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529

Lecture #8 of 17

530

Q: What’s in this set of lectures?

A: B&F Chapters 1, 15 & 4 main concepts:

● Section 1.1: Redox reactions

● Chapter 15: Electrochemical instrumentation

● Section 1.2: Charging interfaces

● Section 1.3: Overview of electrochemical experiments

● Section 1.4: Mass transfer and Semi-empirical treatment of

electrochemical observations

● Chapter 4: Mass transfer

531

Looking forward… Section 1.4 and Chapter 4

● Mass transfer

● Diffusion

● Migration / Drift

● Convection

● Semi-empirical models

● Conductivity

● Transport (Transference) number

● Balance sheets

532… again, at steady-state, ionic migration current must equal electronic current (due to KCL!)

This is a negative current…

}6e– need 6 “+”s,

whose ratios are ti

0.001 M Cu2+/Cu+

… and so is this!

B&F Example #2 (of 3)

migration

533

-5(2) + 7(1) = -3

This is a negative current…

= 3(1) = +3

… neutral!

}6e– need 6 “+”s,

whose ratios are ti

0.001 M Cu2+/Cu+

… and so is this!

… again, at steady-state, counterions move to maintain electroneutrality in diffusion layers


migration

534

5(2) + -7(1) = +3 -5(2) + 7(1) = -3

… and this is a positive current (at that electrode)…

= -3(1) = -3

… also neutral!

}

0.001 M Cu2+/Cu+

… and so is this!

6e– need 6 “+”s,

whose ratios are ti

= 3(1) = +3

… neutral!

… again, at steady-state, counterions move to maintain electroneutrality in diffusion layers


migration

535… RECALL FROM LAST TIME… adding a supporting electrolyte

reduces the original im values

0.001 M Cu2+/Cu+

0.1 M NaClO4

}Na+ClO4

– dominates

because it carries

the most current; its

contribution to σ is

large, and thus the

components have

the largest value of t


migration

536… adding a supporting electrolyte reduces the original im values

0.001 M Cu2+/Cu+

0.1 M NaClO4

But this is actually

much more complex

because of the…

… diffusion effects

here

}Na+ClO4

– dominates

because it carries



large, and thus the

components have



migration


0.001 M Cu2+/Cu+

0.1 M NaClO4


much more complex

because of the…


here }Na+, ClO4

–, and Cl–

should share charge

balance responsibilities

}Na+ClO4

– dominates

because it carries



large, and thus the

components have



migration


0.001 M Cu2+/Cu+

0.1 M NaClO4


much more complex

because of the…


here }Na+, ClO4

–, and Cl–

should share charge


}Na+ClO4

– dominates

because it carries



large, and thus the

components have


CHECK: -5.97(2) + 6.029(1) = -5.911…

migration


0.001 M Cu2+/Cu+

0.1 M NaClO4


much more complex

because of the…


here }Na+, ClO4

–, and Cl–

should share charge


}Na+ClO4

– dominates

because it carries



large, and thus the

components have


CHECK: -5.97(2) + 6.029(1) = -5.911… +2.92(1) + 2.92(1) = 5.84…

migration


0.001 M Cu2+/Cu+

0.1 M NaClO4


much more complex

because of the…


here }Na+, ClO4

–, and Cl–

should share charge


}

CHECK: -5.97(2) + 6.029(1) = -5.911… +2.92(1) + 2.92(1) = 5.84… so, missing +0.071 for Cl– toward the cathode

… and -0.071 for Cl– away from the anode

Na+ClO4– dominates

because it carries



large, and thus the

components have


migration


0.001 M Cu2+/Cu+

0.1 M NaClO4

PRECISION: -5.9709(2) + 6.0291(1) = -5.9127… +2.91(1) + 2.91(1) = 5.82… so, +0.0927 for Cl– toward the cathode

2.91

2.91 2.91

2.91

5.9709 5.9709

6.0291 6.0291

0.0927, so sum is 6

* Balance sheet

now balances

after corrections

here, which are

“underlined”

… and -0.0927 for Cl– away from the anode

* This is a very

challenging sign

convention, so

don’t worry

about it. Just

use absolute

numbers!

migration

542

FeIII(CN)63- + 1e- ⇌ FeII(CN)6

4-

cathode anode

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

?

? ?

* anions move toward the anode

* cations move toward the cathode

Let’s end this discussion of balance sheets with our own example…

Consider a 0.10 M equimolar solution of K3Fe(CN)6 and K4Fe(CN)6:

543

1) Calculate all t’s (for now, assume mobilities are equal, and thus cancel).

1.0

+

2) Convert t’s to migrational ion displacements…

𝒕𝐅𝐞(𝐂𝐍)𝟔

𝟑− =𝟑 𝟎. 𝟏 𝐱 𝟏𝟎−𝟑 𝐦𝐨𝐥/𝐜𝐦𝟑

𝟑 𝟎. 𝟏 𝐱 𝟏𝟎−𝟑 𝐦𝐨𝐥/𝐜𝐦𝟑 + 𝟒 𝟎. 𝟏 𝐱 𝟏𝟎−𝟑 𝐦𝐨𝐥/𝐜𝐦𝟑 + 𝟏 𝟎. 𝟕 𝐱 𝟏𝟎−𝟑 𝐦𝐨𝐥/𝐜𝐦𝟑


𝟑− =𝟑 𝟎. 𝟏

𝟑 𝟎. 𝟏 + 𝟒 𝟎. 𝟏 + 𝟏 𝟎. 𝟕= 𝟎. 𝟐𝟏𝟒 ≈ 𝟎. 𝟐


𝟒− =𝟒 𝟎. 𝟏

𝟑 𝟎. 𝟏 + 𝟒 𝟎. 𝟏 + 𝟏 𝟎. 𝟕= 𝟎. 𝟐𝟖𝟔 ≈ 𝟎. 𝟑

𝒕𝐊+ =𝟏 𝟎. 𝟕

𝟑 𝟎. 𝟏 + 𝟒 𝟎. 𝟏 + 𝟏 𝟎. 𝟕≈ 𝟎. 𝟓

5442) Convert t’s to migrational ion displacements...

3) Calculate diffusional ion displacements by difference:

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-

545

? ?

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-


4-

cathode anode

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

10 e-

10 Fe(CN)64-

10 Fe(CN)63-



5462) Convert t’s to migrational ion displacements...

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-


cathode10 Fe(CN)6

3- (needed) = -2/3 Fe(CN)63- (mig) + 10 2/3 Fe(CN)6

3- (diff)

-10 Fe(CN)64- (needed) = -3/4 Fe(CN)6

4- (mig) – 9 1/4 Fe(CN)64- (diff)

anode-10 Fe(CN)6

3- (needed) = 2/3 Fe(CN)63- (mig) – 10 2/3 Fe(CN)6

3- (diff)

10 Fe(CN)64- (needed) = 3/4 Fe(CN)6

4- (mig) + 9 1/4 Fe(CN)64- (diff)

547

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-


4-

cathode anode

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

10 e-

10 Fe(CN)64-

10 Fe(CN)63-



- 9 1/4 Fe(CN)64-

10 2/3 Fe(CN)63- - 10 2/3 Fe(CN)6

3-

9 1/4 Fe(CN)64-

548

4) Counter ions move to maintain electroneutrality in diffusion layers:

cathode(+10 2/3 Fe(CN)6

3-)(3) + (-9 1/4 Fe(CN)64-)(4)

32 – 37 = -5 (means +5 counterion diffusion… away from the electrode)

anode(+9 1/4 Fe(CN)6

4-)(4) + (-10 2/3 Fe(CN)63-)(3)

37 – 32 = +5 (means -5 counterion diffusion… toward the electrode)


cathode10 Fe(CN)6

3- (needed) = -2/3 Fe(CN)63- (mig) + 10 2/3 Fe(CN)6

3- (diff)

-10 Fe(CN)64- (needed) = -3/4 Fe(CN)6

4- (mig) – 9 1/4 Fe(CN)64- (diff)

anode-10 Fe(CN)6

3- (needed) = 2/3 Fe(CN)63- (mig) – 10 2/3 Fe(CN)6

3- (diff)

10 Fe(CN)64- (needed) = 3/4 Fe(CN)6

4- (mig) + 9 1/4 Fe(CN)64- (diff)

549

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-


4-

cathode anode

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

10 e-

10 Fe(CN)64-

10 Fe(CN)63-



5 K+ - 5 K+

- 9 1/4 Fe(CN)64-

10 2/3 Fe(CN)63- - 10 2/3 Fe(CN)6

3-

9 1/4 Fe(CN)64-

550

so, for the reaction of Fe(CN)63- at the cathode:

5. Finally, we could tally im and id:

𝑖𝑚 =− 2 3

− 2 3 + 10 2 3

= −0.066 (–) means migration will

"reduce" the total current

551


and, for the reaction of Fe(CN)64- at the anode:


(–) means migration will


(+) … the diffusional current

should always be positive

𝑖𝑚 =− 2 3

− 2 3 + 10 2 3

= −0.066

𝑖𝑑 =10 2 3

− 2 3 + 10 2 3

= +1.066

552

10 x 0.5 x 1 = 5 K+

10 x 0.2 x 1/3 = 2/3 Fe(CN)63-

10 x 0.3 x 1/4 = 3/4 Fe(CN)64-


4-

cathode anode

10 e-

10 Fe(CN)64-

10 Fe(CN)63-

10 e-

10 Fe(CN)64-

10 Fe(CN)63-



5 K+ - 5 K+

- 9 1/4 Fe(CN)64-

10 2/3 Fe(CN)63- - 10 2/3 Fe(CN)6

3-

9 1/4 Fe(CN)64-

553




Id will be larger by 6.6% as

compared to its value in the

absence of migration

Id will be smaller by 7.5% as



𝑖𝑚 =− 2 3

− 2 3 + 10 2 3

= −0.066

𝑖𝑑 =10 2 3

− 2 3 + 10 2 3

= +1.066

𝑖𝑚 =+ 3 4

+ 3 4 + 9 1 4

= +0.075

𝑖𝑑 =9 1 4

+ 3 4 + 9 1 4

= +0.925



(+) means migration will

“add to" the total current

554



5. Finally, we could tally im and id: … Boom!… Tough problem analyzed!

𝑖𝑚 =− 2 3

− 2 3 + 10 2 3

= −0.066

𝑖𝑑 =10 2 3

− 2 3 + 10 2 3

= +1.066

𝑖𝑚 =+ 3 4

+ 3 4 + 9 1 4

= +0.075

𝑖𝑑 =9 1 4

+ 3 4 + 9 1 4

= +0.925

Id will be larger by 6.6% as



Id will be smaller by 7.5% as





(+) means migration will

“add to" the total current

555… so supporting electrolyte removes (most) drift for redox species

of interest… but it also removes iRu drop from data (2-for-1!)…

… let’s just look at some figures: 1, 4, 5, 7, 14 ,16



Japp = constant

Problems



Japp = constant

Problems Solutions



Japp = constant

during CV

Problems Solutions

559Day-long 1st exam assigned “tomorrow” and due ≤ Friday

For HW: Box in answers! For B&F 2.4, do not include

junction potentials; for B&F 2.11, Hittorf method is

discussed in JCE 2001 paper; Problem #2 is cumulative

from the class (and does not cover Fick’s 2nd Law)…

Uncompensated resistance:

κ = σ Planar WE

… current is limited by largest resistor in series… which is at WE

… and finally, iRu drop does not always have a typical “R”…

x

Units: Ohm-1 cm-1







κ = σ Planar WE Point WE see B&F,pp. 27–28,and pp. 216–218 for details

Němec, J. Electroanal. Chem., 1964, 8, 166


What happens when x ∞?


x







κ = σ Planar WE Point WE see B&F,pp. 27–28,and pp. 216–218 for details

Němec, J. Electroanal. Chem., 1964, 8, 166


What happens when x ∞?


x

Just this term... with 1/circumference in place of x/A

562

A review of Section 1.4 (and Chapter 4)

● Mass transfer

● Diffusion

● Migration / Drift

● Convection

● Semi-empirical models

● Conductivity

● Transport (Transference) number

● Balance sheets

563

Q: What was in this set of lectures?

A: B&F Chapters 1, 15 & 4 main concepts:

● Section 1.1: Redox reactions

● Chapter 15: Electrochemical instrumentation

● Section 1.2: Charging interfaces

● Section 1.3: Overview of electrochemical experiments

● Section 1.4: Mass transfer and Semi-empirical treatment of

electrochemical observations

● Chapter 4: Mass transfer

564

Time-Dependence in

Electrochemistry

Chapters 4 and 5

565

Q: What’s in this set of lectures?

A: B&F Chapters 4 & 5 main concepts:

● Section 4.4.2: Fick’s Second Law of Diffusion

● Section 5.1: Overview of step experiments

● Section 5.2: Potential step under diffusion controlled

● Sections 5.3 & 5.9: Ultramicroelectrodes

● Sections 5.7 – 5.8: Chronoamperometry/Chronocoulometry

566

Fick’s 1st Law of Diffusion:Adolph Eugen Fick

B&F, pg. 149

… derive it (approximately) in a similar fashion as the diffusion coefficient…get your favorite beverage!

we use both of Fick’s laws of diffusion to derive equations for

transport-controlled electrochemistry...

567


B&F, pg. 149

This is the net flux (correct dimensions)…… with half moving right and half moving left



568


B&F, pg. 149



569


B&F, pg. 149

… derived!Recall…



570

Fick’s 1st Law of Diffusion:

Fick’s 2nd Law of Diffusion:

Adolph Eugen Fick



B&F, pg. 149

… derive it (approximately) in asimilar fashion as Fick’s first law…

571

B&F, pp. 149–150

… derivation is not so bad…

FOR YOUR REFERENCE

572

B&F, pp. 149–150


FOR YOUR REFERENCE

573

(First Law)

… derived!

B&F, pp. 149–150


FOR YOUR REFERENCE

574The experiment we will model is a potential step experiment…

key points: at E1: no reaction (CO(x, 0) = CO*)

at E2: diffusion-controlled reaction (CO(0, t) = 0)

Eeq

> 200 mV

> 200 mV

How to derive expressions for diffusion-controlled current vs. time: 575

1. Solve Fick’s Second Law to get CO(x, t), and in the

process of doing this, you will use boundary conditions

that “customize” the solution for the particular

experiment of interest:






2. Use Fick’s First Law to calculate JO(0, t) from CO(x, t):






3. Calculate the time-dependent diffusion-limited current:

i = nFAJO(0, t)








i = nFAJO(0, t)


… using the… Laplace transform, integration by parts, L’Hôpital’s rule, Schrödinger

equation, complementary error function, Leibniz rule, chain rule… Cool!

579Step 1 is the kicker… we’ll use the Laplace Transform to solve the

linear partial differential equation

The Laplace transform of any function F(t) is:

580

how about F(t) = 1?

Step 1 is the kicker… we’ll use the Laplace Transform to solve the



𝐿 1 =

0

∞

𝑒−𝑠𝑡 1 𝑑𝑡 = 𝑒−𝑠𝑡

−𝑠0

∞

= 0 −1

−𝑠=

1

𝑠

581

how about F(t) = 1?

Step 1 is the kicker… we’ll use the Laplace Transform to solve the



𝐿 1 =

0

∞

𝑒−𝑠𝑡 1 𝑑𝑡 = 𝑒−𝑠𝑡

−𝑠0

∞

= 0 −1

−𝑠=

1

𝑠

𝐿 𝑘𝑡 =

0

∞

𝑒−𝑠𝑡 𝑘𝑡 𝑑𝑡 = 𝑘

0

∞

𝑡𝑒−𝑠𝑡𝑑𝑡 = 𝑘𝑒−𝑠𝑡

𝑠2−𝑠𝑡 − 1

0

∞

how about F(t) = kt?

582Integrated by partshow about F(t) = kt?

𝐿 𝑘𝑡 =

0

∞


0

∞


𝑠2−𝑠𝑡 − 1

0

∞

583Integrated by parts

Used L’Hôpital’s rule


𝐿 𝑘𝑡 =

0

∞


0

∞


𝑠2−𝑠𝑡 − 1

0

∞

= 𝑘 0 −1

𝑠2−1 =

𝑘

𝑠2

584

how about F(t) = e–at?

Integrated by parts



𝐿 𝑘𝑡 =

0

∞


0

∞


𝑠2−𝑠𝑡 − 1

0

∞

= 𝑘 0 −1

𝑠2−1 =

𝑘

𝑠2

𝐿 𝑒−𝑎𝑡 =

0

∞

𝑒−𝑠𝑡𝑒−𝑎𝑡𝑑𝑡 =

0

∞

𝑒− 𝑠+𝑎 𝑡𝑑𝑡 = 𝑒− 𝑠+𝑎 𝑡

− 𝑠 + 𝑎0

∞

585

how about F(t) = e–at?

how about F(t) = kt? Integrated by parts


𝐿 𝑘𝑡 =

0

∞


0

∞


𝑠2−𝑠𝑡 − 1

0

∞

= 𝑘 0 −1

𝑠2−1 =

𝑘

𝑠2

𝐿 𝑒−𝑎𝑡 =

0

∞

𝑒−𝑠𝑡𝑒−𝑎𝑡𝑑𝑡 =

0

∞

𝑒− 𝑠+𝑎 𝑡𝑑𝑡 = 𝑒− 𝑠+𝑎 𝑡

− 𝑠 + 𝑎0

∞

= 0 −1

− 𝑠 + 𝑎=

1

𝑠 + 𝑎

586

OK, now for our case:

Recall, Second Law:

–F(t) =

587

OK, now for our case: –F(t) =

–L{ }= ?

588

well, wait a second, this

term is not so bad…

OK, now for our case: –F(t) =

–L{ }= ?

0

∞

𝑒−𝑠𝑡𝐷𝜕2𝐶 𝑥, 𝑡

𝜕𝑥2𝑑𝑡 = 𝐷

𝜕2

𝜕𝑥2

0

∞

𝑒−𝑠𝑡𝐶 𝑥, 𝑡 𝑑𝑡 = 𝐷𝜕2

𝜕𝑥2 𝐶 𝑥, 𝑠

589

the Laplace transform

of C(x, t)? … Isn’t this

cheating?




–F(t) =

–L{ }= ?

0

∞



𝜕2

𝜕𝑥2

0

∞



590

the Laplace transform

of C(x, t)? … Isn’t this

cheating? Well, ahem,

no!




–F(t) =

–L{ }= ?

0

∞



𝜕2

𝜕𝑥2

0

∞



591

?


not so lucky with

this term…

–F(t) =

–L{ }= ?

−𝐷𝜕2


592

g(x) f '(x)

Integration, by parts, again!

0

∞

𝑒−𝑠𝑡𝜕𝐶𝑂 𝑥, 𝑡

𝜕𝑡𝑑𝑡

𝑎

𝑏

𝑔 𝑥 𝑓′ 𝑥 𝑑𝑥 = 𝑔 𝑥 𝑓 𝑥𝑎

𝑏−

𝑎

𝑏

𝑓 𝑥 𝑔′ 𝑥 𝑑𝑥

593

… and at time = 0, what is the value of C, anywhere?


g(x) f '(x)

0

∞


𝜕𝑡𝑑𝑡

𝑎

𝑏


𝑏−

𝑎

𝑏


= 𝑒−𝑠𝑡𝐶 𝑥, 𝑡0

∞−

0

∞

𝐶 𝑥, 𝑡 −𝑠𝑒−𝑠𝑡 𝑑𝑥

= 0 − 𝐶 𝑥, 0 + 𝑠 𝐶 𝑥, 𝑠

594

… and at time = 0, what is the value of C, anywhere?

… just C*!


g(x) f '(x)

0

∞


𝜕𝑡𝑑𝑡

𝑎

𝑏


𝑏−

𝑎

𝑏


= 𝑒−𝑠𝑡𝐶 𝑥, 𝑡0

∞−

0

∞

𝐶 𝑥, 𝑡 −𝑠𝑒−𝑠𝑡 𝑑𝑥

= 0 − 𝐶 𝑥, 0 + 𝑠 𝐶 𝑥, 𝑠

595L.T. of Fick’s 2nd Law…

now is turns out that the

L.T. of this…

see B&F,pg. 775,for details

is this…

–F(t) =

–L{ }

𝑠 𝐶 𝑥, 𝑠 − 𝐶∗ − 𝐷𝜕2




L.T. of this…

is this…

our equation:

Now what? Well, recall these terms are equal to each other (= 0), then rearrange…

… and what does it look like?

–F(t) =

–L{ }

𝑠 𝐶 𝑥, 𝑠 − 𝐶∗ − 𝐷𝜕2




L.T. of this…

is this…

our equation:

Now what? Well, recall these terms are equal to each other (= 0), then rearrange…

… and what does it look like?

–F(t) =

–L{ }

𝑠 𝐶 𝑥, 𝑠 − 𝐶∗ − 𝐷𝜕2


the time-independent

Schrödinger Eq.

in 1D…

𝑑2

𝑑𝑥2𝜓 𝑥 −

2𝑚

ℏ2𝐸 − 𝑉 𝑥 𝜓 𝑥 = 0

598

the solution of the

Schrödinger Eq. is:

our equation:


Schrödinger Eq.

in 1D…

𝑑2


2𝑚

ℏ2𝐸 − 𝑉 𝑥 𝜓 𝑥 = 0

𝜓 𝑥 = 𝐴′exp− 2𝑚 𝐸 − 𝑉 𝑥

ℏ𝑥 + 𝐵′exp

2𝑚 𝐸 − 𝑉 𝑥

ℏ𝑥

599

… and by analogy, the solution of our equation is:

the solution of the

Schrödinger Eq. is:

our equation:


Schrödinger Eq.

in 1D…

𝑑2


2𝑚

ℏ2𝐸 − 𝑉 𝑥 𝜓 𝑥 = 0

𝜓 𝑥 = 𝐴′exp− 2𝑚 𝐸 − 𝑉 𝑥

ℏ𝑥 + 𝐵′exp

2𝑚 𝐸 − 𝑉 𝑥

ℏ𝑥

𝐶 𝑥, 𝑠 =𝐶∗

𝑠+ 𝐴′ 𝑠 exp −

𝑠

𝐷𝑥 + 𝐵′ 𝑠 exp

𝑠

𝐷𝑥

600

Now, what are A' and B' and how do we get rid of the “s”?

… we need some boundary conditions!



𝑠


𝑠

𝐷𝑥

601

called semi-infinite linear

(because of x) diffusion



1.

L.T.



𝑠


𝑠

𝐷𝑥

lim𝑥→∞


𝑠

602

called semi-infinite linear

(because of x) diffusion



1.



𝑠


𝑠

𝐷𝑥

lim𝑥→∞


𝑠



𝑠


𝑠

𝐷𝑥

What does this do for us?

0… and so B' must be equal to 0

∞

L.T.

603

some more boundary conditions…

2.

L.T.

𝐶 0, 𝑠 = 0

𝐶 0, 𝑡 = 0



𝑠

𝐷𝑥

604

some more boundary conditions…

What does this do for us?



𝑠

𝐷𝑥

2.

L.T.

𝐶 0, 𝑠 = 0

𝐶 0, 𝑡 = 0

1

𝐴′ 𝑠 = −𝐶∗

𝑠… and so



𝑠

𝐷𝑥

605now our solution is fully constrained… and we need “t” back!!

inverse L.T. using Table A.1.1 in B&F

=

606What’s efrc?… Well, first of all, what’s the error function: erf?

607Now… what’s efrc?

608Now… what’s efrc?

Gaussian distribution,

with mean = 0 and std.

dev. = 1/sqrt(2)

609

… well, for large x, erfc = 0 (erf = 1) and so C(x, t) = C* … Check!

… and for x = 0, erfc = 1 (erf = 0) and so C(x, t) = 0 … Check!

… Let’s plot it!

Does this make sense?

610

C* = 1 x 10-6 M

D = 1 x 10-5 cm2 s-1

t = 1s

0.1s0.01s0.0001s

Hey, these look completely reasonable!

(100 µm)

611How large is the diffusion layer? Recall the rms displacement…

root mean square

(rms) displacement

(standard deviation) Δ = 2𝑑 𝐷𝑡, where d is the dimension

D Δ*=

1D

2D

3D

*the rms displacement

… plane

… wire, line, tube

… point, sphere, disk

a characteristic

"diffusion length"

2𝐷𝑡

4𝐷𝑡

6𝐷𝑡

Δ = 2𝑑 𝐷𝑡 =cm2

ss = cm

From a…

612

4 µm

14 µm 40 µm

t = 1s

0.1s

Hey, these look completely reasonable!

0.01s0.0001s

… use the geometric area for calculations

C* = 1 x 10-6 M

D = 1 x 10-5 cm2 s-1

2𝐷𝑡 =(100 µm)

OK… that’s Step #1… 613






i = nFAJO(0, t)


614

… but we just derived CO(x, t):

… now Step #2…

(Fick’s First Law)

… and so we need to evaluate:

−𝐽𝑂 𝑥, 𝑡 = 𝐷𝑂

𝜕

𝜕𝑥𝐶∗erf

𝑥

2 𝐷𝑂𝑡

615… now Step #2…

… we use the Leibniz rule, to get d/dx(erf(x)) as follows:



𝜕

𝜕𝑥𝐶∗erf

𝑥

2 𝐷𝑡

616… now Step #2…



… and using this in conjunction with the chain rule, we get:


𝜕

𝜕𝑥𝐶∗erf

𝑥

2 𝐷𝑡

−𝐽𝑂 𝑥, 𝑡 = 𝐷𝑂𝐶∗1

2 𝐷𝑂𝑡

2

𝜋exp

−𝑥2

4𝐷𝑂𝑡

617


… now Step #2…


… and using this in conjunction with the chain rule, we get:

… and when x = 0 (at the electrode), we have what we wanted…


𝜕

𝜕𝑥𝐶∗erf

𝑥

2 𝐷𝑡

−𝐽𝑂 0, 𝑡 = 𝐶∗𝐷𝑂

𝜋𝑡

−𝐽𝑂 𝑥, 𝑡 = 𝐷𝑂𝐶∗1

2 𝐷𝑂𝑡

2

𝜋exp

−𝑥2

4𝐷𝑂𝑡

618

1. Solve Fick’s Second Law to get CO(x, t). In the



experiment of interest.


i = nFAJ(0, t)

2. Use Fick’s First Law to calculate J0(0, t) from C(x, t):

OK… that’s Steps #1 and 2…

619… and finally, Step #3 using Step #2…


𝜋𝑡

the Cottrell Equation

… and with i = nFAJ(0, t)…

620… and finally, Step #3 using Step #2…

the Cottrell Equation

Frederick Gardner Cottrell, in 1920b. January 10, 1877, Oakland, California, U.S.A.

d. November 16, 1948, Berkeley, California, U.S.A.

… established Research Corp. in 1912

… initial funding from profits on patents for

the electrostatic precipitator, used to clear

smokestacks of charged soot particles

… and with i = nFAJ(0, t)…


𝜋𝑡

621

http://en.wikipedia.org/wiki/Electrostatic_precipitatorhttp://en.wikipedia.org/wiki/Corona_discharge

> 1 kVneutral particles

corona dischargecapture plates

charged particles

Cottrell, then at UC Berkeley, invented the electrostatic precipitator

used to clear smokestacks of charged soot particles…

http://en.wikipedia.org/wiki/Electrostatic_precipitator

623

D = 1.5 x 10-6 cm2 s-1

D = 1 x 10-5 cm2 s-1

… OK, so what does it predict?

the Cottrell

Equation

(10 ms)

624

D = 1.5 x 10-6 cm2 s-1

D = 1 x 10-5 cm2 s-1


the Cottrell

Equation

625… OK, so what does it predict?

the Cottrell

Equation

slope =

nFAπ -1/2D1/2C*

626

slope =

nFAπ -1/2D1/2C*


?????

the Cottrell

Equation

627

1. Huge initial currents… compliance current!

2. Noise.

3. Roughness factor adds to current at early times.

4. RC time limitations yield a negative deviation at short times.

5. Adsorbed (electrolyzable) junk adds to current at short times.

6. Convection, and “edge effects,” impose a “long” time limit on

an experiment.

… use the Cottrell Equation to measure D!

… but what are the problems with this approach?

the Cottrell

Equation

628… use the Cottrell Equation to measure D!

… but what are the problems with this approach?

… solution: Integrate it, with respect to time:

the integrated

Cottrell Equation

1. Huge initial currents… compliance current!

2. Noise.

3. Roughness factor adds to current at early times.

4. RC time limitations yield a negative deviation at short times.

5. Adsorbed (electrolyzable) junk adds to current at short times.

6. Convection, and “edge effects,” impose a “long” time limit on

an experiment.

the Cottrell

Equation

lecture #8 of 17 - uci department of chemistryardo/echem/uci-chem248... · … again, at...

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