1. Electrochemistry Importance of Electrochemistry
starting your car
use a calculator
listen to a radio
corrosion of iron
preparation of important industrial materials
Electrochemistry - the study of the interchange of chemical and electrical energy

2. Galvanic Cells Redox reactions electron transfer reactions
oxidation - loss of electrons, i.e., an increase in oxidation number
reduction - gain electrons, i.e., a decrease in oxidation number

3. Galvanic Cells Ex: 8 H+ + MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3 + 4 H2O
If MnO4- and Fe+2 are present in the same solution, the electrons are transferred directly when the reactants collide.
No useful work is done.
Chemical energy is released as heat.
How can the energy be harnessed?
Separate the half reactions.
Have the electron transfer go through a wire, which can then be directed through a motor or other useful device

4. Galvanic Cells Galvanic cell
Uses a spontaneous redox reaction to produce current to do work
A device in which chemical energy is changed to electrical energy
Also known as a voltaic cell

5. Galvanic Cells Requirements for a Galvanic Cell
Half reactions are in separate compartments
Compartments are connected by a salt bridge
a porous disk connection
a salt bridge (U bridge) containing a strong electrolyte held in a jello-like matrix

6. Galvanic Cells Why a salt bridge?
Without a salt bridge, the reaction will not go
Current starts, but stops due to charge buildup in both compartments
one side becomes negatively charged as electrons are added
one side becomes positively charged as electrons are lost
Creating a charge separation requires a large amount of energy

7. Galvanic Cells
Reaction in a galvanic cell occurs at each of the electrodes
Be able to label the anode, cathode, direction of electron flow, and direction of ion flow

8. Galvanic Cells Cell Potential
The oxidizing agent “pulls” electrons towards itself from the reducing agent
The “pull” or driving force on the electrons is the cell potential, Ecell, aka, the electromotive force (emf) of the cell.
Measured in volts, V
1 V = 1 Joule/Coulomb

9. Galvanic Cell
Cell Potential
Measure with a voltmeter

10. Standard Reduction Potentials
Predict cell potentials if the half cell potentials are known
add half cell potentials to get cell potential
Eox + Ered = Ecell
There is no way to measure half cell potentials

11. Standard Reduction Potentials
So how do we get these half cell potentials if they can’t be measured directly?
Measure a cell potential, assigning one reaction to be the “zero”
2H+ + 2 e- --> H2 Ered = 0.00 V

12. Standard Reduction Potentials
So for 2H+ + Zn --> H2 + Zn+2 the voltage is found to be 0.76 V
If 2 H+ + 2 e- --> H2 Ered = 0.00 V
and Ecell = Eox + Ered,
then Eox = V = 0.76 V
So we can say
Zn --> Zn e- Eox = 0.76 V

13. Standard Reduction Potentials
Combine other half reactions with half reactions with known half cell potentials to complete the table of Standard Reduction Potentials

14. Standard Reduction Potential
Things to know about standard reduction potentials:
Eo values correspond to solutes at 1 M and all gases at 1 atm
When a half reaction is reversed, the sign of Eo is reversed
When a half reaction is multiplied by an integer, Eo remains the same!! Do not multiply Eo by any number!

15. Standard Reduction Potentials
The cell potential is always positive for a galvanic cell
Given Fe+2 + 2e- --> Fe Ered = V
MnO e- + 8 H+--> Mn+2 + 4H2O Ered = 1.51 V
Reverse the reaction that will result in an overall cell potential that is positive, i.e., reverse the rxn involving iron

16. Standard Reduction Potentials
Find the Eo for the galvanic cell based on the reaction
Al+3 + Mg --> Al + Mg+2
Sketch this cell and label the anode, cathode, direction of electron and ion flow, and indicate what reaction occurs at each electrode

17. Standard Reduction Potentials
Calculate Eo and sketch the galvanic cell based on the (unbalanced) reaction:
MnO4- + H+ + ClO3- --> ClO4- + Mn+2 + H2O
Balance the reaction

18. Galvanic Cells Line Notation Used to describe electrochemical cells
Anode components are listed on the left
Cathode components are listed on the right
Separate half cells with double vertical lines: ll
Indicate a phase difference with a single vertical line: l

19. Galvanic Cells Ex: Mg(s) l Mg+2(aq) ll Al+3(aq) l Al(s)
is the line notation for the example in slide # 15
For slide #16, the reactants and products are all ions, and so cannot act as an electrode
an inert electrode is needed - use Pt
Pt(s) l ClO3-(aq),ClO4- ll MnO4-(aq),Mn+2(aq) l Pt(s)

20. Cell Potential, Electrical Work, and Free Energy
Relationship between free energy and cell potential
after some serious derivations:
DGo = -nFE o
n = number of moles of electrons
F = the charge on 1 mole of electrons
= Faraday = 96,485 Coulombs/1 mole e-
this equation provides an experimental means to obtain DGo.
Cell potential is directly related to DGo.
Confirms that a galvanic cell runs in a direction that results in a positive E o, because a positive E o means a negative DGo, which means the reaction is spontaneous

21. Cell Potential and Free Energy
Calculate DGo for the reaction
Cu+2 + Fe --> Cu + Fe+2
Is this reaction spontaneous?

22. Cell Potential and Free Energy
Predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au+3 solution.

23. Dependence of Cell Potential on Concentration
The standard reduction potentials can be used to tell us the cell potential for cells under standard conditions (all concentrations = 1 M)
What will happen to the cell potential when the concentration is not 1 M?

24. Dependence of Cell Potential on Concentration
Cu(s) + 2Ce+4(aq) --> Cu+2(aq) + 2 Ce+3(aq)
Under standard conditions, Eo = 1.36 V. What will the cell potential be if [Ce+4] is greater than 1.0 M?
Use Le Chatelier’s principle.
Increasing the [Ce+4] (when the [Ce+4] > 1 M), then the forward reaction is favored, so the driving force on the electrons is greater, so the cell potential is greater

25. Dependence of Cell Potential on Concentration
2 Al(s) + 3Mn+2(aq) --> 2 Al+3(aq) + 3 Mn(s) Eo = 0.48 V
Predict whether Ecell is larger or smaller than Eocell
a. [Al+3] = 2.0 M, [Mn+2] = 1.0 M
b. [Al+3] = 1.0 M, [Mn+2] = 3.0 M

26. Concentration Cells Cell potential depends on concentration
Make a cell (a concentration cell) in which both compartments contain the same components, but at different concentrations
Usually has a small cell potential
The difference in concentration produces the cell potential

27. Concentration Cells
Given a cell in which both compartments contain aquesous AgNO3
Ag+ + e- ---> Ag Eo = 0.80 V
Left side: 0.10 M Ag+
Right side: 1.0 M Ag+
What will happen?

28. Concentration Cells
There will be a positive cell potential due to the difference in Ag+ concentrations
To reach equilibrium, the driving force is to equalize the Ag+ concentration.
This can be done if the 1.0 M Ag+ could be reduced and the 0.10 M Ag+ could be increased.
The Ag in the 0.10 M Ag+ compartment will dissolve while the Ag in the 1.0 M Ag+ compartment will increase in mass.

29. The Nernst Equation DG = DGo + RT ln Q DGo= -nFEo
-nFE = -nFE o + RT lnQ
E = Eo - RT ln Q Nernst equation nF
The Nernst equation gives the relationship between cell potential and the concentrations of the cell components

30. The Nernst Equation @ 25oC, the Nernst equation can be written as:
E = E o log Q n
Use the Nernst equation to calculate the cell potential when one or more of the components are not in their standard states

31. The Nernst Equation For 2 Al(s) + 3 Mn+2 --> 2 Al+3 + 3 Mn(s)
Calculate E when [Mn+2] = 0.50 M and [Al+3] = 1.50 M
Applying Le Chatelier’s principle, we would predict that the reverse reaction would be favored, so E should be less than Eo.

32. The Nernst Equation
The cell potential calculated by the Nernst equation is the maximum cell potential before any current flows
As an galvanic cell discharges, the concentrations of the components will change, so E will change over time
A cell will spontaneously discharge until equilibrium has been reached

33. The Nernst Equation
Equilibrium has been reached when Q = K, so Ecell = 0
A dead battery is one in which the cell reaction has reached equilibrium
There is no driving force for electrons to be pushed through the wire
DG = 0 at equilibrium, meaning the cell no longer has the ability to do work

34. The Nernst Equation
Describe the cell based on the following half reactions and concentrations:
VO H+ + e- --> VO+2 + H2O Eo = 1.00 V
Zn e- --> Zn Eo = V
T = 25o C
[VO2+] = 2.0 M
[H+] = 0.50 M
[VO+2] = 1.0 x 10-2M
[Zn+2] = 1.0 x 10-1 M

35. Calculating the Keq for Redox Reactions
For a cell at equilibrium,
Ecell = 0, and Q = K
Ecell = Eocell log Q n
0 = Eocell log Keq
Eocell = log Keq
log Keq = nEocell @ 25oC
0.0592

36. Batteries A battery is a galvanic cell
a group of galvanic cells connected in series
cell potentials of the individual cells add up to give the total battery cell potential
a source of direct current

37. Batteries Lead Storage Battery
can function for several years under temperature extremes from -30oF to 120oF
12 Volt storage battery made of six cells
anode = Pb
cathode = Pb coated with PbO2
electrolyte solution = H2SO4 solution

38. Batteries Pb + HSO4- --> PbSO4 + H+ + 2 e-
PbO2 + HSO H+ + 2 e- --> PbSO4 + 2 H2O
___________________________________________________________
Pb + PbO2 + 2HSO H+ --> 2 PbSO4 + 2 H2O
PbSO4 adheres to the electrodes
As the cell discharges, the sulfuric acid is consumed. The condition of the battery can be determined by measuring the density of the solution. As the sulfuric acid concentration decreases, the density decreases.

39. Batteries
The lead storage battery can be recharged because the products adhere to the electrodes, so the alternator can force current through the battery in the opposite direction and reverse the reaction
Even though the battery can be recharged, physical damage from road shock and chemical side reactions (e.g. electrolysis of water)eventually cause battery failure

40. Batteries Dry Cell Battery
invented more than 100 years ago by George Leclanche
Acid Version
anode: Zn
cathode: Carbon rod in contact with a moist paste of solid MnO2, solid NH4Cl, and carbon
2NH MnO2 + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O
produces 1.5 V

41. Batteries Alkaline dry cell anode: Zn
cathode: carbon rod in contact with a moist paste of solid MnO2, KOH or NaOH, and carbon
2MnO2 + H2O + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O
the alkaline dry cell lasts longer because the zinc doesn’t corrode as fast under basic conditions

42. Batteries Nickel-Cadmium battery
Cd + NiO2 + 2 OH- + 2 H2O --> Cd(OH)2 + Ni(OH)2 + 2 OH-
This battery can be recharged because, like the lead storage battery, the products adhere to the electrodes.

43. Fuel Cells
Fuel Cell
a galvanic cell for which the reactants are continuously supplied
used in the U.S. space program
based on the reaction of hydrogen and oxygen to form water:
2 H2(g) + O2(g) --> 2H2O(l)
not exactly a portable power source as this cell weighs about 500 pounds

44. Corrosion
Corrosion is the return of metals to their natural states, i.e., the ores from which they are obtained.
Corrosion involves oxidation of the metal
Corroded metal loses its structural integrity and attractiveness
Metals corrode because they oxidize easily
Metals commonly used for decorative and structural purposes have less positive reduction potentials than oxygen gas

45. Corrosion Noble Metals
copper, gold, silver, and platinum are relatively difficult to oxidize, hence the term noble metals

46. Corrosion
Some metals form a protective oxide coating, preventing the complete corrosion of the metal
Aluminum is the best example, forming Al2O3, which adheres to, and protects the aluminum
Copper forms an external layer of copper carbonate, known as patina
Silver forms silver tarnish which is silver sulfide
Gold does not corrode in air

47. Corrosion of Iron
Important to control the corrosion of iron because it is so important as a structural material
The corrosion of iron is not a direct oxidation process (iron reacting with oxygen), but is actually an electrochemical reaction.

48. Corrosion of Iron Steel has a nonuniform surface
These nonuniform areas are where iron can be more easily oxidized (the anode regions) than at other regions (the cathode regions)
anode: Fe--> Fe e- (the electrons flow through the steel to the cathode)
cathode: O2 + 2 H2O + 4 e- --> 4OH-
The Fe+2 formed in the anodic regions travel through the moisture to the cathodic regions. There, the Fe+2 reacts with oxygen to form rust, which is hydrated iron (III) oxide, Fe2O3.nH2O

49. Corrosion of Iron
For iron to corrode, moisture must be present to act as a salt bridge between the anode and cathode regions.
Steel does not rust in dry air
Salt accelerates the rusting process
Cars rust faster where salt is used on roads to melt ice and snow
Salt on the moist surfaces increases the conductivity of the aqueous solution formed

50. Corrosion Prevention Apply a protective coating
apply paint or metal plating
Chromium or tin are often used to plate steel because they form protective oxides
Zinc can be used to coat steel, a process called galvanizing
Fe--> Fe e- Eo = 0.44 V
Zn --> Zn e- Eo = 0.76 V
Zinc, then, is more likely to be oxidized than iron. It acts as a protective coating because it will react with the oxygen preferentially, so the oxygen and the iron don’t come into contact. The zinc is sacrificed.

51. Corrosion Prevention Alloying is used to prevent corrosion
Stainless steel does not corrode like iron
carbon, chromium and nickel have been added to iron to form stainless steel, an alloy.
These additions to iron form a protective oxide coating that changes steel’s reduction potential to that of a noble metal.

52. Electrolysis Electrolytic Cell
Uses electrical energy to force a nonspontaneous redox reaction to go
Current is forced through a cell for which the Eocell is negative
Importance
charging a battery
producing aluminum metal
chrome plating an object

53. Electrolysis
Compare an electrolytic cell to a galvanic cell. Be able to label the anode, cathode, direction of ion and electron flow, which half reactions take place at each electrode, and where oxidation or reduction are taking place:
Zn+2 + Cu --> Cu+2 + Zn vs.
Zn + Cu+2 --> Zn+2 + Cu

54. Electrolysis Stoichiometry of electrolytic process
1 Ampere = 1 coulomb of charge /sec or
1 A.sec = 1 C
96485 C = the charge on 1 mole of electrons
To plate something is to deposit the neutral metal on the electrode by reducing the metal ions in solution, e.g Cu e---> Cu

55. Electrolysis
How long must a currentof 5.00 A be applied to a solution of Ag+ to produce 10.5 g of silver metal?

56. Electrolysis Electrolysis of a mixture of ions
Which metal will plate out first?
Look at the standard reduction potentials
The more positive the Eo, the more likely the reaction will proceed
Given a solution with Cu+2, Ag+, and Zn+2, which metal will plate out first?

57. Electrolysis
Given a solution containing Ce+4, VO2+, and Fe+3, give the order of oxidizing ability of these species and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage.

58. Electrolysis Overvoltage
a complex phenomenon which results in more voltage needing to be applied to cause a reaction to occur than predicted from the Eo’s, i.e., there will be exceptions!
Ex: Electrolysis of a solution of NaCl
Expect Na+, Cl-, and H2O to be the major species
2Cl- --> Cl2 + 2 e- Eo = V
2H2O --> O2 + 4 H+ + 4 e- Eo = V
It would appear that H2O would be easier to oxidize because Eo is more positive, but in reality, Cl2 is produced. No good explanation, sorry!

59. Commercial Electrolytic Processes
Metals are typically good reducing agents
typically found combined with other substances in ores, mixtures of ionic substances containing oxides, sulfides, and silicate anions
nobles metals, Cu, Ag, Pt, and Au, can be found as pure metals

60. Commercial Electrolytic Processes
Production of aluminum
Al is the third most abundant element on earth
Al is a very active metal found in nature as the oxide in an ore called bauxite (for Les Baux, France where it was discovered)
Charles Hall (U.S.) and Paul Heroult (France) discovered an electrolytic process to produce pure aluminum almost simultaneously - Hall-Heroult process

61. Commercial Electrolytic Processes
Production of Aluminum
Uses molten cryolite, Na3AlF6 as the solvent for aluminum oxide
water is more easily reduced than Al+3, so aluminum oxide could not be dissolved in water to produce aluminum
melting an ionic substance allows for mobility of the ions, but the m.p. of Al2O3 is too high at 2050oC to make melting it practical
Mixing Al2O3 and Na3AlF6 lowers the m.p. to a mere 1000oC, so electrolysis became economically feasible
Price of Aluminum dropped from $100,000/lb to $0.74 /lb!

62. Commercial Electrolytic Processes
Aluminum produced in the Hall-Heroult process is 99.5% pure
To be useful as a structural material, aluminum is alloyed with metals like zinc and manganese.
Production of aluminum uses about 5% of all the electricity used in the U.S.

63. Electrolysis of Sodium Chloride
produces pure sodium metal
melt solid sodium chloride (after mixing with calcium chloride to lower the m.p. from 800oC to 600oC), apply electricity, and sodium is produced

64. Electrolysis of Sodium Chloride
Electrolysis of aqueous sodium chloride
aqueous sodium chloride, aka, brine
process is the second largest consumer of electricity in the U.S.
produces chlorine gas and sodium hydroxide
not a source of sodium metal because water is more easily reduced than Na+
Na+ + e- --> Na Eo = V
2 H2O + 2 e- --> H2 + 2 OH- Eo = -0.83V