Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) ...
-
Upload
jean-warren -
Category
Documents
-
view
221 -
download
4
Transcript of Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) ...
![Page 1: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/1.jpg)
kf kb
v = k C(0,t) = i/nFA
mole/cm2-s = cm/s × mole/cm3 = (coul/s)(mole/coul)(1/cm2)
κ is a transmission coefficient
Electrode Kinetics: Charge transfer control (Activation Polarization)
![Page 2: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/2.jpg)
v = kC(0,t) = i/nFA i = nFAkC(0,t)
inet = nFA[kf CO(0,t) - kbCR(0,t)]
vf = kf CO(0,t)
O + e- = R
vb= kbCR(0,t)
vnet=vf-vb = kf CO(0,t) - kbCR(0,t)
Since v is measured in moles/cm2-s if we multiply by the number of coulombs per mole(nFA) in the rate equation we get a current.
Now consider a 1e- redox reaction
We can write a rate expression for each and the net rate is just the difference.
![Page 3: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/3.jpg)
O+e- R
Apply and E > Eo so that (E-Eo ) > 0. The activation barrier for the anodic reactionhas been reduced by and that for the cathodic reduction reactionHas been increased by .
Reaction coordinate Reaction coordinate
The effect of applied potential on activation free energy
![Page 4: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/4.jpg)
f = F/RT
Application of a potential greaterthan Eo reduces the Act. Barrier forthe oxidation (anodic) process and the activation barrier for the reduction(cathodic) process.
and these Act. Free energies wewrite the forward and backwardrate constants as,
From our general expression for therate constant,
where
![Page 5: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/5.jpg)
Inserting these relations in to inet = nFA[kf CO(0,t) - kbCR(0,t)] yields,
![Page 6: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/6.jpg)
Note that if we set i = 0
Then E = Eequil and we recover the Nernst Equation.
Even though the net current is zero at the equilibrium potential, the oxidation and reduction currents are not – they just add together to yield zero since they are equalIn magnitude. We can express this so-called exchange current (or exchange current density) In terms of either ia or ic,
or
![Page 7: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/7.jpg)
Raising both sides of to the –α power yields
And substitution into the expression for the exchange current yields an expressionFor io in terms of the bulk concentrations of O and R,
dividing
by the expression for io
![Page 8: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/8.jpg)
and after a bit of algebraic manipulation
If the surface and bulk concentrations of O and R are equal then there are NO masstransfer effects. The system is under activation (charge transfer)control.
Butler-Volmer Equation
![Page 9: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/9.jpg)
The symmetry factor, α
α is dependent on the energy landscape in transitioning from O to R and vice versa.This term together with the io term in the B-V equation contains lots of buried physics,much if it poorly understood.
![Page 10: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/10.jpg)
Butler-Volmer Equation
There are two regimes of behavior: High η: At 298 K this occurs when
Tafel Equation
Low η:
Tafel Plot
charge transfer resistance Rct
![Page 11: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/11.jpg)
![Page 12: Kfkf kbkb v = k C(0,t) = i/nFA mole/cm 2 -s = cm/s × mole/cm 3 = (coul/s)(mole/coul) (1/ cm 2 ) Electrode Kinetics: Charge.](https://reader030.fdocuments.net/reader030/viewer/2022032523/56649d745503460f94a54abf/html5/thumbnails/12.jpg)