By Matt Kuhn.

Adapted from “The Efficiency of a Fuel Cell” Heliocentris – Chemistry through Hydrogen

OBJECTIVES:

Students will realize the efficiency benefits of fuel cells compared to internal
combustion engines.
Students will understand the concept of efficiency Ú the ratio of the output to
the input of any system.
Students will learn the basic mechanism of how a fuel cell produces electricity.
Students will recognize how hydrogen can be produced by renewable means
and thus a fuel cell can be used as a zero pollution energy source.
Students will use data to calculate the energy efficiency of a fuel cell and apply
this knowledge to societal energy issues.

MATERIALS:

1) 1 Solar Hydrogen Starter Kit with all components including:

Ø 1 Solar module
Ø 1 Electrolyzer
Ø 1 Fuel cell
Ø 1 Load measurement box or multimeter
Ø 1 Set of connecting leads
Ø 1 Stop watch
Ø 2 Long tubes
Ø 2 Short tubes
Ø 2 Tubing stoppers
2) 1 Lamp (100-150 watt) Available for $200-$300 at
3) 1 Small bottle of distilled water www.fuelcellstore.com or
www.heliocentris.com
INTRODUCTORY DISCUSSION:

Review of Chemical Bonding

Hydrogen and Oxygen bond to form water (H2O) because a single Hydrogen atom has
1 electron in its outer electron cloud, or energy level. It would prefer 2 electrons in
order to be chemically stable. Oxygen has 6 outer electrons but prefers 8 in its outer
energy level. Thus, 4 Hydrogen (2H2) combine in an exothermic reaction with 2
Oxygen (1O2) to form two covalently bonded water molecules. This reaction drives the
fuel cell.

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Review of Electricity
Electricity is the flow of electrons. Electrons flow when a build-up of electric charge on
an anode causes electrons to be attracted to a positive electric charge at the cathode
through a closed circuit. This electron flow can be converted into mechanical energy.

What is a fuel cell and how does it work?

Electrochemical energy conversion device
Fuel cells convert Hydrogen or Hydrogen-reformed fuels directly into electricity
in a reaction with Oxygen
They are quiet, compact, efficient, and clean
The electrolyte membrane must stay hydrated at temperatures where water can
exist as a liquid and vapor (≈80ºC)
Reaction is too slow unless a Platinum catalyst is used
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A single fuel cell produces about 0.7 volts and 1 Amp/cm . Therefore, fuel cells
are stacked together in series to increase the voltage by the number of cells.

Hydrogen is split into a proton and an electron with the aid of a Platinum catalyst. The
protons pass through an electrolyte membrane while the electrons must go around the
membrane through a circuit, doing work along the way. Finally, the protons and
electrons recombine with Oxygen to form water and heat.

Fuel Cell Reactions

Anode (Hydrogen in): Overall Reaction:

2H2 → 4H+ + 4e- (oxidation) 2H2 + 02 → 2H2O (l) + Heat

Cathode (Oxygen in): ?H = - 286 kJ/mole

O 2 + 4e- + 4H + → 2H2O (l)
(reduction)

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Parts of the Fuel Cell

Hydrated polymer electrolyte,

also called the proton
exchange membrane-PEM
(wet plastic wrap)

Anode (-) Cathode (+)

Porous catalyst Grooved Carbon-

layers with Teflon coated, gas
dispersed Platinum diffusion layers

←Fuel Cell
Stack

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Efficiency comparison between fuel cells and internal combustion engines

Internal combustion = 20% compared to fuel cell = 45%

Fuel cells as a renewable energy storage device

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 Electrical Energy Electrical Energy

Water (H2 O)

Renewable
Energy

Oxygen
(O2 )

Electrical Energy Hydrogen (H2 )

Electrolyser
with Hydrogen Storage Tank

From the Congressional Record 1875

“A new source of power... called gasoline has been produced by a Boston

engineer. Instead of burning the fuel under a boiler, it is exploded inside the
cylinder of an engine... The dangers are obvious. Stores of gasoline in the
hands of people interested primarily in profit would constitute a fire and
explosive hazard of the first rank. Horseless carriages propelled by gasoline
might attain speeds of 14, or even 20 miles per hour. The menace to our
people of this type hurtling through our streets and along our roads and
poisoning the atmosphere would call for prompt legislative action even if the
military and economic implications were not so overwhelming... the cost of
producing (gasoline) is far beyond the financial capacity of private industry...
In addition the development of this new power may displace the use of
horses, which would wreck our agriculture.”

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The Experiment

Setting up the fuel cell and purging the system

1. Set up the apparatus as shown in Diagram A . As an alternative to solar power,

you could also use a DC power supply if you want the electrolyser to fill the
tanks more quickly. (The DC voltage and current must not exceed 1.8 V
and 3 A respectively. Check for correct polarity of the electrolyser)

2. Check for correct tube connection between the electrolyser and fuel cell. Keep
the outlet tubes running out of the fuel cell open. Adjust the load measurement
box to “open” circuit.

3. Make sure that both of the electrolyser’s gas storage tanks are filled with distilled
water up to the 0 mL mark. Use the solar panel to provide a current of 700-900
mA to the electrolyser. You should be able to observe gas bubbles discharging
oxygen and hydrogen into the tanks.

4. Run the complete system with the gases produced from the electrolyser for 10
minutes, thus purging the system of unwanted gases. After purging, set the load
resistance to about 2 Ω and check the voltage and current. You should be
getting about 0.6 – 0.9 V and 200 – 600 mA. If this is not the case, your fuel cell
needs another purging.

5. Close both outlet tubes from the fuel cell with stoppers or clamps. Hydrogen and
oxygen gases will now collect in the tanks. Disconnect the power supply to the
electrolyser when 10 mL of hydrogen is stored.

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6. Connect leads between the voltmeter and the fuel cell according to Diagram B.

7. With the load in the “open” circuit setting, and a disconnected electrolyser,
calculate the hydrogen leakage rate by starting with a 10 mL full hydrogen tank
and recording the level of hydrogen left after 180 seconds. Record this rate for
later use.

8. Design a data table to record and measure the following during at least 2 trials:

Initial Voltage in volts – V (EI)

Final Voltage in volts – V (EF)
Initial Current in amperes – A (II)
Final Current in amperes – A (IF)
Initial Volume in milliliters – mL (VolI)
Final Volume in milliliters – mL (VolF)
Elapsed Time in seconds – s (set at 180 s)
Load in ohms – Ω (set at 3 Ω)

9. Reconnect the electrolyser and refill the tanks to 10 mL. Then disconnect the
electrolyser and adjust the load to 3 Ω. The fuel cell is now connected to a load
and will consume the stored hydrogen. Record the time and immediately begin
the first trial.

10. Run the fuel cell for at least two trials and record the measurements on your
data table from step #8. After running the second trial, switch the load to “open”
circuit and open the fuel cell outlet tubes.

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Calculating the Faraday Efficiency and Energy Efficiency

You will need another data table with the following values calculated from the 1st table:

3 Ω for 180 s E I Average H2 Consumed

Average
(VolI – VolF – Step 7 Leakage Rate)

1st Trial V A mL
2nd Trial V A mL
Trial Avg. V A mL

The Faraday Efficiency

The Faraday Efficiency is a percentage that tell us how much of the hydrogen gas is
being used for intended electrical energy production and how much is lost to other
factors. The symbol for efficiency is the Greek letter, eta “η.” Therefore, Faraday
Efficiency’s symbol is “ηF.” It is the ratio between the theoretically calculated volume of
hydrogen consumed by the load at a certain current and the experimentally calculated
volume of hydrogen consumed. Hydrogen naturally exists in pairs, so we use “H 2” to
represent it.
ηF = H2 Vol (th.) / H2 Vol(exp.)

If the fuel cell were perfect and all conditions were ideal, the Faraday Efficiency would
be equal to “1” or 100%. Faraday’s 2nd Law enables us to calculate the theoretical
volume of hydrogen we would expect to be consumed.

H2 Vol(th.) = [Electrical charge in Coulombs (C)] × [H2 Volume per mol]

[Electrical charge delivered by one mol H2]

[Electrical Charge in Coulombs (C)] = I × t à ITrials Average × 180 s = _____ C

[H2 Volume per mol] is a given value. At standard pressure and temperature, the
volume of H2 per mol of H2 is 24,000 mL/mol.

H2 à 2H+ + 2e-, therefore, 1 mole of H2 yields 2 moles of electrons (e-). The

electrical charge of one mole of electrons is given as 96,500 C. Since we have two
moles of electrons,
[Electrical charge delivered by one mol H2] = 2 × 96,500 C = 193,000 C/mol

The experimental volume of H2 can be calculated from the experimental measurements:

H2 Vol(exp.) = Trials Average in milliliters

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Now we can calculate the Faraday Efficiency:

H2 Vol(th.) = (I × t) C × 24,000 mL/mol = ____ mL

193,000 C/mol

ηF = H2 Vol (th.) / H2 Vol(exp.) = ____%

The Energy Efficiency

The energy efficiency of a fuel cell is a percentage that tells us how much energy from
the combination of hydrogen and oxygen resulted in the intended electrical energy and
how much energy was not converted to the intended electrical energy. It is a ratio
between the electricity produced by the consumed hydrogen and the calculated
theoretical energy contained in the consumed hydrogen.

[Electrical energy in Joules H2]

ηE =
[Energy content of consumed H2]Theoretical

[Electrical energy in Joules H2] = ETrail Average × ITrial Average × time (s) = ____ J

The energy content of hydrogen is calculated by using the amount of energy given
off when hydrogen is combusted. This is a given value of 286,000 J/mol
(exothermic) of hydrogen. This is called the molar enthalpy of hydrogen at
standard temperature and pressure.

We also have the given value for the amount of space H2 takes up per mole. H2 has
a molar volume of 24,000 mL/mol. If we divide the average trial volume of
hydrogen used by the molar volume, the millimeters will cancel out and we will
have the number of moles of H2 used.
Trial Volume of H2 used in mL = ____ mol H
2
24,000 mL/mol

[Energy content of consumed H2]Theoretical can now be calculated by multiplying the

amount of energy in Joules/mol H2 by the number of moles used.

____ mol H2 × 286,000J/mol = ____ J = [Energy content of consumed H2]Theoretical

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Now we can calculate the Energy Efficiency:

[Electrical energy in Joules H2]

ηE = = ____%
[Energy content of consumed H2]Theoretical

Optional extended experimentation:

You could vary the resistance to change the current and plot the results to find the
characteristic curve of the fuel cell, and thus its optimal settings for peak efficiency.
You could determine the maximum power output and its efficiency by running trials
with graduated current, voltage, and resistance.

Questions to Ponder

1. How does the Faraday Efficiency compare to the Energy Efficiency?

2. How does the efficiency of fuel cells compare to the efficiency of the internal
combustion engine?
3. Why doesn’t the efficiency equal 100%? What happened to the uncaptured
energy?
4. What are the two main emissions of fuel cells? What are the environmental
benefits of fuel cell powered transportation?
5. What should come first, widespread use of fuels or widespread hydrogen fueling
stations? What are the economic implications of building a hydrogen power
infrastructure?
6. How is a fuel cell similar to a battery?
7. What are some current fuel cell applications in our society?
8. How does the volatility of hydrogen compare to common fossil fuels?
9. Compare the number of moving parts in an internal combustion engine to a fuel
cell. How does this comparison affect their lifespan?
10. What are the price comparisons between automotive fuel cells and automotive
internal combustions engines? What are the primary causes for the difference?
11. How does the storage of a liquid fuel compare to the storage of a gaseous fuel?
12. Why can a fuel cell be considered a renewable energy storage device?
13. Where does the United States get much of its oil? How would the widespread
use of fuel cells affect this?
14. What are the political implications of imported oil compared to domestically
produced hydrogen?

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Teachers guide

The Faraday Efficiency of the fuel cell is less than one for the following reasons:
Competing, simultaneous electrochemical reactions in the fuel cell, which supply
fewer electrons for the same volume of hydrogen consumed.
Chemical reactions between hydrogen and oxygen at the catalysts (catalytic
oxidation/combustion).
Hydrogen and oxygen recombination, or diffusion, by leakage through the
electrolyte membrane. (Common in educational demonstration fuel cells.)

These websites will answer most questions you may have about fuel cells:
http://www.eere.energy.gov/hydrogenandfuelcells/
http://www.crest.org/hydrogen/index.html
http://www.fuelcells.org/

Sample Data and Calculations

Leakage rate calculated at 0.6 mL every 180 s

3 Ω for 180 s E I H2
1st Trial EI = 0.756 V II = 0.210 A VolI = 10 mL
EF = 0.754 V IF = 0.212 A VolF = 3 mL
1st Trial Avg. EAVG1 = 0.755 V IAVG1 = 0.211 A VolI-F – 0.6 = 6.4 mL
2nd Trial EI = 0.752 V II = 0.210 A VolI = 10 mL
EF = 0.746 V IF = 0.208 A VolF = 3 mL
2nd Trial Avg. EAVG2 = 0.749 V IAVG2 = 0.209 A VolI-F – 0.6 = 6.4 mL

3 Ω for 180 s E I Average H2 Consumed

Average (VolI – VolF – Step 7 Leakage Rate)

1st Trial 0.755 V 0.211 A 6.4 mL

2nd Trial 0.749 V 0.209 A 6.4 mL
Trial Avg. 0.752 V 0.210 A 6.4 mL

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The Faraday Efficiency

ηF = H2 Vol (th.) / H2 Vol(exp.)

H2 Vol(th.) = [Electrical charge in Coulombs (C)] × [H2 Volume per mol]

[Electrical charge delivered by one mol H2]

[Electrical Charge in Coulombs (C)] = I × t à 0.210 A × 180 s = 37.8 C

[H2 Volume per mol] is a given value. At standard pressure and temperature, the
volume of H2 per mol of H2 is 24,000 mL/mol.

H2 à 2H+ + 2e-, therefore, 1 mole of H2 yields 2 moles of electrons (e-). The

electrical charge of one mole of electrons is given as 96,500 C. Since we have two
moles of electrons,
[Electrical charge delivered by one mol H2] = 2 × 96,500 C = 193,000 C/mol

The experimental volume of H2 can be calculated from the experimental measurements:

H2 Vol(exp.) = 6.4 mL

Now we can calculate the Faraday Efficiency:

H2 Vol(th.) = 37.8 C × 24,000 mL/mol = 4.7 mL

193,000 C/mol

ηF = H2 Vol (th.) / H2 Vol(exp.) = 4.7 mL / 6.4 mL = 0.73 × 100% = 73% Efficient

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The Energy Efficiency

[Electrical energy in Joules H2]

ηE =
[Energy content of consumed H2]Theoretical

[Electrical energy in Joules H2] = ETrails Average × ITrials Average × elapsed time =
0.752 V × 0.210 A × 180 s = 28.40 J

The energy content of hydrogen is calculated by using the amount of energy given
off when hydrogen is combusted. This is a given value of 286,000 J/mol
(exothermic) of hydrogen. This is called the molar enthalpy of hydrogen at
standard temperature and pressure. (Often given as −285.6kJ/mol, “−” means
exothermic)

We also have the given value for the amount of space H2 takes up per mole. H2 has
a molar volume of 24,000 mL/mol (rounded). If we divide the average trial volume
of hydrogen used by the molar volume, the millimeters will cancel out and we will
have the number of moles of H2 used.
Trial Volume of H2 used in mL = 6.4 mL ÷ 24,000 mL/mol 0.00027 mol H
2
24,000 mL/mol

[Energy content of consumed H2]Theoretical can now be calculated by multiplying the

amount of energy in Joules/mol H2 by the number of moles used.

0.00027 mol H2 × 286,000J/mol = 76.27 J = [Energy content of consumed H2]Theoretical

Now we can calculate the Energy Efficiency:

η E = [Electrical energy in Joules H2] = 28.4 J/76.27 J = 0.37 × 100% = 37% Efficient
[Energy content of consumed H2]Theoretical

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