Syllabus reference R1.3.5
Reactivity 1.3.5 - A fuel cell can be used to convert chemical energy from a fuel directly to electrical energy.
- Deduce half-equations for the electrode reactions in a fuel cell.
Guidance
- Hydrogen and methanol should be covered as fuels for fuel cells.
- The use of proton exchange membranes will not be assessed.
Tools and links
- Reactivity 3.2 - What are the main differences between a fuel cell and a primary (voltaic) cell?
Fuel Cells
Fuel cells are devices that convert chemical energy from a fuel into electrical energy through an electrochemical reaction. Unlike batteries, fuel cells require a continuous supply of fuel and oxidant to operate.
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Types of Fuel Cells
- Proton Exchange Membrane Fuel Cells (PEMFC)
- Fuel: Hydrogen (H2)
- Oxidant: Oxygen (O2)
- Electrolyte: Proton-conducting polymer membrane
- Applications: Transportation, portable power, and stationary power generation
- Solid Oxide Fuel Cells (SOFC)
- Fuel: Hydrogen, carbon monoxide (CO), natural gas
- Oxidant: Oxygen (O2)
- Electrolyte: Solid ceramic material
- Applications: Stationary power generation, combined heat and power (CHP) systems
- Alkaline Fuel Cells (AFC)
- Fuel: Hydrogen (H2)
- Oxidant: Oxygen (O2)
- Electrolyte: Alkaline solution (e.g., potassium hydroxide)
- Applications: Spacecraft, military applications
- Direct Methanol Fuel Cells (DMFC)
- Fuel: Methanol (CH3OH)
- Oxidant: Oxygen (O2)
- Electrolyte: Proton-conducting polymer membrane
- Applications: Portable power devices, such as laptops and mobile phones
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Reactions at the Electrodes
Proton Exchange Membrane Fuel Cells (PEMFC)
- Anode Reaction
- 2H2 → 4H+ + 4e-
- Hydrogen molecules are oxidized, producing protons (H+) and electrons (e-).
- Cathode Reaction
- O2 + 4H+ + 4e- → 2H2O
- Oxygen molecules react with protons and electrons to form water.
- Overall Reaction
- 2H2 + O2 → 2H2O
- The overall reaction produces water and electrical energy.
Solid Oxide Fuel Cells (SOFC)
- Anode Reaction
- H2 + O2- → H2O + 2e-
- Hydrogen reacts with oxide ions to form water and release electrons.
- Cathode Reaction
- O2 + 4e- → 2O2-
- Oxygen molecules gain electrons to form oxide ions.
- Overall Reaction
- H2 + O2 → H2O
- The overall reaction produces water and electrical energy.
Alkaline Fuel Cells (AFC)
- Anode Reaction
- 2H2 + 4OH- → 4H2O + 4e-
- Hydrogen reacts with hydroxide ions to form water and release electrons.
- Cathode Reaction
- O2 + 2H2O + 4e- → 4OH-
- Oxygen reacts with water and electrons to form hydroxide ions.
- Overall Reaction
- 2H2 + O2 → 2H2O
- The overall reaction produces water and electrical energy.
Direct Methanol Fuel Cells (DMFC)
- Anode Reaction
- CH3OH + H2O → CO2 + 6H+ + 6e-
- Methanol reacts with water to produce carbon dioxide, protons, and electrons.
- Cathode Reaction
- 3/2 O2 + 6H+ + 6e- → 3H2O
- Oxygen reacts with protons and electrons to form water.
- Overall Reaction
- CH3OH + 3/2 O2 → CO2 + 2H2O
- The overall reaction produces carbon dioxide, water, and electrical energy.
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Advantages and Disadvantages of Fuel Cells
- Advantages
- High efficiency compared to traditional combustion engines.
- Low emissions, with water being the primary by-product.
- Quiet operation and modular design for various applications.
- Disadvantages
- High cost of fuel cell materials and production.
- Challenges in hydrogen storage and distribution for PEMFCs.
- Durability and stability issues, especially for SOFCs operating at high temperatures.
Conclusion
Fuel cells are promising technologies for clean and efficient energy conversion. Different types of fuel cells operate based on various electrochemical reactions at the electrodes, offering unique advantages and challenges. Continued research and development are essential to overcome current limitations and make fuel cells a viable alternative for widespread energy applications.
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Worked examples
Q461-01 Which of the changes
below occurs with the greatest increase in entropy?
- Na2O(s) + H2O(l) →
2Na+(aq) + 2OH-(aq)
- NH3(g) + HCl(g) → NH4Cl(s)
- H2(g) + I2(g) →
2HI(g)
- C(s) + CO2(g) → 2CO(g)
Answer
|
Entropy can be considered the degree of disorder of a chemical system.
It is increased by the number of particles and their temperature.
In this case it is important to examine the number of moles of free
particles on both sides of the equation.
It may be seen that in equation D there are more moles of gas (maximum
entropy) on the right hand side than on the left hand side. Thus the
entropy increases from left to right. correct
response
Although there are more free ions in A this is not as important in
entropy terms as an increase in the number of moles of gas.
In equation B there is a large decrease in entropy (two gases make
a solid) and in equation C the number of moles of gas on both sides
is equal.
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Q461-02 In which of the following
reactions is the entropy change ( S) closest to zero
- SO2(g) + ½O2(g) →
SO3(g)
- Br2(l) → Br2(g)
- H2(g) + I2(g) →
2HI(g)
- 3Ca(s) + N2 → Ca3N2(s)
Answer
|
Entropy can be considered the degree of disorder of a chemical system.
It is increased by the number of particles and their temperature.
In this case it is important to examine the number of moles of free
particles, i.e. gas, on both sides of the equation.
Equation A the moles of gas decreases
from reactants to products, ΔS is negative.
Equation B the moles of gas increases
from 0 to 1, ΔS is positive.
Equation C the moles of gas stays the
same from reactants to products, ΔS = 0. correct
response
Equation D the moles of gas decreases
from 1 to 0, ΔS is negative.
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Q461-03 Estimate, without
doing a calculation, the magnitude of the entropy change for the following reaction.
Fe2O3(s) + 2Al(s) 
2Fe(s) + Al2O3(s) |
Answer
|
Examination of the equation reveals that the compounds on both sides
of the equation are in the solid state. As solids have very low entropy
it is safe to estimate that the entropy difference between reactants
and products is negligible. Hence ΔS = 0.
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Q461-04 Consider the following
reaction:
The absolute entropy values, S, at 300K for N2(g), H2(g)
and NH3(g) are 193, 131 and 192 JK-1 mol-1
respectively. Calculate ΔSo
for the reaction and explain the sign of So.
Answer
|
On the left hand side there is one mole of nitrogen and three moles
of hydrogen. Their entropy = 193 + (3 x 131) = 586 JK-1
On the right hand side there are two moles of ammonia. Entropy =
(2 x 192) = 384 JK-1
The entropy change, ΔSo,
is 384 - 586 = -202 JK-1
The negative sign indicates that the entropy has decreased from reactants
to products.
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Q461-05 Which reaction has
the greatest positive entropy change?
- CH4(g) + 1½O2(g) →
CO(g) + 2H2O(g)
- CH4(g) + 1½O2(g) →
CO(g) + 2H2O(l)
- CH4(g) + 2O2(g) →
CO2(g) + 2H2O(g)
- CH4(g) + 2O2(g) →
CO2(g) + 2H2O(l)
Answer
|
A positive entropy change means that the products have more entropy
than the reactants. Gases have the largest entropy values, therefore
we are looking for the reaction that produces the greatest positive
change in moles of gas.
reaction 1 2½ moles gas →
3 moles of gas. An increase by ½ mole gas correct
response
reaction 2 2½ moles gas →
1 mole of gas. A decrease of 1½ mole gas
reaction 3 3 moles gas →
3 moles of gas. No change in moles
reaction 4 3 moles gas →
1 mole of gas. A decrease of 2 moles of gas
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Q461-06 Which reaction occurs
with the largest increase in entropy?
- Pb(NO3)2(s) + 2KI(s) →
PbI2(s) + 2KNO3(s)
- CaCO3(s) → CaO(s) + CO2(g)
- 3H2(g) + N2(g) →
2NH3(g)
- H2(g) + I2(g) →
2HI(g)
Answer
|
An increase in entropy change means that the products have more entropy
than the reactants. Gases have the largest entropy values, therefore
we are looking for the reaction that produces the greatest positive
change in moles of gas.
reaction 1 0 moles gas →
0 moles of gas. No change in moles of gas
reaction 2 0 moles gas →
1 mole of gas. A increase of 1 mole of gas correct
response
reaction 3 4 moles gas →
2 moles of gas. A decrease of 2 moles of gas
reaction 4 2 moles gas →
2 mole of gas. No change in moles of gas
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Q461-07 Some chlorine gas
is placed in a flask of fixed volume at room temperature. What change will cause
a decrease in entropy?
- Adding a small amount of hydrogen
- Adding a small amount of chlorine
- Cooling the flask
- Exposing the flask to sunlight
Answer
|
Anything that increases the disorder, such as mixing two gases, or
increasing the temperature, increases the entropy. The reverse is
also tru. Hence decreasing the temperature decreases the entropy,
e.g. Cooling the flask
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Q461-08 Which reaction has
the largest positive value of ΔS
o?
- CO2(g) + 3H2(g) →
CH3OH(g) + H2O(g)
- 2Al(s) + 3S(s) → Al2S3(s)
- CH4(g) + H2O(g) →
3H2(g) + CO(g)
- 2S(s) + 3O2(g) → 2SO3(g)
Answer
|
An increase in entropy change means that the products have more entropy
than the reactants. Gases have the largest entropy values, therefore
we are looking for the reaction that produces the greatest positive
change in moles of gas.
reaction 1 4 moles gas →
2 moles of gas. A decrease of 2 moles of gas, ΔSo
= negative
reaction 2 0 moles gas →
0 mole of gas. No change in moles of gas, ΔSo
= 0 (approx)
reaction 3 2 moles gas →
4 moles of gas. An increase of 2 moles of gas, ΔSo
= positive correct response
reaction 4 3 moles gas →
2 mole of gas. A decrease of 1 mole of gas, ΔSo
= negative
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Q461-09 Which equation represents
a change with a negative value for ΔS?
- 2H2(g) + O2(g) →
2H2O(g)
- H2O(s) → H2O(g)
- H2(g) + Cl2(g) →
2HCl(g)
- 2NH3(g) → N2(g)
+ 3H2(g)
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A negative value for ΔS means that the products have less entropy
than the reactants. There are fewer moles of gas in the products than
in the reactants.
reaction 1 3 moles gas →
2 moles of gas. A decrease of 1 moles of gas, ΔS = negative
correct response
reaction 2 0 moles gas →
1 mole of gas. An increase by 1 mole of gas, ΔS = positive
reaction 3 2 moles gas →
2 moles of gas. No change in moles of gas , ΔS = 0 (approx)
reaction 4 2 moles gas →
4 mole of gas. An increase by 2 moles of gas, ΔS = positive
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Q461-10 Which change does
not lead to an increase in entropy?
- Mixing nitrogen and oxygen gases at room temperature
- Cooling steam so that it condenses to water
- Heating hexane to its boiling point
- Dissolving sugar in water
Answer
|
Entropy is increased by:
- Temperature increase
- Increased number of free particles
- Mixing
From the choices given, only cooling steam
reduces the entropy of the system
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