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Lecture 3.1
Denoting Electrochemical Cells
So, instead of drawing:
Pt
Pt
1 M KCL
KCl (satd.)
Glass
Frit
1 M HCl
Hg2Cl2
10-3 M Fe3+
Hg
10-3 M Fe2+
Pt
SCE
How about a short-hand method?
Pt/Fe3+(10-3M), Fe2+(10-3M), KCl (1M)//KCl(satd.)/Hg2Cl2/Hg/Pt
“/” = phase boundary
“//” = ionically porous material (liquid junction)
+ 0.241 V
vs. NHE
Hg2Cl2 + 2e-
SCE
2Hgo + 2Cl-
Another Example:
Pt/Fe(CN)63-(10-3M), Fe(CN)64-(10-4M), KCl (1M)//KCl(satd.)/Hg2Cl2/Hg/Pt
or
Pt/Fe(CN)63-(10-3M), Fe(CN)64-(10-4M), KCl(0.1 M)//SCE
We find the Ecell is
.010V different. Why?
Lecture 3.2
Look closer:
Cl-
K+
K+ ClCl-
0.1M
or
1.0M
K+
K+ Cl- K+
K+Cl- K+ ClCl- Cl- Cl- K+
Cl- K+ K+
1M
There exists a driving force, DE, due to dC/dx of the ions,
both K+ and ClThe ions want to “equilibrate” but would take a long time to do so.
The potential is known as:
- membrane potential
- diffusion potential
- Liquid Junction Potential
This potential ( E = i Rion motion membrane) will alter E measurement.
Lecture 3.3
Types of L. J. s
All result in
LJP due to
different
chemical
potentials.
1. Same ion / different [ ion ] c+x-llc+x2. Different ions / same [ ion ] c+x-llc+y3. Different ions / different [ ion ] c+x-llc+yWhat else?
4. Different solvent! Different ions!........
…………..Ayy!!!!!
Ecell = ENernst + Ej
Make corrections?
Measure it?
Okay but minimize it!
Use ions with similar ionic
mobilities and [high].
See page 72!!
M+ X-
R
u1 u2
+
d
d
u1 ≈ u2
(KCl)
Lecture 3.4
Let’s move on to more details of what happens when a Polarized
Electrode is placed in a solution with electrolyte and
various redox species.
Pt – WE, SCE – Ref, Pt – Aux
H2O, 1.0 M HCl, 1 mM Fe3+, 1 mM Cr3+
What does voltammogram look like?
1. What reactions can occur at WE?
- What are E 0 of all species?
H2O, Cr3+, H+, Cl-, Fe3+,
Pt

What is E 0 vs. SCE?
- What are electroactive species?
- What is electrode material?
Fe3+/2+ E 0 = + 0.77 V vs. NHE
0
Cr3+/2+ E = - 0.41 V vs. NHE
SCE
0
+ 0.242 V
- 0.242 V
NHE
0
Hg2Cl2 + 2e-
2Hg + 2ClSat’d KCl
So always subtract
+ 0.242 from NHE to
get SCE value.
E
0
Lecture 3.5
Fe
3
/ Fe
2
vs. SCE = +0.53 V
E 0 Cr 3 / Cr 2 vs. SCE = -0.65 V
fast
+eH+ -e- ½ H2
Record Voltammogram:
Pt
e- + Fe3+
(
Fe2+
1000x il of Fe3+/ Fe2+)
No Cr3+
Cr2+!
ilim
ilim ic
0
E vs. SCE
ia
H2O
2H2O
½ O2 + 2H+ + 2eO2 + 4H+ + 4e-
Hmmm… Try Hg drop electrode:
e- + Cr3+
Cr2+
ic
0 ia
Hg
E vs. SCE
Hg2+ + 2eSo, both k and
G limitations!
Lecture 3.6
How do e-s get into molecules?
Well, molecules have orbitals (MO) where either
bonding, non-bonding, or anti-bonding electrons can
reside.
We will remove/add e- to these orbitals where non-bonding
e- can reside.
“effective p-stat”
Potentiostat
and
Energy of
Molecular orbitals
eenergy
(Eelectrode more -
)
So, recall DG = -nFEelectrode. We can make energy of
electrode what we like (within reason).
So, once M is within ~ 10 Å, e- transfers.
Unoccupied
MO
(LUMO)
Occupied
MO
(HOMO)
Lecture 3.7
or
Electrode
Solution (molecule)
Electrode
Solution
(molecule)
But, in order for E – T to occur, must also allow D – L to charge.
What does i – E curve (voltammogram) look like for just HCl?
(Pt WE)
H+ + e-
0
½ H2
E vs. SCE
No E –T occurs! But, some i flows.
qm
+
+
+
+
+
WHY?
-
+
+
+
+
qs
-
+
+
+
+
+
- +
+ +
Lecture 3.8
qm = -qs , Always!!!
Basic Capacitor:
NOT
C
C
D.C. Voltage
Source (fixed)
q
Coulomb

 farad
Eapp
Volt
Depending on the metal, there will be some potential or
some point on the potential axis where no charge will reside
on the metal.
*THIS IS NOT NECESSARILY 0 V vs. REF!!!*
P Z C
qm ~ 0
o of e
i r
n o
t
h
a
r
g
e
bulk
+, - conc
sol’n
bulk
+, - conc
C+, At
PZC
x
C+, -
At
Eapp
x
What is physical structure of the interface ( Double-Layer) ?
+
solvent dipole
-
Solvated or ligated cation of
electrolyte (non-specifically
absorbed)
M+
Z+
Non-solvated cation (electrolyte, etc.)
(typically specifically adsorbed)
Inner
Helmholtz
Plane
(IHP)
qi
Lecture 3.9
Outer Helmholtz Plane
(OHP)
qi + qo = qsoln
qo
E-T occurs at OHP for freely diffusing materials.
EMetal
(~10-200Å depending on [SE])
Energy
ESoln
E
x distance (Å)
IHP
OHP
~3-5Å
DE
0.1V

 2x106V / cm
8
Dx 5x 10 cm
Lecture 3.10
Response from D-L with E step:
E app
i (t ) 
Rs
i(t)
0.37
Eapp
Rs

E app
Rs
e t / RsCDL
= RsCDL
Time to decrease by
0
t
0
0.63 of

E app
Rs
What if we actually scan E?
Rs
CDL
v
i
Scan E at
Eapp
dE
dt
Ei
Rs
~cst i
at
cst v
i(t)
Ei
0
dE
(V/s)
dt
t
t
Lecture 3.11
i(t) = vCd l
If RsCdl is
small and v
is cst
 Ei
  t / Rs C d l 
 

 R  vCdl 
e

 s

i(t)~vCdl
So what if we scan E to Efinal and reverse?
i(t)
Eapp Ef
vCdl
0
Ei
-vCdl
tr
0
tr
t
t
or vs. Eapp
vCdl
i(E)
0
Eapp