Fuel Cell Energy Origin

Understanding fuel cell energy

I’m working through the question of how energy in a fuel cell originates and is released, focusing on redox reactions, free energy changes, and energy conversion.

Understanding energy flow

Hmm, I’m curious about how energy flows through fuel cells. It seems to involve redox reactions, potential differences, free energy changes, and electrical work by charges. Options e, f, and g describe these elements in various combinations.

The energy in a fuel cell originates from redox reactions that result in a potential difference. This potential difference drives the movement of electrons through an external circuit, performing electrical work. The free energy change associated with the redox reactions is directly converted into electrical energy without intermediate steps (like combustion). Therefore, the most comprehensive and accurate description is provided in option (d).

Option (d) includes all the critical elements:

• Redox action: The fundamental chemical reactions in a fuel cell are oxidation-reduction (redox) reactions.
• Resulting potential difference: The redox reactions create a voltage (potential difference) between the anode and cathode.
• Free energy change: The Gibbs free energy change from the redox reactions is the source of the electrical energy.
• Electrical work done by charges: Electrons moving through the external circuit do electrical work.
• Direct energy conversion: The chemical energy is converted directly into electrical energy without intermediate forms.

Other options either omit key components or are less complete. Therefore, only option (d) fully captures the origin and release of energy in a fuel cell.

Answer: Of. Only (d)

Assessing energy release

I’m figuring out the energy released by comparing hydrogen and oxygen bond energies in reactants versus the product water. This involves examining bond strengths to predict the overall energy output.

Calculating energy changes

I’m pulling together the calculation of energy changes in the reaction by tallying the bond energies of bonds broken and formed, focusing on the reactants and products.

Calculating bond energy

I’m listing the bond energies to find the total energy required to break and form bonds. The total energy change, initially minus a sign error, turns out to be 241 kJ/mol.

Calculating bond energy changes

I’m summing bond enthalpies for bonds broken and formed. For bonds broken, the total energy is 679 kJ/mol, and for bonds formed, it’s -920 kJ/mol.

Breaking down the equation

I'm pulling together the bond energies to figure out the reaction's enthalpy change, which comes to -241 kJ/mol, suggesting an exothermic reaction.

Recalculating bond energy

I recalculated the enthalpy change for the reaction using the given bond energies: H-H (432 kJ/mol), O=O (494 kJ/mol), and H-O (460 kJ/mol). The recalculated value is -241 kJ/mol.

Verifying bond enthalpy calculations

I’m thinking through the bond enthalpies of H_2 and O_2, ensuring that all bonds are accurately accounted for to determine the energy released per mole of water formed.

Calculating energy changes

I’m figuring out the energy needed to break and form bonds in the reaction. Breaking bonds costs energy, while forming bonds releases it. This gives us a total enthalpy change of -241 kJ/mol.

Piecing together bond enthalpies

I’m calculating bond enthalpies for H_2 and O_2 reactions, resulting in -241 kJ/mol. This leads to another attempt using standard enthalpies of formation, but the bond enthalpy method seems more accurate overall.

Evaluating bond enthalpy

OK, let me see. I'm looking into reaction H_2(g) + ½O_2(g) → H_2O(l), using the bond enthalpies given for H_2 and O_2 to find the energy released per mole of water formed at NTP.

To calculate the energy released per mole of water produced in the reaction $\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ at NTP using bond enthalpies, we'll follow these steps:

1. List the Bonds Broken and Formed:

   • Bonds Broken (Reactants):

     • One H–H bond in $\text{H}_2$: $1 \times 432 \ \text{kJ/mol}$
     • Half an O=O bond in $\frac{1}{2}\text{O}_2$: $0.5 \times 494 \ \text{kJ/mol}$

   • Bonds Formed (Products):

     • Two O–H bonds in $\text{H}_2\text{O}$: $2 \times 460 \ \text{kJ/mol}$

2. Calculate the Total Energy for Bonds Broken:

   $$\text{Energy to break bonds} = (1 \times 432) + (0.5 \times 494) = 432 + 247 = 679 \ \text{kJ/mol}$$

3. Calculate the Total Energy for Bonds Formed:

   $$\text{Energy to form bonds} = 2 \times 460 = 920 \ \text{kJ/mol}$$

4. Calculate the Enthalpy Change ($\Delta H$):

   $$\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}$$ $$\Delta H = 679 \ \text{kJ/mol} - 920 \ \text{kJ/mol} = -241 \ \text{kJ/mol}$$

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

5. Compare with Given Options:

   • The calculated energy released is 241 kJ/mol (ignoring the negative sign since the question asks for energy released).
   • The provided options are:
     • a. 237 kJ/mol
     • b. 242 kJ/mol
     • c. 237.3 kJ/mol
     • d. 286 kJ/mol
     • e. None of the above

   None of the options exactly match the calculated value of 241 kJ/mol. While option (b) is close, it is not the exact value.

Conclusion:

Since the calculated energy released per mole of water is 241 kJ/mol and none of the given options match this value precisely, the correct choice is:

Answer: e. None of the above

Figuring out interactions

I’m analyzing the question about charge and mass balance in fuel cells, focusing on the interactions between specific chemicals, such as sulfonate and H3O+, H2O and H3O+, H2O and H2O, sulfonate and H2O, and CF2 and H3O+.

Understanding fuel cell dynamics

I’m noting the types of interactions involved in the charge and mass balance processes of fuel cells. These interactions include sulfonate groups, H3O+, water, and CF2 groups.

Breaking down the process

Hydrogen is oxidized at the anode to form protons and electrons. Protons travel through the PEM, while electrons power an external circuit. At the cathode, oxygen, protons, and electrons combine to produce water.

To determine which interactions are involved in the charge and mass balance process during fuel cell operation, let's analyze each option in the context of a typical proton exchange membrane (PEM) fuel cell.

Understanding Fuel Cell Operation:

In a PEM fuel cell, the key processes are:

1. Anode Reaction (Oxidation): $\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$ Hydrogen gas is split into protons ($\text{H}^+$) and electrons ($e^-$).

2. Proton Transport: Protons move through the proton exchange membrane (PEM) to the cathode.

3. Cathode Reaction (Reduction): $\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}$ Oxygen gas combines with protons and electrons to form water.

Components Involved:

• Proton Exchange Membrane (PEM):

  • Typically made of Nafion, a sulfonated tetrafluoroethylene-based fluoropolymer-copolymer.
  • Contains sulfonate groups ($-\text{SO}_3^-$) attached to a fluorocarbon backbone ($-\text{CF}_2-$).

• Ions and Molecules:

  • Hydronium ions ($\text{H}_3\text{O}^+$): Protons associate with water molecules in the membrane.
  • Water molecules ($\text{H}_2\text{O}$): Essential for proton conduction via the Grotthuss mechanism.

Analyzing Each Interaction:

1. Option (a) Sulfonate-H_3O⁺ Interaction:
   • Involved: The fixed sulfonate groups ($-\text{SO}_3^-$) interact with hydronium ions ($\text{H}_3\text{O}^+$), facilitating proton transport.
2. Option (b) H_2O-H_3O⁺ Interaction:
   • Involved: Proton hopping occurs between water molecules and hydronium ions, critical for proton conductivity.
3. Option (c) H_2O-H_2O Interaction:
   • Involved: Hydrogen bonding between water molecules helps maintain membrane hydration and assists in proton transport.
4. Option (d) Sulfonate-H_2O Interaction:
   • Involved: Sulfonate groups attract water molecules, aiding in membrane hydration and proton conduction.
5. Option (e) CF_2-H_3O⁺ Interaction:
   • Not Significantly Involved: The $-\text{CF}_2-$ groups form the hydrophobic backbone of the membrane and do not interact significantly with hydronium ions.

Conclusion:

• Interactions Involved: Options (a), (b), (c), and (d).
• Interaction Not Involved: Option (e).

Therefore, the correct answer is:

Answer: Of. All except (e)

Evaluating factors

I’m looking at the elements influencing reversible voltage in a fuel cell, including coefficients, operating temperature, partial pressures, and the Faraday coefficient.

I’m examining the factors that influence the reversible voltage in a fuel cell, which include standard potential, temperature, and reactant/product activities. The Nernst equation is key to this relationship.

Understanding the Nernst equation

I’m mapping out the Nernst equation, noting how cell potential shifts with standard conditions, temperature, and reaction quotients, and introducing constants like Faraday's constant.

Identifying key factors

I'm comparing the importance of fuel supply, oxidant supply, operating temperature, and partial pressures of fuel and oxidant to determine which factors most significantly influence the reversible voltage. This analysis guides which option best fits the situation.

Mapping out constants

I'm thinking through how Faraday's constant F and the number of electrons n are constants specific to each reaction, meaning the reversible voltage depends inversely on their product nF, as per the Nernst equation.

Assessing the voltage dynamics

OK, let me see. The standard cell potential E° hinges on the Gibbs free energy change ΔG°, which is tied to the reactants and products. This makes me think that E° might be a constant for a given reaction.

Analyzing variables

I'm thinking about how the reversible voltage for a reaction is influenced by factors like standard cell potential, temperature, activities, and partial pressures of reactants and products.

Understanding Reversible Voltage in a Fuel Cell:

The reversible voltage (also known as the open-circuit voltage or equilibrium voltage) of a fuel cell is determined by the Nernst equation, which relates the cell potential to the thermodynamic properties of the cell reaction and the conditions under which the cell operates.

The Nernst equation for a fuel cell reaction is:

Where:

• $E$ is the cell potential under non-standard conditions.
• $E^\circ$ is the standard cell potential.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature (in Kelvin).
• $n$ is the number of moles of electrons transferred in the cell reaction.
• $F$ is Faraday's constant.
• $Q$ is the reaction quotient, which depends on the activities (or effective concentrations) of the reactants and products.

Analyzing Each Option:

1. Option (a): Activity Coefficient of Fuel (H_2) Supply

   • The activity coefficient accounts for deviations from ideal behavior in gases. It affects the effective concentration (activity) of hydrogen gas in the reaction.
   • Relevance: Directly influences the reaction quotient $Q$ in the Nernst equation, thereby affecting the cell potential.

2. Option (b): Activity Coefficient of Oxidant (O_2) Supply

   • Similar to hydrogen, the activity coefficient for oxygen affects its effective concentration in the reaction.
   • Relevance: Also influences $Q$, thus impacting the reversible voltage.

3. Option (c): Operating Temperature (T) of Fuel Cell

   • Temperature appears explicitly in the Nernst equation.
   • Relevance: Changes in temperature affect the cell potential by altering the thermodynamic properties of the reaction.

4. Option (d): Partial Pressure of Fuel & Oxidant

   • The partial pressures of hydrogen and oxygen are directly related to their activities in the gas phase.
   • Relevance: Altering partial pressures changes $Q$, thereby affecting the reversible voltage.

5. Option (e): Faraday Coefficient and Number of Mobile Charges

   • Faraday's constant ($F$) and the number of electrons transferred ($n$) appear in the denominator of the logarithmic term in the Nernst equation.
   • Relevance: They determine how changes in $Q$ translate to changes in cell potential.

Conclusion:

All the factors listed in options (a) through (e) influence the reversible voltage of a fuel cell as per the Nernst equation. They affect either the thermodynamic driving force ($E^\circ$), the reaction quotient ($Q$), or the proportionality constants ($n$ and $F$) in the equation.

Therefore, the reversible voltage in a fuel cell depends on all the factors mentioned.

Answer: g. All of the above

Breaking down fuel cell losses

I’m examining fuel cell losses to identify the origin of power loss. Options include activation loss, Ohmic loss, concentration polarization loss, catalytic losses, electrode-interconnect interface poisoning, or none of the above.

Identifying fuel cell losses

I’m piecing together common fuel cell losses: activation, Ohmic, and concentration losses. Activation is tied to slow surface reactions, Ohmic to internal resistance, and concentration to reactant depletion.

Evaluating the options

I'm thinking through the primary loss mechanisms in fuel cells: activation, ohmic, and concentration polarization losses. Electrode-interconnect interface poisoning and catalytic losses don't fit this context.

Identifying power loss
First, I'll define activation, ohmic, and concentration polarization losses in fuel cells. Then, I'll confirm they are primary causes of power loss.

Understanding Power Losses in Fuel Cells:

Fuel cells convert chemical energy directly into electrical energy through electrochemical reactions. However, during operation, several factors cause losses that reduce the cell's efficiency and power output. The primary origins of power loss in fuel cells are:

1. Activation Losses (Option a):

   • Definition: Activation losses are due to the energy barrier that must be overcome for the electrochemical reactions (oxidation and reduction) to occur at the electrodes.
   • Cause: They result from the slow kinetics of the electrode reactions, especially at the cathode where oxygen reduction is sluggish.
   • Impact: These losses are significant at low current densities and are influenced by the type of catalyst used, temperature, and electrode material.

2. Ohmic Losses (Option b):

   • Definition: Ohmic losses arise from the resistance to the flow of ions in the electrolyte and electrons through the electrodes and interconnects.
   • Cause: They are caused by the inherent electrical resistance of the materials used in the fuel cell components.
   • Impact: Ohmic losses increase linearly with current density and can be minimized by using materials with high electrical conductivity and optimizing the thickness of the electrolyte.

3. Concentration Polarization Losses (Option c):

   • Definition: Also known as mass transport losses, these occur when there's a drop in reactant concentration at the electrode surface.
   • Cause: Due to limitations in the transport of reactants (hydrogen and oxygen) to the reaction sites and removal of products (like water) from the electrodes.
   • Impact: These losses become significant at high current densities when the consumption rate of reactants exceeds the supply rate.

Analyzing Other Options:

4. Catalytic Losses (Option d):

   • Definition: Catalytic losses refer to inefficiencies due to poor catalyst activity or degradation over time.
   • Cause: While catalyst degradation can affect performance, it is often encompassed within activation losses since both relate to the kinetics of the reactions.
   • Impact: Not typically categorized separately in the primary losses but considered part of activation losses.

5. Electrode-Interconnect Interface Poisoning (Option f):

   • Definition: This refers to contamination or degradation at the interfaces between electrodes and interconnects, leading to reduced performance.
   • Cause: Can be due to impurities in the fuel, corrosion, or other chemical interactions.
   • Impact: While it can affect fuel cell operation, it is not considered a primary source of power loss but rather a degradation mechanism over time.

Conclusion:

The primary power losses in fuel cells are due to:

• Activation Losses (a)
• Ohmic Losses (b)
• Concentration Polarization Losses (c)

These three factors are fundamental to understanding and optimizing fuel cell performance. They directly impact the cell's voltage and efficiency during operation.

Therefore, the correct answer is:

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

Evaluating options

I'm working through each parameter to see how it may affect fuel cell output current density. Options span various factors, including reactant concentration and electrode surface area. This helps pinpoint the ones worth considering.

Understanding factors

I’m examining how reactant concentration, reaction temperature, catalysts, surface area, and mixed conducting 3D structural reaction interfaces influence current density in a fuel cell.

Understanding Factors that Optimize Output Current Density in a Fuel Cell:

The output current density of a fuel cell is a measure of the electrical current produced per unit area of the electrode. Optimizing current density is crucial for enhancing the performance and efficiency of fuel cells. Several parameters can be controlled to achieve this optimization. Let's analyze each of the options provided:

Option (a): Increasing Reactant Concentration

• Explanation:
  • Law of Mass Action: According to this principle, increasing the concentration (or partial pressure) of reactants shifts the equilibrium toward the products, enhancing the reaction rate.
  • Impact on Fuel Cell:
    • Anode Side: Increasing hydrogen concentration increases the rate of oxidation reactions.
    • Cathode Side: Increasing oxygen concentration enhances the reduction reactions.
  • Result: Higher reactant concentrations lead to an increase in current density up to the limit where other factors (like mass transport limitations) become significant.

Option (b): Increasing Reaction Temperature

• Explanation:
  • Arrhenius Equation: Reaction rates generally increase with temperature due to higher kinetic energy of molecules, leading to more effective collisions.
  • Impact on Fuel Cell:
    • Enhanced Reaction Kinetics: Faster electrochemical reactions at both anode and cathode.
    • Improved Conductivity: Ionic conductivity of the electrolyte often increases with temperature.
  • Result: Higher temperatures can boost current density but must be balanced against potential material degradation and thermal management issues.

Option (c): Lowering the Activation Barrier via Catalyst

• Explanation:
  • Catalysts: Substances that increase the rate of a chemical reaction without being consumed, by lowering the activation energy required.
  • Impact on Fuel Cell:
    • Anode and Cathode Catalysts: Use of effective catalysts like platinum improves reaction rates for hydrogen oxidation and oxygen reduction.
    • Enhanced Kinetics: Lower activation barriers lead to increased current density at given overpotentials.
  • Result: Improved catalyst activity directly contributes to higher output current density.

Option (d): Scaling Up the Number of Reaction Sites Employing High Surface Area Electrodes

• Explanation:
  • Surface Area Effect: Increasing the electrode surface area provides more active sites for electrochemical reactions.
  • Impact on Fuel Cell:
    • Nanostructured Materials: Use of materials like carbon nanotubes or porous structures increases surface area.
    • Enhanced Mass Transport: More reaction sites facilitate better utilization of reactants.
  • Result: Higher surface area leads to increased current density due to more simultaneous reactions.

Option (e): Employing Mixed Conducting 3D Structural Reaction Interfaces

• Explanation:
  • Mixed Ionic and Electronic Conductors (MIECs): Materials that conduct both ions and electrons enable more efficient charge transport.
  • 3D Structural Interfaces:
    • Benefit: Provide a network where electrochemical reactions can occur throughout the volume, not just on the surface.
    • Enhanced Reaction Zones: Increase the effective area for reactions and improve mass transport.
  • Impact on Fuel Cell:
    • Improved Performance: Allows for higher current densities by facilitating simultaneous ionic and electronic conduction.
  • Result: Utilizing MIECs in a 3D architecture optimizes current density.

Conclusion:

All the options (a) through (e) are valid methods for optimizing the output current density in a fuel cell:

• (a) Increasing reactant concentration enhances reaction rates.
• (b) Increasing temperature accelerates reaction kinetics and ionic conductivity.
• (c) Lowering activation barriers with catalysts improves reaction efficiency.
• (d) Scaling up reaction sites with high surface area electrodes increases the number of simultaneous reactions.
• (e) Employing mixed conducting 3D interfaces maximizes effective reaction zones and improves charge transport.

Therefore, the most comprehensive answer that includes all the parameters is:

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

Evaluating fuel cells

OK, let me see. Fuel cells convert chemical energy into electrical energy, but the key detail is how exactly this conversion occurs. Considering the correct options is crucial for a precise answer.

Mapping fuel cell operations

Fuel cells function through hydrogen electrolysis, separating hydrogen protons and electrons. Oxygen then reacts with electrons to generate an electric current, with ions typically conveyed by polymer membranes.

Examining fuel cells

Fuel cells efficiently convert chemical energy into electrical energy by using externally supplied chemical fuel, without storing it.

Evaluating fuel cell principles

I examined various principles of fuel cell operation, including pressure gradient, volume expansion, and electrochemical action, to determine their accuracy.

Evaluating operation principles

I’m assessing the accuracy of each option by highlighting that fuel cells convert chemical energy to electrical energy, with pressure gradients and volume expansion being less relevant principles for their operation.

Weighing options

I’m gathering info on option b, but it turns out it’s not the best choice after evaluating other options.

Determining the principle

I'm concluding that fuel cells primarily operate on the principle of chemical energy conversion and electrical energy release, which aligns with option b.

Evaluating options

I'm thinking about options b and e, but leaning towards b as it seems more accurate.

Identifying the principle

Assess the significance of each option to determine which best describes how fuel cells operate, emphasizing their conversion of chemical energy into electrical energy.

Distributing components

I’m piecing together the role of fuel cells and batteries. Batteries lock energy internally, but fuel cells need ongoing reactants. And while pressure affects performance, it’s not a core operational principle.

Evaluating energy dynamics

Considering fuel cells and their efficiency, I found they excel in transforming chemical energy directly into electrical energy through electrochemical reactions.

Elaborating on energy conversion

Highlighting that fuel cells involve both chemical energy conversion and electrical energy release, with the chemical energy of reactants being transformed into electrical energy through electrochemical reactions.

To determine the principle(s) on which fuel cells operate, let's analyze each option:

Option a: Chemical energy storage and electrical energy transfer

• Analysis: Fuel cells do not store chemical energy. Instead, they continuously convert chemical energy from externally supplied fuels (like hydrogen) and oxidants (like oxygen) into electrical energy. Energy storage devices, such as batteries, store chemical energy internally and release it when needed, which is not the case with fuel cells.
• Conclusion: This option is incorrect because fuel cells are energy conversion devices, not energy storage devices.

Option b: Chemical energy conversion and electrical energy release

• Analysis: This accurately describes the fundamental operation of fuel cells. They convert the chemical energy of the fuel and oxidant directly into electrical energy through electrochemical reactions.
• Conclusion: This option correctly states the principles of fuel cell operation.

Option c: Chemical energy conversion under a pressure gradient and electrical energy release

• Analysis: While operating pressure can influence fuel cell performance, a pressure gradient is not a fundamental principle of their operation. Fuel cells do not require a pressure gradient to function.
• Conclusion: This option is not a primary principle of fuel cell operation.

Option d: Chemical energy storage and electrical energy release under volume expansion

• Analysis: Fuel cells do not store chemical energy, and volume expansion is not a key aspect of their operation. There is no significant volume change contributing to energy release in fuel cells.
• Conclusion: This option is incorrect.

Option e: Electrochemical action only

• Analysis: Fuel cells operate based on electrochemical reactions. While this is true, stating "electrochemical action only" overlooks the essential aspect of chemical energy conversion into electrical energy.
• Conclusion: This option is partially correct but incomplete.

Option f: (a), (d), and (e)

• Analysis: Since options (a) and (d) are incorrect and option (e) is incomplete, this combination is not correct.
• Conclusion: This option is incorrect.

Option g: (c) and (e)

• Analysis: Given that option (c) is not fundamental and option (e) is incomplete, this combination is not the best choice.
• Conclusion: This option is not correct.

Final Conclusion:

• The most accurate description is provided by Option b: "Chemical energy conversion and electrical energy release."
• Fuel cells operate by converting chemical energy from the fuel and oxidant into electrical energy through electrochemical reactions.

Answer: Option b

Analyzing fuel cell dynamics

I’m thinking through the oxygen output flux and fuel input flux needed to sustain a 1000 kA current in a H_2-Air fuel cell, given a reactant input flow of 20 moles per second.

Piecing together the components

I’m analyzing a H_2-Air fuel cell to calculate oxygen and hydrogen consumption rates for a 1000 kA current, considering air supplied at 20 moles s⁻¹.

Probing molecular interactions

I’m digging into the reactions in a hydrogen fuel cell, focusing on electron flow: two electrons from H_2 at the anode and four electrons from O_2 at the cathode, creating H_2O ultimately.

Piecing together electron transfer

First, I’m mapping out the number of electrons transferred per mole of reactant. For H_2, 2 electrons per mole are involved. Considering oxygen's reaction, it shows 2 electrons per half mole of O_2.

Calculating consumption rates

First, I determine the consumption rate of H_2 by applying the formula $v = \frac{I}{nF}$. Plugging in the given values, I calculate $v_{\text{H}_2} \approx 5.182 \text{ mol/s}$.

Calculating oxygen usage

I'm thinking through the calculation of the oxygen supply, given air at 20 mol/s and oxygen making up 21% by volume. The oxygen consumption is approximately 2.591 mol/s.

Determining oxygen output

I calculated the oxygen output flux by subtracting the oxygen consumed from the oxygen supplied, arriving at a flux of approximately 1.6 mol/s. Additionally, the fuel input flux was determined to be approximately 5.19 mol/s.

Identifying discrepancies

I evaluated all options, confirming the correct answer is option (a), which was calculated to be approximately 5.182 mol/s.

To solve this problem, we'll follow these steps:

1. Understand the Fuel Cell Reaction and Electron Transfer:

   • The overall reaction in a hydrogen-air (oxygen) fuel cell is: $$\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$$
   • Electrons Transferred:
     • Anode Reaction (Oxidation of Hydrogen): $$\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$$
       • Electrons per mole of H_2: $n_{\text{H}_2} = 2$
     • Cathode Reaction (Reduction of Oxygen): $$\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}$$
       • Electrons per mole of O_2 consumed: $n_{\text{O}_2} = 4$ (since one mole of O_2 consumes 4 moles of electrons)

2. Use the Relation $I = nFv$:

   • $I$ is the current (in Amperes)
   • $n$ is the number of electrons transferred per mole of reactant
   • $F$ is Faraday's constant ($F = 96485 \ \text{C/mol}$)
   • $v$ is the molar flow rate of the reactant (in moles per second)

3. Calculate the Molar Flow Rate of Hydrogen ($v_{\text{H}_2}$):

   • Given current $I = 1000 \ \text{kA} = 1,000,000 \ \text{A}$
   • For hydrogen: $$v_{\text{H}_2} = \frac{I}{n_{\text{H}_2} F} = \frac{1,000,000 \ \text{A}}{2 \times 96485 \ \text{C/mol}} \approx 5.182 \ \text{mol/s}$$

4. Calculate the Molar Flow Rate of Oxygen Consumed ($v_{\text{O}_2}$):

   • From the stoichiometry, $\text{H}_2 : \text{O}_2 = 2 : 1$
   • Therefore: $$v_{\text{O}_2} = \frac{v_{\text{H}_2}}{2} = \frac{5.182}{2} \approx 2.591 \ \text{mol/s}$$

5. Determine the Oxygen Supplied and Oxygen Output Flux:

   • Air Composition: Air is approximately 21% oxygen by volume (and moles).
   • Total Air Supplied: $20 \ \text{mol/s}$
   • Oxygen Supplied: $$\text{O}_2 \text{ supplied} = 0.21 \times 20 \ \text{mol/s} = 4.2 \ \text{mol/s}$$
   • Oxygen Output Flux: $$\text{O}_2 \text{ output flux} = \text{O}_2 \text{ supplied} - v_{\text{O}_2} = 4.2 - 2.591 \approx 1.609 \ \text{mol/s}$$

6. Compare Calculated Values with Given Options:

   Option a: 1.6 mol/s (O_2 output flux) and 5.19 mol/s (H_2 input flux)

   • Our calculations:
     • Oxygen output flux ≈ 1.609 mol/s (matches 1.6 mol/s)
     • Hydrogen input flux ≈ 5.182 mol/s (matches 5.19 mol/s)
   • Conclusion: Matches our calculated values.

   Option b: 1.6 mol/s and 3.2 mol/s respectively

   • Hydrogen input flux of 3.2 mol/s does not match our calculated value of approximately 5.182 mol/s.
   • Conclusion: Does not match.

   Option c: None of the above

   • Our calculations match Option a.
   • Conclusion: Not correct.

   Option d: 5.19 mol/s and 1.6 mol/s respectively

   • The values are correct but the order is reversed (should be O_2 output flux first, then H_2 input flux).
   • Conclusion: Order does not match the question's requirement ("oxygen output flux and fuel input flux respectively").

   Option e: 2.5 mol/s and 5.7 mol/s respectively

   • Oxygen output flux of 2.5 mol/s does not match our calculated value of approximately 1.6 mol/s.
   • Conclusion: Does not match.

7. Final Answer:

   • Option a is the only one that aligns with our calculations.

Answer: Option a. 1.6 moles s⁻¹ and 5.19 moles s⁻¹ respectively

Summary:

• Oxygen Output Flux: Approximately 1.6 mol/s
• Fuel (Hydrogen) Input Flux: Approximately 5.19 mol/s
• Correct Option: a

Choosing the right criteria

I'm working through the question of which requirements are essential for an effective fuel cell catalyst. The options include activity, conductivity, chemical stability, thermal stability, and thermodynamic stability.

Evaluating requirements

I start by identifying the key elements for an effective fuel cell catalyst: activity for electrochemical reactions, conductivity for electron transfer, chemical stability, and thermal stability to withstand operating conditions.

Evaluating stability needs

I'm noting that thermodynamic stability might be redundant since chemical and thermal stability already cover it. Thermal stability is crucial, especially in high-temperature fuel cells.

Breaking down requirements

First, I analyze activity, which is crucial for facilitating electrochemical reactions efficiently.

Assessing catalyst requirements

I’m mapping out the essential traits of a fuel cell catalyst: lowering activation energy, ensuring electronic conductivity, and maintaining chemical and thermal stability under demanding conditions.

Evaluating catalyst requirements

I'm piecing together the necessary qualities for a high temperature catalyst, like structural integrity and thermal stability. Ensuring these traits is crucial for maintaining efficiency and preventing damage.

To determine the prime requirements for an effective fuel cell catalyst, let's examine each option:

Option (a): Activity

• Explanation:
  • Catalytic Activity is crucial because the catalyst must efficiently facilitate the electrochemical reactions at the electrodes.
  • It should lower the activation energy, increasing the reaction rate without being consumed.
• Conclusion: Essential requirement.

Option (b): Conductivity

• Explanation:
  • Electrical Conductivity is important because the catalyst often forms part of the electrode structure.
  • It must allow for efficient electron transfer between the reactants and the external circuit.
• Conclusion: Essential requirement.

Option (c): Chemical Stability

• Explanation:
  • Chemical Stability ensures the catalyst maintains its integrity and activity over time.
  • It must resist corrosion, oxidation, and reactions with fuel cell components or reactants.
• Conclusion: Essential requirement.

Option (d): Thermal Stability

• Explanation:
  • Thermal Stability is vital because fuel cells may operate at elevated temperatures.
  • The catalyst should maintain its structure and activity without degradation or sintering at operational temperatures.
• Conclusion: Essential requirement, especially for high-temperature fuel cells.

Option (e): Thermodynamic Stability

• Explanation:
  • Thermodynamic Stability refers to the catalyst's ability to remain in its lowest energy state under operational conditions.
  • While important, it overlaps with chemical and thermal stability.
• Conclusion: Important but not distinct from chemical and thermal stability.

Analyzing the Options:

• Option (f): (a), (b), (c), and (d)
  • Includes all essential requirements: activity, conductivity, chemical stability, and thermal stability.
• Option (g): (a), (b), (c), and (e)
  • Replaces thermal stability with thermodynamic stability, which is less specific and overlaps with chemical stability.

Final Conclusion:

The prime requirements for an effective fuel cell catalyst are:

1. Activity (a): To facilitate reactions efficiently.
2. Conductivity (b): For effective electron transfer.
3. Chemical Stability (c): To resist degradation and maintain performance.
4. Thermal Stability (d): To operate effectively at fuel cell temperatures.

Thermodynamic stability (e), while important, is encompassed within chemical and thermal stability and is not a separate prime requirement.

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

Evaluating fuel cell statements

I'm working through multiple-choice questions on fuel cells, examining statements to pinpoint the incorrect one. This involves verifying each statement against known fuel cell principles and characteristics.

Evaluating fuel cell potential

Fuel cells emerge as a clean, green energy source, producing DC power through electrochemical reactions, unlike batteries, which store and release energy.

Understanding fuel cells

Fuel cells and batteries both generate electricity through electrochemical reactions. Fuel cells continuously supply reactants from external sources, while batteries contain internal reactants. The principles of fuel cells align with reversible thermodynamics.

Evaluating fuel cell principles

I’m thinking through how fuel cells operate under reversible thermodynamics, their maximum efficiency tied to Gibbs free energy change, and the electrochemical reactions integral to their function.

To determine which statement is not correct about fuel cells, let's analyze each option carefully:

Option a: Fuel cell is a clean and green energy alternative

• Explanation:
  • Fuel cells produce electricity through electrochemical reactions, typically combining hydrogen and oxygen to form water.
  • Emissions: The primary byproduct is water vapor, making fuel cells a clean energy source with minimal pollutants.
• Conclusion: Correct. Fuel cells are indeed a clean and green energy alternative.

Option b: Fuel cells produce DC power

• Explanation:
  • Fuel cells generate electricity directly through electrochemical reactions without combustion.
  • Output: The electricity produced is in the form of direct current (DC).
• Conclusion: Correct. Fuel cells produce DC power.

Option c: Fuel cell operating principle is like that of the batteries

• Explanation:
  • Similarities:
    • Both fuel cells and batteries convert chemical energy into electrical energy through electrochemical reactions.
    • Both involve redox reactions occurring at electrodes with an electrolyte facilitating ion transport.
  • Differences:
    • Fuel Cells: Require a continuous supply of fuel and oxidant from external sources.
    • Batteries: Contain a fixed amount of reactants stored internally; once depleted, they need recharging or replacement.
• Conclusion: Partially Correct. While the fundamental electrochemical principles are similar, fuel cells differ from batteries in operation (continuous fuel supply vs. stored reactants). However, the statement emphasizes the operating principle, which is similar.

Option d: Fuel cell action complies with the principles of reversible thermodynamics

• Explanation:
  • Reversible Thermodynamics: Refers to processes that occur without entropy production, maximizing efficiency.
  • Fuel Cells:
    • Operate based on thermodynamic principles where the maximum efficiency is determined by the Gibbs free energy change.
    • Aim for reversible operation to achieve high efficiencies.
• Conclusion: Correct. Fuel cell operation aligns with the principles of reversible thermodynamics.

Option e: Electrochemical action is the root of fuel cell working

• Explanation:
  • Fuel cells rely on electrochemical reactions (oxidation and reduction) to generate electricity.
  • Core Mechanism: The electrochemical conversion of fuel and oxidant into electricity, heat, and water.
• Conclusion: Correct. Electrochemical action is fundamental to fuel cell operation.

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

• Explanation:
  • Internal Combustion Engines (ICEs):
    • Typically have thermal efficiencies ranging from 20% to 40%.
    • Limited by the Carnot efficiency due to heat-to-work conversion.
  • Fuel Cells:
    • Have electrical efficiencies ranging from 40% to 60%, with potential for higher efficiencies when waste heat is utilized (combined heat and power systems).
    • Not limited by Carnot efficiency since they convert chemical energy directly to electrical energy.
• Conclusion: Incorrect. Fuel cells generally have higher efficiencies than internal combustion engines.

Option g: Fuel cell has the potential to substitute and replace fossil fuel

• Explanation:
  • Fuel Cells:
    • Can run on hydrogen produced from renewable sources, reducing reliance on fossil fuels.
    • Offer a cleaner alternative for energy generation in transportation, stationary power, and portable applications.
  • Potential: Seen as a promising technology for transitioning to sustainable energy systems.
• Conclusion: Correct. Fuel cells have the potential to substitute and replace fossil fuels in various applications.

Final Answer:

• The statement that is not correct is Option f: "Fuel cell efficiency is lower than that of the internal combustion engines (ICEs)."
• This is incorrect because fuel cells generally have higher efficiencies compared to ICEs.

Answer: Option f

Breaking down the problem

I’m working through calculating the ohmic voltage loss in a PEMFC by examining the membrane's thickness and conductivity at a specific current density. This involves assessing the membrane's contribution to cell resistance, leading to a voltage drop.

Mapping out the process

I’m calculating the ohmic voltage loss by first figuring out the membrane resistance (R_membrane), then the total resistance (R_total), and lastly, the voltage loss (V_ohmic) using the given data.

Crafting an equation

I’m building the membrane resistance equation by dividing its thickness by the product of conductivity and area, leading to an expression that combines 0.01 cm and 0.10 Ω⁻¹ cm⁻¹ × 10 cm².

Piecing together resistance

I’m calculating the total resistance by summing the electronic and membrane slips, then figuring out the ohmic voltage loss by multiplying the total current by this resistance.

To determine the ohmic voltage loss for the given PEM fuel cell (PEMFC) under the specified conditions, we'll follow these steps:

1. Calculate the Resistance of the NAFION Membrane ($R_{\text{membrane}}$)
2. Calculate the Total Resistance ($R_{\text{total}}$)
3. Determine the Total Current ($I$)
4. Calculate the Ohmic Voltage Loss ($V_{\text{ohmic}}$)
5. Select the Correct Option

Given Data:

• Electrode Area ($A$): 10 cm²
• Membrane Conductivity ($\sigma$): 0.10 Ω⁻¹ cm⁻¹
• Membrane Thickness ($l$): 100 µm
• Electronic Resistance ($R_{\text{electronic}}$): 0.005 Ω
• Current Density ($j$): 1 A cm⁻²

Step 1: Calculate the Resistance of the NAFION Membrane ($R_{\text{membrane}}$)

First, convert membrane thickness to centimeters:

Use the formula for resistance:

Plug in the values:

Calculate the denominator:

Calculate $R_{\text{membrane}}$:

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

Step 3: Determine the Total Current ($I$)

Given the current density ($j$) and electrode area ($A$):

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

Step 5: Select the Correct Option

The calculated ohmic voltage loss is 0.15 V.

Therefore, the correct option is:

• Option a. 0.15 V

Answer: a. 0.15 V