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Chem 106 Tuesday 19 April 2011
A few OWL problems
Chapter 20: Electrolysis
Interchapter: Chemistry of Nitrogen
Makeup of the atmosphere
Aurora borealis: Role of N2 and O2
Nitrogen oxides – Major air pollution players
4/19/2011 1
ElectrolysisUsing electrical energy to produce chemical change.
Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)
⊕
4
BATTERY
+
Na+Cl-
Anode Cathode
electrons
BATTERY
+
Na+Cl-
Anode Cathode
electrons
• NOTE: Polarity (+/-) of
electrodes is reversed from
electrochemical cells.Sn2+
• Electrolysis of aqueous SnCl2.
• Here, a battery pushes electrons
“uphill” from Cl- to Sn2+ in the non-
spontaneous direction.
5
e-e-
BATTERY
+
Na+Cl-
Anode Cathode
electrons
BATTERY
+
Na+Cl-
Anode Cathode
electrons
Sn2+
6
Why is the cathode labeled NEGATIVE in an electrolysis cell, but
POSITIVE in electrochemical cell?
The NEGATIVE terminal is where the ELECTRONS “PILE UP”.
This occurs “naturally” at the anode of an electrochemical cell.
But, in an electrolysis cell, electrons are driven there by the
battery voltage.
e-e-
e-e-e-e-e-e-e-e-e-e-
e-e-e-e-e-e-e-e-e-
The difference is the driving
force for the production of
electrons:
Electrochemical c.
Spontaneous linked half-
reactions.
Electrolysis c. Voltage from the
attached battery.
4/12/2011 7
Cu2+(aq) ionsAg+(aq) ions
Cu Ag
Electrons-->
Cu ����
Cu2+ + 2e-
Ag+ + e-
���� Ag
Oxidation
Anode
Negative
Reduction
Cathode
Positive
Salt bridge
e-e-e-e-e-e-e-e-e-e-
e-
e-e-e-e-e-
e-e-e-e-e-
e-
Electrochemical cell
Anode (+)
2 Cl- ---> Cl2(g) + 2e-
Cathode (-)
Sn2+ + 2e- ---> Sn(s)
BATTERY
+
Na+Cl-
Anode Cathode
electrons
BATTERY
+
Na+Cl-
Anode Cathode
electrons
Sn2+
Eo for cell = E˚cathode - E˚anode (in the electrolysis direction)
= - 0.14 V – (+1.36 V)
= –1.50 V
This much external voltage is required to reverse the electrochemical cell.
9
e-e-
Electrolysis of aqueous sodium iodide NaI (aq) produces I2 and OH- ion.
Anode (+): 2 I- ---> I2(g) + 2e- (-0.54 V)
Cathode (-): 2 H2O + 2e- ---> H2 + 2 OH- (-0.83 V)
Eo for cell = -1.45 V
10
4/19/2011 11
Na metal is not produced because the reduction potential Eo of Na+ (-2.71 V) is more
negative than that of WATER (-0.83 V). So the water molecules get the electrons,
not the Na+ ions.
Say you electrolyzed
aqueous solutions of FeI2,
ZnI2, and AlI3, which one
would not form a METAL
at the cathode?
4/19/2011 12
FeI2
ZnI2
AlI3
9
25
4
1. FeI2
2. ZnI2
3. AlI3
4/19/2011 13
Say you electrolyzed
aqueous solutions of FeI2,
ZnI2, and AlI3, which one
would not form a METAL
at the cathode?
1. FeI2
2. ZnI2
3. AlI3
Stoichiometry of Electrolysis
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- ---> Ag(s)1 mol e- ---> 1 mol Ag
If we could measure the moles of e-, we could know the quantity of Ag formed.
But how to measure moles of e-?
14
time(sec)
lomb)charge(couofAmount)( =ampCurrent
Current (Coul/s) x time (s) = Charge (Coulomb)
mole e- x (mol species/1 mol e-) = mol species
mol species x atom or molecular weight (g/mol) = mass (g)
15
Stoichiometry of Electrolysis
−
−
−
=
=
emoleeoleCoulombs/m96,500
Coulombs
eoleCoulombs/m96,500Faraday1
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass
of Ag metal is deposited?
Solution
(a)Calculate charge
Charge (C) = current (A) x time (t)
= (1.5 amp)(1Cou/s-amp)(15.0 min)(60 s/min) = 1350 C
seconds
coulombs = (amps) I
seconds
coulombs = (amps) I
16
Stoichiometry of Electrolysis
Solution
(a) Charge = 1.50 Coul/sec x 15.0 min x 60 sec/min = 1350 Coulombs
(b) Calculate moles of e- used
seconds
coulombs = (amps) I
seconds
coulombs = (amps) I
1350 C • 1 mol e -
96,500 C==== 0.0140 mol e -1350 C •
1 mol e -
96,500 C==== 0.0140 mol e -
0.0140 mol e - • 1 mol Ag
1 mol e -==== 0.0140 mol Ag or 1.51 g Ag0.0140 mol e - •
1 mol Ag
1 mol e -==== 0.0140 mol Ag or 1.51 g Ag
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass
of Ag metal is deposited?
(c) Calculate quantity of Ag
17
* = an excited state (electron in a
high energy atomic or molecular
orbital)
N2 + e- � N2+* + 2e-
N2+* � N2
+ + UV photon
UV photon + O2 � O2*
O2* � 2 O* (O excited states)
O* � O + red, green photon
O atom fluorescence
4/19/2011 22
Fluorescence can only occur at high altitudes (> 300 km) where there are about
10 atoms/cm3.
At low altitudes:
-No electrons
-O* excited states lose energy as heat in collisions with other molecules
(“quenching” of fluorescence).
(There are two different excited states for O atoms, one yielding the red photons, the
other green. The red ones occur only at higher altitudes. This state of O requires a longer
time interval before it can “spit out” its red photon, and the thinner atmosphere up there
allows fewer intermolecular collisions that would otherwise quench the red fluorescence.)
(What is the number density at 1 atmosphere?)
1 mole of gas molecules/22.4 L
= 6.022 x 1023 molecules/2.24 x 104 cm3
= 2.29 x 1019 molecules/cm3
4/19/2011 23
Aurora studies verify the chemistry of the upper
atmosphere. This is a type of remote spectroscopy
(with the eyes!) that can also be used to identify
molecules on distant planets and stars.
4/19/2011 24
N2 originated from
outgassing of molten
rock. N2 is a common
molecule throughout
the universe.
N2 is one of more
than 100 molecules
observed so far in
studies of interstellar
dust clouds.
4/19/2011 25
Nitrogen fixation by the root nodule bacterium azotobacter vinelandii
Nitrogenase enzyme from protein data bank (www.rcsb.org/pdb/, 1M34.pdb)
4/19/2011 28
Iron-sulfur clusters
Molybdenum atom
Adenosine
diphosphate (ADP)
Two identical halves
N2 + 6 H+ + 6 H:- + nATP ���� 2 NH3 + nADP + nPO4
Nitrogen oxides N2O, NO, and NO2 are produced
by reaction of N atoms or N2 with O atoms or O2
N2 + O � N2O
O2 � 2 O
N2 + 2 O2 � 2 NO2
N2 + O2 � 2 NOProduced in any high
temperature combustion using
air as O2 source.
These are not removed by
automobile catalytic converters.
In upper atmosphere
4/19/2011 30
N2O is a stable molecule that contains several π-bonds.
The necessity of drawing resonance forms in the
Lewis formula points to multi-atom π-bonds.
4/19/2011 31
Agree or disagree: N2O4 is resonance-
stabilized.
4/19/2011 33
Yes
No
Abst
ain
18
3
141. Yes
2. No
3. Abstain
colorless brown
N2O4 � 2 NO2
∆Ho = +57.1 kJ/mol
∆So= +176 J/mol-K
∆Go = ∆Ho - T ∆So
Cold Warm 4/19/2011 36
T (°C) T (Kelvins) ∆∆∆∆G° (kJ/mol)
K=
[NO2]2/[N2O4]
-30 243.15 14.306 0.0008
-20 253.15 12.546 0.0026
-10 263.15 10.786 0.0072
0 273.15 9.026 0.0188
10 283.15 7.266 0.0457
20 293.15 5.506 0.1045
30 303.15 3.746 0.2262
40 313.15 1.986 0.4664
50 323.15 0.226 0.9195
60 333.15 -1.534 1.7402