Ohmic Voltage Loss Calculation

To determine the ohmic voltage loss in the fuel cell, we need to calculate the total resistance and then use Ohm's Law to find the voltage loss.

1. Calculate the Membrane Resistance ($R_{\text{membrane}}$)

The resistance of the membrane can be calculated using the formula:

Where:

• $d$ is the membrane thickness (in cm)
• $\sigma$ is the conductivity (in $\Omega^{-1} \text{cm}^{-1}$)
• $A$ is the area (in $\text{cm}^2$)

Given:

• $d = 100 \, \mu\text{m} = 0.01 \, \text{cm}$
• $\sigma = 0.10 \, \Omega^{-1} \text{cm}^{-1}$
• $A = 10 \, \text{cm}^2$

Plugging in the values:

2. Calculate the Total Resistance ($R_{\text{total}}$)

Add the electronic resistance:

3. Calculate the Total Current ($I$)

Using the current density and area:

4. Calculate the Ohmic Voltage Loss ($V_{\text{ohmic}}$)

Using Ohm's Law:

Answer: 0.15 V

To calculate the total number of moles of hydrogen gas (H_2) consumed by the fuel cell during its operation, we'll use Faraday's laws of electrolysis, which relate the amount of substance consumed or produced at an electrode to the quantity of electricity passed through the electrolyte.

1. Calculate the Total Charge Passed During Each Period

First Period (1 hour at 2 A):

• Time ($t_1$) = 1 hour = 3600 seconds
• Current ($I_1$) = 2 A
• Charge ($Q_1$) = Current × Time

Second Period (2 hours at 5 A):

• Time ($t_2$) = 2 hours = 7200 seconds
• Current ($I_2$) = 5 A
• Charge ($Q_2$) = Current × Time

Total Charge Passed:

2. Relate Charge to Moles of Electrons

Using Faraday's constant ($F = 96,485 \, \text{C/mol e}^-$):

3. Determine Moles of Hydrogen Consumed

In a hydrogen fuel cell, the oxidation of hydrogen at the anode is:

This means that 1 mole of H_2 produces 2 moles of electrons. Therefore:

Answer: Approximately 0.224 moles of hydrogen gas are consumed.

To calculate the oxygen output flux and the hydrogen fuel input flux required to produce a current of 1000 kA (1,000,000 A) in a hydrogen-air fuel cell supplied with air at 20 moles per second, we'll use the given relation and stoichiometry of the fuel cell reactions.

Given:

• I (Current) = 1,000,000 A
• Air supply rate = 20 moles/s (which contains oxygen)
• Relation: $I = n F v$
  • I = Current in Amperes (Coulombs per second)
  • n = Number of electrons transferred per mole of reactant
  • F = Faraday's constant ($96,485 \, \text{C/mol}$)
  • v = Rate of consumption of reactant (moles per second)

Step 1: Determine the Moles of Oxygen in Air Supply

Air is composed of approximately 21% oxygen by volume (and mole fraction). Therefore:

Step 2: Calculate the Required Hydrogen Input Flux ($v_{\text{H}_2}$)

Anode Reaction (Hydrogen Oxidation):

• n = 2 electrons per mole of H_2

Using the relation:

Step 3: Calculate the Required Oxygen Consumption Flux ($v_{\text{O}_2,\text{consumed}}$)

Cathode Reaction (Oxygen Reduction):

• n = 4 electrons per mole of O_2 (since O_2 is diatomic and accepts 4 electrons)

Using the relation:

Step 4: Calculate the Oxygen Output Flux ($v_{\text{O}_2,\text{out}}$)

Final Answers:

• Hydrogen Fuel Input Flux: $v_{\text{H}_2} \approx 5.182 \, \text{mol/s}$
• Oxygen Output Flux: $v_{\text{O}_2,\text{out}} \approx 1.609 \, \text{mol/s}$

Summary:

To produce a current of 1000 kA:

• The fuel cell must consume approximately 5.182 moles per second of hydrogen.
• The fuel cell consumes approximately 2.591 moles per second of oxygen from the supplied air.
• The unused oxygen exits the fuel cell at a rate of approximately 1.609 moles per second.

Answer: The oxygen output flux is approximately 1.609 mol/s, and the hydrogen fuel input flux is approximately 5.182 mol/s to produce a current of 1000 kA.

To calculate the rate of water production in grams per hour by the 20-cell fuel cell stack, we'll follow these steps:

Given Data:

• Number of cells in series ($N$): 20
• Active area per electrode ($A$): 100 cm^2
• Current density ($J$): 0.8 A/cm^2
• Time ($t$): 1 hour (3600 seconds)
• Faraday's constant ($F$): 96,485 C/mol $\text{e}^-$
• Molar mass of water ($M_{\text{H}_2\text{O}}$): 18.015 g/mol

Step 1: Calculate the Total Current

Current per cell ($I_{\text{cell}}$):

Since the cells are connected in series, the total current ($I_{\text{total}}$) remains the same:

Step 2: Calculate the Total Charge Passed in One Hour

Step 3: Calculate the Moles of Electrons Transferred

Step 4: Determine the Moles of Water Produced

In the fuel cell reaction, the formation of water involves electron transfer:

Overall Reaction:

• Electrons transferred per mole of $\text{H}_2\text{O}$: 2 mol e^-

Step 5: Calculate the Mass of Water Produced

Answer:

The rate of water production by the 20-cell fuel cell stack is approximately 26.87 grams per hour.

Final Answer: Approximately 27 grams per hour of water are produced by the fuel cell stack.

To calculate the energy released per mole of water ($\text{H}_2\text{O}$) formed in the reaction:

we will use the concept of bond enthalpies. The energy change in a reaction ($\Delta H_{\text{rxn}}$) can be calculated using the bond energies of the reactants and products:

Given Bond Enthalpies:

• H–H bond: $432 \, \text{kJ/mol}$
• O=O bond: $494 \, \text{kJ/mol}$
• H–O bond: $460 \, \text{kJ/mol}$

Step 1: Identify Bonds Broken and Formed

Bonds Broken (Reactants):

1. One H–H bond in $\text{H}_2$
2. Half an O=O bond in $\frac{1}{2}\text{O}_2$

Bonds Formed (Products):

1. Two H–O bonds in $\text{H}_2\text{O}$ (since each water molecule has two H–O bonds)

Step 2: Calculate Energy Required to Break Bonds

Step 3: Calculate Energy Released in Forming Bonds

Step 4: Calculate the Net Energy Change

The negative sign indicates that the reaction is exothermic (energy is released).

Answer:

The energy released per mole of water produced is approximately 241 kJ.

Final Answer: Approximately 241 kJ of energy is released per mole of water formed in the reaction H_2 + O_2 → H_2O.

To determine which interactions are involved in the charge and mass balance process during fuel cell operation, especially in a Proton Exchange Membrane Fuel Cell (PEMFC), let's analyze each option.

Understanding the Components:

1. Sulfonate Groups (-SO_3^-):

   • Fixed anionic sites on the proton exchange membrane (e.g., Nafion).
   • Play a crucial role in facilitating proton conduction.

2. Hydronium Ions (H_3O^+):

   • Protons (H^+) associate with water molecules to form hydronium ions.
   • Essential carriers of positive charge through the membrane.

3. Water Molecules (H_2O):

   • Act as a medium for proton transport via the Grotthuss mechanism.
   • Participate in hydration and facilitate proton hopping.

4. Perfluorinated Backbone (e.g., CF_2 Groups):

   • Hydrophobic component of the membrane polymer.
   • Generally inert in proton conduction.

Analyzing Each Interaction:

a. Sulfonate-H_3O^+ Interaction

• Explanation:
  • Sulfonate groups are negatively charged and attract hydronium ions.
  • This interaction anchors hydronium ions, aiding in proton conductivity.
• Conclusion:
  • Yes, this interaction is crucial for charge balance.

b. H_2O-H_3O^+ Interaction

• Explanation:
  • Proton transfer occurs between hydronium and water molecules.
  • Facilitates the Grotthuss mechanism for efficient proton transport.
• Conclusion:
  • Yes, essential for both charge and mass transport.

c. H_2O-H_2O Interaction

• Explanation:
  • Hydrogen bonding between water molecules forms a conductive network.
  • Supports the movement of protons and maintains membrane hydration.
• Conclusion:
  • Yes, vital for maintaining the proton conduction pathway.

d. Sulfonate-H_2O Interaction

• Explanation:
  • Sulfonate groups interact with water, enhancing membrane hydration.
  • A hydrated membrane improves proton conductivity.
• Conclusion:
  • Yes, important for mass balance and membrane functionality.

e. CF_2-H_3O^+ Interaction

• Explanation:
  • CF_2 groups are part of the hydrophobic backbone of the membrane polymer.
  • They do not interact significantly with hydronium ions.
• Conclusion:
  • No, this interaction is negligible in charge and mass balance.

Final Decision:

• Interactions Involved:
  • All interactions except for (e) are involved in the charge and mass balance process during fuel cell operation.

Answer:

f. All except (e)

Answer: The correct option is f. All except (e).

Answer: b. All of the above

Explanation:

In a proton exchange membrane fuel cell (PEMFC), several key properties and mechanisms enable efficient operation:

1. Presence of H^+ Ions with Adequate Available Sites in Its Backbone Structure for Mobility (Option a):

   • The proton exchange membrane (often made of Nafion) contains sulfonic acid groups (-SO_3H) that can dissociate to release H^+ ions.
   • These H^+ ions are mobile and move through the membrane via the Grotthuss mechanism.
   • The membrane's backbone provides sites that facilitate proton conduction.

2. Adequate Porosity Permitting Ionic Mobility (Option c):

   • While the membrane is not porous in the traditional sense, it has a microstructure with hydrophilic channels formed by the aggregation of sulfonic acid groups.
   • These channels are filled with water and allow H^+ ions to move efficiently.
   • This internal "porosity" or network of channels is crucial for ionic mobility.

3. Selective Ion Permeation Properties Permitting Only H^+ Ion Mobility (Option d):

   • The membrane is selectively permeable to protons (H^+ ions) and blocks other ions.
   • This selectivity ensures that only protons are transported from the anode to the cathode, maintaining the efficiency of the cell.

4. Block Electronic Charge Mobility (Option e):

   • The membrane acts as an electrical insulator for electrons.
   • By blocking electronic conductivity, it prevents short circuits and ensures that electrons travel through the external circuit, generating electrical work.

5. Hydrophilic Properties (Option f):

   • The sulfonic acid groups confer hydrophilic characteristics to the membrane.
   • Hydrophilicity allows the membrane to absorb and retain water, which is essential for maintaining proton conductivity.
   • Adequate hydration is necessary to keep the proton-conducting channels functional.

Given that all the listed properties are true and essential for the operation of a PEMFC, the correct choice is:

b. All of the above

Answer: a. Alkaline fuel cell (AFC)

Explanation:

The reversible (thermodynamic) efficiency of a fuel cell is determined by the ratio of the Gibbs free energy change ($\Delta G$) to the enthalpy change ($\Delta H$) of the electrochemical reaction:

This ratio decreases with increasing temperature because $\Delta G$ becomes less negative due to the $T\Delta S$ term in the Gibbs free energy equation ($\Delta G = \Delta H - T\Delta S$).

Fuel cells operating at lower temperatures have higher reversible efficiencies because the Gibbs free energy change is a larger fraction of the enthalpy change at lower temperatures.

Let's analyze each fuel cell type based on their operating temperatures:

1. Alkaline Fuel Cell (AFC)

   • Operating Temperature: Low (~60–90°C)
   • Fuel and Oxidant: Hydrogen and Oxygen
   • Characteristics:
     • High electrochemical reaction rates due to favorable kinetics in alkaline electrolytes.
     • Lower activation overpotentials.
     • Highest reversible efficiency due to low operating temperature and efficient hydrogen oxidation.

2. Phosphoric Acid Fuel Cell (PAFC)

   • Operating Temperature: Medium (~150–220°C)
   • Efficiency: Lower reversible efficiency than AFC due to higher operating temperature.

3. Molten Carbonate Fuel Cell (MCFC)

   • Operating Temperature: High (~600–700°C)
   • Efficiency: Lower reversible efficiency because high temperatures decrease $\Delta G/\Delta H$.

4. Proton Exchange Membrane Fuel Cell (PEMFC)

   • Operating Temperature: Low (~60–80°C)
   • Fuel and Oxidant: Hydrogen and Oxygen
   • Characteristics:
     • High efficiency at low temperatures.
     • Slightly lower reversible efficiency compared to AFC due to different electrolyte properties and possible issues with water management.

5. Solid Oxide Fuel Cell (SOFC)

   • Operating Temperature: Very High (~700–1000°C)
   • Efficiency: Lower reversible efficiency due to high operating temperatures.

6. Direct Methanol Fuel Cell (DMFC)

   • Operating Temperature: Low (~60–120°C)
   • Fuel: Methanol
   • Characteristics:
     • Uses liquid methanol as fuel, which has lower electrochemical efficiency than hydrogen.
     • Lower reversible efficiency compared to hydrogen fuel cells (AFC and PEMFC).

Conclusion:

• AFCs have the highest reversible efficiency among the listed options due to their low operating temperature and efficient hydrogen oxidation in an alkaline environment.
• PEMFCs also operate at low temperatures but may have slightly lower reversible efficiency than AFCs due to differences in electrolyte properties and water management challenges.
• MCFCs and SOFCs operate at high temperatures, resulting in lower reversible efficiencies.

Therefore, the correct answer is:

a. Alkaline fuel cell (AFC)

Answer: f. Only (d)

Explanation:

In a fuel cell, the origin of energy and its subsequent release for applications involve several key processes:

1. Redox Reactions:

   • Oxidation-reduction (redox) reactions occur at the electrodes.
   • At the anode, the fuel (e.g., hydrogen) is oxidized, releasing electrons.
   • At the cathode, the oxidant (e.g., oxygen) is reduced, accepting electrons.

2. Resulting Potential Difference:

   • The redox reactions create a potential difference (voltage) between the anode and cathode.
   • This potential difference is due to the difference in the chemical potential of the reactants and products.

3. Free Energy Change ($\Delta G$):

   • The Gibbs free energy change of the reaction represents the maximum useful work obtainable from the process.
   • In a fuel cell, this free energy is directly converted into electrical energy.

4. Electrical Work Done by Charges:

   • Electrons flow from the anode to the cathode through an external circuit, performing electrical work.
   • The movement of charges generates an electric current that can be harnessed for applications.

5. Direct Energy Conversion:

   • Fuel cells convert chemical energy directly into electrical energy without intermediate steps (e.g., combustion or thermal processes).
   • This direct conversion increases efficiency and reduces energy losses.

Option Analysis:

• Option (a): Mentions energy storage, which is not a primary function of fuel cells. Fuel cells are energy converters, not storage devices.
• Option (b): Partially correct but lacks mention of potential difference and electrical work.
• Option (c): Includes energy conversion and release but still omits the critical aspect of potential difference and the role of charge movement.
• Option (d): Accurately encompasses all essential aspects:
  • Redox action
  • Resulting potential difference
  • Free energy change
  • Electrical work done by charges
  • Direct energy conversion
• Options (e) and (g): Include options that are incomplete or contain inaccuracies.
• Option (f): Correctly identifies that only option (d) fully describes the origin and release of energy in a fuel cell.

Therefore, the most comprehensive and accurate description is provided by option (d), and the correct choice is f. Only (d).

Answer: g. All of the above, i.e., (a) to (e)

Explanation:

Optimizing the output current density in a fuel cell involves controlling various parameters that enhance the rate of electrochemical reactions and improve overall cell performance. Here's how each option contributes:

a. Increasing Reactant Concentration

• Effect: Higher concentrations of fuel (e.g., hydrogen) and oxidant (e.g., oxygen) increase the availability of reactants at the electrode surfaces.
• Result: Enhances the reaction rate according to the principles of chemical kinetics, leading to higher current densities.

b. Increasing Reaction Temperature

• Effect: Elevated temperatures increase the kinetic energy of molecules.
• Result: Speeds up the electrochemical reactions, reduces activation losses, and improves ion conductivity within the electrolyte.

c. Lowering the Activation Barrier via Catalyst

• Effect: Catalysts reduce the activation energy required for the reactions at the electrodes.
• Result: Increases the reaction rate without being consumed in the process, leading to higher current outputs.

d. Scaling Up the Number of Reaction Sites Employing High Surface Area Electrodes

• Effect: Using electrodes with larger surface areas provides more active sites for the electrochemical reactions.
• Result: Enhances the overall reaction rate and current density by allowing more simultaneous reactions.

e. Employing Mixed Conducting 3D Structural Reaction Interfaces

• Effect: Materials that conduct both ions and electrons (mixed conductors) in a three-dimensional structure facilitate better transport properties.
• Result: Improves the efficiency of charge transfer and mass transport within the fuel cell, leading to higher current densities.

Since all the options (a) to (e) are valid methods to optimize the output current density in a fuel cell, the correct choice is:

g. All of the above, i.e., (a) to (e)

Answer: g. All of the above

Explanation:

The reversible voltage of a fuel cell, also known as the open-circuit voltage (OCV), is determined by the Nernst equation, which relates the cell potential to the standard electrode potential and the activities (or effective concentrations) of the reactants and products involved in the electrochemical reaction. The Nernst equation is given by:

Where:

• $E$ = Reversible cell voltage
• $E^\circ$ = Standard cell potential
• $R$ = Universal gas constant ($8.314\, \text{J/mol·K}$)
• $T$ = Absolute temperature (in Kelvin)
• $n$ = Number of electrons transferred in the reaction
• $F$ = Faraday's constant ($96,485\, \text{C/mol}$)
• $Q$ = Reaction quotient, which depends on the activities (or partial pressures) of the reactants and products

Let's analyze each option to see how it influences the reversible voltage:

a. Activity Coefficient of Fuel (H_2) Supply

• Impact: The activity coefficient adjusts the effective concentration (activity) of hydrogen gas. In non-ideal conditions, the activity coefficient accounts for deviations from ideal behavior, affecting the reaction quotient $Q$ in the Nernst equation.
• Conclusion: It directly influences the reversible voltage.

b. Activity Coefficient of Oxidant (O_2) Supply

• Impact: Similar to hydrogen, the activity coefficient of oxygen adjusts its effective concentration. It affects the reaction quotient $Q$ and thus the cell voltage.
• Conclusion: It directly influences the reversible voltage.

c. Operating Temperature (T) of Fuel Cell

• Impact: Temperature appears explicitly in the Nernst equation in the term $\frac{RT}{nF}$. Changes in temperature alter this term, affecting the cell voltage.
• Conclusion: It directly influences the reversible voltage.

d. Partial Pressure of Fuel & Oxidant

• Impact: The partial pressures of hydrogen and oxygen gases determine their activities when considering ideal gases (activity ≈ partial pressure). They are components of the reaction quotient $Q$ in the Nernst equation.
• Conclusion: They directly influence the reversible voltage.

e. Faraday Coefficient and Number of Mobile Charge(s)

• Impact:
  • Faraday's Constant ($F$): Represents the charge per mole of electrons. It's a fundamental constant in the Nernst equation.
  • Number of Electrons Transferred ($n$): Represents the number of electrons involved in the electrochemical reaction. It affects the magnitude of the voltage change per unit activity change.
• Conclusion: Both $F$ and $n$ directly influence the reversible voltage through the Nernst equation.

Since all the factors listed from (a) to (e) are components that the reversible voltage depends on, the correct choice is:

g. All of the above

Answer:

The correct sequence of processes in the operation of a fuel cell is:

1. Fuel Transport
2. Oxidant Transport
3. Catalytic Action
4. Electrochemical Reaction
5. Charge (Ion and Electron) Transport
6. Energy Release
7. Product Removal

Explanation:

Fuel cells operate through a series of interconnected processes that convert chemical energy directly into electrical energy. Here's the detailed sequence:

1. Fuel Transport (Step 1):

   • Process: The fuel (e.g., hydrogen gas) is delivered to the anode side of the fuel cell.
   • Importance: Adequate fuel supply is essential for continuous operation and maintaining the desired power output.

2. Oxidant Transport (Step 2):

   • Process: The oxidant (e.g., oxygen from air) is supplied to the cathode side of the fuel cell.
   • Importance: The availability of the oxidant is crucial for the reduction reaction at the cathode.

3. Catalytic Action (Step 3):

   • Process: Catalysts at both the anode and cathode facilitate the electrochemical reactions.
     • Anode Catalyst: Accelerates the oxidation of the fuel, releasing electrons.
     • Cathode Catalyst: Enhances the reduction of the oxidant by accepting electrons.
   • Importance: Catalysts lower the activation energy, increasing reaction rates and fuel cell efficiency.

4. Electrochemical Reaction (Step 4):

   • Process: Chemical reactions occur at the electrodes.
     • Anode Reaction (Oxidation): Fuel molecules lose electrons. $$\text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^-$$
     • Cathode Reaction (Reduction): Oxidant molecules gain electrons. $$\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}$$
   • Importance: These reactions are the heart of the fuel cell's operation, converting chemical energy into electrical energy.

5. Charge (Ion and Electron) Transport (Step 5):

   • Process:
     • Electron Transport: Electrons released at the anode travel through the external circuit to the cathode, creating an electric current that can perform work.
     • Ion Transport: Ions (e.g., protons, $\text{H}^+$) move through the electrolyte from the anode to the cathode.
   • Importance: The movement of charges maintains electrical neutrality and allows continuous operation.

6. Energy Release (Step 6):

   • Process: The flow of electrons through the external circuit releases electrical energy.
   • Importance: This is the usable energy output of the fuel cell, harnessed to power devices or systems.

7. Product Removal (Step 7):

   • Process: Reaction products (e.g., water in a hydrogen fuel cell) are removed from the electrode sites.
   • Importance: Removing products prevents blockage of active sites, maintaining efficiency and preventing damage.

Summary:

The correct sequence ensures the continuous and efficient operation of the fuel cell by:

• Supplying reactants (fuel and oxidant).
• Facilitating reactions through catalysts.
• Conducting electrochemical reactions to generate charges.
• Transporting charges to produce electrical energy.
• Releasing energy for external use.
• Removing products to sustain ongoing reactions.

Note: The energy release (Step 6) occurs as a direct result of the charge transport (Step 5) because the movement of electrons through an external load generates electrical power. Product removal (Step 7) is essential after energy release to maintain the reaction sites for ongoing operation.

Therefore, the correct sequence is:

1. Fuel transport
2. Oxidant transport
3. Catalytic action
4. Electrochemical reaction
5. Charge (ion and electron) transport
6. Energy release
7. Product removal

Answer: f. (a), (c), and (e)

Explanation:

The power and capacity rating of a fuel cell are determined by several key parameters. Let's analyze each option to understand how they contribute to the fuel cell's performance.

Understanding Fuel Cell Power and Capacity:

1. Power Rating:

   • Definition: The maximum electrical power output a fuel cell can deliver.
   • Determined by:
     • Size of the Cell: Larger active electrode area allows for higher current densities, increasing power output.
     • Redox Reaction Kinetics: Faster reaction rates at the electrodes enhance current generation.
     • Rate of Fuel and Oxidant Flow: Adequate supply ensures that the reactants are available at the required rates to sustain high power output.

2. Capacity Rating:

   • Definition: The total amount of energy a fuel cell can produce before refueling.
   • Determined by:
     • Size of the Fuel Reservoir: A larger reservoir holds more fuel, allowing the fuel cell to operate longer before needing a refill.

Analyzing Each Option:

Option a. Size of cell and fuel reservoir respectively

• Explanation:
  • Size of Cell: Determines the power rating because a larger cell can generate more power due to increased active area.
  • Fuel Reservoir Size: Determines the capacity rating as it affects how long the fuel cell can operate before refueling.
• Conclusion: Correct.

Option b. Size of fuel reservoir and cell respectively

• Explanation:
  • This reverses the logical relationship.
  • Fuel Reservoir Size: While important for capacity, it doesn't directly determine power.
  • Size of Cell: Influences power output, not capacity.
• Conclusion: Incorrect.

Option c. Redox action kinetics only

• Explanation:
  • Redox Reaction Kinetics: Faster kinetics enhance the rate at which electrons are transferred, increasing power output.
  • However, it doesn't affect capacity unless it impacts fuel utilization efficiency significantly.
• Conclusion: Partially correct (affects power but not capacity).

Option d. Catalytic action at interconnect metal-electrode interface

• Explanation:
  • Interconnects: Typically serve to electrically connect cells in a stack and are designed to have minimal resistance.
  • Catalytic Action at Interconnects: Not a significant factor in power or capacity rating.
• Conclusion: Incorrect.

Option e. Rate of fuel and oxidant flow

• Explanation:
  • Rate of Flow: Directly affects the power rating by ensuring that reactants are supplied at rates matching the demand for electricity.
  • Insufficient flow rates can limit the maximum achievable power.
• Conclusion: Correct.

Option f. (a), (c), and (e)

• Combines:
  • (a): Correct for power (size of cell) and capacity (size of fuel reservoir).
  • (c): Correct for power (redox kinetics).
  • (e): Correct for power (rate of fuel and oxidant flow).
• Conclusion: This option includes all correct parameters affecting power and capacity ratings.

Option g. (b), (d), and (e)

• Combines:
  • (b): Incorrect relationship between size and ratings.
  • (d): Incorrect, as catalytic action at interconnects doesn't significantly impact ratings.
  • (e): Correct for power rating.
• Conclusion: This option includes incorrect parameters.

Final Conclusion:

The parameters that determine fuel cell power and capacity rating are:

• (a): Size of cell and fuel reservoir respectively
• (c): Redox action kinetics only
• (e): Rate of fuel and oxidant flow

Therefore, the correct answer is:

f. (a), (c), and (e)

Answer: f. Fuel cell efficiency is lower than that of the internal combustion engines (ICEs)

Explanation:

In the context of fuel cells, most of the statements provided are correct except for option (f). Let's analyze each option to understand why option (f) is not correct.

a. Fuel cell has the potential to substitute and replace fossil fuel

• Explanation:
  • Fuel cells can use hydrogen or other fuels derived from renewable sources.
  • They offer a clean alternative to fossil fuels in transportation, stationary power generation, and portable power applications.
  • Conclusion: Correct statement.

b. Fuel cells produce DC power

• Explanation:
  • Fuel cells generate direct current (DC) electricity as a result of electrochemical reactions.
  • The DC power can be used directly or converted to alternating current (AC) for various applications.
  • Conclusion: Correct statement.

c. Fuel cell is a clean and green energy alternative

• Explanation:
  • Fuel cells produce electricity with minimal pollutants.
  • When hydrogen is used as fuel, the only byproduct is water.
  • They reduce greenhouse gas emissions compared to fossil fuel-based technologies.
  • Conclusion: Correct statement.

d. Electrochemical action is the root of fuel cell working

• Explanation:
  • Fuel cells operate based on electrochemical reactions where chemical energy is directly converted into electrical energy.
  • No combustion process is involved.
  • Conclusion: Correct statement.

e. Fuel cell action complies with the principles of reversible thermodynamics

• Explanation:
  • Fuel cells are governed by thermodynamic principles, particularly the concepts of Gibbs free energy and reversible processes.
  • They aim to operate as close as possible to the reversible (ideal) conditions to maximize efficiency.
  • Conclusion: Correct statement.

f. Fuel cell efficiency is lower than that of the internal combustion engines (ICEs)

• Explanation:
  • Fuel cells generally have higher efficiencies compared to internal combustion engines.
  • Fuel Cell Efficiency:
    • Typically ranges from 40% to 60% for fuel cells operating on hydrogen.
    • Can be even higher (up to 85%) when waste heat is utilized in combined heat and power (CHP) systems.
  • Internal Combustion Engine Efficiency:
    • Usually around 20% to 35% for gasoline and diesel engines.
    • Limited by the Carnot efficiency due to thermal conversion processes.
  • Fuel cells are not constrained by the Carnot cycle since they convert chemical energy directly to electrical energy without combustion.
  • Conclusion: Incorrect statement.

g. Fuel cell operating principle is like that of the batteries

• Explanation:
  • Both fuel cells and batteries convert chemical energy into electrical energy through electrochemical reactions.
  • Similarities:
    • Involve oxidation and reduction reactions.
    • Have electrodes (anode and cathode) and an electrolyte.
  • Differences:
    • Fuel Cells:
      • Require a continuous supply of fuel and oxidant.
      • Do not store energy internally.
    • Batteries:
      • Store energy chemically within the cell.
      • Operate in a closed system until recharged.
  • Despite differences, the fundamental operating principle (electrochemical conversion) is similar.
  • Conclusion: Correct statement.

Conclusion:

• Option (f) is the only incorrect statement because it inaccurately claims that fuel cell efficiency is lower than that of internal combustion engines, whereas in reality, fuel cells are more efficient.
• All other options are correct descriptions of fuel cells.

Therefore, the correct answer is:

f. Fuel cell efficiency is lower than that of the internal combustion engines (ICEs)

Answer: f. 2

Explanation:

In a fuel cell reaction involving molar concentrations of fuel and oxidant with charge and mass balance, the total exchange of electrons in the redox (reduction-oxidation) process can be determined by analyzing the half-reactions occurring at the anode and cathode.

Typical Fuel Cell Reaction (Hydrogen-Oxygen Fuel Cell):

Anode Reaction (Oxidation):

• Process:
  • Hydrogen gas ($\text{H}_2$) is oxidized.
  • Each molecule of hydrogen releases 2 electrons.
• Electrons Produced: 2 electrons per molecule of $\text{H}_2$.

Cathode Reaction (Reduction):

• Process:
  • Oxygen gas ($\text{O}_2$) is reduced.
  • Consumes 2 electrons per molecule of water formed.
• Electrons Consumed: 2 electrons per molecule of $\text{H}_2\text{O}$.

Overall Reaction:

• Net Electron Exchange:
  • Although electrons are transferred internally between the anode and cathode, they cancel out in the overall reaction.
  • However, the total number of electrons exchanged in the redox process is 2 electrons per mole of hydrogen gas reacted.

Conclusion:

• Total Exchange of Electrons: 2 electrons
  • 2 electrons are released at the anode and 2 electrons are consumed at the cathode.
• The options mentioning different numbers of electrons at the anode and cathode interfaces do not accurately represent the balanced electron exchange in the fuel cell reaction.

Therefore, the correct answer is:

f. 2

Answer: f. (a), (b), (c), and (d)

Explanation:

An effective fuel cell catalyst must meet several critical requirements to facilitate efficient electrochemical reactions while maintaining durability and performance over time. Let's evaluate each option to determine which parameters are essential.

(a) Activity

• Importance:
  • The catalyst must be highly active to accelerate the electrochemical reactions at the anode and cathode.
  • High activity reduces activation energy barriers, increasing reaction rates and improving overall fuel cell efficiency.

(b) Conductivity

• Importance:
  • The catalyst should possess good electrical conductivity.
  • Conductive catalysts facilitate the efficient transfer of electrons during redox reactions, minimizing resistance losses.

(c) Chemical Stability

• Importance:
  • Chemical stability ensures the catalyst remains unchanged and active in the corrosive environment of a fuel cell.
  • It must resist degradation due to exposure to reactants, intermediates, products, and potential contaminants.

(d) Thermal Stability

• Importance:
  • The catalyst must withstand the operating temperatures of the fuel cell without loss of structure or activity.
  • Thermal stability prevents sintering, agglomeration, or phase changes that could reduce catalytic effectiveness.

(e) Thermodynamic Stability

• Importance:
  • While thermodynamic stability (i.e., stability against spontaneous decomposition or reaction) is generally desirable, it's not always a primary concern.
  • Some catalysts may be thermodynamically unstable but kinetically stable under operating conditions, allowing them to function effectively.

Conclusion:

• Options (a), (b), (c), and (d) are prime requirements for an effective fuel cell catalyst.

  • Activity ensures efficient reaction rates.
  • Conductivity allows for effective electron transfer.
  • Chemical Stability maintains catalyst integrity in the chemical environment.
  • Thermal Stability ensures performance at operating temperatures.

• Option (e) (Thermodynamic Stability) is less critical in practical fuel cell applications compared to chemical and thermal stability.

Therefore, the correct choice is:

f. (a), (b), (c), and (d)

Final Answer: f. (a), (b), (c), and (d)

Answer: g. (a) and (c)

Answer: e. (a), (b), and (c)

Explanation:

The primary sources of power loss in fuel cells arise due to three main types of overpotentials or polarizations:

a. Activation Losses (Option a):

• Description:
  • Activation losses are associated with the energy barrier that must be overcome for the electrochemical reactions to occur at the electrodes.
  • They represent the voltage loss due to the kinetics of the electrode reactions, especially the sluggish oxygen reduction reaction at the cathode.
• Factors Influencing Activation Losses:
  • Catalyst Activity: Low catalyst activity increases activation overpotentials.
  • Temperature: Higher temperatures can reduce activation losses by increasing reaction rates.
  • Reactant Concentration: Higher concentrations can decrease activation losses.
• Impact on Fuel Cell Performance:
  • Significant at low current densities.
  • They cause a rapid drop in voltage at the beginning of the polarization curve.

b. Ohmic Losses (Option b):

• Description:
  • Ohmic losses result from the resistance to the flow of ions in the electrolyte and electrons through the electrodes and other cell components.
  • They cause a voltage drop proportional to the current density according to Ohm's Law ($V = IR$).
• Factors Influencing Ohmic Losses:
  • Electrolyte Conductivity: Lower ionic conductivity increases ohmic losses.
  • Membrane Thickness: Thicker membranes can lead to higher resistance.
  • Electrode and Bipolar Plate Materials: Materials with higher electrical resistance contribute to ohmic losses.
• Impact on Fuel Cell Performance:
  • Increase linearly with current density.
  • Represent the middle region of the polarization curve where voltage drops steadily.

c. Concentration Polarization Losses (Option c):

• Description:
  • Concentration losses, also known as mass transport losses, occur due to limitations in the transport of reactants to the electrode surfaces and the removal of products from the reaction sites.
  • They are associated with the depletion of reactant concentration at the catalyst sites.
• Factors Influencing Concentration Losses:
  • Reactant Flow Rates: Insufficient flow rates lead to depletion zones.
  • Gas Diffusion Layer Properties: Poorly designed layers impede mass transport.
  • Operating Conditions: High current densities exacerbate concentration losses.
• Impact on Fuel Cell Performance:
  • Become significant at high current densities.
  • Cause a sharp voltage drop at the end of the polarization curve.

Other Options:

• d. Catalytic Losses:

  • While catalyst degradation can affect fuel cell performance, the term "catalytic losses" is not commonly used as a primary category of power loss.
  • Issues with catalysts typically manifest as increased activation losses.

• f. Electrode-Interconnect Interface Poisoning:

  • This refers to degradation mechanisms over time, such as contamination or poisoning of catalysts.
  • Not considered a primary cause of power loss during normal operation.

• g. None of the Above:

  • Incorrect, as options (a), (b), and (c) are the primary causes of power loss.

Conclusion:

The origin of loss of power in fuel cells arises primarily due to:

1. Activation Losses (kinetic barriers to the electrochemical reactions).
2. Ohmic Losses (resistive losses in ionic and electronic conduction).
3. Concentration Polarization Losses (mass transport limitations of reactants and products).

Final Answer: e. (a), (b), and (c)

Answer: g. (c) and (f)

Explanation:

We are asked to identify the statements that are NOT true in the case of a fuel cell. Let's evaluate each statement individually.

a. EMF in a fuel cell varies with activity of the reactants and products

• True
  • The electromotive force (EMF) or cell voltage of a fuel cell is governed by the Nernst equation, which relates the cell potential to the activities (or effective concentrations/partial pressures) of the reactants and products.
  • As the activities or partial pressures of the reactants and products change, the EMF varies accordingly.

b. Fuel cell efficiency decreases with rise in temperature

• True
  • The theoretical efficiency of a fuel cell is given by the ratio of Gibbs free energy change to enthalpy change ($\eta = \Delta G / \Delta H$).
  • Gibbs free energy ($\Delta G$) decreases with increasing temperature (since $\Delta G = \Delta H - T\Delta S$), leading to a decrease in efficiency.
  • Therefore, as temperature rises, fuel cell efficiency decreases.

c. Fuel cell efficiency remains constant irrespective of temperature

• Not True
  • As explained above, fuel cell efficiency depends on temperature and generally decreases with increasing temperature.
  • Therefore, the statement that efficiency remains constant irrespective of temperature is false.

d. Fuel cell efficiency is always less than ideal thermodynamic efficiency

• True
  • In practical fuel cells, there are losses due to overpotentials, ohmic resistances, and mass transport limitations.
  • These losses mean that the actual efficiency is always less than the ideal thermodynamic efficiency.
  • The ideal efficiency is based on reversible processes without any losses, which is unattainable in real systems.

e. For stable fuel cell, both charge and mass balance is essential

• True
  • Charge balance ensures that the flow of electrons and ions is maintained, preventing charge buildup that could disrupt cell operation.
  • Mass balance ensures that reactants are supplied and products are removed at rates that sustain continuous operation.
  • Both balances are crucial for the stable and efficient functioning of a fuel cell.

f. Fuel cell voltage is independent of pressure

• Not True
  • The cell voltage depends on the partial pressures of the reactants and products, as described by the Nernst equation.
  • Changes in pressure affect the activities (partial pressures) of gases, thereby influencing the EMF.
  • Therefore, fuel cell voltage is dependent on pressure.

Conclusion:

• The statements that are NOT true are:
  • (c) Fuel cell efficiency remains constant irrespective of temperature.
  • (f) Fuel cell voltage is independent of pressure.

Therefore, the correct answer is:

g. (c) and (f)

Answer: b. Chemical energy conversion and electrical energy release

Explanation:

Fuel cells operate on the fundamental principle of converting chemical energy directly into electrical energy through electrochemical reactions. Let's analyze each option to determine which statement accurately describes this principle.

Option Analysis:

a. Chemical energy storage and electrical energy transfer

• Explanation:
  • Fuel cells do not store chemical energy; instead, they continuously convert the chemical energy of supplied fuel and oxidant into electrical energy.
  • Electrical energy transfer alone does not encompass the generation of electrical energy from chemical reactions.
• Conclusion: Not correct.

b. Chemical energy conversion and electrical energy release

• Explanation:
  • Fuel cells convert chemical energy from the reactants (fuel and oxidant) into electrical energy that is released and can be harnessed to perform work.
  • This direct conversion is the core operating principle of fuel cells.
• Conclusion: Correct.

c. Chemical energy conversion under a pressure gradient and electrical energy release

• Explanation:
  • While pressure can influence fuel cell performance (e.g., gas partial pressures affect reaction rates and cell voltage via the Nernst equation), pressure gradients are not a fundamental principle of fuel cell operation.
  • The essential principle is the conversion of chemical energy to electrical energy, not the reliance on pressure gradients.
• Conclusion: Partially correct but includes unnecessary condition (pressure gradient).

d. Chemical energy storage and electrical energy release under volume expansion

• Explanation:
  • Again, fuel cells do not store chemical energy; they convert it as long as fuel is supplied.
  • Volume expansion is not a primary aspect of fuel cell operation.
• Conclusion: Not correct.

e. Electrochemical action only

• Explanation:
  • Fuel cells indeed operate through electrochemical reactions, involving oxidation at the anode and reduction at the cathode.
  • However, stating "only" limits the description and omits the crucial aspect of energy conversion and release.
• Conclusion: Partially correct but incomplete.

f. (a), (d), and (e)

• Explanation:
  • As previously determined, options (a) and (d) are not correct, and (e) is partially correct.
  • Combining incorrect and incomplete options does not yield a correct statement.
• Conclusion: Not correct.

g. (c) and (e)

• Explanation:
  • Option (c) is partially correct but includes unnecessary specifics (pressure gradient).
  • Option (e) is partially correct but incomplete.
  • Combining these does not fully and accurately describe the operating principle of fuel cells.
• Conclusion: Not the best choice.

Summary:

• Fuel cells operate by converting chemical energy directly into electrical energy through electrochemical reactions.

• The most accurate and complete description among the options is:

  b. Chemical energy conversion and electrical energy release