Electroanalytical Chemistry

1
Electroanalytical Chemistry

• Lecture 3 Electrolyte - the Unassuming and
  Unappreciated Electrochemical Workhorse

2
Junction Potentials

• Whenever we change electrolyte or solvent there
  is a cost in that we create a double layer
  (electrode) which has a small but often
  measureable potential
• We call this potential a junction potential

3
General - Junction Potentials

• To solve this we must know 3 things
• How the concentration of each ion changes
• How the ion activity varies with concentration
• How the transport number varies with the
  concentration of the ions

4
The Henderson Equation

• We cannot solve analytically so we make 2
  simplifying assumptions
• concentration of each ion varies linearly from
  region 1 to region 2 (1)
• ?i ci, i.e., ? i ? 1 for each ion (2)
• Then we obtain the Henderson equation

5
Classification of Junction Potentials

• 3 Categories of Junction Potentials
• Type 1 2 solutions of same electrolyte with
  different concentrations in contact
• Type 2 2 solutions of same concentrations of
  electrolytes in contact which share a common
  univalent (z1) ion
• Type 3 2 different solutions containing
  different electrolytes and/or different
  concentrations in contact

6
Type I Liquid Junctions

• For Type 1, potential, in V, given by
• Note RT/F 0.02569 V at 298 K
• Checks
• When c2c1, EJ 0
• When tt-, EJ 0 (good electrolyte!)

7
Observations about Type I Liquid Junction
Potentials

• Consider
• If t gt t- ? EJ _______
• If c1 gt c2 ? EJ _______
• So, this tells us if we want to minimize EJ we
  must
• Keep cs ___________________
• Use electrolytes with t______t-

8
EXAMPLE (BF 2.4e)
0.01 M KCl
0.1 M KCl

• Ag/AgCl(s)/K,Cl-(1 M)/K, Cl-(0.1 M)/AgCl(s)/Ag
• Left (1), right (2) so c2 0.1 and c1 1.0
• t 73.52/(73.5276.34) 0.49
• EJ - 0.05916 ((2 0.49)-1)log(0.1) -
  0.05916 (- 0.02) (-1) - 0.001 V
• Note this represents one approach to
  measurement of junction potentials

K
Cl-
9
EXAMPLE 2
0.01 M HCl
0.1 M HCl

• Same junction -HCl not KClAg/AgCl(s)/H,Cl-(1
  M)/H, Cl-(0.1 M)/AgCl(s)/Ag
• Left (1), right (2) so c2 0.1 and c1 1.0
• t 349.82/(349.8276.34) 0.82
• EJ - 0.05916 ((2 0.82)-1)log(0.1) -
  0.05916 (- 0.64) (-1) - 0.038 V
• Note junction potential much larger
  significant perturbation on Ecell

H
Cl-
10
Type 2 Liquid Junctions - the Lewis Sargent
Relation

• Type 2 solutions of two different electrolytes
  having common anion/cation and same concentration
  in contact with each other

11
EXAMPLE

• Calculate the liquid junction potential for
  contact of 0.1 M HCl and 0.1 M KCl at 250C.
• Common anion so negative andEJ - 0.025916 ln
  ((349.8276.34)/(73.5276.34)) - 0.0259 ln
  2.84 - 27 mV

12
Type 3 - the General Case

• 2 solutions of different electrolyte at different
  concentrations in contact with each other
• From table 2.3.3uH 36.25uCl- 7.912uK
  7.619uNO3- 7.404

Region 1
Region 2
0.1 M HCl
0.05 M KNO3
H
Cl-
K
NO3-
13
Mass Transport - Revisited

• Recall 3 modes
• Diffusion
• Migration
• Convection
• If use short times ( lt 10 s) and dont stir
  (quiescent), then i id im

14
Consider Sign of id and im

• Consider current for quiescent solutions of
  (a)Cu2 and (b) Cu(CN)42- at negatively charged
  electrodes
• Conclusion id and im do not have to be in same
  direction

id
_
_
id
Cu(CN)42-
Cu2
im
im
15
EXAMPLE

• Identify and give sign for all contributions to
  current for quiescent solutions of Cu(CN)2 at a
  negatively charged electrode

_
Id ??
Cu(CN)2
Im ??
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Significance

• If our goal is to set up experiments such that
  results are easily quantified one way to do this
  is eliminate/minimize im such that itot id
• We must now consider how we can accomplish this
  experimentally

17
When/How Can We Do This?

• Lets begin by examining what happens in
  different regions of the solution
• A. In bulk solution
• B. Near/at the electrode surface

18
A. In Bulk Solution

• Here concentration gradients are small, so i im
  (i.e., id ? 0)

19
B. At/near the Electrode - The Balance Sheet
Approach

• Steps
• 1. Write reaction
• 2. Identify the two half reactions
• 3. Determine t and t-
• 4. Define total current 10 electrons and
  calculate imj
• 5. Determine moles of product at each electrode
• 6. Determine diffusion contributions at each
  electrode

20
EXAMPLE

• Consider electrolysis of 1 mM HCl at Pt
  electrodes
• (Hint what kind of electrodes are Pt?)

21
B. At/near the Electrode - The Balance Sheet
Approach

• Steps
• 1. Write reaction
• 2. Identify the two half reactions
• 3. Determine t and t-
• 4. Define total current 10 electrons and
  calculate imj
• 5. Determine moles of product at each electrode
• 6. Determine diffusion contributions at each
  electrode

22
Step 1 Write Reaction

• Electrolysis so
• 2HClaq H2? Cl2?
• i.e.,
• 2H 2Cl- H2? Cl2?

23
Step 2 Identify Half-Reactions

• Cathode 2H 2e- H2 ? 0 V
• Anode 2Cl- -2e- Cl2 ? -1.36 V
• Net 2H 2Cl- H2 ? Cl2 ? -1.36 V
• Note -1.36 V Ecell consistent with
  electrolytic process

24
Step 3 Determine t and t-

• Two species present H and Cl-
• Recall
• So, t 349.82/(349.8276.34) 0.82t-
  76.34/(349.8276.34) 0.18

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Step 4 Define Total Current ? 10 e- and
Calculate imj (Bulk)

• Given
• So, imH 10 0.82 8 e- andimCl- 10 0.18
  2 e-
• This means8H 8e- 4H2 at cathodeCl- - 2e-
  Cl2 at anode

26
Putting This into Our Balance Sheet
Pt/H, Cl-(aq, 1 mM)/Pt
Cathode
Anode
10 e-
10 e-
10H 10e- 5H2
10Cl- - 10e- 5Cl2
10Cl-
10H
8 H
2 Cl-
27
Step 5 Determine Moles of Product at Each
Electrode

• If 10 e- involved and if n2 then
• Cathode produces 5H2
• Anode produces 5Cl2
• i.e., 10 H 10 Cl- 5H2? 5Cl2?

28
Step 6 Determine Diffusion Contributions

• Recall, I im id
• So, at cathodeIf I 10 e- and im 8e- then id
  2e- (2H) (note need 2Cl- for
  electroneutrality)
• So, at anodeIf I 10 e- and im 2e- then id
  8e- (8H) (note need 8Cl- for
  electroneutrality)
• Putting this into our Balance Sheet...

29
Putting This into Our Balance Sheet
Pt/H, Cl-(aq, 1 mM)/Pt
Cathode
Anode
10 e-
10 e-
10H 10e- 5H2
10Cl- - 10e- 5Cl2
10Cl-
10H
migration
8 H
2 Cl-
2 H
8 Cl-
diffusion
8 H
2 Cl-
30
EXAMPLE 2

• Consider electrolysis at Pt electrodes of 1 mM
  HCl to which 0.1 M KNO3 is added

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B. At/near the Electrode - The Balance Sheet
Approach

• Steps
• 1. Write reaction
• 2. Identify the two half reactions
• 3. Determine t and t-
• 4. Define total current 10 electrons
• 5. Determine moles of product at each electrode
• 6. Determine diffusion contributions at each
  electrode

32
Step 1 Write Reaction

• Electrolysis so
• 2HClaq H2? Cl2?
• i.e., 2H 2Cl- H2? Cl2?
• Reaction is the same as in previous example
• Electrolyte does NOT participate in redox process!

33
Step 2 Identify Half-Reactions

• Cathode 2H 2e- H2 ? 0 V
• Anode 2Cl- -2e- Cl2 ? -1.36 V
• Net 2H 2Cl- H2 ? Cl2 ? -1.36 V
• Note -1.36 V Ecell consistent with
  electrolytic process
• Same comment as on previous slide

34
Step 3 Determine t and t-

• Four species present H, K, NO3-, and Cl-
• Recall
• Denom (10-3349.82)(10-376.34)(0.173.52)(0.
  171.44) 14.92
• t(H) 10-3349.82/14.92 0.023
• t-(Cl-) 10-376.34/ 14.92 0.005
• t(K) 0.173.52/14.92 0.493
• t-(NO3-) 0.171.44/14.92 0.479

35
Step 4 Define Total Current ? 10 e- and
Calculate imj (Bulk)

• Given
• So, imH 0.023 10 e- 0.23 e-
• imCl- 0.005 10 e- 0.05 e-
• imK 0.493 10 e- 4.93 e-
• imNO3- 0.479 10 e- 4.79 e-

Really small!
Represents bulkof migration I
36
Step 5 Determine Moles of Product at Each
Electrode

• If 10 e- involved and if n2 then
• Cathode produces 5H2
• Anode produces 5Cl2
• i.e., 10 H 10 Cl- 5H2? 5Cl2?
• Note same as first example

37
Step 6 Determine Diffusion Contributions

• Recall, I im id
• So, at cathodeIf I 10 e-, id(H) 10 - 0.23
  9.77 e-
• At anode id(Cl-) 10 - 0.05 9.95 e-
• Electroneutrality id(K) 4.93 e- (?)
  id(NO3-) 4.93 e- ?)
• Putting this into our Balance Sheet...

38
Putting This into Our Balance Sheet
Pt/H, Cl-(aq, 1 mM)/Pt
Cathode
Anode
10 e-
10 e-
10H 10e- 5H2
10Cl- - 10e- 5Cl2
10 e-
10H
10Cl-
0.23 H
0.05 Cl-
migration
4.93 K
4.79 NO3-
9.77 H
9.77 H
diffusion
9.95 Cl-
9.95 Cl-
4.93 K
4.93 K
4.79 NO3-
4.79 NO3-
39
Conclusions About Electrolyte

• Electrolyte dominates migrational contribution
• Note comparatively high concentration of
  electrolyte is required to do this
• Therefore, it simplifies the mass transport
  picture for our analyte

40
Nernst-Planck Equation
Ji(x) flux of species i at distance x from
electrode (mole/cm2 s) Di diffusion coefficient
(cm2/s) ?Ci(x)/?x concentration gradient at
distance x from electrode ??(x)/?x potential
gradient at distance x from electrode ?(x)
velocity at which species i moves (cm/s)
41
If We Use High Concentration of Good Electrolyte
in Quiescent Solution...

• Nernst-Planck Equation reduces to...

42
Ficks First Law

• Flux of substance is proportional to its
  concentration
• i.e., flux ? concentration

43
Ficks Second Law

• Describes how concentration of substance varies
  with time due to diffusion
• is readily solved for wide variety of boundary
  conditions
• requires use of LaPlace transforms
• solution of this equation represents basis of
  many electrochemical experiments

44
What is D0?

• Fourth equivalent measure of the transport
  properties of an ion in solution (10-8 - 10-10
  m2/s)
• Einstein Equation
• Nernst Einstein Equation