(s)|Co2+(aq)||Au3+(aq)|Au(s)> E°cell = E°cathode – E°anode = 1.50 V – (–0.28 V) = 1.78 V b. What is the standard reaction free energy for this cell? ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜6 mol rxn ⎟⎜96485 molC e – ⎟(1.78 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= –1030.46 molkJrxn € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. ΔG˚= –RT lnK ΔG˚= –1030.46 molkJrxn = –1030460 molJ rxn –1030460 molJ rxn ΔG˚ lnK = – =− = 415.7 J RT (8.314 mol rxn⋅K )(298.15K) K = e 415.7 This is a very large K, this reflects the spontaneous nature of this reaction. € d. Calculate the emf at 25°C of this cell in which the concentration of Co2+ ions is 0.30 mol/L and that of the Au3+ ions is 0.0010 mol/L. Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant: 0.05916 E cell = E˚− logQ n [Co 2+ ]3 [0.30M]3 Q= = = 27000 [Au 3+ ]2 [0.0010M]2 0.05916 E cell = 1.78V − log(27000) 6mol e – E cell = 1.74V e. Assume you use this cell as a battery to power one of your electric devices that draws 1€ mA of current. If you ran the battery at a constant 1 mA for 3000 hours, how many grams of Co would be consumed? To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Co ! grams of Co 1mA = 0.001 Cs ⎛ 0.001C ⎞ ⎛ 3600s ⎞ C used = current × time = ⎜ ⎟ × ⎜ ⎟ × ( 3000hr) ⎝ 1s ⎠ ⎝ 1hr ⎠ C used = 10800C ⎛ 1mol e – ⎞ ⎛ 1mol Co ⎞ ⎛ 59g Co ⎞ ⎟ ⎜ ⎟⎟ ⎟ × ⎜⎜ mass Co = (10800C) × ⎜ – ⎟ × ⎜ ⎝ 96485C ⎠ ⎝ 2mol e ⎠ ⎝1mol Co ⎠ mass Co = 3.30gCo So 3.30 grams of cobalt would be consumed. € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ 2. Consider the electrolysis of molten calcium chloride with inert electrodes. The following experimental observations were noted: (1) Bubbles of pale green chlorine gas are produced at one electrode, and (2) Silvery white molten metallic calcium is produced at the other electrode. a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? <Pt(s)|Cl-‐(aq)|Cl2(g)||Ca2+(aq)|Ca(s)|Pt(s)> E°cell = E°cathode – E°anode = –2.76 V – (1.36 V) = –4.12 V b. What is the standard reaction free energy for this cell? ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜2 mol rxn ⎟⎜96485 molC e – ⎟( –4.12 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= 795 molkJrxn c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ ΔG˚= –RT lnK ΔG˚= 795 molkJrxn = 795000 molJ rxn 795000 molJ rxn ΔG˚ lnK = – =− = –320.7 J RT (8.314 mol rxn⋅K )(298.15K) K = e –320.7 This is a very small K, this reflects the non-‐spontaneous nature of this reaction. € d. How many hours are required to plate 12.00 g of metallic calcium from 1.00 M CaCl2 (aq) by using a current of 3.00 A? To figure this problem out, we will do a series of conversions: grams of Ca ! moles of Ca ! moles of e– ! charge ! charge and current ! time ⎛ 1mol Ca ⎞ ⎛ 2mol e – ⎞ ⎛ 96485C ⎞ ⎛ 1s ⎞ ⎛ 1hr ⎞ ⎟⎟ × ⎜⎜ ⎟⎟ × ⎜⎜ ⎟ time = (12.00gCa) × ⎜⎜ ⎟ × ⎜ ⎟ – ⎟ × ⎜ ⎝ 40.1g Ca ⎠ ⎝1mol Ca ⎠ ⎝ 1mol e ⎠ ⎝ 3C ⎠ ⎝ 3600s ⎠ time = 5.347hr It would require 5.3 hours to plate 12.00 g of metallic calcium from 1.00 M CaCl2 (aq) by using a current of 3.00 A. € e. Determine the volume (in liters, at STP) of chlorine gas that can be produced in this cell by using a current of 7.30 mA for 2.11 hours. To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e-‐ ! moles of Co ! liters of Cl2 current = 7.30mA = 0.0073 Cs ⎛ 3600s ⎞ time = 2.11hr × ⎜ ⎟ = 7596s ⎝ 1hr ⎠ ⎛ 0.0073C ⎞ C used = current × time = ⎜ ⎟ × (7596s) ⎝ 1s ⎠ C used = 55.45C ⎛ 1mol e – ⎞ ⎛1mol Cl2 ⎞ ⎛ 22.4L Cl2 ⎞ ⎟ ⎜ ⎟⎟ ⎟ × ⎜⎜ volume Cl2 = (55.45C) × ⎜ – ⎟ × ⎜ ⎝ 96485C ⎠ ⎝ 2mol e ⎠ ⎝ 1mol Cl2 ⎠ volume Cl2 = 6.44 × 10 −3 L In this situation 6.44x10-‐3L or 6.44 mL Cl2 would be produced. € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ Short Hand Notation of Electrochemical Cells 3. Consider the following cell <Ni(s)|Ni2+(aq)||Ag+(aq)|Ag(s)> a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? E°cell = E°cathode – E°anode = 0.80 V – (–0.23 V) = 1.03 V b. What is the standard reaction free energy for this cell? ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜2 mol rxn ⎟⎜96485 molC e – ⎟(1.03 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= −198.76 molkJrxn c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. € ΔG˚= –RT lnK ΔG˚= −198.76 molkJrxn = −198760 molJ rxn −198760 molJ rxn ΔG˚ lnK = – =− = 80.18 J RT (8.314 mol rxn⋅K )(298.15K) K = e 80.18 = 6.65 × 10 34 This is a very large K, this reflects the spontaneous nature of this reaction. Revised CR 1/16/14 € © LaBrake & Vanden Bout 2014 Name:____________________ d. What is the concentration of Ni2+ ions if the emf at 25°C of this cell is +1.68 V and the concentration of Ag+ ions is 0.0280 mol/L? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant: 0.05916 E cell = E˚− logQ n [Ni 2+ ] x Q= + 2 = [Ag ] [0.0280M]2 ⎞ 0.05916 ⎛ x 1.68 = 1.03V − – log⎜ 2 ⎟ 2mol e ⎝ [0.0280M] ⎠ ⎛ ⎞ x log⎜ 2 ⎟ = −21.97 ⎝ [0.0280M] ⎠ x −22 2 = 1.06 × 10 [0.0280M] x = [Ni 2+ ] = 8.32 × 10 −26 M e. How many hours would it take for this galvanic cell to plate 30.00 g of 2+ + metallic silver € from 1.00 M solutions of Ni (aq) and Ag (aq) assuming it produces a constant current of 2.25 A? To figure this problem out, we will do a series of conversions: grams of Ag ! moles of Ag ! moles of e-‐ ! charge ! current ! time 2.25A = 2.25 Cs ⎛ 1mol Ag ⎞ ⎛ 1mol e – ⎞ ⎛ 96485C ⎞ ⎛ 1s ⎞ ⎛ 1hr ⎞ ⎟⎟ × ⎜⎜ ⎟⎟ × ⎜⎜ ⎟ time = 30.00g Ag × ⎜⎜ ⎟ × ⎜ ⎟ – ⎟ × ⎜ ⎝107.87g Ag ⎠ ⎝1mol Ag ⎠ ⎝ 1mol e ⎠ ⎝ 2.25C ⎠ ⎝ 3600s ⎠ ( ) time = 3.3hours € It would require 3.3 hours to plate 30.00 g of metallic silver from 1.00 M solutions of Ni2+(aq) and Ag+(aq) assuming it produces a constant current of 2.25 A. Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ 4. Consider the following cell <Pt(s)| Ce3+(aq)|Ce4+(aq)||Cu2+(aq) |Cu(s)> a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? E°cell = E°cathode – E°anode = 0.34 V – (1.70 V) = –1.36 V b. What is the standard reaction free energy for this cell? ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜2 mol rxn ⎟⎜96485 molC e – ⎟( −1.36 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= 262.44 molkJrxn c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. € ΔG˚= –RT lnK ΔG˚= 262.44 molkJrxn = 262440 molJ rxn 262440 molJ rxn ΔG˚ lnK = – =− = –105.87 J RT (8.314 mol rxn⋅K )(298.15K) K = e –105.87 ≈ 1.0 × 10 −46 This is a very small K, this reflects the non-‐spontaneous nature of this reaction. € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ d. What is the concentration of Ce4+ ions if the emf at 25°C of this cell is –1.20 V, the concentration of Cu2+ ions is 0.60 mol/L and the concentration of Ce3+ ions is 0.30 mol/L? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant: 0.05916 E cell = E˚− logQ n [Ce 4 + ]2 x2 x2 Q= = = [Cu 2+ ][Ce 3+ ]2 [0.60M][0.30M]2 0.054 0.05916 ⎛ x 2 ⎞ −1.20V = –1.36V − log⎜ ⎟ 2mol e – ⎝ 0.054 ⎠ 2 ⎛ x ⎞ log⎜ ⎟ = −5.409 ⎝ 0.054 ⎠ x2 = 3.90 × 10 −6 0.054 x = [Ce 4 + ] = 4.59 × 10 −4 M e. Determine the mass (in grams) of metal copper that can be produced in this cell by € using a current of 5.0 A for 2.7 days. To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Cu ! grams of Cu 5.0A = 5.0 Cs ⎛ 5.0C ⎞ ⎛ 3600s ⎞ ⎛ 24hr ⎞ C used = current × time = ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × (2.7days) ⎝ 1s ⎠ ⎝ 1hr ⎠ ⎝ 1day ⎠ C used = 1166400C ⎛ 1mol e – ⎞ ⎛ 1mol Cu ⎞ ⎛ 63.55g Cu ⎞ ⎟ ⎜ ⎟⎟ ⎟ × ⎜⎜ mass Cu = (1166400C) × ⎜ – ⎟ × ⎜ ⎝ 96485C ⎠ ⎝ 2mol e ⎠ ⎝ 1mol Cu ⎠ mass Cu = 384gCu So 384 g Cu would be produced. € Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ Electrochemical Cell Diagrams 5. Consider the following cell: a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? <Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)> E°cell = E°cathode – E°anode = –0.76 V – (0.34 V) = –1.10 V b. What is the standard reaction free energy for this cell? Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜2 mol rxn ⎟⎜96485 molC e – ⎟( −1.10 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= 212.27 molkJrxn c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. € ΔG˚= –RT lnK ΔG˚= 212.27 molkJrxn = 212270 molJ rxn 212270 molJ rxn ΔG˚ lnK = – =− = –85.63 J RT (8.314 mol rxn⋅K )(298.15K) K = e –85.63 ≈ 6.5 × 10 −38 This is a very small K, this reflects the non-‐spontaneous nature of this reaction. € d. What is the potential for this cell if the Cu2+ concentration is 5.8 x 10-‐3M and the Zn2+ concentration is 1.3 x 10-‐1 M ? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant: 0.05916 E cell = E˚− logQ n [Cu 2+ ] [5.8 × 10 −3 M] Q= = = 0.0446 [Zn 2+ ] [1.3 × 10 −1 M] 0.05916 E cell = –1.10V − log(0.0446) 2mol e – E cell = −1.06V e. Determine the mass (in grams) of zinc metal that can be produced in this cell by using a current of 6.0 A for 1.5 days. € To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Zn ! grams of Zn Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ 6.0A = 6.0 Cs ⎛ 6.0C ⎞ ⎛ 3600s ⎞ ⎛ 24hr ⎞ C used = current × time = ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × (1.5days) ⎝ 1s ⎠ ⎝ 1hr ⎠ ⎝ 1day ⎠ C used = 777600C ⎛ 1mol e – ⎞ ⎛ 1mol Zn ⎞ ⎛ 65.38g Zn ⎞ ⎟ ⎜ ⎟⎟ ⎟ × ⎜⎜ mass Zn = (777600C) × ⎜ – ⎟ × ⎜ ⎝ 96485C ⎠ ⎝ 2mol e ⎠ ⎝ 1mol Zn ⎠ mass Zn = 263.46gZn So 263.5 g Zn would be produced 6. Consider the following cell: € a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell? <Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)> E°cell = E°cathode – E°anode = 0.34 V – (–0.76 V) = 1.10 V Revised CR 1/16/14 © LaBrake & Vanden Bout 2014 Name:____________________ b. What is the standard reaction free energy for this cell? ΔG˚= –nFE˚ ⎛ mol e – ⎞⎛ ⎞ ΔG˚= –⎜2 mol rxn ⎟⎜96485 molC e – ⎟(1.10 CJ ) ⎝ ⎠⎝ ⎠ ΔG˚= −212.27 molkJrxn c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C. € ΔG˚= –RT lnK ΔG˚= −212.27 molkJrxn = −212270 molJ rxn −212270 molJ rxn ΔG˚ lnK = – =− = 85.63 J RT (8.314 mol rxn⋅K )(298.15K) K = e 85.63 ≈ 1.5 × 10 37 This is a very large K, this reflects the spontaneous nature of this reaction. d. Calculate cell in which the concentration of Cu2+ ions is € the emf at 25°C of this 2+ 0.1250 mol/L and that of the Zn ions is 0.0020 mol/L. Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant: 0.05916 E cell = E˚− logQ n [Zn 2+ ] [0.0020M] Q= = = 0.016 [Cu 2+ ] [0.1250M] 0.05916 E cell = –1.10V − log(0.016) 2mol e – E cell = −1.15V e. How many hours would it take for this galvanic cell to plate 15.00 g of 2+ 2+ metallic copper € from 1.00 M solutions of Cu (aq) and Zn (aq) assuming it produces a constant current of 5.0 A? To figure this problem out, we will do a series of conversions: grams of Cu ! moles of Cu ! moles of e-‐ ! charge ! current ! time 5.0A = 5.0 Cs ⎛ 1mol Cu ⎞ ⎛ 2mol e – ⎞ ⎛ 96485C ⎞ ⎛ 1s ⎞ ⎛ 1hr ⎞ ⎟⎟ × ⎜⎜ ⎟⎟ × ⎜⎜ ⎟ time = 15.00g Cu × ⎜⎜ ⎟ × ⎜ ⎟ – ⎟ × ⎜ ⎝ 63.55g Cu ⎠ ⎝1mol Cu ⎠ ⎝ 1mol e ⎠ ⎝ 5.0C ⎠ ⎝ 3600s ⎠ time = 2.53hours It would require 2.53 hours to plate 15.00 g of metallic copper from 1.00 M solutions of Cu2+(aq) and Zn2+(aq) assuming it produces a constant current of 5.0 A. ( € Revised CR 1/16/14 ) © LaBrake & Vanden Bout 2014

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