Pre-Lab Questions/Answers Experiment 6 Part I 1. Based on Ohm s Law, calculate the current (ma) flow through a 1.00Ω resistor when the voltage across the resister is 5.00 mv. 2. Calculate the standard reaction ΔH O rxn and ΔG O rxn for the following reaction from their standard ΔH O f and ΔG O f. (see Zumdahl Appendix four, or Tro, Appendix II B)) Species CH 3 OH (l)+ 3/2 O 2 (g)à CO 2 (g)+2h 2 O (l) ΔG O f at 25 o C (KJ/mol) ΔH O f at 25 o C (KJ/mol) CH 3OH - 166-239 O 2 0 0 CO 2-394 - 394 H 2O - 237-286 3. The overall methanol fuel cell reaction is CH 3 OH + 3/2 O 2 à CO 2 +2H 2 O. Two half reactions and the standard half cell reduction potentials are given (see Zumdahl Appendix five, or Tro, Appendix II, D for common standard reduction half cell potentials): Oxidation (anode): CH 3 OH+6OH - à CO 2 + 5H 2 O +6e- Reduction (cathode): ½ O 2 + H 2 O+ 2e- à 2OH - E o anode= -0.81V E o cathode = 0.40 V Calculate the standard fuel cell potential E o (or open circuit potential) from their standard halfcell reduction potentials. 4. For per mole of CH 3 OH, calculate the standard open circuit voltage E o open using ΔG O rxn from question 2 and the maximum efficiency of the fuel cell in terms of ΔH O rxn. 5. When the methanol fuel cell is connected to a mini motor, the potential of the cell is decreased to 0.80V. Calculate the electric work in kj/mol CH 3 OH provided by the fuel cell and the efficiency of the fuel cell when running a mini motor (Faraday s constant is 96,485 Coulombs/mol) Part II 1. Write the half-cell reactions for the following electrochemical processes: (a) the waterelectrolysis cell containing sodium hydroxide and (b) the cell containing copper sulfate. 2. The following data were collected using the procedure described in this experiment. Data are given here only for calculating the volume of oxygen gas, but calculations for the volume of hydrogen gas are done in the same way.
Meniscus of oxygen read on buret: at the 24.90 ml mark Void volume from stopcock to 50 ml mark: 2.62 ml Height of NaOH column: 27.70 cm Temperature: 22.8 o C Barometric pressure: 773 mm Hg Initial weight of Cu cathode: 20.8816 g Final weight of Cu cathode: 21.0195 g Time, min: 2 6 10 14 18 19.5 Current, ma: 352 357 361 363 366 368 Total time of electrolysis: 19.5 min a) Calculate the volume of dry oxygen produced, corrected to 0 o C and 1.00 atm pressure. b) Calculate the volume of dry oxygen produced per mole of copper metal deposited. Use atomic weight of Cu = 63.55 g/mol c) Calculate the coulombs of charge carried in the circuit during the experiment and the value of Faraday s constant. The current to be used for this calculation is the average of the listed numbers: (1/6)(352 + 357 + 361 + 363 + 366 + 368) = 361ma. The time is the total elapsed time (19.5 min). (361 ma) (1 Amp/ 1000 ma) = 0.361 a Recall: Coulombs of charge = amps x seconds 3. Explain why: a) A solution of NaOH, instead of pure water, is used in the water electrolysis. b) The volume of the void in the buret, between the stopcock and the 50 ml mark, is measured. c) The copper strips are cleaned with abrasive until all corrosion and dirt are removed. d) The current is measured at regular time intervals during the electrolysis, rather than only once at the beginning or the end of the run. e) The height of the NaOH column in each buret, from the meniscus to the level in the beaker, is measured. f) The temperature and barometric pressure are measured. g) The presence of water vapor in the gases in the burets must be considered.
Answers Part I 1. Based on Ohm s Law, calculate the current (ma) flow through a 1.00Ω resistor when the voltage across the resister is 5.00 mv. I (A) = V (V)/R (Ω) = 0.00500(V)/1.00 (Ω) =5.00 ma 2. Calculate the standard reaction ΔH O rxn and ΔG O rxn for the following reaction from their standard ΔH O f and ΔG O f. (see Zumdahl Appendix four, or Tro, Appendix II B) CH 3 OH + 3/2 O 2 à CO 2 +2H 2 O ΔG Species O f at 25 o C ΔH O f at 25 o C (kj/mol) (k/mol) CH 3OH - 166-239 O 2 0 0 CO 2-394 - 394 H 2O - 237-286 ΔG O rxn at 25 o C= (- 394)+2(- 237)- (- 166)- 3/2 (0) = - 702 kj/mol of CH 3 OH ΔH O rxn at 25 o C= (- 394)+2(- 286)- (- 239)- 3/2 (0)= - 727 kj/mol of CH 3 OH 3. The overall methanol fuel cell reaction is CH 3 OH + 3/2 O 2 à CO 2 +2H 2 O. Two half reactions and their standard half cell reduction potential are given (common standard reduction half cell potentials can be found from Zumdahl Appendix five, or Tro, Appendix II, D): Oxidation (anode): CH 3 OH+6OH - à CO 2 + 5H 2 O +6e- Reduction (cathode): ½ O 2 + H 2 O+ 2e- à 2OH - E o anode= -0.81V E o cathode = 0.40 V Calculate the standard fuel cell potential E o (or open circuit potential) from their standard halfcell reduction potentials. E o cell = E o open = 0.40 (-0.81V) = 1.21 V 4. For per mole of CH 3 OH, calculate the standard open circuit voltage E o open using ΔG O rxn from question 2 and the maximum efficiency of the fuel cell in terms of ΔH O rxn. E o open = -ΔG O rxn/nf = -(-702) KJ/mol / [(6mol e-/mol CH 3 OH )* 96485C] = 1.21 V This value is the same if it is calculated using half-cell potentials from question 3. ŋ max = ΔG O /ΔH O = 702kJ/mol /727kJ/mol = 97%
5. When the methanol fuel cell is connected to a mini motor, the potential of the cell is decreased to 0.80V. Calculate the electric work in kj/mol CH 3 OH provided by the fuel cell and the efficiency of the fuel cell when running a mini motor (Faraday s constant is 96,485 Coulombs/mol) W el = -E q = -E n F= 0.80V * (6mol e-/mol CH 3 OH) * 96485C/mol e- = -463 kj/mol CH 3 OH ŋ = W el /ΔH = 463kJ/mol / 727kJ/mol= 64 % Part II 1. Write the half-cell reactions for the following electrochemical processes: (a) the waterelectrolysis cell containing sodium hydroxide and (b) the cell containing copper sulfate. (a) cathode: 2 H 2 O + 2 e - H 2 + 2 OH - (Eq. 1) (a) anode 2 H 2 O O 2 + 4 H + + 4 e - (Eq. 2) (b) cathode Cu 2+ (aq) + 2 e - Cu (s) (Eq. 4) (b) anode Cu(s) Cu 2+ (aq) + 2 e - (Eq. 5) 2. The following data were collected using the procedure described in this experiment. Data are given here only for calculating the volume of oxygen gas, but calculations for the volume of hydrogen gas are done in the same way. Meniscus of oxygen read on buret: at the 24.90 ml mark Void volume from stopcock to 50 ml mark: 2.62 ml Height of NaOH column: 27.70 cm Temperature: 22.8 o C Barometric pressure: 773 mm Hg Initial weight of Cu cathode: 20.8816 g Final weight of Cu cathode: 21.0195 g Time, min: 2 6 10 14 18 19.5 Current, ma: 352 357 361 363 366 368 Total time of electrolysis: 19.5 min a) Calculate the volume of dry oxygen produced, corrected to 0 o C and 1.00 atm pressure. volume = 50.00-24.90 +2.62 = 27.72 ml (uncorrected for P w or to STP) Total pressure = 773 mm Hg Pressure of NaOH column: 277.0 mm NaOH X (1.076 mm Hg/13.55 mm NaOH) =22.00 mm Hg (Eq. 10) ln P w = 1.613 + 0.06227T = 3.0328; P w = 20.75 mm Hg (Eq. 9) P dry O2 = (773 22.00 20.75) mm Hg = 730.2 mm Hg (extra sig. fig. carried) V STP = (27.72 ml) (730.2 mm/760mm STP ) (273.15 K STP /(273.15+22.8)K) = 24.58 ml b) Calculate the volume of dry oxygen produced per mole of copper metal deposited. Use atomic weight of Cu = 63.55 g/mol Cu = 21.0195-20.8816 = 0.1379 g = 0.1379 g/63.55 g/mol = 0.002170 mol
V dry O2, STP /mol Cu = 0.02458 L/0.002170 mol = 11.3 L/mol Cu (3 sig. fig. because pressure is known only to 3 sig. fig) c) Calculate the coulombs of charge carried in the circuit during the experiment and the value of Faraday s constant. The current to be used for this calculation is the average of the listed numbers: (1/6)(352 + 357 + 361 + 363 + 366 + 368) = 361ma. The time is the total elapsed time (19.5 min). (361 ma) (1 Amp/ 1000 ma) = 0.361 a Recall: Coulombs of charge = amps x seconds q = (0.361 Amp) (19.5 min) (60 sec/min) = 422 coulombs Faraday s constant = coulombs/mole electrons = C/(mol Cu x 2mol e - /mol Cu) = C/(mol O 2 x 4 mol e - /mol O 2 ) based on Cu: F = 422 C/(0.002170 mol Cu x 2 mol e - /1mol Cu) = 97,200 = 9.72 x 10 4 C/mol e - based on O 2 : moles O 2 = (0.02458 L)(1 atm)/((0.08206 L-atm/mol K)(273K)) = 0.001096 F = 422 C/(0.001096 mol O 2 x 4 mol e - /1 mol O 2 ) = 96,200 = 9.62 x 10 4 C/mol e - 3. Explain why: a) A solution of NaOH, instead of pure water, is used in the water electrolysis. Water is a poor electrolyte, and base protects the steel electrodes from oxidation. b) The volume of the void in the buret, between the stopcock and the 50 ml mark, is measured. This volume will be filled with gas, part of the total volume that must be known. c) The copper strips are cleaned with abrasive until all corrosion and dirt are removed. Copper plate will not adhere to a corroded or dirty surface. d) The current is measured at regular time intervals during the electrolysis, rather than only once at the beginning or the end of the run. The current may not stay constant during the experiment, but an average of measurements taken throughout the time will reflect the total flow of electrons. e) The height of the NaOH column in each buret, from the meniscus to the level in the beaker, is measured. The NaOH column contributes to the internal pressure that balances the atmospheric pressure. f) The temperature and barometric pressure are measured. These values must be known to convert the volumes of gas to STP; and the temperature must be known to calculate the vapor pressure of water inside the burets. The measurements and calculations are necessary to calculate the number of moles of gas. g) The presence of water vapor in the gases in the burets must be considered. The pressure of the water vapor contributes to the total pressure inside the burets.