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 ΔHOrxn and ΔGOrxn for the following reaction from their standard ΔHOf and ΔGOf. (see Zumdahl Appendix four, or Tro, Appendix II B)) CH3OH (l)+ 3/2 O2 (g)à CO2 (g)+2H2O (l)
CH3OH ΔGOf at 25oC (KJ/mol)
-­‐166 ΔHOf at 25oC (KJ/mol)
-­‐239 O2 0 0 CO2 -­‐394 -­‐394 H2O -­‐237 -­‐286 Species 3. The overall methanol fuel cell reaction is CH3OH + 3/2 O2 à CO2+2H2O. 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): CH3OH+6OH- à CO2+ 5H2O +6e-
Eoanode= -0.81V
Reduction (cathode): ½ O2+ H2O+ 2e- à2OH-
Eocathode = 0.40 V
Calculate the standard fuel cell potential Eo (or open circuit potential) from their standard halfcell reduction potentials.
4. For per mole of CH3OH, calculate the standard open circuit voltage Eoopen using ΔGOrxn from
question 2 and the maximum efficiency of the fuel cell in terms of ΔHOrxn.
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 CH3OH 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: 22.8oC
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
Current, ma:
352
357
361
363
366
Total time of electrolysis: 19.5 min
19.5
368
a) Calculate the volume of dry oxygen produced, corrected to 0oC 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 ΔHOrxn and ΔGOrxn for the following reaction from their standard ΔHOf and ΔGOf. (see Zumdahl Appendix four, or Tro, Appendix II B) CH3OH + 3/2 O2 à CO2+2H2O
CH3OH ΔGOf at 25oC (kJ/mol)
-­‐166 ΔHOf at 25oC (k/mol)
-­‐239 O2 0 0 CO2 -­‐394 -­‐394 H2O -­‐237 -­‐286 Species ΔGOrxn at 25oC= (-­‐394)+2(-­‐237)-­‐(-­‐166)-­‐ 3/2 (0) = -­‐702 kJ/mol of CH3OH
ΔHOrxn at 25oC= (-­‐394)+2(-­‐286)-­‐(-­‐239)-­‐ 3/2 (0)= -­‐727 kJ/mol of CH3OH
3. The overall methanol fuel cell reaction is CH3OH + 3/2 O2 à CO2+2H2O. 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): CH3OH+6OH- à CO2+ 5H2O +6e-
Eoanode= -0.81V
Reduction (cathode): ½ O2+ H2O+ 2e- à2OH-
Eocathode = 0.40 V
Calculate the standard fuel cell potential Eo (or open circuit potential) from their standard halfcell reduction potentials.
Eocell = Eo open = 0.40 – (-0.81V) = 1.21 V
4. For per mole of CH3OH, calculate the standard open circuit voltage Eoopen using ΔGOrxn from
question 2 and the maximum efficiency of the fuel cell in terms of ΔHOrxn.
Eo open = -ΔGOrxn/nF = -(-702) KJ/mol / [(6mol e-/mol CH3OH )* 96485C] = 1.21 V
This value is the same if it is calculated using half-cell potentials from question 3.
ŋ max= |ΔGO /ΔHO |= |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 CH3OH 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)
Wel= -E q = -E n F= 0.80V * (6mol e-/mol CH3OH) * 96485C/mol e- = -463 kJ/mol CH3OH
ŋ = |Wel /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 H2O + 2 e-→ H2 + 2 OH-
(Eq. 1)
(a) anode
2 H 2O → O 2 + 4 H + + 4 e -
(Eq. 2)
(b) cathode
(b) anode
2+
-
Cu (aq) + 2 e → Cu (s)
2+
Cu(s) → Cu (aq) + 2 e
(Eq. 4)
-
(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: 22.8oC
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
Current, ma:
352
357
361
363
366
Total time of electrolysis: 19.5 min
19.5
368
a) Calculate the volume of dry oxygen produced, corrected to 0oC and 1.00 atm pressure.
volume = 50.00 -24.90 +2.62 = 27.72 mL (uncorrected for Pw 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 Pw = 1.613 + 0.06227T = 3.0328; Pw = 20.75 mm Hg (Eq. 9)
Pdry O2 = (773 – 22.00 – 20.75) mm Hg = 730.2 mm Hg (extra sig. fig. carried)
VSTP = (27.72 mL) (730.2 mm/760mmSTP) (273.15 KSTP/(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
Vdry 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 O2 x 4 mol e-/mol O2)
based on Cu: F = 422 C/(0.002170 mol Cu x 2 mol e-/1mol Cu) = 97,200 = 9.72 x 104 C/mol ebased on O2: moles O2 = (0.02458 L)(1 atm)/((0.08206 L-atm/mol K)(273K)) = 0.001096
F = 422 C/(0.001096 mol O2 x 4 mol e-/1 mol O2) = 96,200 = 9.62 x 104 C/mol e3. 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.