Houghton Mifflin Company and G. Hall. All rights r eserved. 1 Lecture 25 Electrolysis Define electrolysis? Some examples. What are the values of G and E cell ? Electrolysis of water. Some industrial applications. Corrosion
Dec 20, 2015
Houghton Mifflin Company and G. Hall. All rights reserved.
1
Lecture 25 Electrolysis
Define electrolysis?Some examples.What are the values of G and
Ecell?Electrolysis of water.Some industrial applications.Corrosion
Houghton Mifflin Company and G. Hall. All rights reserved.
2
Two Types of Cells
Cell 1: does work by releasing free energy from a spontaneous reaction to produce electricity such as a battery.
Cell 2: does work by absorbing free energy from a source of electricity to drive a non-spontaneous reaction such as electroplating.
Houghton Mifflin Company and G. Hall. All rights reserved.
3
A voltaic (Galvanic) cell can power an electrolytic cell
Houghton Mifflin Company and G. Hall. All rights reserved.
4
Electrolysis
The splitting (lysing) of a substance or decomposing by forcing a current through a cell to produce a chemical change for which the cell potential is negative.
Houghton Mifflin Company and G. Hall. All rights reserved.
5
Electrolysis
Previously our lectures on electrochemistry were involved with voltaic cells i.e. cells with Ecell > 0 and G < 0 that were spontaneous reactions.
Today we discuss electrochemical cells where Ecell < 0 and G > 0 that are non-spontaneous reactions and require electricity for the reactions to take place. We can take a voltaic cell and reverse the electrodes to make an electrochemical cell.
Houghton Mifflin Company and G. Hall. All rights reserved.
6
VoltaicElectrolytic
Houghton Mifflin Company and G. Hall. All rights reserved.
7
Houghton Mifflin Company and G. Hall. All rights reserved.
8
Houghton Mifflin Company and G. Hall. All rights reserved.
9
Fig. 21.17
Houghton Mifflin Company and G. Hall. All rights reserved.
10
Fig. 21.18: Car battery, both voltaic and electrochemical cell.
Houghton Mifflin Company and G. Hall. All rights reserved.
11
Increase oxidizing power
Increase reducing power
Houghton Mifflin Company and G. Hall. All rights reserved.
12
A standard electrolytic cell. A power source forces the opposite reaction
Houghton Mifflin Company and G. Hall. All rights reserved.
13
Electrolysis
Houghton Mifflin Company and G. Hall. All rights reserved.
14
(a) A silver-plated teapot. (b) Schematic of the electroplating of a
spoon.
Houghton Mifflin Company and G. Hall. All rights reserved.
15
Schematic of the electroplating of a
spoon.
AgNO3(aq)
Houghton Mifflin Company and G. Hall. All rights reserved.
16
The electrolysis of water produces hydrogen gas at the cathode (on the right) and oxygen gas at the anode
(on the left).
Houghton Mifflin Company and G. Hall. All rights reserved.
17
Fig. 21.19 Electrolysis of
water
Houghton Mifflin Company and G. Hall. All rights reserved.
18
Electrolysis of water
At the anode (oxidation):2H2O(l) + 2e- = H2(g) + 2OH-(aq) E=-0.42V
At the cathode (reduction):2H2O(l) = O2(g) + 4H+(aq) + 4e- E= 0.82VOverall reaction after multiplying anode
reaction by 2,2H2O(l) = 2H2(g) + O2(g) Eo
cell = -0.42 -0.82 = -1.24 V
Houghton Mifflin Company and G. Hall. All rights reserved.
19
Electrolysis: Consider the electrolysis of a solution that is 1.00 M in each of CuSO4(aq) and NaCl(aq) Oxidation possibilities follow.2Cl–(aq) = Cl2(g) + 2e– E° = –1.358 V2SO4
2–(aq) = S2O82–(aq) + 2e– E° = –2.01 V
2H2O = 4H+(aq) + O2(g) + 4e– E° = –1.229 V
Reduction possibilities follow:Na+(aq) + e– = Na(s) E° = –2.713 VCu2+(aq) + 2e– = Cu(s) E° = +0.337 V2H2O + 2e– = H2(g) + 2OH–(aq) E° = +0.828 V
Houghton Mifflin Company and G. Hall. All rights reserved.
20
Electrolysis
We would choose the production of O2(g) and Cu(s). But the voltage for producing O2(g) from solution is
considerably higher than the standard potential, because of the high activation energy needed to form O2(g).
The voltage for this half cell seems to be closer to –1.5 V in reality.
The result then is the production of Cl2(g) and Cu(s).anode, oxidation: 2Cl–(aq) = Cl2(g) + 2e– E° = –1.358 V
cathode, reduction: Cu2+(aq) + 2e– : Cu(s) E° = +0.337 V
overall: CuCl2(aq) : Cu(s) + Cl2(g) E = –1.021 V We must apply a voltage of more than +1.021 V to
cause this reaction to occur.
Houghton Mifflin Company and G. Hall. All rights reserved.
21
E = -2.37 V
Houghton Mifflin Company and G. Hall. All rights reserved.
22
E = 1.07 V
Ecell = -2.37-1.07 = -3.44V
Houghton Mifflin Company and G. Hall. All rights reserved.
23
Houghton Mifflin Company and G. Hall. All rights reserved.
24
Prob. 21.9
Houghton Mifflin Company and G. Hall. All rights reserved.
25
Stoichiometry of electrolysis: Relation between amounts of charge and
productFaraday’s law of electrolysis
relates to the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell (half reaction).
Each balanced half-cell shows the relationship between moles of electrons and the product.
Houghton Mifflin Company and G. Hall. All rights reserved.
26
Application of Faraday’s law
1. First balance the half-reactions to find number of moles of electrons needed per mole of product.
2. Use Faraday constant (F = 9.65E4 C/mol e-) to find corresponding charge.
3. Use the molar mass of substance to find the charge needed for a given mass of product.1 ampere = 1 coulomb/second or 1 A = 1 C/sA x s = C
Houghton Mifflin Company and G. Hall. All rights reserved.
27
Stoichiometry of Electrolysis
How much chemical change occurs with the flow of a given current for a specified time?
current and time quantity of charge
moles of electrons moles of analyte
grams of analyte
Houghton Mifflin Company and G. Hall. All rights reserved.
28
Fig. 21.20
Houghton Mifflin Company and G. Hall. All rights reserved.
29
Doing work with electricity.
Houghton Mifflin Company and G. Hall. All rights reserved.
30
Houghton Mifflin Company and G. Hall. All rights reserved.
31
Industrial Applications of Electrolysis
http://academic.pgcc.edu/~ssinex/E_cells.ppt.. All rights reserved.
32
What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)?
Na+ Cl-
Let’s examine the electrolytic cell for molten NaCl.
All rights reserved. http://academic.pgcc.edu/~ssinex/E_cells.ppt.
33
+-battery
Na (l)
electrode half-cell
electrode half-cell
Molten NaCl
Na+
Cl-
Cl-
Na+
Na+
Na+ + e- Na 2Cl- Cl2 + 2e-
Cl2 (g) escapes
Observe the reactions at the electrodes
NaCl (l)
(-)
Cl-
(+)
http://academic.pgcc.edu/~ssinex/E_cells.ppt. All rights reserved.
34
+-battery
e-
e-
NaCl (l)
(-) (+)
cathode anode
Molten NaCl
Na+
Cl-
Cl-
Cl-
Na+
Na+
Na+ + e- Na 2Cl- Cl2 + 2e-
cationsmigrate toward
(-) electrode
anionsmigrate toward
(+) electrode
At the microscopic level
http://academic.pgcc.edu/~ssinex/E_cells.ppt. All rights reserved.
35
Molten NaCl Electrolytic Cell
cathode half-cell (-)REDUCTION Na+ + e- Na
anode half-cell (+)OXIDATION 2Cl- Cl2 + 2e-
overall cell reaction2Na+ + 2Cl- 2Na + Cl2
X 2
Non-spontaneous reaction!
Houghton Mifflin Company and G. Hall. All rights reserved.
36
The Downs Cell for the Electrolysis of Molten Sodium Chloride
Houghton Mifflin Company and G. Hall. All rights reserved.
37
If the products are mixed, the result is household bleach.2 NaOH(aq) + Cl2(g) = NaCl(aq) + NaOCl(aq) + H2O
Houghton Mifflin Company and G. Hall. All rights reserved.
38
The Mercury Cell for Production of Chlorine and Sodium Hydroxide
Houghton Mifflin Company and G. Hall. All rights reserved.
39
A schematic diagram of an electrolytic cell for producing aluminum by the
Hall-Heroult process.
Houghton Mifflin Company and G. Hall. All rights reserved.
40
Fig. 22.19 A schematic diagram of an electrolytic cell for producing aluminum by the Hall-Heroult
process.
http://academic.pgcc.edu/~ssinex/E_cells.ppt.
41
The Hall Process for Aluminum
Electrolysis of molten Al2O3 mixed with cryolite – lowers melting point
Cell operates at high temperature – 1000oC
Aluminum was a precious metal in 1886.
A block of aluminum is at the tip of the Washington Monument!
42
carbon-lined steel vesselacts as cathode
CO2 bubbles
Al (l)Al2O3 (l)
Drawoff Al (l)
-
+
Cathode: Al+3 + 3e- Al (l)
Anode: 2 O-2 + C (s) CO2 (g) + 4e-
frompowersource
Al+3
O-2O-2
Al+3
O-2
graphite anodes
e-
e- From: http://academic.pgcc.edu/~ssinex/E_cells.ppt.
http://academic.pgcc.edu/~ssinex/E_cells.ppt.
43
The Hall Process
Cathode: Al+3 + 3e- Al (l)
Anode: 2 O-2 + C (s) CO2 (g) + 4e-
4 Al+3 + 6 O-2 + 3 C (s) 4 Al (l) + 3 CO2 (g)
x 4
x 3
The graphite anode is consumed in the process.
Houghton Mifflin Company and G. Hall. All rights reserved.
44
Fig. 22.21: Production of solid Mg
Houghton Mifflin Company and G. Hall. All rights reserved.
45
Corrosion
Electrochemistry plays a major role in corrosion and protection against it.
46
Calculating the cell potential, Eocell, at
standard conditions
Fe+2 + 2e- Fe Eo = -0.44 v
O2 (g) + 2H2O + 4e- 4 OH- Eo = +0.40 v
This is corrosion or the oxidation of a metal.
Consider a drop of oxygenated water on an iron objectFe
H2O with O2
Fe Fe+2 + 2e- -Eo = +0.44 v2x
2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 v
reverse
From: http://academic.pgcc.edu/~ssinex/E_cells.ppt.
Houghton Mifflin Company and G. Hall. All rights reserved.
47
Cathodic Protection Against Corrosion
Underground steel pipes offer the strength to transport fluids at high pressures, but they are vulnerable to corrosion driven by electrochemical processes. A measure of protection can be offered by driving a magnesium rod into the ground near the pipe and providing an electrical connection to the pipe. Since the magnesium has a standard potential of -2.38 volts compared to -.41 volts for iron, it can act as a anode of a voltaic cell with the steel pipe acting as the cathode. With damp soil serving as the electrolyte, a small current can flow in the wire connected to the pipe. The magnesium rod will be eventually consumed by the reactionMg(s) -> + Mg2+(aq) + 2e-
while the steel pipe as the cathode will be protected by the reactionO2(g) + 2H2O(l) + 4e- -> 4OH-(aq).
From: http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/corrosion.html#c2
Houghton Mifflin Company and G. Hall. All rights reserved.
48
Lecture summary
Electrolysis is often the reverse of voltaic cell in that Ecell < 0, and G >0 and reaction is non-spontaneous.
Electrolysis of water and to produce O2 and H2.
Faraday’s law allows us to determine how much current is needed to produce a certain amount of an element.
Industrial applications are numerous for producing a variety of solid elements (Al, Mg, Na, etc).