11
Electrochemistry and Society I. Corrosion = oxidation of pure metals to their oxides A. Corrosion Basics 1) Metals (M o ) are easily oxidized to cationic forms (M n+ ) [Table 18.1] 2) o 1/2 of O 2 gas reduction > oxidation of most metals a) O 2 + 4H + + 4e - 2H 2 O o 1/2 = +1.23 V b) This leads to an o cell that is positive for this process = spontaneous c) M o + O 2 M x O y o cell = + 3) Most metals don’t completely decompose because M x O y protects the vulnerable M o underneath from further corrosion 4) Aluminum Example a) Al 3+ + 3e - Al o o = -1.70 V b) O 2 + 4H + + 4e - 2H 2 O o = +1.23 V c) Al 2 O 3 produced has o 1/2 = -0.60 V, resulting in a much less favorable corrosion process, once the aluminum underneath is covered. o cell = +2.93 V

Electrochemistry and Society

Embed Size (px)

DESCRIPTION

Electrochemistry and Society. Corrosion = oxidation of pure metals to their oxides Corrosion Basics Metals (M o ) are easily oxidized to cationic forms (M n+ ) [Table 18.1] e o 1/2 of O 2 gas reduction > oxidation of most metals - PowerPoint PPT Presentation

Citation preview

Page 1: Electrochemistry and Society

Electrochemistry and Society

I. Corrosion = oxidation of pure metals to their oxidesA. Corrosion Basics

1) Metals (Mo) are easily oxidized to cationic forms (Mn+) [Table 18.1]

2) o1/2 of O2 gas reduction > oxidation of most metals

a) O2 + 4H+ + 4e- 2H2O o1/2 = +1.23 V

b) This leads to an ocell that is positive for this process = spontaneous

c) Mo + O2 MxOy ocell = +

3) Most metals don’t completely decompose because MxOy protects the vulnerable Mo underneath from further corrosion

4) Aluminum Example

a) Al3+ + 3e- Alo o = -1.70 V

b) O2 + 4H+ + 4e- 2H2O o = +1.23 V

c) Al2O3 produced has o1/2 = -0.60 V, resulting in a much less favorable

corrosion process, once the aluminum underneath is covered.

ocell = +2.93 V

Page 2: Electrochemistry and Society

5) The “Noble Metals” (Ag, Au, Cu, Pt) do not react with oxygen as easily

a) Auo o1/2 = +1.50 V no corrosion

b) Ago o1/2 = +0.80 V Ag2S tarnish formed instead of the oxide

c) Cuo o1/2 = +0.16 V Cu2CO3 forms green “patina”

B. The Corrosion of Iron

1) This is the most economically important corrosion process due to structural steel

2) Steel has a non-uniform surface due to physical stress

a) Anodic Region: Fe Fe2+ + 2e-

b) Cathodic Region: O2 + 2H2O + 4e- 4OH-

3) Fe2+ then acts as the salt bridge electrolyte if wet (added salt speeds up corrosion)

Cathode: 4Fe2+ + O2 (4 + 2n)H2O 2Fe2O3 • nH2O + 8H+

Page 3: Electrochemistry and Society

C. Preventing Corrosion

1) Paint covers the surface to prevent the contact of oxygen and the metal

2) Plating steel with Cr or Sn to produce very stable oxides

3) Galvanizing = coating with Zinc

a) Fe Fe2+ + 2e- -o1/2 = +0.44 V

b) Zn Zn2+ + 2e- -o1/2 = +0.76 V

c) Corrosion occurs on Zn rather than Fe (sacrificial coating)

4) Stainless Steel = Fe + Cr + Ni -o1/2 ~ Noble metal

5) Cathodic Protection = protects buried steel or ships with a sacrificial reactant

a) Active metal (Mg) connected to pipe by a wire -o1/2 = +2.37 V

b) Bars of Ti attached to ship -o1/2 = +1.60 V

Page 4: Electrochemistry and Society

II. Electrolysis = using electric energy to produce chemical change (opposite of cell)A. Example

1) Consider the Cu/Zn Galvanic Cell

a) Anode: Zn Zn2+ + 2e-

b) Cathode: Cu2+ + 2e- Cu ocell = +1.10 V

2) If we attach a power source of o > +1.10 V, we can force e- to go the other way

a) Anode: Cu Cu2+ + 2e-

b) Cathode: Zn2+ + 2e- Zn

c) Called an Electrolytic Cell

Page 5: Electrochemistry and Society

B. Calculations with Electrolytic Cells1) How much Chemical Change? Is usually the question.2) Find mass of Cuo plated out by passing 10 amps (10 C/s) through Cu2+ solution.

a) Cu2+ + 2e- Cuo(s) b) Steps: current/time, charge (C), moles e-, moles Cu, grams Cu

3) Example: How long must a current of 5.00 amps be applied to a Ag+ solution to produce 10.5 g of silver metal?

C. Electrolysis of Water

1) Galvanic: 2H2 + O2 2H2O (Fuel Cell)2) Electrolytic Cell:

a) Anode: 2H2O O2 + 4H+ + 4e- -o = -1.23 V

b) Cathode: 4H2O + 4e- 2H2 + 4OH- o = -0.83 V

c) Overall: 6H2O 2H2 + O2 + 4(H+ + OH-)

2H2O 2H2 + O2 ocell = -2.06 V

3) We must add a salt to increase the conductance of pure water [H+] = [OH-] = 10-7

Cu g 5.94 g/mol) Cu)(63.546 mol 0935.0(

Cu mol 0935.0-e mol 2

Cu mol 1e-) mol (0.187

-e mol 0.187 C 96,485

-e mol 1C) (18,000

C 18,000 s/min) min)(60 C/s)(30 10(

Page 6: Electrochemistry and Society
Page 7: Electrochemistry and Society

D. Electrolysis of Mixtures

1) Mixture of Cu2+, Ag+, Zn2+; What is the order of plating out?

a) Ag+ + e- Ag o1/2 = +0.80 V

b) Cu2+ + 2e- Cu o1/2 = +0.34 V

c) Zn2+ + 2e- Zn o1/2 = -0.76 V

2) Reduction of Ag+ is easiest (o = most positive) followed by Cu, then Zn

3) Example: Ce4+ (o1/2 = +1.70 V), VO2+ (o

1/2 = +1.00 V), Fe3+ (o1/2 = +0.77 V)

III. Commercial Electrolytic ProcessesA. Production of Aluminum

1) Most metals are found naturally as their oxides, MxOy

2) Only the noble metals are typically found as the pure metal

3) Bauxite = aluminum ore; Al2O3

4) Aluminum is the third most abundant element on crust (oxygen and silicon)

5) No commercial process for pure Al until 1854 (o1/2 = 1.66 V)

6) Al was more valuable than gold or silver

Page 8: Electrochemistry and Society

8) Hall—Heroult Process

a) Al3+ + 3 e- Al o1/2 = -1.66 V

b) 2H2O + 2e- H2 + 2OH- o1/2 = -0.83 V

c) Can’t make Al in water because water gets reduced before Al3+

d) Use molten Al2O3/Na3AlF6 mixture at 1000 oC

9) Aluminum alloys with Zn, Mn are most useful because they are stronger

10) Aluminum production uses 5% of the electricity consumed in the U.S.

Page 9: Electrochemistry and Society

B. Electrorefining = purification of metals

1) Impure Cuo anode: Cuo Cu2+ + 2e-

2) Pure Cuo cathode: Cu2+ + 2e- Cuo (99.95% pure)

3) Also useful for purification of Zn, Fe

4) Gold, Silver, Platinum fall to the bottom of the tank as sludge (won’t plate out)

C. Metal Plating

1) Coat easily corrodable metal object with a noble metal

2) Ag+ + e- Ago on a spoon

D. Electrolysis of NaCl

1) Production of Na metal from NaCl in a “Downs Cell”

a) Anode: 2Cl- Cl2 + 2e-

b) Cathode: Na+ + e- Nao

c) Cell is designed to to keep products apart so they can’t reform NaCl

2) Production of Cl2, OH- in a Mercury Cell

a) Water is reduced to OH- (o1/2 = -0.83 V) before Na+ (o

1/2 = -2.71 V)

b) Anode: 2Cl- Cl2 + 2e-

c) Cathode: 2H2O + 2e- H2 + 2OH-

d) Chlor-Alkali Process = second largest electricity user in U.S. (after Al)

Page 10: Electrochemistry and Society

Downs Cell

Page 11: Electrochemistry and Society

Mercury Cell for Chor-Alkali Process