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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
Lecture 4. Fuel Cell Reaction Kinetics
  • Introduction to Electrode Kinetics
  • Activation Energy
  • Exchange Current Density
  • Galvani Potential & Butler-Volmer Equation
  • Tafel Equation
  • Electrode Catalysts
slide2

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)

electrochemistry basics
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)
current is a rate
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

current is a rate1
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)

charge is an amount
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)

current density
Current Density

Current density (j)

j = i/A (A/cm2)

Reaction rate per unit area (J)

why charge transfer rxns have an activation energy
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) ]
calculating net reaction rate
Calculating Net Reaction Rate

Forward reaction: M    H  (M + e-) + H+

Reverse reaction: (M + e-) + H+ M   H

-

-

reaction rate at equilibrium
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.

slide17

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

slide20

Galvani Potential

a andc are Galvani potentials.

E0 = a+ c

slide21

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.

slide23

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

slide24

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?

slide25

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.

slide26

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’)
slide27

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

slide29

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.

slide30

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.

slide31

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 (

slide32

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:

slide33

Tafel Equation

Tafel Equation:

slide34

Tafel Equation

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

slide35

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

slide37

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)