Intersection 15: The End 12/12/06. The Final Exam Questions 1.Electron configurations, atomic...

Intersection 15: The End

12/12/06

The Final Exam Questions

1. Electron configurations, atomic structure, periodic trends, atomic orbitals.

2. Quantitative calculations.3. Lewis & VSEPR Structures and polarity.4. Thermochemistry5. Thermochemistry6. Thermochemistry7. Acid/Base chemistry8. Acid/Base chemistry9. Oxidation number10. Redox reactions11. Redox reactions12. Electrochemistry

Research at the University of Michigan

A unique opportunity in your life

• Women in Science and Engineering (WISE)– http://www.wise.umich.edu/

• Undergraduate Research Opportunities Program– http://www.lsa.umich.edu/urop/

• Your future class professors • Dr. Gottfried and Dr. Banaszak Holl

• Send specific request with information about yourself and interests (CV). Use contacts. Blanket emails not great approach.

For Help Finding Research Labs

Outline

• Vitamin C note

• Electrochemistry– Nernst– Ampere– Electrolytic cells– Environmental – Hybrid cars

Vitamin C

• Which fruit keeps best over a long period of time?

• Fruits known to sailors– Orange, lemon ~1400– Limes 1638– Shaddock (grapefruit) 1700

• Used lemon or lime juice preserved in brandy

Picture from: www.corrosion-doctors.org/ Biographies/images/

Nernst

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Is potential always the same?

Standard conditions: 1 atm, 25oC, 1 M

What will influence the potential of a cell?

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Mathematical Relationships: Nernst

The Nernst Equation:  Eo = standard potential of the cellR = Universal gas constant = 8.3145 J/mol*KT = temperature in Kelvinn = number of electrons transferredF = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode)

E = Eo - RT ln Q nF

Cu+2 + Zn(s) → Zn+2 + Cu(s) Q =

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Applying the Nernst Equation

This cell is operating at 25oC with 1.00x10-5M Zn2+ and 0.100M Cu2+?

Predict if the voltage will be higher or lower than the standard potential

E = Eo - RT ln Q nF

Cu+2 + Zn(s) → Zn+2 + Cu(s)

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Eo = standard potential of the cell

R = Universal gas constant = 8.3145 J/mol*K

T = temperature in Kelvin

n = number of electrons transferred

F = Faraday’s constant = 96,483.4 C/mol

Q = reaction quotient (concentration of anode divided by the concentration of the cathode)

E = Eo - RT ln Q nF Zn+2 + 2e- -> Zn -0.76 V

Cu+2 + 2e- -> Cu 0.34 V

25oC + 273 = K

n =

1.00x10-5M Zn2+ and 0.100M Cu2

Cu+2 + Zn(s) → Zn+2 + Cu(s)

Q = [Zn+2]/[Cu+2]

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Were your predictions correct?

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Picture from: musee-ampere.univ-lyon1.fr/

Ampere

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The Units of Electrochemistry

• Coulomb

– 1 coulomb equals 2.998 x 109 electrostatic units (eu)

– eu is amount of charge needed to repel an identical charge 1 cm away with unit force

– Charge on one electron is -1.602 x 10-19 coulomb

Problem: An aluminum ion has a +3 charge. What is this value in coulombs?

magnitude of charge is same at that of e-, opposite sign3 x 1.602 x10-19 = 4.806 x 10-19 coulomb

Key Point: electrons or ions charges can be measured in coloumbs

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The Units of Electrochemistry

• Ampere

– Amount of current flowing when 1 coulomb passes a given point in 1 second

– Units of Amperes are Coulombs per second

– Current (I) x time (C/s x s) gives an amount of charge.

Problem: How much current is flowing in a wire in which 5.0 x 1016 electrons are flowing per second?

The charge transferred each second = (5.0 x 1016 electrons/sec) x (1.602 x 10-19 coulomb/electron)= 8.0 x 10-3 coulombs/sec = amps

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The Units of Electrochemistry

• Ampere

– Amount of current flowing when 1 coulomb passes a given point in 1 second

– Units of Amperes are Coulombs per second

– Current (I) x time (C/s x s) gives an amount of charge.

– We can express electron or ION flow in amps!

Problem: If 1 mol Al+3 ions passes a given point in one hour, what is the current flow?

1 mol Al +3 ions 6.022 x 1023 Al +3 ions 4.806 x 10-19 coulomb 1 hour

Hour 1 mol Al +3 ions 1 Al+3 ion 3600 sec= 80 C/s = 80 A

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From: Moore, Stanitski, and Jurs Chemistry: The Molecular Science 2nd Edition/

p.941

Batteries

Primary Cellsnon-reversible, non-rechargeable electrochemical cell

"dry" cell & alkaline cell 1.5 v/cellmercury cell 1.34 v/cellfuel cell 1.23v/cell

Secondary Cells reversible, rechargeable electrochemical cell

Lead-acid (automobile battery) 2 v/cellNiCad 1.25 v/cellLithium batteries

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“Flash Light” Batteries

Primary Cells

"Dry" Cell

Zn(s) + 2 MnO2(s) + 2 NH4+

Zn+2(aq) + 2 MnO(OH)(s) + 2 NH3

Alkaline Cell

Zn(s) + 2 MnO2(s) ZnO(s) + Mn2O3(s)

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Leclanche “Dry” Cell

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Mercury BatteryPrimary Cells

Zn(s) + HgO(s) ZnO(aq) + Hg(l)

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Fuel Cells

anode

H2(g) + 2 OH-(aq) → 2 H2O(l) + 2 e-

cathode

O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)

Picture from: http://www.bpa.gov/Energy/N/projects/fuel_cell/education/fuelcellcar/

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Lead-Acid (Automobile Battery)

Pb(s) + PbO2(s) + 2 H2SO4 2 PbSO4(s) + 2 H2O

2 v/cell thus 12 volt battery = 6-2 volt cells

Secondary Cell

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Nickel-Cadmium (Ni-Cad)

Secondary CellCd(s) + 2 Ni(OH)3(s) ↔ Cd(OH)2(s) + Ni(OH)2(s)

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Lithium Batteries

• Lithium batters (developed in 1970; used in watches, pacemakers, etc.)

• The 4-volt lithium battery, which has up to 33 percent higher energy density and 60 less weight than a nickel-metal hydride battery of the same size, has made possible the miniaturization of the current generation of electronic devices

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Lithium chemistry

• What is the ½ reaction involving lithium?

• Is this a reduction or oxidation reaction?

• Does it take place at the anode or cathode?

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The other ½ of the battery:

• Co+3 + e- → Co+2 (+1.92 V)

• What is the standard potential of the cell?

[Li+ + e- → Li (-3.045 V)]

• A lithium battery has a potential listed at 4V. Does this differ from the number you calculated? Why?

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Alternative Anodes

• Which elements would you predict to have oxidation potentials similar to lithium? (no peeking at the table of standard reduction

potentials!)

Is what way is lithium preferable to these elements?

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Alternative Cathodes

• Co+3 + e- → Co+2 • Disadvantage: Use of cobalt oxide can lead to

thermal runaway. Increased temperature and pressure cause case of battery to break and fumes are released. Lithium is exposed to oxygen and hydrogen in the air.

• What are the alternatives and why aren’t they being used in conjunction with lithium to build a battery with higher potential?

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Table of Standard Reduction Potentials

Eo Eo

F2 + 2e- --> 2F- 2.87 Fe3+ + 3e- ---> Fe -0.04Co3+ + e- --> Co2+ 1.80 Pb2+ + 2e- ---> Pb -0.13Cl2 + 2e- ---> 2Cl- 1.36 Ni2+ + 2e- ---> Ni -0.25O2 + 4H+ + 4e- -> 2H2O 1.23 Co2+ + 2e- ---> Co -0.29Hg2+ + 2e- ---> Hg 0.85 Cr3+ + e- ---> Cr2 -0.40Ag+ + e- ---> Ag 0.80 Fe2+ + 2e- ---> Fe -0.41I2 + 2e- ---> 2I- 0.54 Zn2+ + 2e- ---> Zn -0.76Cu+ + e- ---> Cu 0.52 Mn2+ + 2e- ---> Mn -1.182H+ + 2e- ---> H2 0.00 Al3+ + 3e- ---> Al -1.66

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Other commercially viable batteries

• Besides cobalt, two other reduction reactions used in lithium batteries are shown below. What is the potential of these batteries?

• Fe+2 + 2e- → Fe (-0.41 V)

• ½ I2 + e- → I- (0.54 V)

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Other cathode reactions being explored

• Al and Mg are also being explored as materials for the cathode. What are the advantages and disadvantages of these materials?

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Fig. 19-25, p.953

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Corrosion: A Case of Environmental Electrochemistry

• Facts about formation of rust– Iron does not rust in dry air: moisture must be present– Iron does not rust in air-free water: O2 must be present– The loss of iron and the deposition of rust often occur

at different places on the same object– Iron rusts more under acidic conditions (low pH)– Iron rusts more quickly in contact with ionic solutions– Iron rusts more quickly in contact with a less active

metal (such as Cu) and more slowly in contact with a more reactive metal (such as Zn)

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Balance the Redox Reaction

Fe(s) + O2 (g) Fe+2(aq) + H2O(l) in acid

Corrosion: A Case of Environmental Electrochemistry

O2(g) + 4 H+(aq) + 4 e- => 2 H2O(l) Eo = 1.23 V

Fe(s) => Fe+2(aq) + 2 e- Eo = 0.44

V

2 Fe(s) + O2(g) + 4 H+(aq) => 2 H2O(l) + Fe+2

(aq)

Eo = 1.67 V

2Fe2+(aq) + ½O2(g) + (2+n)H2O(l) Fe2O3.nH2O(s) + 4H+(aq)

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Corrosion: A Case of Environmental Electrochemistry

• Sacrificial cathode

Fig. 19-27, p.955

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During the reconstruction of the Statue of Liberty, Teflon spacers were placed between the iron skeleton and the copper plates that cover the statue. What purpose do these spacers serve?

Cu+2 + 2e- Cu 0.153 VFe+2 + 2e- Fe -0.44 V

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Hybrid Cars

2004 Toyota PriusGasoline and Electric Engines

2004 Honda Civic Gasoline and Electric Engines

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Electrochemistry and the Honda Civic

The battery in the 2004 Honda Civic is rated for 6.0 Ampere-hours. If the electric motor draws 42 amps to accelerate from 0 to 60 mph, and the time required is 15 seconds.

The Honda Civic battery is a nickel metal hydride battery. The half-reaction for generating the electrical current is:

MH + OH- → M + H2O + e-

How much metal (assume its nickel) is produce on accelerating from zero to 60?

42 C 15 sec 1 mole e- 1 mol Ni 58.69 g Ni 0.38 g Ni

sec 96, 487 C 1 mole e- 1 mol Ni

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Hybrid Cars: Can We PowerUsing Sunlight and Fuel Cells?

SpectruM2005 Michigan Solar Car Team

Photovoltaics and Battery

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Some Needed Data• The Toyota Prius can generate ~ 100 Amps

– Let’s assume our commute requires on average at ~ 20% max current for 1 hour.

• A 1 m2 solar cell array can generate 3 amps in full summer sunlight

Questions:1. If the electrical power was generated with a fuel cell, how

much hydrogen would be consumed? How much volume would be required at 1 atm pressure to hold the hydrogen?

2. Assume you can place 10 m2 of solar cells on your car. How long would it take to generate the needed hydrogen assuming full summer sunlight?

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1. Hydrogen Consumption

At the anode (where oxidation occurs): 2 H2 (g) + 4 H2O (l) → 4 H3O+ (aq) + 4 e-

At the cathode (where reduction occurs) O2 (g) + 2 H2O (l) + 4e- → 4 OH- (aq)

Overall reaction: 2 H2(g) + O2 → 2 H2O (l)

Average 20% of maximum power = 20 amps20 amps x 3600 secs = 72,000 coulombs

72,000 C 1 Faraday 1 mol electrons 2 mol H2 0.37 mol H2

96,487 C 1 Faraday 4 mol electrons

V = nRT/P = (0.37 x 0.08206 x 300) / 1 = 9 Liters

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2. Hydrogen Generation

A 1 x 1 m2 solar panel generates a peak current of 3 amps.Our car with 10 m2 of solar panels could generate 30 amps.

Electrolysis of water generates hydrogen with the following half-reaction:

At the cathode (where reduction occurs): 4 H3O+ (aq) + 4 e- → 2 H2 (g) + 4 H2O (l)

0.37 mol H2 4 mol e- 1 Faraday 96, 487 C sec 1 hour 0.66 h

2 mol H2 1 mol e- 1 Faraday 30 C 3600 sec

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Issues Arising

• Can generate hydrogen from water quickly enough to supply average demand by fuel cell.

• Important since solar cells cannot supply peak current flow used in Prius of ~ 100 amps.

• Where to store hydrogen??– As water is good. High density storage.– Molarity??– ~110 M

• As gas is bad, requires too much space (~ 9 L)– Need good storage materials!! (Active research in Yaghi group)

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Evaluations

• Please fill out both an evaluation for Dr. Banaszak Holl and Dr. Gottfried and put them in the correct envelopes!

• Constructive comments very helpful!

Thanks for a great semester!