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Lecture 4 . Fuel Cell Reaction Kinetics

Lecture 4 . Fuel Cell Reaction Kinetics . Introduction to Electrode Kinetics Activation Energy Exchange Current Density Galvani Potential & Butler- Volmer Equation Tafel Equation Electrode Catalysts. Fuel Cell Performance Curve. Reversible Voltage (Chapter 2).

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Lecture 4 . Fuel Cell Reaction Kinetics

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  1. Lecture 4. Fuel Cell Reaction Kinetics • Introduction to Electrode Kinetics • Activation Energy • Exchange Current Density • Galvani Potential & Butler-Volmer Equation • Tafel Equation • Electrode Catalysts

  2. Fuel Cell Performance Curve Reversible Voltage (Chapter 2) Activation Loss (Lecture 4) Ohmic Loss (Lecture 5) Cell voltage(V) Cell voltage(V) Cell voltage(V) Current Density (A/cm2) Current Density (A/cm2) Current Density (A/cm2) Concentration Loss (Lecture 6) Net Fuel Cell Performance Cell voltage(V) Cell voltage(V) Current Density (A/cm2) Current Density (A/cm2)

  3. Electrochemistry Basics • Electrochemical rxns DIFFER from chemical rxns • Electrochemical rxns are HETEROGENEOUS • Current is a RATE (i = dQ/dt = zF (dN/dt) = zFv ) • Charge is an AMOUNT ( • Current density is more important than current • An ACTIVATION BARRIER (DG‡) determines rxn rate (and therefore current)

  4. Electrochemical Processes Are Heterogeneous

  5. Current Is a Rate Example 3-1. Assuming 100% fuel utilization, how much current can a fuel cell produce if provisioned with 5 sccm H2 gas at STP? (sccm = standard cubic centimeter per minute) Assume sufficient oxidant is supplied. i = dQ/dt = zF (dN/dt) = zFv

  6. Current Is a Rate v= dN/dt = p(dV/dt)/RT == 2.05x10-4 (mol/min) v = 2.05x10-4 (mol/min) i = zFv = (2) (96485) (2.05x10-4)/60 = 0.659 (A)

  7. Charge Is An Amount Example 3-2. A fuel cell operates for 1 h at 2A current load and then operates for 2 more h at 5A current load. Calculate the total amount of H2 consumed by the fuel cell. Assume 90% fuel utilization. Q = (i1t1 +i2t2) = (2A)(3600s) + (5A)(7200S)=43,200 C NH2 = Q/(zF)/90% = 43200/(2*96485)/0.9 =0.248 (mole)

  8. Current Density Current density (j) j = i/A (A/cm2) Reaction rate per unit area (J)

  9. Potential Controls Electron Energy

  10. Reaction Rates Are Finite

  11. Why Charge Transfer RXNs Have An Activation Energy ? Five steps for H2 oxidation • H2 (bulk)  H2 (near electrode) • H2 (near electrode) +M  M  H2 • M   H2+M  2 (M   H) • 2 x [M   H  (M+e-) + H+(near electrode)] • 2 x [ H+(near electrode) H+(bulkelectrolyte) ]

  12. Why Charge Transfer RXNs Have An Activation Energy ?

  13. Why Charge Transfer RXNs Have An Activation Energy ?

  14. Activation Energy Determines Reaction Rate

  15. Calculating Net Reaction Rate Forward reaction: M    H  (M + e-) + H+ Reverse reaction: (M + e-) + H+ M   H - -

  16. Reaction Rate at Equilibrium is the exchange current density. Although at equilibrium the net reaction rate is zero, both forward and reverse Reactions are taking place at a rate which is characterized by j0; this is called dynamic equilibrium.

  17. Galvani Potential a) Chemical Free Energy (M…H) DGorxn + (M + e-) + H+ Distance From Interface b) Electrical Energy -nFf = Distance From Interface c) DG‡ Chemical + Electrical Energy j0 Distance From Interface

  18. Galvani Potential

  19. Galvani Potential

  20. Galvani Potential a andc are Galvani potentials. E0 = a+ c

  21. Bulter-Volmer Equation For symmetric reaction, the transfer coefficient α=0.5. For most electrochemical reactions, α=0.2 – 0.5. is a voltage loss to the Galvani potential.

  22. Bulter-Volmer Equation j

  23. 1.2 Cell voltage(V) 0.5 Current density (mA/cm2) 1000 Activation Overvoltage actrepresents a voltage loss due to activation, also called overvoltage. Theoretical EMF or Ideal voltage hact j0 = 10-2 j0 = 10-5 j0 = 10-8

  24. Activation Overvoltage Example 3.3. If a fuel cell reaction exhibits α=0.5 and z=2 at rom temperature, what activation overvoltage is required to increase the forward current density by one order of magnitude and decrease the reverse current density by one order of magnitude?

  25. Activation Overvoltage Solution: Since α=0.5, the reaction is symmetric. We can look at either the forward or the reverse term in the Butler-Volmer equation to calculate the overvoltage necessary to cause an order-of-magnitude change in current density.

  26. How to Improve Kinetic Performance? Improving kinetic performance = increasing jo There are four major ways to increase jo • Increase CR • Decrease DG‡ • Increase T • Increase Reaction Sites (A/A’)

  27. How a PEMFC Works? Flow Structure Electrode/Catalyst Ion Membrane Proton Flow Electron Flow 2H2→ 4H+ + 4e- H2 Anode e- “MEA” H+ Membrane Cathode H2O O2 O2 + 4H+ + 4e-→ 2H2O

  28. How a SOFC Works?

  29. Increase Reactant Concentration The benefit to increase reactant concentration for increasing the voltage is limited (due to the logarithmic form in the Nernst equation). In contrast, the kinetic benefit to increasing reactant concentration is significant, with a linear rather than logarithmic impact. Unfortunately, the penalty of decreasing the reactant concentration is also significant.

  30. Decrease Activation Barrier A decrease in the activation barrier,, will increase the current density (j0). This is usually achieved by changing the free energy surface of the reaction on different electrode catalysts. Using different catalysts also affects the α.

  31. Increase T and Reaction Sites Increasing the T could accelerate the reaction and therefore the current density (j0). In actual fuel cell, the T effect is more complicated. Its effect is decided by the sign of { }. = C exp (

  32. Tafel Equation When act is very small (<15 mV at 298K, or j<<j0) 2. When act is large (>50-100 mV at 298K, or j>>j0) Tafel Equation:

  33. Tafel Equation Tafel Equation:

  34. Tafel Equation Example 3.4. Calculate j0 and α from the hypothetical reaction in the following figure. Assume T=298 and z=2.

  35. Tafel Equation j0= e-10 =4.54x10-5 (A/cm2). From the figure the slope of the Tafel equation is about (0.25-0.10)/(-5-(-8)) = 0.05. From the slope we can calculate α = [RT/(slope * zF)] = (8.314*298/(0.05*2*96485) α= 0.257

  36. Tafel Equation

  37. Fuel Cell Kinetic Problems • Given hact = 200mV at j = 1 A/cm2 (a=0.5, T=300K, z=2), calculate jo. (0.436 mA/cm2) • If jo = 1x10-2 A/cm2 when T is increased to 350K, calculate DG‡. (52.0 kJ/mol) • What is hact at j = 1 A/cm2 now that T=350K? ( 138.9 mV)

  38. Electrode Catalysts

  39. Electrode Catalysts

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