Electrochemistry Electrochemistry is a very diverse area. It can be broadly divided into a) analytical electrochemistry which is concerned with methods of measurement.

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Transcript Electrochemistry Electrochemistry is a very diverse area. It can be broadly divided into a) analytical electrochemistry which is concerned with methods of measurement.

Electrochemistry
Electrochemistry is a very diverse area. It can be broadly divided into
a)
analytical electrochemistry which is concerned with methods of measurement involving
potentiometry (pH meters etc), voltammetry, and modern sensors (generally voltammetric
in nature)
b)
physical electrochemistry is the basis of analytical electrochemistry, but is generally
concerned with the effect of electric fields, charge balance, and diffusion.
c)
Chemical electrochemistry usually is less interested in quantitative analysis but is devoted
to understanding the mechanisms of electron transfer based on chemical structure.
d)
Biologic electrochemistry can be understood as a form of physical electrochemistry
(charge and fields around biomolecules) and of chemical electrochemistry (electron transfer
events in biological systems)
e)
Geologic and environmental electrochemistry is concerned with all of the above as they
take place in the natural environment . Of particular interest are the oxidation reduction
reactions of iron, manganese, chromium, arsenic, sulfur, as these set the parameters in
which life can grow and, as the oxidation state of iron has substantial implications for the
stability of various minerals and their dissolution/formation.
f)
Technical - in this field fall the major areas of batteries, solar energy, fuel cells, and
corrosion sciences.
Electrochemistry
Fundamental concepts
Electron flow, nomenclature
Energy in the electron flow
Kinetics of the electron transfer event as
influenced by
a) energy
b) homogeneous vs heterogeneous system
Electrochemistry
Energy :
1.
2.
3.
Heat out/Heat in = enthalpy, H
Equilibrium Constants, K
Universal Entropy ~ Free energy, G
Use total free energy – not just enthalpy

aq

4 ,aq
8H  MnO
2 
aq 
 5Fe
2
aq
Mn
3
aq
 5Fe
 4 H2 O
This reaction is exothermic.
How much?
 Hrxo 
species
8H+(aq)
4H2O(liquid)
MnO4-(aq)
Mn2+(aq)
5Fe2+(aq)
5Fe3+(aq)
o
n

H
 i , products f ,i , products 
Hfo (kJ/mol)
0
-286
-543
-218.8
-47.69
-87.86
o
n

H
 i ,reac tan ts f ,i ,reac tan ts
 H rxo  1  218.8  5  87.86  4  286
 8 0  1  543  5  47.69
Hrxo    2126    78145
. 
Hrxo  1344kJ
This reaction could do a lot of work for
us if we could get the energy

aq
2
aq
8H  5Fe


4 ,aq 
 MnO
Fe2+
2
aq
Mn
3
aq
 5Fe
MnO4-
Hrxo  1344kJ
 4 H2 O
Heat water to convert energy?
inefficient
clumsy
heat loss
Water Bath
Fe2+
MnO4-
Water circulation
steam
drive pistons
Make
electricity
Alternative Strategy: Capture electrons directly
Split into 2 partial reactions
Fe2+
MnO4e
Mn2+
Fe3+
To do this
Need to balance
Redox reactions
Writing the Net Ionic Reaction of an Oxidation-Reduction Reaction
Split into Reactions to be balanced
Write oxidation half reaction
Write reduction half reaction
A. Balance atoms of element
oxidized
b. Balance Rx sides by adding e
c. Balance charge by adding H+
or OHd. Balance hydrogen by adding
H2O
e. Balance oxygen
A. Balance atoms of element
reduced
b. Balance Rx sides by adding e
c. Balance charge by adding H+
or OHd. Balance hydrogen by adding
H2O
e. Balance oxygen
Combine and balance electrons
Writing the Net Ionic Reaction of an Oxidation-Reduction Reaction
Reaction to be balanced
Write oxidation half reaction
Write reduction half reaction
A. Balance atoms of element
A. Balance atoms of element
reduced
oxidized
b. Balance Ox number with e
b. Balance Ox number with e
c. Balance charge by adding H+
c. Balance charge by adding H+
or OHor OHProtons
are added d. Balance hydrogen by adding
d. Balance hydrogen
by adding
in an acidic solution
H2O
H2O
and hydroxyls are e. Balance oxygen
e. Balance oxygen
added in a basic
solution.
Combine and balance electrons
When in doubt add protons
Example: Balance the following redox equation in an acidic solution
2
( aq )
Fe

4( aq )
 MnO
3
( aq )
 Fe
Ox #balance
2+
3+
2
( aq )
Fe
2
( aq )
+7
 Mn

4( aq )
MnO
3
( aq )
 Fe
+2
2
( aq )
 Mn
Mass balance
Mass balance
2
( aq )
Fe
3
( aq )
 Fe
ox# balance
e
+2
-1 = 4(-2) +?
8-1 = ?
7=?

4( aq )
MnO
2
( aq )
 5e  Mn
Example: Balance the following redox equation in an acidic solution
2
( aq )
Fe
2+
2
( aq )
Fe

4( aq )
 MnO
3
( aq )
2
( aq )
 Fe
3+
+7
 Mn

4( aq )
MnO
3
( aq )
 Fe
+2
2
( aq )
 Mn
Mass balanced
Mass balanced
2
( aq )
Fe
3
( aq )
 Fe
e

Ox numbers 2
4( aq )
( aq )
MnO
 5e  Mn
+2
-1 + -5 = -6

4( aq )
MnO
Charge balance

2
( aq )
 5e  8Haq  Mn
Example: Balance the following redox equation in an acidic solution
2
( aq )
Fe
2
( aq )
Fe
3
( aq )
 Fe

4( aq )
 MnO
 Fe

4( aq )
e
MnO
3
( aq )
2
( aq )
 Mn

( aq )
 5e  8H
2
( aq )
 Mn
Balance hydrogen by adding water

4( aq )
MnO

( aq )
 5e  8H
Check oxygen balance
4O
2
( aq )
 Mn
=+2
Check total charge balance: (-1)+5(-1)+8(+1)=+2
 4H2 O
=4O
Example: Balance the following redox equation in an acidic solution
2
( aq )
Fe
2
( aq )
Fe
3
( aq )
 Fe

4( aq )
 MnO
3
( aq )
 Fe
2
( aq )
 Mn
e

4( aq )
MnO

( aq )
 5e  8H
 Mn
Recombine while balancing electrons
2
( aq )
5Fe
3
( aq )
 5Fe
 5e
2
( aq )
 4H2 O
Example: Balance the following redox equation in an acidic solution
2
( aq )
Fe
2
( aq )
Fe
3
( aq )
 Fe

4( aq )
 MnO
3
( aq )
 Fe
 Mn
e

4( aq )
MnO

( aq )
 5e  8H
2
( aq )
5Fe

4( aq )
MnO
2
( aq )
2
( aq )
 5Fe

( aq )
 8H
 4H2 O
3
( aq )
 5e
2
( aq )
 4H2 O
 Mn
 5Fe
3
( aq )
 5Fe
2
( aq )
 Mn
Final equation does not show any electrons
because electrons “do not exist” in solution
OJO: Important Point
The reactions

4( aq )
MnO

( aq )
 5e  8H
2
( aq )
5Fe
1.
2.
3.
2
( aq )
 Mn
3
( aq )
 5Fe
 4H2 O
 5e
do not really occur by themselves
they are linked through the electrons
DEFINITION: = 1/2 reactions
If all reactions are considered
Half reactions where does the
Electron go?


aq
e  nH2 O  e

aq

e  nH3O  H   H2 O

eaq
 H2 O  H   HO 
10
1 1
k , 2.3x10 M s
k , 19
. x101 M 1 s 1
Hydrated electrons not only react with water
But with other species including biological,
Hence it is a good way to sterilize water

aq

e  RSH  HS  R 



eaq
 O2  O2  HOO   HO 
pK HA 4.9
H2 O
10
1 1
k , 11
. x10 M s
k , ~ 1010 M 1 s 1
Hydrated electrons in aerobic biology will produce
Finite fluxes of the soft radical HOO
Not only have to consider rate
But energy


aq
e  nH2 O  e

eaq
 nH3 O   H   H2 O
E , 30
. VvsNHE
o
k , 2.3x1010 M 1 s 1
E o ,   2.10V vs NHE

eaq
 H2 O  H   HO 
H-O bond
Water
464 kJ
Hydronium 301 kJ
k , 19
. x101 M 1 s 1
E ,   2.930V vs NHE
o
Some other practical considerations
Do you think the Mn/Fe reaction
will continue for long?
e
+
anion
cation
MnO4( aq )  5e  8H(aq )
 Mn(2aq )  4 H2 O
5Fe(2aq )  5Fe(3aq )  5e
1MnO4-
5Fe2+
Net charge
=0
Net charge
=10(-1)+5(+3)
=+5
-
-
-
-
-
-
-+
-
+
Net charge
=0
e
Fe3+
Mn2+
Net charge = (+2)+2(+1
+(-1)=+3
Will want to let spectator ions flow
(but not the reactants!)
e (current)
Fe3+
5+
-
+1
“jelly” (salt bridge) retards motion of Fe3+/2+ MnO4“jelly” allows motion of spectators which produces
Charge balance
Weird Grammar Rules: Those Italians!
Volta discovered this process
1.
2.
3.
Always make electrons flow to right
Electrons flow down to the cathode
(cat = Greek for down).
Electrons flow up into the anode
(an = Greek for up)
Count Alessandro Volta,
Italy
~1800, first battery
e (current)
Rsb
anode
Oxidation
electrons taken
Out (up = anode)
oxidation
Cl-
There is
Resistance
In the system,
We will come
Back to this
cathode
Rsoln
Reduction
electrons accepted
In (down = cathode)
reduction
An
An ox
ox jumped
jumped over
over aa red
red cat
Vocabulary for Work when using
electrons instead of heat.
Ohm’s Law: Voltage = current x resistance
V  IR
Voltage = energy required to move charge
= Joules/Coulomb
Georg Simon Ohm, 1789-1854
German physicist having a good hair day
a coulomb is a unit of charge
F=Faraday = 96,485 coulombs of charge/mole of e
Coulomb  Joule
Coulomb
Joule 
Coulomb

 mole e  


Coulomb
mole
mole
e
e






Coulomb


molese
FFF FF VVV V   J
nn molese
molese
nnn molese
molese
FF
V



 





 


Ohm
nFV  J
Joule
Coulomb
Nernst
neg sign accounts for negative electron
 nFV   G
V directly relates to free energy because we are not
Separately the work terms into heat and changing surrounding randomness with the heat
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
For standard conditions (1 mole, 1 atm, 25C):
 nFV   G
o
Greek
G
Komodo (Indonesia) K
Vamale (Polynesia) Vo
o
 nFV   G   RT ln K
o
o
RT
V 
ln K
nF
o
Different languages, same information. Represent
Total energy (heat + entropy) associated with a
reaction
Relationship G, K, V Example Problem 1 :
What are K and the standard voltage associated with the
Fire oxidation of lead given tabulated free energies?
 Grxo 
 n G
2 Pbs  O2( g)  2 PbOs
0
f , products
  n G of ,reac tan ts
 0kJ 
 0kJ  

  187.9 kJ   
  1molO2 

 G   2molPbOs 
    2molPbs 
 molPbO   
 molPbs 
 molO2  

o
rx
Grxo  357.8kJ
 G   RT ln K
o
Ke
 G 0
RT
kJ 

   357 .8


mol 
Ke

kJ 
 8.314 x 10  3
 298 K
mol K 

K  5.23x10 62
Relationship G, K, V Example Problem 1 :
What are K and the standard voltage associated with the
Fire oxidation of lead given tabulated free energies?
2 Pbs  O2( g)  2 PbOs
G  357.8kJ
o
rx
K  5.23x10
-nFV0 = Go
?
62
Compound neutral
O usually -2
+4
Ox # = 0
So Pb = +2
2(+2)=4
=-4electrons
Relationship G, K, V Example Problem 1 :
What are K and the standard voltage associated with the
Fire oxidation of lead given tabulated free energies?
2 Pbs  O2( g)  2 PbOs
G  357.8kJ
o
rx
K  5.23x10
62
-nFV0 = Go
? n=4
 nFV o   G o   357.8


 357.8kJ   1000 J   


 
 

o


mol
rx
as
written
kJ




G
1
V


  0.92V
Vo 


J
nF

  4mol electrons   96487coulombs    

  mol reaction   mol electron     coulomb  


2 Pbs  O2( g)  2 PbOs
G  357.8kJ
o
rx
K  5.23x10
62
V  0.92V
o
All tell us that reaction
Will spontaneously
Proceed to the right
Favoring products
How will we conveniently store info?
Vo values for 1/2 reactions
Compared to protons
Reaction
Cs+ + e
K+ + e
Na+ + e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Br2 + 2e
Cl2 + 2e
F2 + 2e
Cs
K
Na
Fe
Pb
H2(gas)
Cu
4OHH2O2
2Br2Cl2F-
Observations, Please!!!!
Vo
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
1.
2.
3.
4.
5.
6.
7.
8.
9.
What seems to be the “grammar” for
the reactions?
What is the zero point?
What do you expect the value for Cs to be?
How do the values for the halogens compare
to the group I elements?
Is there a trend in the halogens?
How does this relate to the periodic chart?
How does this relate to “charge density”?
Who wants the electrons?
Where are the guys that want the electrons
located on the chart?
We said that electrons are rapidly
Aquated and, rapidly react
What about their energy?


aq
e  nH2 O  e

eaq
 nH3 O   H   H2 O
H-O bond
Hydronium 301 kJ
Water
464 kJ

eaq
 H2 O  H   HO 
E , 30
. VvsNHE
o
k , 2.3x1010 M 1 s 1
E o ,   2.10V vs NHE
k , 19
. x101 M 1 s 1
E o ,   2.930V vs NHE
don’t have e
Reaction
e + H2O
Cs+ + e
K+ + e
Na+ + e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Br2 + 2e
Cl2 + 2e
F2 + 2e
want most
have e
want least
eaq
Cs
K
Na
Fe
Pb
H2(gas)
Cu
4OHH2O2
2Br2Cl2F-
Vo
-3.0
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
Start arrow on right hand side and end on left hand
Reaction
Cs+ + e
K+ + e
Na+ + e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Br2 + 2e
Cl2 + 2e
F2 + 2e
Don’t have
want most
Have e
want least
Cs
K
Na
Fe
Pb
H2(gas)
Cu
4OHH2O2
2Br2Cl2F-
Vo
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
electrons
flow
down hill
away from
negative
voltage
Think of
A water
tower
Start arrow on right hand side and end on left hand
Uphill reactions: not probable
Reaction
Cs+ + e
K+ + e
Na+ + e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Br2 + 2e
Cl2 + 2e
F2 + 2e
Cs
K
Na
Fe
Pb
H2(gas)
Cu
4OHH2O2
2Br2Cl2F-
Vo
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
Start arrow on right hand side and end on left hand
Can I react F2 with K+?
Reaction
Cs+ + e
Cs
K+ + e
K
Na+ + e
Na
No, there
Fe2+ + 2e
Fe is nobody
electrons,
Pb2+ + 2e
Pb
2H+ + 2e
H2(gas) source!
no
electron
Cu2+ + 2e
Cu
O2 + 2H2O + 4e
4OHO2 + 2H+ + 2e
H2O2
Br2 + 2e
2BrCl2 + 2e
2ClF2 + 2e
2F-
to
Vo
?
-2.95
-2.71
give
-0.44 away
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
Start arrow on right hand side and end on left hand
Can I exchange e between Cs with Pb?
Reaction
Cs+ + e
K+ + e
Na+ + e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Br2 + 2e
Cl2 + 2e
F2 + 2e
Cs
K
Na
Fe
Pb
H2(gas)
Cu
4OHH2O2
2Br2Cl2F-
Vo
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
There is nobody to accept electrons!
Example problem Standard V (good exam prototypes)
Which reactions will go?
a)
b)
c)
d)
Cs metal plus KBr?
F2 gas plus PbCl2
Na metal plus chlorine gas
Na+ + Cl-
Strategy:
1. Pick one who has electrons
2. Pick one who doesn’t
3. Draw an arrow, starting where the electron
is.
4. Is it up or downhill?
Reaction
e + H2O
K+ + e
Na+ + e
NCl3_4H+ + 6e
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
N2(g) + 8H+ + 6e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Ag+ + e
NO3- + 4H+ + 3e
Br2 + 2e
2NO3- + 12H+ + 10e
Cl2 + 2e
Au+ + e
F2 + 2e
eaq
K
Na
3Cl- + NH4+
Fe
Pb
H2(gas)
2NH4+
Cu
4OHH2O2
Ag
NO(g) +2H2O
2BrN2(g) +6H2O
2ClAu
2F-
Vo
-3.0
-2.95
-2.71
-1.37
-0.44
-0.13
0
0.275
0.34
0.40
0.68
0.799
0.957
1.09
1.246
1.36
1.83
2.87
Who rusts most?
Pb
Fe
Cu
Ag
Au
Why?
What do we use for plumbing?
Why was gold considered the sacred material?
Reaction
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Sn4++2e
N2(g) + 8H+ + 6e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Fe3+ +e
Hg22+ +2e
Ag+ + e
NO3- + 4H+ + 3e
Br2 + 2e
2NO3- + 12H+ + 10e
Cl2 + 2e
Au+ + e
F2 + 2e
Fe
Pb
H2(gas)
Sn2+
2NH4+
Cu
4OHH2O2
Fe2+
2Hg(l)
Ag
NO(g) +2H2O
2BrN2(g) +6H2O
2ClAu
2F-
Vo
-0.44
-0.13
0
0.154
0.275
0.34
0.40
0.68
0.769
0.796
0.799
0.957
1.09
1.246
1.36
1.83
2.87
Saturday
Friday
Thursday
Tuesday
Wednesday
Monday
Sunday
~100 B.C
(Context Slide 1)
Change in metal
sequence
occurs at time
that acids were
developed (1100-1400AD)
(Islamic Chemists)
xM solid  2 O2 
 M 2 Oy , solid
y
G
Air oxidation
n
MLnx,aqueous 
 M aqueous  xLaqueous1/K
n
M aqueous
 ne 
 M solid
~1300A.D
f
V
MLnx,aqueous  ne 
 M solid  xLaqueous
Chemical oxidation
(Context Slide 2)
1.
2.
3.
Some Rules
Voltages sum
Reversed reactions =change of sign
Don’t worry about #electrons (n)
since V = Joule/coulomb of charge
Example Calculation: Summing V equations
What is the voltage for the reaction:
2 A 
 A B
V?
Given that

A  e
A

A  e 
B
A


A e
A  e 
B
2 A 
 A B
Vao
Vao/b
 Vao
Vao/b
V  Vao/b  Vao
Example Summing V equations: If your lab partner
attempts to add fluorine gas to a beaker containing
potassium metal what should you do? Justify by
calculating the reaction voltage and the free energy

F2 , g  2e 
 2F
K  e
K

2 K  2e 
 2K

2K 
2
K
 2e


F2, g  2e 
2
F



2 K  F2 
2
K

2
F

V   2.87
V o   2.95
o
Say your prayers and duck.
V   2.95
o
V o     2.95
V o   2.87
V o   582
.
 G o   nFV o
 Go
 2 F   2.95
 2 F  2.87
 2 F   582
. 
 29.648x104   582
.    1123kJ
How does concentration fit In?
 G   G  RT ln Q
o
 nFV   G
 G   nFV
0
o
 nFV   nFV  RT ln Q
o
 nFV o
RT
V

ln Q
 nF
 nF
RT
V V 
ln Q
nF
o
RT  C  D
V V 
ln
a
b
nF  A  B
c
o
d
Nernst Equation:
RT  C  D
V V 
ln
nF  A a  B b
c
d
o
At 25 oC
 C  D
0.0592
V V 
log
a
b
n
 A  B
c
d
o
When the reaction favors products, it is
Spontaneous, or Galvanic
Luigi
Galvani:
“Frog leg Guy”
1780
More accurately:
M   a
a
Activity coefficient
   CM a 
concentration
activity
M a
log     0509
. z
u
1
2
CZ
i
2
i
2
u
1 u
Ionic strength
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
Electrochemistry
Electrochemistry is a very diverse area. It can be broadly divided into
a)
analytical electrochemistry which is concerned with methods of measurement involving
potentiometry (pH meters etc), voltammetry, and modern sensors (generally voltammetric
in nature)
b)
physical electrochemistry is the basis of analytical electrochemistry, but is generally
concerned with the effect of electric fields, charge balance, and diffusion.
c)
Chemical electrochemistry usually is less interested in quantitative analysis but is devoted
to understanding the mechanisms of electron transfer based on chemical structure.
d)
Biologic electrochemistry can be understood as a form of physical electrochemistry
(charge and fields around biomolecules) and of chemical electrochemistry (electron transfer
events in biological systems)
e)
Geologic and environmental electrochemistry is concerned with all of the above as they
take place in the natural environment . Of particular interest are the oxidation reduction
reactions of iron, manganese, chromium, arsenic, sulfur, as these set the parameters in
which life can grow and, as the oxidation state of iron has substantial implications for the
stability of various minerals and their dissolution/formation.
f)
Technical - in this field fall the major areas of batteries, solar energy, fuel cells, and
corrosion sciences.
A preview……
http://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb1/part2/redox.htm
See also:Awesome site
Biological “Galvanic”
(Spontaneous) Cell:
Respiration
Note the negative
To positive
Arrangement of
Voltages.
Electrons flow away
From the
Negative sign.
Note also very
Small voltage steps, 0.01 V is a large driver!
CoQ = Coenzyme Q
What is the role of the long tail?
Ubiquinone, Q
2e, 2H+
Ubiquinol, QH2

Q  2 H  2e H2 Q


~ 0.6999V
0
Open browser to see and rotate molecule
http://www.reciprocalnet.org/recipnet/showsample.jsp?sampleId=27344188&sampleHistoryId=13823
Biological
Electrochemistry
Cytochrome C
Cytochrome c oxidase
Fe
Containing
Heme group

in
8H  O2,g
4H
Membrane

out
Hemeglobin:
Oxygen carrier
Fe is
square planar
with 2 more
coordination
sites
top and
bottom.
One is
used for
oxygen
transport
http://www.elmhurst.edu/~chm/vchembook/568globularprotein.html
Review: Module 18: Complex Ions
Review: Module 17B: Acid Bases
Biological “Electrolytic” or
Non-spontaneous cell:
Photosynthesis
Electrons are “pumped” up towards
More negative voltage
The pump chemistry is
Similar (but not identical)
to metal ligand crystal
Field splitting
light
700nm
680nm
Photosystems I and II
(Context Slide 1)
Electrochemistry in Mining
The Conquest of Mexico
In 1550 the Viceroy wrote to the King
“In just a few years a large area of forest has been destroyed [near the
Taxco silver mines], and it appears that the wood supply will be
depleted sooner than the ore. Ordinances have been made regarding
the conservation of the forest, and likewise regarding the paths that the
Indian workers use for making charcoal, cutting wood, and on the
maximum loads that may carry.”
Requires a less fuel
Intensive method
Mercury consumed in New World Spanish silver mines
(1560-1820):170,000 tons; USA gold rush (1850-1900): 70,000 tons
Amalgamation was introduced in the 1550s in M exico by a Spanish immigrant, Bartolome de
M edina, who wrote Dec. 29, 1555 (1):
1.
(Context Slide 2)
I, Bartolome de Medina do declare that I learned in Spain through discussion with a
German, that silver can be extracted from ore without the necessity for s melting it, or
refining it, or incurring any other considerable expense. With this information I
resolved to come to New Spain. Leaving my home, my wife and my children in Spain, I
came to test it, knowing that if I were successful, I would render a great service to Our
Lord, and to his Majesty and to all this realm. And having spent much time and money
and suffered mental anguish, and seeing that I was not going to be able to make it
work, I commended myself to Our Lady and I begged Her to enlighten me and guide
me, so that I might be successful and it pleased Our Lady to enlighten me and put me
on the right path so that I could make it work.
Probert, A. Bartolome de Medina: The Patio P rocess and the Sixteenth Century Silver Crisis. In M ines
of Silver and Gold in the Americas.
A description of the process 1555.
Grind the ore fine. Steep it in strong brine. Add mercury and mix thoroughly. Repeat
mixing daily for several weeks. Every day take a pinch of ore mud and examine the
mercury. See? It is bright and glistening. As times passes, it should darken as silver
minerals are decomposed by salt and the silver forms an alloy with mercury.
Amalgam is pasty. Wash out the spent ore in water. Retort residual amalgam;
mercury is driven off and silver remains.
Solubility

Ag2 S  2 Ag   S 2
2
3
2 Ag  4S2 O
 2 Ag S2 O3  2
3
K 
insoluble
  8.41x1026  1026.9
Could drive solubility
f
Complexation
Ksp  1051
2
 2.9 x10
13 2
8SO42  16e  32 H   8H2 SO3  8H2 O
V10  0158
.
8H2 SO3  16e  8H   4S2 O32  12 H2 O V2o  0.400
S 2  Ss  2e
 V3o     0.447
Reduction
Of S
8SO  30e  40H
(how to
log K
Get the ligand)
2
4

 4S2 O32  20H2 O  Ss
o
Vnet
 1005
.
n
o

V
net
0.0592 net
30
1002
log Knet 
.   507.7
0.0592
Kvoltage,net  10507.7
(Context Slide 2)
2
4
Ag2 S  8SO
 30e  40H  2 Ag S2 O3  2  20H2 O  S s
3

 K
Ktotal  Knet K f
2
sp
 10507.7 1026.9 1050.1   10484.5
Couple Reactions Example
• Most native silver has long since been used:
2. but we still mine silver dust.
3. How is this economically feasible?
4. How could we get rich with a new process involving CN
extraction?
What is the voltage, free energy, and K
Associated with this reaction?
Ag s  CN

aq
O

2,g 
Ag CN 

2 ,aq
Ag s  CN

aq
Ag s  CN
O

2,g 
 
aq 



Ag CN 2 ,aq 1. Balance the equation
a. Split into ½ reactions
b. Balance each ½ reaction
c. Recombine
Ag CN 

2,aq
Ag s  2CN aq 
 Ag CN 
O2, g 
?
O2 , g 
 2 H2 O

2,aq
Ag s  2CN aq 
 Ag CN  2,aq  1e

4 Ag s  8CN
 
aq 
4 Ag CN 

2 ,aq
4 Ag s  8CN

aq
 4e
 
aq 
O2 , g  4 Haq 
 2 H 2 O
O2 , g  4 Haq  4e 
 2 H2 O



4 Ag CN 2 ,aq  4e
V o     0.31
V o  123
.
O2, g  4 H  4e 2 H2 O


o
4 Ag s  8CN aq  O2 , g  4 Haq 
 4 Ag CN  2 ,aq  2 H2 O V  154
.
rx

The free energy for the reaction is a mere:
G = -nFVorx = -4(96485)(1.53) = -5.9x105 J
G = -RTln K
K = e(-G/RT) = e(-(-590000/(298x8.314)) = e238 = 10238/2.3
= 10103
all you need is:
CN (cheap)
O2 (air is cheap (an aerator))
Hypothetical Modern Silver/Gold Mine
O2
Bulldozer
(Context Slide)
CNaerator
Tibor Kocsis
Same process used to recover
silver at photography studios,
in silver plating.
Major cyanide spills: Czech, Elbe River, Jan. 2006; Romania, Tisza
River, Nov. 2005; Laos, June, 2005; Ghana River Kubreko, Jan, 2005;
China, Papua New Guinea, Ghana, Romania (10 tons Danube River,
Mar. 2004), Ghana, Honduras, Nicaragua, China, 2002: Nevada, USA
(Context Slide) Using Bugs to Mine Cu
from CuS
Biomining for
Gold and
Copper in Botswana
collect
Ksp=10-36
Cu 2 S2  2  s  8Fe 3  4 H2 O  8Fe 2  Cu 2  2SO42  16H 
-14e
8Fe
2

8Fe
bugs
3
Catalytic reagent,
supplied courtesy of bugs, Thiobacillus ferridoxin
(Context Slide)
Coupled Chemical Equation Example Calculate the formal
potential for the reaction to form the initial corrosion product,
Fe(OH)3,s reaction at pH 7


2 Fe 2 Fe
2
V     0.44
 4e

O2  2 H2 O  4e 
4OH 
VOo2  0.40
V    0.771
3
2 Fe 2 
2
Fe
 2e

1
2

O2  H2 O  2e 
2OH 
Fe 3  3OH   Fe(OH ) 3
2Fe(OH) 3 
3H2 O  Fe2 O3
slow
VOo2  0.40
Ksp  6.3x10  38
Introduction: Key Concepts
1. Grammer Rules: write all reactions as
reductions
2. Half reactions
3. Aquated electrons (carry electrons from
electrode to solution species) (disinfectants)
4. Voltage is an energy term
5. Applications
1. Disinfectants
2. Biology
3. Geology
4. Industry