Hydrogen-oxygen fuel cells

1. Overview

Hydrogen-oxygen fuel cells are electrochemical cells that convert the chemical energy of a fuel (hydrogen) and an oxidizing agent (oxygen) directly into electrical energy through chemical reactions. They represent a significant advancement in "green" technology because they provide a clean alternative to the internal combustion engine.

Key Definitions

• Fuel Cell: An electrochemical device that continuously converts chemical energy into electrical energy as long as fuel and oxidant are supplied.
• Electrolyte: A substance (often a liquid or gel) that allows ions to move between the electrodes but does not conduct electrons.
• Exothermic Reaction: A reaction that releases energy to the surroundings; the reaction inside a fuel cell is highly exothermic, but the energy is harnessed as electricity.

Core Content

A hydrogen-oxygen fuel cell uses hydrogen gas and oxygen gas to generate an electric current. Unlike a battery, which stores a finite amount of energy, a fuel cell will produce electricity as long as it has a steady supply of these gases.

How it works:

• Hydrogen gas is supplied to the negative electrode (anode).
• Oxygen gas is supplied to the positive electrode (cathode).
• The overall chemical reaction produces water as the only chemical product.
• This process is "clean" because no carbon dioxide ($CO_2$), nitrogen oxides, or sulfur dioxide are emitted at the point of use.

Equations for the overall reaction:

• Word Equation: hydrogen + oxygen → water
• Symbol Equation: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

Extended Content (Extended Only)

When comparing hydrogen-oxygen fuel cells to traditional gasoline (petrol) engines in vehicles, there are several trade-offs regarding efficiency and practicality.

Advantages of Fuel Cells (vs. Gasoline Engines):

• Environmental Impact: Fuel cells produce only water ($H_2O(l)$), whereas gasoline engines produce greenhouse gases like carbon dioxide ($CO_2(g)$), and pollutants like carbon monoxide ($CO(g)$) and nitrogen oxides ($NO_x(g)$).
• Efficiency: Fuel cells are more efficient at converting chemical energy into kinetic energy (movement) than internal combustion engines, which lose a lot of energy as heat.
• Maintenance: Fuel cells have no moving parts, making them quieter and potentially more reliable than complex mechanical engines.

Disadvantages of Fuel Cells (vs. Gasoline Engines):

• Storage and Transport: Hydrogen is a gas at room temperature and has a low energy density by volume. It must be stored under very high pressure in heavy tanks, which takes up significant space in a vehicle.
• Infrastructure: There is a widespread network of petrol stations globally, but very few hydrogen refueling stations currently exist.
• Production: While the fuel cell itself is clean, most hydrogen is currently produced from fossil fuels (methane), which releases $CO_2$. Producing hydrogen by electrolysis of water is expensive and requires large amounts of electricity.
• Safety: Hydrogen is highly flammable and difficult to detect if a leak occurs.

Key Equations

Type	Equation
Overall Reaction	$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

Components of the equation:

• $H_2$: Hydrogen gas (fuel)
• $O_2$: Oxygen gas (oxidant)
• $H_2O$: Water (product)
• $(g)$: Gas state symbol
• $(l)$: Liquid state symbol

Common Mistakes to Avoid

• ❌ Wrong: Stating that hydrogen fuel cells produce "smoke" or "exhaust gases."
• ✓ Right: Stating that water is the only chemical product of the reaction.
• ❌ Wrong: Forgetting state symbols in the equation ($2H_2 + O_2 \rightarrow 2H_2O$).
• ✓ Right: Always include $(g)$ for the reactants and $(l)$ or $(g)$ for the water product: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$.
• ❌ Wrong: Thinking that fuel cells "run out" of charge like a battery.
• ✓ Right: Fuel cells work continuously as long as fuel (hydrogen) is supplied.

Exam Tips

• Command Words: If an exam question asks you to "State" the product, simply write "water." If it asks you to "Describe the advantages," ensure you mention both the environmental aspect (no $CO_2$) and the efficiency aspect.
• Contextual Questions: You may be asked why fuel cells are used in spacecraft. The answer is that they provide both electricity and a source of drinking water for astronauts.
• State Symbols: Examiners frequently award a specific mark just for the correct state symbols $(g)$ and $(l)$ in this topic.
• The "Only" Product: When describing the reaction, always emphasize that water is the only product. This is a key phrase in mark schemes to distinguish fuel cells from combustion engines.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) State the reactants and the product of a hydrogen-oxygen fuel cell. [3]

(b) State two advantages of using hydrogen-oxygen fuel cells in vehicles, compared to gasoline/petrol engines. [2]

Worked Solution:

(a)

1. Reactants: Hydrogen and oxygen. Hydrogen and oxygen are consumed in the fuel cell reaction.
2. Product: Water. Water is the only chemical product generated by the fuel cell.

How to earn full marks:

• One mark each for correctly identifying hydrogen and oxygen as the reactants.
• One mark for correctly identifying water as the product.

(b)

1. Advantage 1: Hydrogen-oxygen fuel cells produce only water as a byproduct, whereas gasoline/petrol engines produce greenhouse gases and pollutants. This highlights the environmental benefit of fuel cells.
2. Advantage 2: Hydrogen-oxygen fuel cells are more energy efficient than gasoline/petrol engines. Fuel cells convert a higher percentage of the fuel's energy into useful work.

How to earn full marks:

• One mark for each valid advantage stated.
• Answers must clearly compare the fuel cell to gasoline/petrol engines.

Common Pitfall: Students sometimes confuse the inputs and outputs of a fuel cell. Remember that hydrogen and oxygen are the reactants that go into the fuel cell, and water is the product that comes out. Also, make sure your advantages are clearly comparative to gasoline/petrol engines.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) Describe, in terms of energy transfer, how a hydrogen-oxygen fuel cell generates electricity. [3]

(b) Suggest three reasons why hydrogen-oxygen fuel cells are not yet widely used in vehicles, despite their advantages. [3]

Worked Solution:

(a)

1. Hydrogen and oxygen react chemically. This is the initial chemical reaction.
2. Chemical energy from the reaction is converted directly into electrical energy. This energy conversion is the core principle of the fuel cell.
3. This electrical energy is used to drive an electric motor. The electric motor then propels the vehicle.

How to earn full marks:

• One mark for mentioning the chemical reaction between hydrogen and oxygen.
• One mark for stating that chemical energy is converted to electrical energy.
• One mark for describing the use of electrical energy to drive a motor.

(b)

1. Hydrogen is difficult and expensive to store safely in vehicles. This refers to the challenges of hydrogen storage.
2. The infrastructure for hydrogen refueling is not widely available. This addresses the lack of refueling stations.
3. The production of hydrogen can be energy-intensive and may rely on fossil fuels. This acknowledges the potential environmental impact of hydrogen production.

How to earn full marks:

• One mark for each valid reason suggested.
• Answers must address the practical limitations of fuel cell technology.

Common Pitfall: When discussing energy transfer, be specific about the types of energy involved. Don't just say "energy is transferred"; say "chemical energy is converted to electrical energy." Also, focus on practical reasons for limited fuel cell adoption, such as cost and infrastructure.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A hydrogen-oxygen fuel cell powers a small electric car.

(a) Write the balanced chemical equation for the overall reaction in the hydrogen-oxygen fuel cell. [2]

(b) Describe the process by which hydrogen and oxygen react within the fuel cell to generate electricity. Include the role of the electrolyte and electrodes. [4]

(c) The electric car consumes 0.4 kg of hydrogen in one hour of driving. The energy released by the combustion of 1 kg of hydrogen is 142 MJ. Calculate the power output of the fuel cell in kilowatts (kW), assuming 60% efficiency. [2]

Worked Solution:

(a)

1. $2H_2 + O_2 \rightarrow 2H_2O$ This is the balanced equation showing the formation of water from hydrogen and oxygen.

How to earn full marks:

• One mark for correct formulas of reactants and products.
• One mark for correct balancing.

(b)

1. Hydrogen molecules are supplied to the anode, where they are oxidised, releasing electrons ($H_2 \rightarrow 2H^+ + 2e^-$). Describes the oxidation of hydrogen at the anode.
2. Oxygen molecules are supplied to the cathode, where they are reduced, accepting electrons ($O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$). Describes the reduction of oxygen at the cathode.
3. The electrolyte allows the passage of ions (H+). Explains the function of the electrolyte.
4. The flow of electrons through an external circuit generates electricity. Describes the generation of current.

How to earn full marks:

• One mark for describing the oxidation of hydrogen at the anode.
• One mark for describing the reduction of oxygen at the cathode.
• One mark for mentioning the role of the electrolyte in ion transport.
• One mark for explaining the flow of electrons generating electricity.

(c)

1. Energy released per hour: $0.4 \text{ kg} \times 142 \text{ MJ/kg} = 56.8 \text{ MJ}$. Calculates the total energy released by the consumed hydrogen.
2. Actual energy output (60% efficiency): $56.8 \text{ MJ} \times 0.6 = 34.08 \text{ MJ} = 34.08 \times 10^6 \text{ J}$. Calculates the energy output considering the efficiency.
3. Power output: $P = \frac{E}{t} = \frac{34.08 \times 10^6 \text{ J}}{3600 \text{ s}} = 9466.67 \text{ W} \approx 9.47 \text{ kW}$. Calculates the power using the formula P=E/t.

How to earn full marks:

• One mark for calculating the total energy released (56.8 MJ).
• One mark for calculating the power output with correct units. $\boxed{9.47 \text{ kW}}$

Common Pitfall: Make sure you understand the half-equations occurring at the anode and cathode. Also, pay close attention to units in the power calculation. Remember to convert MJ to J and hours to seconds to get the power in Watts, then convert to kW.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A research team is testing a new hydrogen-oxygen fuel cell for use in powering a remote weather station.

(a) State the type of energy conversion that occurs within a hydrogen-oxygen fuel cell. [1]

(b) The fuel cell operates with an electrolyte that is acidic. Suggest what effect this may have on the materials used in the construction of the fuel cell, and how this could be mitigated. [3]

(c) The fuel cell produces a voltage of 0.8 V and delivers a current of 6.0 A to the weather station's instruments. Calculate the electrical power output of the fuel cell. [2]

(d) The weather station requires a continuous power supply of 4.2 W. Calculate the percentage of the fuel cell's power output that is actually used by the weather station. [1]

(e) Suggest two reasons why the percentage calculated in part (d) is less than 100%. [2]

Worked Solution:

(a)

1. Chemical energy to electrical energy. This is the fundamental energy conversion in a fuel cell.

How to earn full marks:

• One mark for stating the conversion from chemical to electrical energy.

(b)

1. The acidic electrolyte may corrode the metal components of the fuel cell. Acids are corrosive, especially to metals.
2. Use corrosion-resistant materials such as platinum, gold, or specific alloys. These materials are less reactive and more resistant to corrosion.
3. Apply a protective coating to the metal surfaces to prevent contact with the electrolyte. Protective coatings act as a barrier.

How to earn full marks:

• One mark for identifying corrosion as a potential problem.
• One mark for suggesting a corrosion-resistant material.
• One mark for suggesting a protective coating.

(c)

1. Power = Voltage x Current = $0.8 \text{ V} \times 6.0 \text{ A} = 4.8 \text{ W}$. Applies the formula P=VI.

How to earn full marks:

• One mark for using the correct formula (P=VI).
• One mark for the correct answer and unit. $\boxed{4.8 \text{ W}}$

(d)

1. Percentage used = $\frac{\text{Power used}}{\text{Power output}} \times 100 = \frac{4.2 \text{ W}}{4.8 \text{ W}} \times 100 = 87.5%$. Calculates the percentage of power used.

How to earn full marks:

• One mark for the correct calculation and unit. $\boxed{87.5%}$

(e)

1. Energy losses due to heat generation in the fuel cell. Some energy is lost as heat due to internal resistance.
2. Inefficiencies in the voltage conversion or regulation circuitry. Voltage converters and regulators are not perfectly efficient.

How to earn full marks:

• One mark for each valid reason suggested.
• Answers must explain why the power output is not fully utilized.

Common Pitfall: Remember the basic power equation, P=VI. Also, when calculating percentages, make sure you're dividing the smaller value by the larger value to get a percentage less than 100%. Don't forget the percentage sign (%) in your final answer.