“A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th- F 2-3:30 pm Module #21 Electrochemistry.

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Transcript “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th- F 2-3:30 pm Module #21 Electrochemistry.

“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Chemistry
General
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
 qq 
qq
C3. Size Matters
 or  k 
E  k
 d
r r 

C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
el
1
2
1 2



Properties and Measurements
Property
Size
Volume
Weight
Temperature
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
electronic states in atom
Electronegativity
Heat flow measurements
Standard Molar Enthalpy
Energy of electron in vacuum
F
constant pressure, define system vs surroundings
per mole basis (intensive)
25 oC, 1 atm, from stable state
Hfo Haq+ =0
We ended our last module considering the two following reactions:
2 Pbs  O2( g )  2 PbO
 G   355kJ
kJ 

   355


mol 
Ke
 G 0
RT
e

kH 
 8.314 x 10  3
 298 K
mol K 

K  e143  127 x1062
We also had looked up K for the reaction of Fe rusting
K for rusting of Fe = 10261
Will these reactions proceed the same way in solution as in air oxidation?
Could we “capture” the energy of these reactions more efficiently than a fire?
Do “bugs” make fires to get their energy?, If not can we mimic them?
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Some review of relevant
Energy concepts
Thermochemistry and work
Free energy and maximum work
Chemical reactions involve Zn(s)  2 H  (aq)  Zn 2 (aq)  H2 ( g)
1. heat exchange
As a review:
Heat exchange Constant
At constant
Atm.pressure
Pressure
who is oxidized?
who is reduced?
what is the oxidation number on H2?
Who is an oxidizing agent?
thalpein – to heat
en - in
H for (?) heat
H = Greek:
enthalpy
qP   H
Subscript
Reminds us that
Pressure is constant
1 atm pressure = constant pressure
This means heat flow, q, is enthalpy change
Chemical reactions involve Zn(s)  2 H  (aq)  Zn 2 (aq)  H2 ( g)
1. heat exchange
constant Atm. pressure
2. work
V
PressureVolume
work
w   P V
I feel a Noble
Prize coming on
Maximize work!!!
Solve Global Warming
five Navy Avengers
disappeared in the Bermuda
Triangle on Dec. 5, 194
One possible source of energy
=?????
3 to 8 standard cubic feet of
biogas per pound of manure.
The biogas usually contains 60
to 70% methane.
Methane
Gas
Recovery
At landfills
How much PV work occurs for 1 atm constant T burning of methane as a free volume fire?
CH4( g )  2O2( g )  CO2( g )  2 H2 O( l )
PV  nRT
Work is done by the system on the reaction
(compression)
At constant T, 1 atm P:
P  V     n RT


P  V   n gas final  n gas initial RT
L  atm 

P  V   1  3moles  0.0821
 298K

mol  K 
P  V    48.9316 L  atm
%&$*! Conversions
.
kJ 
 01013
P  V     48.9316 L  atm 
   4.9kJ
 L  atm 
2mole change
Consider the contribution of volume change for water in this reaction
CH4( g )  2O2( g )  CO2( g )  2 H2 O( l )
2moleH O 
2
(l )
3

18
g
1
cm
water   1L 


*
*
*  3 3   0.036 L


 mol   1gwater   10 cm 
%&$*! Conversions
01013
.
kJ
PV  1atm 0.036 L
 0.0036kJ
L  atm
Most reactions total (q):
PV 1 mole gas
PV 2mole liquid water
Energy in kJ
~ 1000
kJ
~
2.5
kJ
~
0.0036 kJ
Sig fig tells us that PV energy small compared to q
H   890kJ
CH4( g )  2O2( g )  CO2( g )  2 H2 O(l )
 
w    PV    4.9kJ   0.0036kJ    4.90036kJ
Enclosing the reaction
In a combustion engine
Allows us to capture work
Total INTERNAL energy Enthalpy
Of an isolated system
Or heat exchange work
Constant P
There is a sign change which
Indicates we get to use the work
ΔE
Bunsen burner (open)
-885kJ
Automobile engine (closed)
-885kJ
Fuel Cell, theoretical no heat transfer -885 kJ
ΔH
-890kJ
-665kJ
-67kJ
Maximum non expansion work at constant pressure and temperature is
G  wnon,exp ansion,max,cons tan tT , P
w
+5kJ
-220kJ
-818kJ
 Grx   Hrx  T Srx
If
T or  Srx  0
Then
 Grx   Hrx
If
T  0 or  Srx  0 Then
 Grx   Hrx
K does not go below 0
So this is impossible
H   890kJ
Maximum work less
Than total internal
energy
CH4( g )  2O2( g )  CO2( g )  2 H2 O(l )
More organized phases
Maximum work less than internal energy because energy used to create a
more organized phase
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Some review of relevant
Energy concepts
Units of energy and work
energy
force dis tan ce
work
Power 

time
time
d  212 ft 
12 ft
Wikepedia
32,572 ft lbforce
 2.4 
1hp  180lbs
 212 ft  
 min utes 
min
 32,572 ft lbforce   min  542.8667 fl lbf

  60s  
min
s


Nm
J
 542.8667 ft lbforce   0.3048m  4.44822 N 


 745.699
  745.699


s
lbf
s
s


  1 ft  
J
1hp  745.699
s
average horse peak
average horse average
hp
14.9 R.D. Stevenson and R. J. Wasserzug
<1 hp Nature, 364, 195-195, Jul 1993
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
kg  m
2
F ma
kg
P

 s2 
A
A
ms
m
Energy, General
Animal hp
BTU
Heat
Gram Calorie
horse on tread mill, 745 J/s
lb water oF
g water oC
British Thermal Unit (>1700 AD) Energy required to raise one lb of water at it’s
maximum density (39.1 oF) 1 oF
Energy to raise 1 g of water by 1 oC
Means “defined as”
Kinetic energy
kg  m2
1
 1J oule
2
s
1 2
E k  mv
2
2 kg mass moving at 1 m/s
1
 1m
E k  2kg  
 s
2
kg  m
Ek  1
s2
2
2
James Joule (1818-1889)
English
Physicist who related
Heat energy to animal work
(to sell steam engines)
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
kg  m
2
F ma
kg
P

 s2 
A
A
ms
m
Energy, General
Animal hp
horse on tread mill, 745 J/s
heat
BTU
1 lb water 1 oF
calorie
1 g water 1 oC
Kinetic J
2 kg mass moving at 1 m/s
Electrostatic
1 electric charge of 1 coulomb in a 1 V field
Chemistry Rule #1 = it’s all about Charge
nFV  J
J =Work required to move one electric charge of one coulomb through an
Electric potential difference of 1 V
J  VC
 kQ1 
E el  
 Q2
 d 
Na+
Cl-
d
Electric potential, V, Exerted by Q1
over distance d on the charge Q2 of
object 2
8.99 x10 J  m
Charge,q
And size,
D matter!
9
k
C
oulomb

2
J
V 
C
Same old, same old, new names
a coulomb is a unit of charge
F=Faraday = 96,485 coulombs of charge/mole of e
 coulombs   Joules 
nmolese molee   Coulomb   Joules
nFV  J
 nFV   G
neg sign accounts for negative electron
V directly relates to free energy
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
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Some example problems
Relating V, free energy,
And K
For standard conditions (1 mole, 1 atm, 25C):
 nFV   G
o
o
Language
Komodo (Indonesia) K
o
o
Greek
G  nFV   G   RT ln K
The Rosetta Stone
Vamale (Polynesia) Vo
Different languages, same information. Represent
Total free energy (maximum work) associated with a
reaction
Relationship G, K, V Example Problem 1 :
What are K and the standard voltage associated with the
oxidation of lead given tabulated standard free energies
of formation? (1 atm, 298 K)
G 
o
rx
 n G
0
f , products
  n G
o
f ,reac tan ts
2 Pbs  O2( g)  2 PbOs
 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
Ke

kJ 
 8.314 x10  3
  273 K
mol K 

158
K  2.89 x1068
Relationship G, K, V Example Problem 1 :
What are K and the standard voltage associated with the
oxidation of lead given tabulated standard free energies
of formation? (1 atm, 298 K)
2 Pbs  O2( g)  2 PbOs
Grxo  357.8kJ
-nFV0 = Go
?
K  2.89 x1068
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
oxidation of lead given tabulated standard free energies
of formation? (1 atm, 298 K)
2 Pbs  O2( g)  2 PbOs
G  357.8kJ
o
rx
K  2.89 x1068
-nFV0 = Go
? n=4
 nFV o   G o   357.8

 357.8kJ    1000 J   

     357
.8kJ
1000
J   







oo






mol
rx
as
written
kJ




G
1
V


o
  0.92V
V ooo    G    mol rx as written   kJ  

V  nF    4mol electrons   96487coulombs
   
J


nF


     coulomb  
  mol reaction   mol electron
 


2 Pbs  O2( g)  2 PbOs
G  357.8kJ
o
rx
K  2.89 x1068
V  0.92V
o
All tell us that reaction
Will spontaneously
Proceed to the right
Favoring products
So….V>0 is spontaneous
How will we conveniently store info?
What will be the reference point?
T = boiling water
E of electrons – vacuum, far away from the nucleus etc.
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
½ reactions, standard voltages
and
Electrochemical
“cells”
1. Create a body of reference reactions
2. with a SCALE defined to one reference rx
3. Make an instrument to calibrate or measure
the scale.
+4e
2 Pbs  O2( g)  2 PbOs
-4e

s
2 Pb
2
2 Pb  4e
Simplify for the tables

s
Pb
2
Pb  2e
Could imagine oxygen reacting
Similarly with Fe, CH4!
Divide reaction into 2 parts
O2 g  4e 2O


2
Set a common “grammar” – all
comparisons are for reductions
2
Pb  2e  Pbs
Standard conditions
1 atm, 298 K,
1 M or 1 atm
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-
Vo
?
-2.95
-2.71
-0.44
-0.13
0
0.34
0.40
0.68
1.09
1.36
2.87
“Medicine is the Art of Observation” (ABB, III, M.D.)
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?
don’t have e
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
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
Vo < -2.95 (e.g. -3,
-4…)
Obeys
Rule #2
Everybody
Wants to
Be Like
Mike
(get to Group
18 e configuration)
1. Create a body of reference reactions
2. with a SCALE defined to one reference rx
3. Make an instrument to calibrate or
measure the scale.
Measure flow of e
1atmH2,g
Unknown
But wait, isn’t there a problem?
1M Haq
Reference ½ reaction
2 Haq ,1 M  2e  H2 ,1atm
Do you think this reaction will continue for long?
Here’s a case
where spectator
ions are
importantsolution net
nuetrality!!!!
5e
MnO4( aq )  5e  8H(aq )
 Mn(2aq )  4 H2 O
5Fe(2aq )  5Fe(3aq )  5e
10 X- 5Fe2+
Net charge
=0
Fe3+
Net charge
=10(-1)+5(+3)
=+5
1MnO4-, 8H+
-
-
-
-
-
-
-
+ +- +
+
+
+
++
Net charge
=0
8
7 X-
e
Mn2+
Net charge =+2 +7(-1)
= +5
Charge build up stops reaction
Will want to let spectator ions flow
(but not the reactants!)
e (current)
Fe3+
5+
-
-5
“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
OIL
Oxidation is Loss
e (current)
anode
Oxidation
electrons taken
Out (up = anode)
Rig
Reduction is Gain
Cl-
cathode
Reduction
electrons accepted
In (down = cathode)
oxidation
reduction
An
An ox
ox jumped
jumped over
over aa red
red cat
There 2 kinds of electrochemical (or Voltaic) “cells”
1. Spontaneous (Galvanic) Electrochemical cell (V +; free energy -)
2. Non-spontaneous (re-charging or Electrolytic) (V -; free energy +)
Prez Bush II’s 2006 State of the Union
The coming of the hydrogen economy
(Considered much more likely by scientists:
Methane based fuel Cell)
 Hrxo   150.7kJ
 Hrxo  0
 Grxo   2501
. kJ
 Grxo  0
CH4( g )  H2 O  CO( g )  3H2 ( g )
3H2( g)  6Haq
Resistive energy losses in the external circuit
 Hrxo   711kJ
 Grxo   857.4kJ
o
 Hnet
rx   560.9 kJ
o
 Gnet
rx   607.3kJ
3
2
O2,g  6Haq  3H2 O
CH4 ( g )  2 O2  2 H2 O( l )  CO( g )
3
Feed hydrogen gas to fuel cell anode
To use methane
need to a pre-step
conversion to
hydrogen over
a nickel oxide
catalyst.
CO+3H2
work
Ni
3H2
Ni
Ni
Advantage:
methane exists
Disadvantage”
CH4(g)
Less energy
H2O
recycle
Feed air
To fuel cell
cathode
3/2O2
H+
H+ 6e
6H+ 6e H+
3H2O
H+
H+
H+
Proton conducting polymer to reduce weight
Scientific American, Sept 2006 Issue on Energy2005 – big research
Bucks from the public
Hydrogen Economy
1. H2 production
spigot
1.
2.
Coal/gasoline/methane conversion = 110gCO2/km driven
current cars 150gCO2/km driven
1. Need to capture CO2 and dump and/or
2. Reduce CO2
1.
coal/gasoline/methane capture CO2 and pump to ground
2.
solar
3.
wind
4.
nuclear
Autos (= ~ 40% U.S. energy consumption)
• Proton-exchange membrane – membranes degrade with use fuel cells
last only ~2,000 <1/2 necessary for commercial vehicle
• Cost of fuel cells – current cost 1M$, could drop to $125/kw ($30/kw
combustion engine)
• Storage of H2 on the car
compression/super cooled
metal hydride
4. Safety of compressed H2
5. Timeline – earliest predicted
commercialization 20 years
Green and red are
hydrogen gas on an
organic lattice to serve
as fuel source
Hydrogen gas
http://www.physorg.com/news11458.html
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Using Standard
Voltages to predict
Spontaneous reactions
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
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
K
Na
3Cl- + NH4+
Fe
Pb
H2(gas)
2NH4+
Cu
4OHH2O2
Ag
NO(g) +2H2O
2BrN2(g) +6H2O
2ClAu
2F-
Vo
-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
Consider 6 of the 7 earliest
known pure elements:
Au, Ag, Cu, Pb, Sn, Fe
Who rusts (reacts with O2)
more spontaneously?
Why is Au considered
sacred or valuable in
many cultures across
history?
Why was Pb used for
plumbing?
Reaction
Fe2+ + 2e
Pb2+ + 2e
2H+ + 2e
Sn4++2e
N2(g) + 8H+ + 6e
Cu2+ + 2e
O2 + 2H2O + 4e
O2 + 2H+ + 2e
Fe+3 +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-
Medicine is the Art of Observation
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)
Days of weeks related to 7
planets and “7” metals
Less spontaneous
Not really
Known so
Assigned as alloy
Changes occur at the time
that acids were developed
(1100-1400AD)
(Islamic Chemists)
xM solid  2 O2 
 M 2 Oy , solid
y
Less likely to lose e
G
Air oxidation
New assignments
M solid


n
M aqueous
 ne
-V
Kf
n
n
M aqueous
 xLaqueous 
 MLx ,aqueous
Change in order here
~1300A.D
n
M solid  xLaqueous 
 MLx ,aqueous  ne
Chemical oxidation in HCl
(Context Slide 2)
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Summing various kinds
Of reactions together:
Calculations
Some Rules
1.Voltages sum
2.Reversed reactions =change of sign
3.When summing voltages only don’t worry
about #electrons (n) since V = Joule/coulomb
of charge
4.To sum a Voltage and a K:
a. first convert both to free energy (now worry
about # electrons n)
b.Sum the free energies
c. Convert the summed free energy back to a
Voltage (worry about #electrons n here also)
Example Calculation: Summing V equations
Given that

A  e
A

Vao
A  e 
B
Vao/b
What is the voltage for the reaction:
2 A 
 A B
A


A e
A  e 
B
2 A 
 A B
V?
 Vao
Vao/b
V  Vao/b  Vao
Standard V for
REDUCTION
½ reactions
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
V   2.87
V o   2.95
2 K   2e 
 2K
V o   2.95
o
Say your prayers and duck.
Notice here multiplying does not affect V
Reversal switches sign, tho

2K 
2
K
 2e

F2, g  2e 2 F





2 K  F2 
2
K

2
F

V o     2.95
V   2.87
o
V o   582
.
 G o   nFV o
 2 F   582
. 
 29.648x104   582
.    1123kJ
Example Summing Voltages and Ks
We saw that the relative rank for easy of oxidation of metals changed
Around 1300 A.D. when technology from the Islamic world was
developed that allowed for the production of strong acids. What is the
standard potential for the oxidation of gold in the presence of 1 M HCl
given the following information:

Auaq
 e
 Aus
V o  183
.

1
Auaq
 2Cl,aqs 
 AuCl2 ,aq
K f  2.71x10 11
Is gold easier or harder to oxidize in 1 M HCl?
Organize information into one oxidation reaction

Aus 
 Auaq  e
V o   183
.

1
Auaq
 2Cl,aqs 
 AuCl2 ,aq
K f  2.71x1011

Aus  2Claq 
AuCl

2,aq  e
Sign changes
?
Example Summing Voltages and Ks
What is the standard potential for the oxidation of gold in the presence
of 1 M HCl? Is gold easier or harder to oxidize in 1 M HCl?
 nFV   G   RT ln K
o

Aus 
 Auaq  e
V o   183
.

1
Auaq
 2Cl,aqs 
AuCl

2 ,aq
Aus  2Cl
 
aq 
AuCl

2,aq
K f  2.71x1011
e
?.
V  115
o
net
o
 G o   177
. x10 5
 G o  6.52 x104
o
Gnet
 111
. x105
Easier to
Oxidize in
o
o
5
G   nFV    1 96487  183
.    177
. x10
Presence of
 298 ln2.71x1011    652
Go   RT ln K   83141
.
. x104 HCl
a. Convert to free energies
b. Sum
c. Convert to Vnet
5
5
111
.
x
10
111
.
x
10
Vneto 

  115
.
 nF
 1 96487
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Voltage and Concentrations
How does concentration fit In?
 G   G  RT ln Q
o
 nFV   G
 G   nFV
0
o
 nFV   nFV  RT ln Q
o
o

 nFV   RT ln Q 
  nFV 

 


  nF    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
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
Example Concentration Cell: Calculate the cell
potential for a spontaneous (galvanic) cell based on
the reaction where [Mn2+] = 0.50 M, [Al3+] = 1.50 M
At 25 oC, 1 atm
2 
aq 
2 Als  3 Mn
2 Al
3
aq
Know
 C  D
0.0592
V V 
log
n
 A a  B b
c
d
o
concentrations
25 oC
1 atm
 3 Mns
Don’t know
n
Vo
Standard conditions
Means we can use tables of standard voltages
Example Concentration Cell: Calculate the cell
potential for a spontaneous (galvanic) cell based on
the reaction where [Mn2+] = 0.50 M, [Al3+] = 1.50 M
At 25 oC, 1 atm
2 
aq 
2 Als  3 Mn
2 Al
3
aq
Don’t know
n =6
Vo
Know
 C  D
0.0592
V V 
log
n
 A a  B b
c
 3 Mns
d
o
concentrations
25 oC
1 atm
3
2 Als 
2
Al

aq  6e
3 Mn
2
aq
 6e 
 3 Mn s
2 
aq 
2 Als  3 Mn
2 Al
3
aq
2
aq
3 Mn
 6e 
 2 Al s
 6e 
 3 Mns
3
2 Als 
2
Al

aq  6e
2
aq
3 Mn
 6e 3 Mns


2 Al
3
aq
 3 Mns
Std
V   166
. grammer
o
V   118
.
o
V o     166
. 
V   118
.
o
+0.48
Calculate the cell potential for a spontaneous
(galvanic) cell based on the reaction where [Mn2+] = 0.50 M
[Al3+] = 1.50 M, at25 oC, 1 atm
Example Conc. Cell: Calculate the cell potential for a
spontaneous (galvanic) cell based on the reaction
where [Mn2+] = 0.50 M, [Al3+] = 1.50 M at 25 oC.
1atm
2
3
2 Als  3 Mn

aq 
Know
c
d




0.0592
C D
o
V V 
log
n
 A a  B b
concentrations
n=6 Vo=+0.48
2 Alaq  3 Mns
Don’t know
 
 
3
aq
Al
0.0592
V  0.48 
log
2
6
Mnaq
What happened to Als and Mns?
150
0.0592
. 
V  0.48 
log
3
6
 05
.
2
V  048
.  001
.  047
.
2
3
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Voltage and Biology
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 Example Calc. 1: What
would be the standard potential of the ½ reaction at a
physiologically appropriate pH (7.4)?
Q  2 H   2e  H2 Q
~ 0.6999V 0


 H2 Q 
0
.
0592

V  0.6999V 0 
log
2
2
 Q H   
 1 
0
.
0592
0

V  0.6999V 
log
2
2
 1 H   
0.0592
0
 2
V  0.6999V 
log H 
2
For standard conditions (1 mole, 1 atm, 25C):
Biological Electrochemistry Example Calc. 1
What would be the standard potential of the ½
reaction at a physiologically appropriate pH (7.4)?
0.0592
 2
V  0.6999V 
log H 
2
0
V  06999
.
V 0  00592
.
log H  
V  06999
.
V 0  00592
.
log107.4 
V  0.6999V   0.0592  7.4
0
V  0.6999V 0    0.43808
V  02619
.
Notice that not only
Does pH control
Structure of proteins
It controls the total
Energy associated with
Many reactions
Which is why we emphasize acid base
Chemistry over.
Which is why we emphasize acid base
Chemistry over and over.
Which is why we emphasize acid base
Chemistry over and over and over.
Biological
Electrochemistry
Example
Calculation
2
Watch also the protons
+4H+
Cytochrome C
Cytochrome c oxidase
Fe
Containing
Heme group
H2 O

in
4H  O2,g
1
2
2e
Pump
Membrane
2H

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 Electrochemistry Example Calc. 2: Cytochrome, CyFe2+,
reacts with the air we breathe to supply energy required to synthesize
adenosine triphosphate (ATP). The body uses ATP as an energy source
to drive other reactions. At pH 7.0 the following reduction potentials
pertain to this oxidation of CyFe2+

4e  4 H  O

2,g 
 2 H2 O
CyFeaq3  e  CyFeaq2
V   082
. V
o
V 0  0.22V
a) What is ΔG for the oxidation of CyFe2+ by air?
b) If the synthesis of 1.0 mol ofATP from adenosine diphosphate
(ADP) requires a ΔG of 37.7 kJ, how many moles of ATP are
synthesized per mole of O2?
Biological Electrochemistry Example Calc. 2: At pH 7.0 the
following reduction potentials pertain to this oxidation of CyFe2+
4e  4 H   O2, g   2 H2 O
CyFeaq3  e  CyFeaq2
V o   082
. V
V 0  0.22V
a) What is ΔG for the oxidation of CyFe2+ by air?
4e  4 H   O2, g   2 H2 O
2 
aq  4
4CyFe
3
aq
CyFe  4e
4 H   O2, g  4CyFeaq2   2 H2 O  4CyFeaq3
V o   082
. V
V     0.22V 
0
V o   0.60V
Biological Electrochemistry Example Calc. 2: At pH 7.0 the
following reduction potentials pertain to this oxidation of CyFe2+
4 H   O2, g  4CyFeaq2   2 H2 O  4CyFeaq3
V o   0.60V
n  4e
a) What is ΔG for the oxidation of CyFe2+ by air?
 G   nFV
0
o
 G   4 96485 0.60   231564
,
J   2316
. kJ
0
b) If the synthesis of 1.0 mol ofATP from adenosine diphosphate (ADP)
requires a ΔG of 37.7 kJ, how many moles of ATP are synthesized
per mole of O2?
. molATP    2316
. kJ  615
. molATP
 10
 


 37.7 kJ   1molO2 
1molO2
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
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours W – F 2-3 pm
Module #21
Electrochemistry
Reduction/Oxidation
Reaction Coupled to
Complexation and
Precipitation, more
challenging examples
(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 
f
2
Ksp  1051
 2.9 x10

13 2
 8.41x1026  1026.9
Complexation
8SO42  16e  32 H   8H2 SO3  8H2 O
Reduction
Of S
8SO
( to make
Complexing
agent)
8H2 SO3  16e  8H   4S2 O32  12 H2 O V2o  0.400
S 2  Ss  2e
 V3o     0.447
2
4
 30e  40H   4S2 O32  20H2 O  Ss
n
o
log Knet 
Vnet
0.0592
30
1002
log Knet 
.   507.7
0.0592
Ag2 S  8SO
NET RX
o
Vnet
 1005
.
Too
complicat
d for an
exam,
Kvoltage,net  10507.7
(Context Slide 2)
2
4
V10  0158
.
 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
Ag2 S  2 Ag   S 2
2
4
Ag2 S  8SO
Ksp  1051
Insoluble
 30e  40H  2 Ag S2 O3  2  20H2 O  S s

3
Soluble: driven by oxidation/reduction
and complexation
 K
Ktotal  Knet K f
2
sp
 10507.7 1026.9 1050.1   10484.5
Couple Reactions Example Calculation 1
Good for
• Most native silver has long since been used:
an exam
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
could be an exam
question either
math or written
one
Biomining for
Gold and
Copper in Botswana
collect
Ksp=10-36
S ox # = -2
CuSs  8Fe 3  4 H2 O  8Fe 2  Cu2  SO42  8H 
-2 to +6=-8
Need 8 e
8Fe
2

8Fe
bugs
3
4x(-2)=-8
S have to have +6
 8e
Catalytic reagent, supplied courtesy
of bugs, Thiobacillus ferridoxin
http://www.learner.org/channel/courses/biology/textbook/microb/microb_14.htm
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
Reduction/Oxidation
Reactions: Corrosion
The Billion $ Question
Or: why your taxes will have
To always go up
Coupled Chemical Equation Challenge Calculation 2
Calculate the formal potential for the reaction to form the
initial corrosion product, Fe(OH)3,s reaction at pH 7, 1 atm,
298 K from iron metal and oxygen in the presence of water
Given the following information.
Fe(OH ) 3  Fe 3  3OH 
Ksp  6.3x10  38
Fe 2  2e 
 Fes
V o   0.44
2
Fe 3  e 
 Fe
V  0.771
OH 3
OO 22HHOO44ee 44OH
22
22



o
O2
V
 0.40
Want a reaction that looks
sort of like:
Fes  O2 , g  H2 O 
 Fe OH  3
Balance electrons
Start with the oxidation/reduction first
V o     0.44
2
Fes 
 2e
 Fe
3
Fe 2  
e
 Fe



33
Fes s
Fe
Fe
Fe
3e3e4

V    0.771
V o     0.44    0.771)
Coupled Chemical Equation Challenge Calculation 2
Calculate the formal potential for the reaction to form the
initial corrosion product, Fe(OH)3,s reaction at pH 7, 1 atm,
298 K from iron metal and oxygen in the presence of water
Given the following information.
 nFV o   G o   RT ln K
Fe(OH ) 3  Fe 3  3OH 
Ksp  6.3x10  38
4
4 Fe 3  12OH   4 Fe(OH ) 3

3O2  6H2 O  12e 
12OH 
3
4 Fes 
 12e
 4 Fe
Reverse and quadruple
4
 1   1

   
 38 
K
6
.
3
x
10


 sp 
VOo2  0.40
J 
1 


 G    8.3145
  298 K  ln


 6.3x10  38 
mol  K 
4
o
  8.49 x105 J
 129.648x104  040
.    Go   4.63x105 J
V o     0.44    0.771)
 129.648x104  0441
.  .771   Go   383
. x105 J
4 Fes  3O2, g  6H2 O  4 Fe(OH ) 3
Gnet  170
. x106 J
Now we have the reaction need total voltage: Use Rosetta Stone
Coupled Chemical Equation Challenge Calculation 2
Calculate the formal potential for the reaction to form the
initial corrosion product, Fe(OH)3,s reaction at pH 7, 1 atm,
298 K from iron metal and oxygen in the presence of water
Given the following information.
 nFV o   G o   RT ln K
4
4 Fe 3  12OH   4 Fe(OH ) 3
4
 1   1

   
 38 
 Ksp   6.3x10 
VOo2  0.40
 4.63x105 J
V o   0.331
 383
. x105 J

3O2  6H2 O  12e 
12OH 
3
4 Fes 
4
Fe
 12e

Gnet  170
. x106 J
4 Fes  3O2, g  6H2 O  4 Fe(OH ) 3
G
 170
. x10
V 

 146
.
4
 nF   129.648x10 
o
net
o
net
 8.49 x105 J
6
Why do we care if this is
A highly favorable reaction?
Why use it as an example?
(Context Slide)
Chicago Tribune Week of April 11, 2007
Corrosion Involves several Reactions which result in iron
OO removal
OO
OO
Begins at a pit or hole
Na+
O H
H
ClO H
H
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
5 neighbors
Which atom
is more likely
to be removed
from the crystal?
WHY?
Fe
FeFe
Fe
Fe
Fe
Fe
Fe
Fe
Fe
FeFe
Fe
FeFe
Fe
Fe
Fe
4 neighbors
Less bonds
to break
(Context Slide)
Corrosion Involves several Reactions
OO
OO
OO
Na+
Cl-
O H
H
O H
H
OO
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe
Fe Fe Fe Fe Fe Fe Fe
Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe2+
2+
2
2 Fe 
2
Fe
 4e

2
aq
Fe

aq
 xCl  FeCl
O2 , gas 
 O2 ,aqueous
o
 VFE
n
x ,aq
Kf
Khenry
(Context Slide)
Corrosion Involves several Reactions
OO
OO
Na+
Fe2+ Cl
O H
O H H
H OO
Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
2+


2
aq
2 Fe 2 Fe
Fe
2
 4e

aq
 xCl  FeCl
V
n
x ,aq
O2 , gas 
 O2 ,aqueous

O2  2 H2 O  4e 
4OH 
2
Feaq
 2OH aq  Fe OH  2 ,s
o
FE
Kf
Khenry
VOo2
To stop this process
prevent access of
1. water
2. NaCl
3. oxygen
K sp ,(Context
Fe  OH 2
Slide)
Varnish - need to dry from inside out - requires initial O2 entrance
and then sealing otherwise you get a “bubble” OO
OO
Cl
O H
O H
+
H
H
Na
Dry, polymerized
unpolymerized
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe FeFe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Use Pb3O4
Fe Fe Fe Fe FeFe Fe
Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe
To carry O2
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
OO
toxic
OO
Cl
O H
O H
+
Dense
H
H
Na
barrier
CryOy
toxic
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe FeFe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe FeFe Fe
Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
(Context Slide)
Third alternative
(Context Slide)
Corrosion surface coated unevenly with various and multiple salts manipulate to get an even layer of Fe2O3
1. Increase production of Fe3+
2. Prevent loss of Fe2+ by random walk
Electron
conducting
composite
film containing
clay
orients
polymer
increases path
length of O2
Na+
OO
OO
O H
H
Cl-
O H
H
Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe FeFe Fe Fe Fe
Fe
Fe Fe Fe Fe FeFe Fe
Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
2+
2+
Third alternative
(Context Slide)
Corrosion surface coated unevenly with various and multiple salts manipulate to get an even layer of Fe2O3


2 Fe 2 Fe
1. Increase production of Fe3+
2. Prevent loss of Fe2+ by random walk
2
 4e

PAN ox  4 H  4e 
 PAN red
PAN red ,1  PAN ox ,2 
 PAN ox ,1  PAN red , 2
Electron
conducting
composite
film containing
clay
orients
polymer
increases path
length of O2
Na+
OO
OO
O H
H
Cl-
O H
H
Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe FeFe Fe Fe Fe
Fe
Fe Fe Fe Fe FeFe Fe
Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
2+
2+
Third alternative
(Context Slide)
Corrosion surface coated unevenly with various and multiple salts manipulate to get an even layer of Fe2O3


2 Fe 2 Fe
1. Increase production of Fe3+
2. Prevent loss of Fe2+ by random walk
2
 4e

PAN ox  4 H  4e 
 PAN red
10-85
Fe2O3 Ksp
Fe(OH)2 Ksp 10-38.8
PAN red ,1  PAN ox ,2 
 PAN ox ,1  PAN red , 2
2 Fe
2
 O2  2 H2 O 2 Fe


Fe  3OH
 4OH
Na+
More insoluble,
denser
3
3
  Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe3 FeFe Fe Fe Fe

Cl-
OO
OO
O H
H
O H
H
Fe Fe Fe Fe Fe Fe Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe
Fe
Fe
Fe Fe Fe Fe FeFe Fe
Fe Fe
Fe Fe Fe Fe Fe Fe Fe Fe Fe
slow Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
2 Fe(OH ) 3   Fe
 3Fe
H2Fe
O FeFeFe
2 O3Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe Fe
Fe(OH )
2+
Fe3+ Fe
3+
2+
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th- F 2-3:30 pm
Module #21
Electrochemistry
What you should know
1. Balance oxidation/reduction reactions
2. Convert between V, free energy, and K
3. Calculate standard V of a reaction cell from standard ½
reactions
4. Determine if the cell (reaction) is Galvanic or
Electrolytic
5. Calculate V of a reaction at non-standard conditions
(change in conc.)
6. Add various electrochemical reactions together (flip
reaction change sign)
7. Add various electrochemical and solubility and/or
complexation rxs together: This is tricky – need to add
in terms of free energy not voltages
8. Be able to explain one example ( energy, biological,
environmental, corrosion) of electrochemical reactions.