Electroanalytical Chemistry
Lecture 3,Electrolyte - the
Unassuming and Unappreciated
Electrochemical Workhorse
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
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
? i
n
i i
i
J
d
F
RT
z
t
E ln
2
1R e g i o n
1
? ?
?
??
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:
? ?
? ? ?
?
?
?
?
?
?
?
?
?
?
n
i
iii
n
i
iii
n
i
iiii
n
i
ii
i
ii
J
cuz
cuz
ccuz
cc
z
uz
E
F
RT
1
1
1
1
)1(
)2(
ln
)1()2(
)1()2(
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 (z=1) ion
? Type 3,2 different solutions containing
different electrolytes and/or different
concentrations in contact
Type I Liquid Junctions
? For Type 1,potential,in V,given by:
? Note,RT/F = 0.02569 V at 298 K
? Checks,
? When c2=c1,EJ = 0
? When t+=t-,EJ = 0 (good electrolyte!)
? ? cc
c
ctE w h e r e
zF
RT
J 21
1
2 _ln12 ????
?
? ?
c
ctE
z
VKat J
1
2lo g120 5 9 1 6.0)(,2 9 8_ ???
?
Observations about Type I Liquid
Junction Potentials
? Consider:
? If t+ > t- ? EJ _______
? If c1 > c2 ? EJ _______
? So,this tells us if we want to minimize EJ
we must:
? Keep c’s ___________________
? Use electrolytes with t+______t-
EXAMPLE (B&F 2.4e)
? 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/(73.52+76.34) = 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
0.01 M
KCl
0.1 M
KCl
K+Cl-
? Same junction -HCl not KCl:
Ag/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/(349.82+76.34) = 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
EXAMPLE 2 0.01 M
HCl
0.1 M
HCl
H+ Cl-
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
c a t i o nc o m m o nif
M
M
F
RT
X
X
JE
__;ln
1
2 ???
?
?
EXAMPLE
? Calculate the liquid junction potential for
contact of 0.1 M HCl and 0.1 M KCl at
250C.
? Common anion so negative and
EJ = - 0.025916 ln
((349.82+76.34)/(73.52+76.34)) = - 0.0259
* ln 2.84 = - 27 mV
Type 3 - the General Case
? 2 solutions of different electrolyte at
different concentrations in contact with each
other
? From table 2.3.3:
uH+ = 36.25
uCl- = 7.912
uK+ = 7.619
uNO3- = 7.404
0.1 M
HCl
0.05 M
KNO3
H+
Cl-
K+
NO3-
Region 1 Region 2
Mass Transport - Revisited
? Recall 3 modes:
? Diffusion
? Migration
? Convection
? If use short times ( < 10 s) and don’t stir
(quiescent),then i = id + 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
Consider Sign of id and im
_
id
im im
id
Cu2+ Cu(CN)4
2-
_
? Identify and give sign for all contributions
to current for quiescent solutions of
Cu(CN)2 at a negatively charged electrode:
EXAMPLE
_
Id
Im
Cu(CN)2
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
When/How Can We Do This?
? Let’s begin by examining what happens in
different regions of the solution:
A,In bulk solution
B,Near/at the electrode surface
A,In Bulk Solution
? Here concentration gradients are small,so i
= im (i.e.,id ? 0)
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
EXAMPLE
? Consider electrolysis of 1 mM HCl at Pt
electrodes
? (Hint,what kind of electrodes are Pt?)
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
Step 1,Write Reaction
Electrolysis so:
2HClaq = H2? + Cl2?
i.e.,
2H+ + 2Cl- = H2? + Cl2?
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
Step 3,Determine t+ and t-
? Two species present,H+ and Cl-
? Recall:
? So,t+ = 349.82/(349.82+76.34) = 0.82
t- = 76.34/(349.82+76.34) = 0.18
?
?
j
jjj
iii
i
cz
cz
t
?
?
Step 4,Define Total Current ?
10 e- and Calculate imj (Bulk)
? Given:
? So,imH+ = 10 * 0.82 = 8 e- and
imCl- = 10 * 0.18 = 2 e-
? This means:
8H+ + 8e- = 4H2 at cathode
Cl- - 2e- = Cl2 at anode
c u r r e n tt o t a lIw h e r e
I
z
ti
j
j
mj
__ ??
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
10H+ 10Cl-
8 H+
2 Cl-
Step 5,Determine Moles of
Product at Each Electrode
? If 10 e- involved and if n=2 then:
? Cathode,produces 5H2
? Anode,produces 5Cl2
? i.e.,10 H+ + 10 Cl- = 5H2? + 5Cl2?
Step 6,Determine Diffusion
Contributions
? Recall,I = im + id
? So,at cathode:
If I = 10 e- and im = 8e- then id = 2e- (2H+)
(note,need 2Cl- for electroneutrality)
? So,at anode:
If I = 10 e- and im = 2e- then id = 8e- (8H+)
(note,need 8Cl- for electroneutrality)
? Putting this into our Balance Sheet...
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
10H+ 10Cl-
8 H+
2 Cl-
migration
diffusion
2 H+
8 H+2 Cl-
8 Cl-
EXAMPLE 2
? Consider electrolysis at Pt electrodes of
1 mM HCl to which 0.1 M KNO3 is added
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
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!
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
Step 3,Determine t+ and t-
? Four species present,H+,K+,NO3-,and Cl-
? Recall:
Denom = (10-3*349.82)+(10-3*76.34)+(0.1*73.52)+(0.1*71.44)
= 14.92
t+(H+) = 10-3*349.82/14.92 = 0.023
t-(Cl-) = 10-3*76.34/ 14.92 = 0.005
t+(K+) = 0.1*73.52/14.92 = 0.493
t-(NO3-) = 0.1*71.44/14.92 = 0.479
?
?
j
jjj
iii
i cz
czt
?
?
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-
c u r r e n tt o t a lIw h e r e
I
z
ti
j
j
mj
__ ??
Really small!
Represents bulk
of migration I
Step 5,Determine Moles of
Product at Each Electrode
? If 10 e- involved and if n=2 then:
? Cathode,produces 5H2
? Anode,produces 5Cl2
? i.e.,10 H+ + 10 Cl- = 5H2? + 5Cl2?
? Note,same as first example
Step 6,Determine Diffusion
Contributions
? Recall,I = im + id
? So,at cathode:
If 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...
Putting This into Our Balance Sheet
Pt/H+,Cl-(aq,1 mM)/PtCathode
Anode
10 e-
10 e-
10H++ 10e- = 5H2 10Cl- - 10e- = 5Cl2
10H+ 10Cl-
0.23 H+
0.05 Cl- migration
diffusion9.77 H
+
9.95 Cl-9.95 Cl-
9.77 H+
4.93 K+
4.79 NO3-
10 e-
4.93 K+
4.79 NO3- 4.79 NO3-
4.93 K+
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
Nernst-Planck Equation
? ? ? ? ? ? ? ?xx xRT Fx xx CCDzCDJ iiiiiii ?? ???????? ?
Diffusion Migration Convection
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)
If We Use High Concentration of
Good Electrolyte in Quiescent
Solution...
Nernst-Planck Equation reduces to...
Fick’s First Law
? Flux of substance is proportional to its
concentration:
? i.e.,flux ? concentration
? ? ? ?
x
txtx CDJ o
o ?
???,,
0
Fick’s 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
? ? ? ?
x
CDC tx
t
tx
o
o
o
2
2
,,
?
?
?
? ?
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
s
F
RT m
z
uD
i
i
oi /;
2?
?
??
?
??
zeD i
i
oi
kT
22
?
?
?
?
?
?
?
?
?
?
Lecture 3,Electrolyte - the
Unassuming and Unappreciated
Electrochemical Workhorse
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
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
? i
n
i i
i
J
d
F
RT
z
t
E ln
2
1R e g i o n
1
? ?
?
??
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:
? ?
? ? ?
?
?
?
?
?
?
?
?
?
?
n
i
iii
n
i
iii
n
i
iiii
n
i
ii
i
ii
J
cuz
cuz
ccuz
cc
z
uz
E
F
RT
1
1
1
1
)1(
)2(
ln
)1()2(
)1()2(
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 (z=1) ion
? Type 3,2 different solutions containing
different electrolytes and/or different
concentrations in contact
Type I Liquid Junctions
? For Type 1,potential,in V,given by:
? Note,RT/F = 0.02569 V at 298 K
? Checks,
? When c2=c1,EJ = 0
? When t+=t-,EJ = 0 (good electrolyte!)
? ? cc
c
ctE w h e r e
zF
RT
J 21
1
2 _ln12 ????
?
? ?
c
ctE
z
VKat J
1
2lo g120 5 9 1 6.0)(,2 9 8_ ???
?
Observations about Type I Liquid
Junction Potentials
? Consider:
? If t+ > t- ? EJ _______
? If c1 > c2 ? EJ _______
? So,this tells us if we want to minimize EJ
we must:
? Keep c’s ___________________
? Use electrolytes with t+______t-
EXAMPLE (B&F 2.4e)
? 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/(73.52+76.34) = 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
0.01 M
KCl
0.1 M
KCl
K+Cl-
? Same junction -HCl not KCl:
Ag/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/(349.82+76.34) = 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
EXAMPLE 2 0.01 M
HCl
0.1 M
HCl
H+ Cl-
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
c a t i o nc o m m o nif
M
M
F
RT
X
X
JE
__;ln
1
2 ???
?
?
EXAMPLE
? Calculate the liquid junction potential for
contact of 0.1 M HCl and 0.1 M KCl at
250C.
? Common anion so negative and
EJ = - 0.025916 ln
((349.82+76.34)/(73.52+76.34)) = - 0.0259
* ln 2.84 = - 27 mV
Type 3 - the General Case
? 2 solutions of different electrolyte at
different concentrations in contact with each
other
? From table 2.3.3:
uH+ = 36.25
uCl- = 7.912
uK+ = 7.619
uNO3- = 7.404
0.1 M
HCl
0.05 M
KNO3
H+
Cl-
K+
NO3-
Region 1 Region 2
Mass Transport - Revisited
? Recall 3 modes:
? Diffusion
? Migration
? Convection
? If use short times ( < 10 s) and don’t stir
(quiescent),then i = id + 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
Consider Sign of id and im
_
id
im im
id
Cu2+ Cu(CN)4
2-
_
? Identify and give sign for all contributions
to current for quiescent solutions of
Cu(CN)2 at a negatively charged electrode:
EXAMPLE
_
Id
Im
Cu(CN)2
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
When/How Can We Do This?
? Let’s begin by examining what happens in
different regions of the solution:
A,In bulk solution
B,Near/at the electrode surface
A,In Bulk Solution
? Here concentration gradients are small,so i
= im (i.e.,id ? 0)
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
EXAMPLE
? Consider electrolysis of 1 mM HCl at Pt
electrodes
? (Hint,what kind of electrodes are Pt?)
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
Step 1,Write Reaction
Electrolysis so:
2HClaq = H2? + Cl2?
i.e.,
2H+ + 2Cl- = H2? + Cl2?
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
Step 3,Determine t+ and t-
? Two species present,H+ and Cl-
? Recall:
? So,t+ = 349.82/(349.82+76.34) = 0.82
t- = 76.34/(349.82+76.34) = 0.18
?
?
j
jjj
iii
i
cz
cz
t
?
?
Step 4,Define Total Current ?
10 e- and Calculate imj (Bulk)
? Given:
? So,imH+ = 10 * 0.82 = 8 e- and
imCl- = 10 * 0.18 = 2 e-
? This means:
8H+ + 8e- = 4H2 at cathode
Cl- - 2e- = Cl2 at anode
c u r r e n tt o t a lIw h e r e
I
z
ti
j
j
mj
__ ??
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
10H+ 10Cl-
8 H+
2 Cl-
Step 5,Determine Moles of
Product at Each Electrode
? If 10 e- involved and if n=2 then:
? Cathode,produces 5H2
? Anode,produces 5Cl2
? i.e.,10 H+ + 10 Cl- = 5H2? + 5Cl2?
Step 6,Determine Diffusion
Contributions
? Recall,I = im + id
? So,at cathode:
If I = 10 e- and im = 8e- then id = 2e- (2H+)
(note,need 2Cl- for electroneutrality)
? So,at anode:
If I = 10 e- and im = 2e- then id = 8e- (8H+)
(note,need 8Cl- for electroneutrality)
? Putting this into our Balance Sheet...
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
10H+ 10Cl-
8 H+
2 Cl-
migration
diffusion
2 H+
8 H+2 Cl-
8 Cl-
EXAMPLE 2
? Consider electrolysis at Pt electrodes of
1 mM HCl to which 0.1 M KNO3 is added
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
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!
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
Step 3,Determine t+ and t-
? Four species present,H+,K+,NO3-,and Cl-
? Recall:
Denom = (10-3*349.82)+(10-3*76.34)+(0.1*73.52)+(0.1*71.44)
= 14.92
t+(H+) = 10-3*349.82/14.92 = 0.023
t-(Cl-) = 10-3*76.34/ 14.92 = 0.005
t+(K+) = 0.1*73.52/14.92 = 0.493
t-(NO3-) = 0.1*71.44/14.92 = 0.479
?
?
j
jjj
iii
i cz
czt
?
?
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-
c u r r e n tt o t a lIw h e r e
I
z
ti
j
j
mj
__ ??
Really small!
Represents bulk
of migration I
Step 5,Determine Moles of
Product at Each Electrode
? If 10 e- involved and if n=2 then:
? Cathode,produces 5H2
? Anode,produces 5Cl2
? i.e.,10 H+ + 10 Cl- = 5H2? + 5Cl2?
? Note,same as first example
Step 6,Determine Diffusion
Contributions
? Recall,I = im + id
? So,at cathode:
If 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...
Putting This into Our Balance Sheet
Pt/H+,Cl-(aq,1 mM)/PtCathode
Anode
10 e-
10 e-
10H++ 10e- = 5H2 10Cl- - 10e- = 5Cl2
10H+ 10Cl-
0.23 H+
0.05 Cl- migration
diffusion9.77 H
+
9.95 Cl-9.95 Cl-
9.77 H+
4.93 K+
4.79 NO3-
10 e-
4.93 K+
4.79 NO3- 4.79 NO3-
4.93 K+
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
Nernst-Planck Equation
? ? ? ? ? ? ? ?xx xRT Fx xx CCDzCDJ iiiiiii ?? ???????? ?
Diffusion Migration Convection
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)
If We Use High Concentration of
Good Electrolyte in Quiescent
Solution...
Nernst-Planck Equation reduces to...
Fick’s First Law
? Flux of substance is proportional to its
concentration:
? i.e.,flux ? concentration
? ? ? ?
x
txtx CDJ o
o ?
???,,
0
Fick’s 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
? ? ? ?
x
CDC tx
t
tx
o
o
o
2
2
,,
?
?
?
? ?
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
s
F
RT m
z
uD
i
i
oi /;
2?
?
??
?
??
zeD i
i
oi
kT
22
?
?
?
?
?
?
?
?
?
?
