問題

A 10 cm2 PEMFC employs an electrolyte membrane with a conductivity of 0.10 Ω-1 Cm-1. R electronic has been determined to be 0.005 Ω for this fuel cell. Assuming only other contribution to cell resistance arising from the NAFION membrane, what would be the value of ohmic voltage loss for this fuel cell at a current density of 1 Acm-2 if the membrane thickness is 100 µm. Select correct option from the following;

思考

Analyzing fuel cell parameters

I'm working through a PEMFC problem, considering electrolyte membrane conductivity and thickness, electronic resistance, and current density to compute the ohmic voltage loss.

Mapping the process

First, I’m gathering info on calculating the ohmic voltage loss for the fuel cell, starting with the membrane resistance. This sets the stage for a detailed analysis.

Calculating resistance

I found the membrane resistance by dividing the membrane thickness by the product of membrane conductivity and area. Adding the electronic resistance, I determined the total resistance to be 0.015 Ω.

回答

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 (RmembraneR_{\text{membrane}})

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

Rmembrane=dσ×AR_{\text{membrane}} = \frac{d}{\sigma \times A}

Where:

  • dd is the membrane thickness (in cm)
  • σ\sigma is the conductivity (in Ω1cm1\Omega^{-1} \text{cm}^{-1})
  • AA is the area (in cm2\text{cm}^2)

Given:

  • d=100μm=0.01cmd = 100 \, \mu\text{m} = 0.01 \, \text{cm}
  • σ=0.10Ω1cm1\sigma = 0.10 \, \Omega^{-1} \text{cm}^{-1}
  • A=10cm2A = 10 \, \text{cm}^2

Plugging in the values:

Rmembrane=0.010.10×10=0.011=0.01ΩR_{\text{membrane}} = \frac{0.01}{0.10 \times 10} = \frac{0.01}{1} = 0.01 \, \Omega

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

Add the electronic resistance:

Rtotal=Relectronic+Rmembrane=0.005Ω+0.01Ω=0.015ΩR_{\text{total}} = R_{\text{electronic}} + R_{\text{membrane}} = 0.005 \, \Omega + 0.01 \, \Omega = 0.015 \, \Omega

3. Calculate the Total Current (II)

Using the current density and area:

I=Current Density×A=1A/cm2×10cm2=10AI = \text{Current Density} \times A = 1 \, \text{A/cm}^2 \times 10 \, \text{cm}^2 = 10 \, \text{A}

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

Using Ohm's Law:

Vohmic=I×Rtotal=10A×0.015Ω=0.15VV_{\text{ohmic}} = I \times R_{\text{total}} = 10 \, \text{A} \times 0.015 \, \Omega = 0.15 \, \text{V}

Answer: 0.15 V

問題

A fuel cell operates for 1 h @ 2A current load and then for another 2 h @ 5 A current load. Assuming 100% fuel utilization, calculate the total number of moles of fuel (H2) consumed by the fuel cell during this operation. Select the right answer from the following;

思考

Understanding the task

OK, let's break this down: the fuel cell is consuming fuel over three hours at specified currents. We need to figure out the total fuel consumption for this period, considering 100% fuel utilization.

Calculating consumption

I’m first gathering info on the total charge passed through the fuel cell for each period. Then, I’ll use Faraday's laws to determine the moles of hydrogen consumed.

Assessing hydrogen fuel cell charges

I'm calculating the total charge passed during two periods. First, 7200 C is determined for 1 hour. Then, 36,000 C is added for 2 hours. The grand total is 43,200 C.

Determining total moles consumed

I calculated the total moles of electrons passed using the charge and Faraday's constant, then divided by 2 to find the total moles of H2 consumed, rounding to 0.224 mol.

回答

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 (t1t_1) = 1 hour = 3600 seconds
  • Current (I1I_1) = 2 A
  • Charge (Q1Q_1) = Current × Time
Q1=I1×t1=2A×3600s=7200CQ_1 = I_1 \times t_1 = 2 \, \text{A} \times 3600 \, \text{s} = 7200 \, \text{C}

Second Period (2 hours at 5 A):

  • Time (t2t_2) = 2 hours = 7200 seconds
  • Current (I2I_2) = 5 A
  • Charge (Q2Q_2) = Current × Time
Q2=I2×t2=5A×7200s=36,000CQ_2 = I_2 \times t_2 = 5 \, \text{A} \times 7200 \, \text{s} = 36,000 \, \text{C}

Total Charge Passed:

Qtotal=Q1+Q2=7200C+36,000C=43,200CQ_{\text{total}} = Q_1 + Q_2 = 7200 \, \text{C} + 36,000 \, \text{C} = 43,200 \, \text{C}

2. Relate Charge to Moles of Electrons

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

Moles of electrons=QtotalF=43,200C96,485C/mol0.448mol e\text{Moles of electrons} = \frac{Q_{\text{total}}}{F} = \frac{43,200 \, \text{C}}{96,485 \, \text{C/mol}} \approx 0.448 \, \text{mol e}^-

3. Determine Moles of Hydrogen Consumed

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

H22H++2e\text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^-

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

Moles of H2=Moles of electrons2=0.448mol e20.224mol H2\text{Moles of H}_2 = \frac{\text{Moles of electrons}}{2} = \frac{0.448 \, \text{mol e}^-}{2} \approx 0.224 \, \text{mol H}_2

Answer: Approximately 0.224 moles of hydrogen gas are consumed.

問題

Considering the input and output flow rate of reactants can be converted to equivalent current, in a H2-Air fuel cell, in accordance with the relation; I = Q s-1 = nFv where v refers to the rate of consumption of reactants in mole s-1, calculate the oxygen output flux and fuel input flux respectively to produce a current ~1000 kA by the fuel cell when supplied with air @ 20 mole s-1.

思考

Calculating fuel cell dynamics

I'm working through the oxygen output flux and fuel input flux needed for a 1000 kA current in a H2-Air fuel cell, given air at 20 mole s-1.

Determining key fluxes

I'm figuring out the oxygen output flux and fuel input flux for a H2-Air fuel cell, ensuring units and context are accurately captured. This requires calculating voltage and moles consumed.

Understanding fuel cell reactions

I'm thinking through the fuel cell reactions: at the anode, H2 splits into H+ and electrons; at the cathode, O2 is reduced to H2O.

Estimating oxygen intake

OK, let me see. I’m calculating the oxygen plasma consumed to generate a 1000 kA current. Given 20 moles per second of air flow at 21% oxygen, this yields 4.2 moles per second.

Calculating required voltages

I’m working out the voltages needed for hydrogen and oxygen at the anode and cathode, respectively, to generate 1000 kA. Using the flux equation, I calculated the desired fluxes based on the stoichiometry and current values.

Calculating flux rates

I’m mapping out the hydrogen and oxygen flux rates using the given current, the Faraday constant, and the number of moles. For hydrogen, the flux is around 5.182 mol/s, while for oxygen, it’s approximately 2.591 mol/s.

Balancing input and output

I’m focusing on the oxygen input of 4.2 mol/s and pondering the surplus oxygen. The oxygen output flux will reflect the oxygen left in the fuel cell after reaction.

回答

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=nFvI = n F v
    • I = Current in Amperes (Coulombs per second)
    • n = Number of electrons transferred per mole of reactant
    • F = Faraday's constant (96,485C/mol96,485 \, \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:

Oxygen input flux ( vO2,in )=0.21×20mol/s=4.2mol/s\text{Oxygen input flux ( $ v_{\text{O}_2,\text{in}} $ )} = 0.21 \times 20 \, \text{mol/s} = 4.2 \, \text{mol/s}

Step 2: Calculate the Required Hydrogen Input Flux (vH2v_{\text{H}_2})

Anode Reaction (Hydrogen Oxidation):

H22H++2e\text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^-
  • n = 2 electrons per mole of H_2

Using the relation:

vH2=InF=1,000,000A2×96,485C/mol1,000,000192,9705.182mol/sv_{\text{H}_2} = \frac{I}{n F} = \frac{1,000,000 \, \text{A}}{2 \times 96,485 \, \text{C/mol}} \approx \frac{1,000,000}{192,970} \approx 5.182 \, \text{mol/s}

Step 3: Calculate the Required Oxygen Consumption Flux (vO2,consumedv_{\text{O}_2,\text{consumed}})

Cathode Reaction (Oxygen Reduction):

12O2+2H++2eH2O\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}
  • n = 4 electrons per mole of O_2 (since O_2 is diatomic and accepts 4 electrons)

Using the relation:

vO2,consumed=InF=1,000,000A4×96,485C/mol1,000,000385,9402.591mol/sv_{\text{O}_2,\text{consumed}} = \frac{I}{n F} = \frac{1,000,000 \, \text{A}}{4 \times 96,485 \, \text{C/mol}} \approx \frac{1,000,000}{385,940} \approx 2.591 \, \text{mol/s}

Step 4: Calculate the Oxygen Output Flux (vO2,outv_{\text{O}_2,\text{out}})

vO2,out=vO2,invO2,consumed=4.2mol/s2.591mol/s=1.609mol/sv_{\text{O}_2,\text{out}} = v_{\text{O}_2,\text{in}} - v_{\text{O}_2,\text{consumed}} = 4.2 \, \text{mol/s} - 2.591 \, \text{mol/s} = 1.609 \, \text{mol/s}

Final Answers:

  • Hydrogen Fuel Input Flux: vH25.182mol/sv_{\text{H}_2} \approx 5.182 \, \text{mol/s}
  • Oxygen Output Flux: vO2,out1.609mol/sv_{\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.

問題

Consider a 20-cell stack in series operating under steady condition of fuel & oxidant supply with 100 cm2 active area per electrode with a current density of 0.8 Acm-2. The rate of water production in grams per hour by this fuel cell stack is;

思考

Understanding the problem

OK, let me see. The setup involves a 20-cell hydrogen fuel cell stack. Hydrogen oxidizes at the anode, producing protons and electrons, while oxygen at the cathode reduces to form water. The overall reaction is 2 H2 + O2 -> 2 H2O.

Calculating total current

I'm working through the total current produced by the stack. With a current density of 0.8 A/cm^2, an active electrode area of 100 cm^2, and 20 series cells, I can estimate it using the formula I=J×A×NI = J \times A \times N.

Piecing together the data

I’m gathering info on the total current per cell, calculating total charge per hour, and then figuring out the moles of electrons transferred using Faraday's constant.

Determining water output

I determined that 4 moles of electrons produce 2 moles of water. With 2.984 moles of electrons, approximately 1.492 moles of water are produced. Converting this to mass, I calculated around 26.875 grams of water per hour.

Assessing current flow

I’m piecing together whether each cell in a series circuit produces water independently or if it’s the total current across all cells that matters. This leads me to think through the mechanics of a series circuit.

Mapping the chemical reaction

I started by calculating the total charge and then converted it to moles of electrons. Using the reaction stoichiometry, I determined the moles and mass of water produced per hour.

回答

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 (NN): 20
  • Active area per electrode (AA): 100 cm²
  • Current density (JJ): 0.8 A/cm²
  • Time (tt): 1 hour (3600 seconds)
  • Faraday's constant (FF): 96,485 C/mol e\text{e}^-
  • Molar mass of water (MH2OM_{\text{H}_2\text{O}}): 18.015 g/mol

Step 1: Calculate the Total Current

Current per cell (IcellI_{\text{cell}}):

Icell=J×A=0.8A/cm2×100cm2=80AI_{\text{cell}} = J \times A = 0.8 \, \text{A/cm}^2 \times 100 \, \text{cm}^2 = 80 \, \text{A}

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

Itotal=Icell=80AI_{\text{total}} = I_{\text{cell}} = 80 \, \text{A}

Step 2: Calculate the Total Charge Passed in One Hour

Q=Itotal×t=80A×3600s=288,000CQ = I_{\text{total}} \times t = 80 \, \text{A} \times 3600 \, \text{s} = 288,000 \, \text{C}

Step 3: Calculate the Moles of Electrons Transferred

Moles of electrons(ne)=QF=288,000C96,485C/mol2.984mol e\text{Moles of electrons} \, (n_{\text{e}^-}) = \frac{Q}{F} = \frac{288,000 \, \text{C}}{96,485 \, \text{C/mol}} \approx 2.984 \, \text{mol e}^-

Step 4: Determine the Moles of Water Produced

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

Overall Reaction:

2H2+O22H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}
  • Electrons transferred per mole of H2O\text{H}_2\text{O}: 2 mol e⁻
Moles of H2O(nH2O)=ne2=2.984mol e21.492mol\text{Moles of } \text{H}_2\text{O} \, (n_{\text{H}_2\text{O}}) = \frac{n_{\text{e}^-}}{2} = \frac{2.984 \, \text{mol e}^-}{2} \approx 1.492 \, \text{mol}

Step 5: Calculate the Mass of Water Produced

Mass of H2O=nH2O×MH2O=1.492mol×18.015g/mol26.87g\text{Mass of } \text{H}_2\text{O} = n_{\text{H}_2\text{O}} \times M_{\text{H}_2\text{O}} = 1.492 \, \text{mol} \times 18.015 \, \text{g/mol} \approx 26.87 \, \text{g}

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.

問題

In a fuel cell, energy is released when fuel (H2) and oxidant (O2) react in molar ratio producing water (H2O). Assuming origin of energy to be the difference in the energetics of the hydrogen (H)-Oxygen (O) bonds in the product vis-à-vis original hydrogen-hydrogen and oxygen-oxygen bonds in the reactants, calculate the energy released per mole of the product in the reaction; H2 + O2 ↔ H2O at NTP. Given; Standard bond enthalpies at NTP are; H-H bond = 432 kJ mole-1, O = O bond = 494 kJ mole-1 and H-O bond = 460 kJ mole-1.

思考

Beginning the reaction

OK, let me see. Balancing the chemical equation involves ensuring an equal number of each type of atom on both sides of the equation, leading me to balance it correctly.

Breaking down the chemical reaction

I’m working through the balanced equation for the reaction of hydrogen and oxygen to form water, focusing on calculating energy released per mole of product using bond enthalpies.

Analyzing bond transformations

I’m examining the reactants and products, counting the bonds broken and formed, and noting the bond structure in H2O.

Calculating energy changes

I’m curious about the energy released in the formation of water. By totaling the bond energies, the reaction releases 241 kJ/mol.

回答

To calculate the energy released per mole of water (H2O\text{H}_2\text{O}) formed in the reaction:

H2+12O2H2O\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}

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

ΔHrxn=(Bond energies of bonds broken)(Bond energies of bonds formed)\Delta H_{\text{rxn}} = \sum \text{(Bond energies of bonds broken)} - \sum \text{(Bond energies of bonds formed)}

Given Bond Enthalpies:

  • H–H bond: 432kJ/mol432 \, \text{kJ/mol}
  • O=O bond: 494kJ/mol494 \, \text{kJ/mol}
  • H–O bond: 460kJ/mol460 \, \text{kJ/mol}

Step 1: Identify Bonds Broken and Formed

Bonds Broken (Reactants):

  1. One H–H bond in H2\text{H}_2
  2. Half an O=O bond in 12O2\frac{1}{2}\text{O}_2

Bonds Formed (Products):

  1. Two H–O bonds in H2O\text{H}_2\text{O} (since each water molecule has two H–O bonds)

Step 2: Calculate Energy Required to Break Bonds

Energy to break H–H bond=1×432kJ/mol=432kJ\text{Energy to break H–H bond} = 1 \times 432 \, \text{kJ/mol} = 432 \, \text{kJ} Energy to break O=O bond=12×494kJ/mol=247kJ\text{Energy to break O=O bond} = \frac{1}{2} \times 494 \, \text{kJ/mol} = 247 \, \text{kJ} Total energy to break bonds=432kJ+247kJ=679kJ\text{Total energy to break bonds} = 432 \, \text{kJ} + 247 \, \text{kJ} = 679 \, \text{kJ}

Step 3: Calculate Energy Released in Forming Bonds

Energy released in forming H–O bonds=2×460kJ/mol=920kJ\text{Energy released in forming H–O bonds} = 2 \times 460 \, \text{kJ/mol} = 920 \, \text{kJ}

Step 4: Calculate the Net Energy Change

ΔHrxn=Energy to break bondsEnergy to form bonds\Delta H_{\text{rxn}} = \text{Energy to break bonds} - \text{Energy to form bonds} ΔHrxn=679kJ920kJ=241kJ/mol\Delta H_{\text{rxn}} = 679 \, \text{kJ} - 920 \, \text{kJ} = -241 \, \text{kJ/mol}

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.

問題

The charge and mass balance process during fuel cell operation comprise of the following interactions only;

Question 20Select one:

a.
Sulfonate-H3O+ interaction


b.
H2O-H3O+ interaction


c.
H2O-H2O interaction


d.
Sulfonate-H2O interaction


e.
CF2-H3O+ interaction


f.
All except (e)

思考

Evaluating interaction dynamics

I’m analyzing the charge and mass balance process in fuel cells, considering interactions like Sulfonate-H3O+, H2O-H3O+, and others to determine which are essential.

Understanding PEM fuel cells

I'm digging into how proton conduction works in PEM fuel cells, using Nafion, a polymer with fixed sulfonate groups and fluorinated backbones. Hydrated protons move through, involving water molecules.

Assessing interactions

I’m evaluating key interactions essential for proton conduction in a polymer electrolyte membrane. Notably, sulfonate-H3O+ and H2O-H3O+ interactions are vital, while CF2-H3O+ interaction is less critical.

回答

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)

問題

waht is the ans

思考

Answering a question

The assistant determines that the charge and mass balance process in fuel cells involves all interactions except "all except (e)". The user repeats their query likely due to a typographical error, presumably meaning "what is the answer?".

Clarifying the response

OK, I’m working through the user's request and ensuring the answer is both correct and clear. The correct option is f. All except (e).

回答

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

問題

Which of the following is TRUE in a proton exchange membrane fuel cell (PEMFC)?

Question 1Select one:

a.
Presence of H+ ion with adequate available sites in its backbone structure for mobility


b.
All of the above


c.
Adequate porosity permitting ionic mobility


d.
Selective ion permeation properties permitting only H+ ion mobility


e.
Block electronic charge mobility


f.
Hydrophilic properties

思考

Figuring out PEMFC truth

OK, let me see. I’m piecing together the options to pinpoint the most accurate statement about proton exchange membrane fuel cells (PEMFC).

Categorizing the options

I’m gathering info on each statement to find the best truth about proton exchange membrane fuel cells (PEMFC). This involves assessing areas like ionic mobility, selective ion permeation, hydrophilic properties, and influences on electronic charge movement.

Understanding PEMFC רוצה грамотное понимание. Membrane selective to H+ ions, hydrated and blocking electrons is key.

Assessing membrane characteristics

I’m evaluating the presence of H+ ions, adequate porosity, selective ion permeation, and blocking electronic charge mobility to confirm the membrane's functionality.

Assessing membrane properties

I'm noting that the proton exchange membrane fuel cell (PEMFC) leans on hydrophilic traits from sulfonic acid groups, which absorb water crucial for proton conductivity. This points me to the veracity of options a, d, e, and f but casts doubt on option c.

Evaluating correct options

OK, let me see. Options a, d, e, and f align with the PEMFC attributes. Selecting just one is tricky, but option f, addressing hydrophilic properties, could be a key choice.

Evaluating options

I'm thinking through the options, examining how the membrane's hydrated channels or microstructural features enable proton conductivity and ion mobility. It seems clear "All of the above" could align with these concepts.

回答

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

問題

Which fuel cell among the following has the highest reversible efficiency;

Question 2Select one:

a.
Alkaline fuel cell (AFC)


b.
Phosphoric acid fuel cell (PAFC)


c.
Molten carbonate fuel cell (MCFC)


d.
Proton exchange membrane fuel cell (PEMFC)


e.
Solid oxide fuel cell (SOFC)


f.
Direct methanol fuel cell (DMFC)


g.
(a) and (c)

思考

Analyzing fuel cell options

I’m examining various fuel cells to pinpoint the one with the highest reversible efficiency. Each choice, from alkaline to direct methanol, needs careful evaluation.

Evaluating reversible efficiency

I’m digging into reversible efficiency in fuel cells, focusing on thermodynamics and reaction conditions. It’s interesting to see that lower temperatures enhance efficiency by aligning thermodynamic changes of enthalpy and Gibbs free energy.

Temperature constraints

AFC, PEMFC, and DMFC operate at low temperatures (~70-80°C), while MCFC and SOFC function at higher temperatures (~650-1000°C).

Weighing fuel cell types

I’m mapping out the efficiencies of AFC, PEMFC, and DMFC. While hydrogen fuel cells (AFC and PEMFC) tend to perform better in reversible efficiency, AFC slightly edges out PEMFC due to its higher reaction kinetics and lower overpotentials in alkaline environments.

Assessing fuel efficiency

I’m scratching my head over the Alkaline Fuel Cell (AFC) for its simple reaction and lower operating temperature, suggesting it could have a slight edge in reversible efficiency compared to others.

回答

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 (ΔG\Delta G) to the enthalpy change (ΔH\Delta H) of the electrochemical reaction:

Reversible Efficiency (ηrev)=ΔGΔH\text{Reversible Efficiency (} \eta_{\text{rev}} \text{)} = \frac{\Delta G}{\Delta H}

This ratio decreases with increasing temperature because ΔG\Delta G becomes less negative due to the TΔST\Delta S term in the Gibbs free energy equation (ΔG=ΔHTΔS\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 ΔG/ΔH\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)

問題

Origin of energy in a fuel cell followed by its release for applications lies in;

Question 3Select one:

a.
Redox action, energy storage, and release


b.
Redox action and free energy change


c.
Redox action, free energy change, energy conversion, and release


d.
Redox action, resulting potential difference, free energy change, and electrical work done by charge(s) and direct energy conversion


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


f.
Only (d)


g.
(b), (c), and (d)

思考

Evaluating hypothesis options

I’m really mapping out the best energy origin and release mechanisms for fuel cells. The answer needs a solid explanation and aligns with OpenAI's guidance.

Understanding fuel cell energy triggers

I’m focusing on deciphering the origin of energy in fuel cells through redox reactions, free energy change, energy conversion, and achieving potential difference.

Connecting redox reactions

Interestingly enough, the free energy change (ΔG) from redox reactions transitions into electrical energy. Electrons surge from anode to cathode, driving external circuit work and creating a substantial potential difference.

I’m thinking through the completeness of the explanation and determining the best answer.

回答

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 (ΔG\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).

問題

The output current density in a fuel cell can be optimized by control of which of the following parameters;

Question 8Select one:

a.
Increasing reactant concentration


b.
Increasing reaction temperature


c.
Lowering the activation barrier via catalyst


d.
Scaling up the number of reaction sites employing high surface area electrodes


e.
Employing mixed conducting 3 D structural reaction interfaces


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


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

思考

Identifying key features

The question focuses on understanding the unique characteristics of a proton exchange membrane fuel cell (PEMFC), such as H+ ion presence, adequate porosity, selective ion permeation, electronic charge mobility, and hydrophilic properties.

Identifying fuel cell types

I’m mapping out specific fuel cell types:

  • a. Alkaline fuel cell (AFC)
  • g. Combining solid oxide fuel cell (SOFC) with another type.

Figuring out the options

We’re working through the best answer for optimizing fuel cell output current density. Considering reactant concentration, reaction temperature, catalyst enhancement, reaction sites, and mixed conducting structures. To find the optimal choice, further analysis is needed.

Examining fuel cell parameters

I'm gathering info on parameters that can boost fuel cell output current density. This includes reactant concentration, reaction temperature, catalyst activation barrier, high surface area electrodes, and mixed conducting 3D structural reactions.

Evaluating factors

I’m examining options (a) to (e) to optimize fuel cell output current density, considering reactant concentration, temperature, and reaction kinetics.

Evaluating each option

I reviewed methods to boost current density, focusing on catalysts, high surface area electrodes, and mixed conducting interfaces to enhance reaction rates and active sites.

Confirming validity

I’m noting that option (e), which involves employing mixed conducting 3D structural reaction interfaces, is valid. This method enhances electrochemical reactions in Solid Oxide Fuel Cells (SOFCs), supporting the conclusion.

Considering all parameters

I’m mapping out all parameters of a fuel cell to optimize output current density. Each parameter plays a crucial role in achieving the desired efficiency.

Adjusting variables

Increasing reactant concentration, boosting temperature, using catalysts, and adding high surface area electrodes and mixed conducting materials all improve fuel cell performance.

回答

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)

問題

Reversible voltage in a fuel cell depends solely on;

Question 9Select one:

a.
Activity coefficient of fuel (H2) supply


b.
Activity coefficient of oxidant (O2) supply


c.
Operating temperature (T) of fuel cell


d.
Partial pressure of fuel & oxidant


e.
Faraday coefficient and number of mobile charge(s)


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


g.
All of the above

思考

Assessing the influences

I'm analyzing the factors affecting the reversible voltage of a fuel cell, including the activity coefficient of fuel and oxidant, operating temperature, and partial pressures of fuels and oxidants.

Understanding the Nernst equation

Taking a closer look at the Nernst equation, which ties the reversible voltage of a fuel cell to the standard potential and the reaction quotient, influenced by reactant and product concentrations or partial pressures.

Piecing together

I’m thinking about standard cell potential, temperature, and partial pressures or activities of reactants and products in the Nernst equation, and how each option affects these factors.

Examining the factors

I'm curious about the factors influencing the reversible voltage in a fuel cell, like partial pressures and Faraday's constant. They seem crucial to the Nernst equation and thus analyze the voltage.

回答

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:

E=ERTnFlnQE = E^\circ - \frac{RT}{nF} \ln Q

Where:

  • EE = Reversible cell voltage
  • EE^\circ = Standard cell potential
  • RR = Universal gas constant (8.314J/mol\cdotpK8.314\, \text{J/mol·K})
  • TT = Absolute temperature (in Kelvin)
  • nn = Number of electrons transferred in the reaction
  • FF = Faraday's constant (96,485C/mol96,485\, \text{C/mol})
  • QQ = 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 QQ 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 QQ 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 RTnF\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 QQ in the Nernst equation.
  • Conclusion: They directly influence the reversible voltage.

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

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

問題

Fuel cell working depends on a chain of processes comprising; 1. Fuel transport, 2. Oxidant transport, 3. Catalytic action, 4. Electrochemical reaction, 5. Charge (ion and electron) transport, 6. Product removal, 7. Energy release. Which of the following sequence is correct?

思考

Piecing together essential processes

OK, let me see. I’m analyzing the critical stages of fuel cell operation, like fuel transport, catalytic action, and energy release. This helps outline the correct sequence to answer the user's question.

Starting with fuel and oxidant transport, catalytic action and electrochemical reactions follow.

Breaking down the process step-by-step. Hm, I’m thinking about the sequence: fuel transport to anode, oxidant transport to cathode, electrochemical reaction at anode, charge transport via electrons and ions, product removal.

Mapping the sequence

OK, let me see. I’m piecing together the process: fuel transport, oxident transport, catalytic action, electrochemical reaction, charge transport, energy release, and product removal.

回答

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. H22H++2e\text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^-
      • Cathode Reaction (Reduction): Oxidant molecules gain electrons. 12O2+2H++2eH2O\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, H+\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

問題

Which of the following parameters determine fuel cell power and capacity rating?

Question 12Select one:

a.
Size of cell and fuel reservoir respectively


b.
Size of fuel reservoir and cell respectively


c.
Redox action kinetics only


d.
Catalytic action at interconnect metal-electrode interface


e.
Rate of fuel and oxidant flow


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


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

思考

Analyzing fuel cell parameters

I’m mapping out which parameters impact fuel cell power and capacity rating. The choices range from size criteria to chemical actions and flow rates, so I’m considering their relevance to power and capacity.

Assessing fuel cell performance

Considering how factors like cell size, reaction kinetics, catalytic action, reactant flow rate, and design impact the maximum power and energy capacity of fuel cells.

Evaluating options

I'm weighing various choices to pinpoint the right method for assessing fuel cell performance. Each option presents a distinct approach, contrasting methods and adding complexity.

Weighing the options

Noticing the link between cell size and power, I’m thinking fuel reservoir size gauges capacity. The rationale behind these choices aligns well with the concepts.

Analyzing performance factors

I'm looking at how redox kinetic effects and catalytic actions at interfaces impact fuel cell performance, particularly in maximizing power output by influencing current density and efficiency.

回答

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)

問題

Which of the following is not correct in the case of fuel cells?

Question 13Select one:

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


b.
Fuel cells produce DC power


c.
Fuel cell is a clean and green energy alternative


d.
Electrochemical action is the root of fuel cell working


e.
Fuel cell action complies the principles of reversible thermodynamics


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


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

思考

Gathering info

I’m listing the choices to identify if any statement about fuel cells is incorrect, covering energy potential, power type, environmental impact, electrochemical action, thermodynamics, efficiency, and operational principles.

Evaluating fuel cell statements

I’m confirming the accuracy of statements about fuel cells, including their ability to replace fossil fuels, their DC power production, and their classification as a clean and green energy alternative. Each option is being meticulously reviewed to ensure precision.

Assessing fuel cell claims

I’m thinking through the statement that fuel cells' efficiency is lower than internal combustion engines. Fuel cells typically have higher efficiency, making this claim inaccurate.

回答

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)

問題

In a fuel cell reaction comprising molar concentration of fuel and oxidants with charge and mass balance, the total exchange of electrons in the REDOX process is;

Question 14Select one:

a.
4 at anode interface and none at cathode interface


b.
4


c.
None of the above


d.
2 at anode interface and 4 at cathode interface


e.
0


f.
2


g.
3

思考

Understanding fuel cell reactions and exploring electron transfer

Thinking through electron exchange

I'm considering how many electrons move in a fuel cell's redox reaction, focusing on the overall reaction and the quantity exchanged per mole of fuel.

Identifying electron transfer

I noticed that in the hydrogen fuel cell reaction, 2 electrons transfer per H2 molecule: 2 at the anode and 2 at the cathode.

Choosing the most appropriate option

I'm figuring out the correct answer by eliminating unlikely choices and noting the option with the least electrons exchanged during the fuel cell reaction.

回答

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):

At the anode:H22H++2e\text{At the anode:} \quad \text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^-
  • Process:
    • Hydrogen gas (H2\text{H}_2) is oxidized.
    • Each molecule of hydrogen releases 2 electrons.
  • Electrons Produced: 2 electrons per molecule of H2\text{H}_2.

Cathode Reaction (Reduction):

At the cathode:12O2+2H++2eH2O\text{At the cathode:} \quad \frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}
  • Process:
    • Oxygen gas (O2\text{O}_2) is reduced.
    • Consumes 2 electrons per molecule of water formed.
  • Electrons Consumed: 2 electrons per molecule of H2O\text{H}_2\text{O}.

Overall Reaction:

H2+12O2H2O\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}
  • 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

問題

The prime requirements for an effective fuel cell catalyst are;

Question 15Select one:

a.
Activity


b.
Conductivity


c.
Chemical stability


d.
Thermal stability


e.
Thermodynamic stability


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


g.
(a), (b), (c), and (e)

思考

OK, let's break it down: considering the elements typical of proton exchange membrane fuel cells, the ideal answer seems to encompass all presented characteristics.

Highlighting crucial traits

I'm mapping out the essential characteristics of an effective fuel cell catalyst: activity, conductivity, chemical stability, and thermal stability.

Piecing together catalyst requirements

When evaluating fuel cell catalyst needs, I’m focusing on activity, conductivity, chemical stability, and thermal stability. Thermodynamic stability, while beneficial, seems less crucial here.

Weighing key catalyst attributes

I'm evaluating the prime attributes for a fuel cell catalyst, emphasizing activity, conductivity, chemical stability, and thermal stability. The inclusion of thermodynamic stability appears irrelevant here.

Opting for thermal stability feels practical due to its straightforward application in everyday scenarios.

Evaluating catalyst criteria

I'm mapping out the prime criteria for fuel cell catalysts: activity, conductivity, chemical stability, and thermal stability. Each aspect is crucial for efficient catalysis and longevity.

回答

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)

問題

NAFION membrane, the trademark product by Dupont chemicals, in a PEMFC is basically;

Question 7Select one:

a.
An ionomer comprising polymer and ionized components in its backbone structure


b.
A copolymer of tetrafluoroethylene and per fluorinated vinyl ether


c.
A copolymer of tetrafluoroethylene & sulfonic acid coupled per fluorinated vinyl ether


d.
A copolymer of PVDF-TrFE and per fluorinated vinyl ether


e.
(b), (c), and (d)


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


g.
(a) and (c)

思考

Figuring out Nafion

I’m choosing the best option that accurately describes Nafion in terms of its ionomer and copolymer composition. The options need to clarify Nafion's key characteristics.

Clarifying the question

The assistant is determining the correct response by understanding the question's elements, making sure the answer is accurate and compliant with OpenAI's policies.

Validating the details

I’m checking the copolymer of PVDF-TrFE with perfluorinated vinyl ether. Nafion includes PVDF-conjointly with sulfonic acid-based polymer.

Evaluating options

I’m examining various answer choices to find the most accurate ones. Since option d is incorrect, options e and f are off the table. Options a, b, and c seem promising.

Clarifying Nafion's structure

Nafion is an ionomer composed of polymer and ionized components, specifically a copolymer of tetrafluoroethylene and sulfonic acid-coupled perfluorinated vinyl ether, making options (a), (b), and (c) accurate.

Evaluating options

I’m grappling with the dilemma in selecting the optimal response for Nafion. Option f (a), (b), (c), and (d) seems viable despite the incorrect option d. Options a, b, and c lack a perfect match.

回答

Answer: g. (a) and (c)

問題

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

Question 6Select one:

a.
Activation loss


b.
Ohmic loss


c.
Concentration polarization loss


d.
Catalytic losses


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


f.
Electrode-interconnect interface poisoning


g.
None of the above

思考

Assessing fuel cell power losses

I'm thinking through the types of power loss in fuel cells—activation, ohmic, concentration polarization, and catalytic—which the assistant should identify and explain.

Analyzing copolymer choices

I’m mapping out the options for copolymers. Options (a), (b), and (c) involve fluorinated compounds, leading me to conclude that (a) and (c) are the most suitable choices.

Mapping out power loss factors

First, I’m identifying primary sources of power loss in a fuel cell: activation, ohmic, and concentration polarization losses. Catalytic losses might fall under activation, and electrode-interconnect interface poisoning is less common. Option e is the best choice.

Identifying power loss sources

I’m examining the sources of power loss in fuel cells, highlighting activation losses (linked to electrochemical reaction kinetics) and ohmic losses (arising from resistance in the flow paths).

回答

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=IRV = 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)

問題

Which of the following statement is NOT True in the case of fuel cell;

Question 18Select one:

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


b.
Fuel cell efficiency decreases with rise in temperature


c.
Fuel cell efficiency remains constant irrespective of temperature


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


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


f.
Fuel cell voltage is independent of pressure


g.
(c) and (f)

思考

Troubleshooting fuel cell statements

I'm piecing together the true and false statements about fuel cells. It’s interesting to see that fuel cell efficiency changes with temperature, unlike the claim.

Verifying fuel cell statements

I’m checking the truth of statements about fuel cells. EMF depends on reactant and product activity. Fuel cell efficiency drops with rising temperature due to the relationship between ΔG and ΔH.

Evaluating fuel cell claims

Fuel cell efficiency is not constant with temperature (option c). It’s always less than ideal due to irreversibilities and losses (option d). Stable cells require charge and mass balance (option e). Voltage depends on pressure, as per the Nernst equation (option f).

回答

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 (η=ΔG/ΔH\eta = \Delta G / \Delta H).
    • Gibbs free energy (ΔG\Delta G) decreases with increasing temperature (since ΔG=ΔHTΔS\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)

問題

Fuel cells operate on the principle(s) of

Question 19Select one:

a.
Chemical energy storage and electrical energy transfer


b.
Chemical energy conversion and electrical energy release


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


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


e.
Electrochemical action only


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


g.
(c) and (e)

思考

Identifying core principles

I reviewed the question about fuel cell principles. After analyzing available options, I selected the one that accurately reflects fuel cells' operation mechanisms.

Evaluating options

OK, let me see. I’m assessing each option, noting that fuel cells convert chemical energy to electrical energy without storing it. So, options with "chemical energy storage" seem unsuitable.

Pinpointing the core

I’m analyzing fuel cells' principles: converting chemical to electrical energy. Options b and c are on my radar, but it seems "b" is more precise.

Assessing fuel cell properties

I'm thinking through options c and e, analyzing the role of pressure and electrochemical reactions. Options f and g blend other options, guiding me towards the best answer being option b.

Piecing together fuel cell dynamics

Breaking down how a fuel cell leverages redox reactions, electrical work, and free energy to convert chemical energy into direct electrical energy.

回答

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


分享這個問答