Download Oxidation-Reduction and Electrochemistry

Oxidation‐Reduction
and
Electrochemistry
David A. Katz
Department of Chemistry
Pima Community College
Oxidation‐Reduction Reactions
In an oxidation‐reduction (Redox) reaction, electrons are transferred from one species to another.
For example, in a single replacement reaction
Cu (s) + 2 AgNO3 (aq)  2 Ag (s) + Cu(NO3)2 (aq) The Cu atoms lose electrons to form Cu2+ in the
Cu(NO3)2 (aq) and the Ag+ gains electrons to form metallic Ag
Oxidation‐Reduction Reactions
• This can be more easily observed by writing the net ionic equation for the reaction:
Cu (s) + 2 Ag+ (aq)  2 Ag (s) + Cu2+ (aq) • The metallic Cu atoms are uncombined, so they are considered to have an oxidation number of zero.
• The initial combined Ag+ ions are in a +1 oxidation state.
• Each Cu atom will lose 2 electrons to 2 Ag+ ions
• The resulting Ag atoms are considered to have an oxidation number of zero
Oxidation‐Reduction Reactions
Cu (s) + 2 Ag+ (aq)  2 Ag (s) + Cu2+ (aq)
Oxidation Numbers
In order to keep track of what loses electrons and what gains them, we list the oxidation numbers of each element.
Oxidation and Reduction
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.
Oxidation and Reduction
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron and they combine to form H2.
Oxidation and Reduction
• The species that contains the element that is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• The species that contains the element that is oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons.
Assigning Oxidation Numbers
1. Elements in their elemental form have an oxidation number of 0.
2. The oxidation number of a monatomic ion is the same as its charge.
Assigning Oxidation Numbers
3. The oxidation number of metals depends on their position in the periodic table
•
•
•
•
•
Group IA elements are +1
Group IIA elements are +2
Group IIIA elements are +3
Group IVA metals are usually +2 or +4
Group VA metals are usually +3 or +5
Assigning Oxidation Numbers
4. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
– Oxygen always has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.
– Hydrogen is always −1 when bonded to a metal
– Hydrogen is +1 when bonded to a nonmetal.
Assigning Oxidation Numbers
4. Nonmetals (continued).
– Fluorine always has an oxidation number of −1.
– The halogens (Cl, Br, and I)have an oxidation number of −1 when they are negative
– The halogens (Cl, Br, and I) will have positive oxidation numbers in oxyanions
(ClO‐, ClO2‐, ClO3‐, etc.)
Assigning Oxidation Numbers
5. The sum of the oxidation numbers in a neutral compound is 0.
6. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
Balancing Oxidation‐Reduction Equations
Oxidation‐reduction equations may be difficult to balance.
Generally, the easiest way to balance the equation of an oxidation‐reduction reaction is via the half‐
reaction method.
This involves treating the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.
Balancing Redox Equations by the Half‐Reaction Method
1. Assign oxidation numbers to the elements in the equation
2. Determine what is oxidized and what is reduced.
3. Write the oxidation and reduction half‐
reactions.
Balancing Redox Equations by the Half‐Reaction Method
4. Balance each half‐reaction.
a.
b.
c.
d.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
5. Multiply the half‐reactions by integers so that the electrons gained and lost are the same.
Balancing Redox Equations by the Half‐Reaction Method
6. Add the half‐reactions, subtracting things that appear on both sides.
7. Make sure the equation is balanced according to mass.
8. Make sure the equation is balanced according to charge.
Balancing Redox Equations by the Half‐Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) (acidic solution)
Balancing Redox Equations by the Half‐Reaction Method
First, assign oxidation numbers (remember oxygen is ‐2)
+7
+3
+2
+4
MnO4− + C2O42‐  Mn2+ + CO2
The oxidation number of manganese goes from +7 to +2, it is reduced.
The oxidation number of carbon goes from +3 to +4, it is oxidized.
Balancing Redox Equations by the Half‐Reaction Method
Determine the oxidizing and reducing agents
+7
+3
+2
+4
MnO4− + C2O42‐  Mn2+ + CO2
The MnO4‐ is the oxidizing agent
The C2O42‐ is the reducing agent
Balancing Redox Equations by the Half‐Reaction Method
The oxidation half‐reaction is
C2O42−  CO2
Balance the half reaction
C2O42−  2 CO2
This balances both the carbon atoms and the oxygen atoms
Balancing Redox Equations by the Half‐Reaction Method
Balance the charge by adding 2 electrons to the right side.
C2O42−  2 CO2 + 2 e−
Balancing Redox Equations by the Half‐Reaction Method
The reduction half‐reaction is
MnO4−  Mn2+
The manganese is balanced
In order to balance the 4 oxygens, we add 4 water molecules to the right side.
MnO4−  Mn2+ + 4 H2O
Balancing Redox Equations by the Half‐Reaction Method
MnO4−  Mn2+ + 4 H2O
The addition of water on the right side of the equation included hydrogen atoms.
To balance the 8 hydrogens, add 8 H+ to the left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
Balancing Redox Equations by the Half‐Reaction Method
8 H+ + MnO4−  Mn2+ + 4 H2O
Balance the charges, add 5 e− to the left side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
Balancing Redox Equations by the Half‐Reaction Method
Before the two half‐reactions can be added together, the number of electrons lost must be equal to the electrons gained:
C2O42−  2 CO2 + 2 e− (2 e‐ lost)
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O (5 e‐ gained)
To attain the same number of electrons on each side, multiply the first reaction by 5 and the second by 2.
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
Balancing Redox Equations by the Half‐Reaction Method
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
Add these half‐reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e− Cancel out the electrons from both sides to get
16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2
Balancing Redox Equations
in Basic Solution
• If a reaction occurs in basic solution, use OH‐
and H2O instead of the H+ and H2O used in acid solution
• The easiest method is to balance the equation as if it occurred in acid.
• Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.
• If this results in water on both sides of the equation, subtract water from each side as a final step.
Balancing Redox Equations
in Basic Solution
Complete and balance the following redox equation that takes place in basic solution
CN‐ (aq) + MnO4‐ (aq)  CNO‐ (aq) + MnO2 (s) (basic solution) First, assign oxidation numbers (remember oxygen is ‐2)
+2 ‐3 +7 ‐2 +4 ‐3 ‐2 +4 ‐2
CN‐ (aq) + MnO4‐ (aq)  CNO‐ (aq) + MnO2 (s)
Note: In an anion such as CN‐, C, which comes first, would be positive and N would be negative. If negative, N would be ‐3.
Balancing Redox Equations
in Basic Solution
+2 ‐3 +7 ‐2 +4 ‐3 ‐2 +4 ‐2
CN‐ (aq) + MnO4‐ (aq)  CNO‐ (aq) + MnO2 (s)
The oxidation number of carbon goes from +2 to +4, it is oxidized. (C lost 2 e‐)
The oxidation number of manganese goes from +7 to +4, it is reduced. (Mn gained 3 e‐)
MnO4‐ is the oxidizing agent
CN‐ is the reducing agent
Balancing Redox Equations
in Basic Solution
The reduction half‐reaction is
MnO4−  MnO2
Balance the half reaction
MnO4−  MnO2
Balance the oxygen atoms, add H+ and H2O
4 H+ + MnO4−  MnO2 + 2 H2O
Balancing Redox Equations
in Basic Solution
Balance the charges, add 3 e‐ to the left side 3 e‐ + 4 H+ + MnO4−  MnO2 + 2 H2O
Balancing Redox Equations
in Basic Solution
The oxidation half‐reaction is
CN−  CNO‐
Since the C atoms are balanced, balance the oxygen. Add H2O to the left side and H+ to the right side.
H2O + CN−  CNO‐ + 2 H+
Balancing Redox Equations
in Basic Solution
H2O + CN−  CNO‐ + 2 H+
Balance the charge, add 2 e‐ to the right side
H2O + CN−  CNO ̶ + 2 H+ + 2 e‐
Balancing Redox Equations
in Basic Solution
Before the two half‐reactions can be added together, the number of electrons lost must be equal to the electrons gained:
H2O + CN−  CNO‐ + 2 H+ + 2 e‐ (2 e‐ lost)
3 e‐ + 4 H+ + MnO4−  MnO2 + 2 H2O (3 e‐ gained)
To attain the same number of electrons on each side, multiply the first equation by 3 and the second by 2.
3 H2O + 3 CN−  3 CNO‐ + 6 H+ + 6 e‐
6 e‐ + 8 H+ + 2 MnO4−  2 MnO2 + 4 H2O
Balancing Redox Equations
in Basic Solution
Add the half‐reactions
3 H2O + 3 CN− + 8 H+ + 2 MnO4− 
3 CNO‐ + 6 H+ + 2 MnO2 + 4 H2O
Cancel out the H+ and H2O to get
2 H+ + 3 CN− + 2 MnO4−  3 CNO‐ + 2 MnO2 + H2O
Balancing Redox Equations
in Basic Solution
This reaction, however, takes place in basic solution
2 H+ + 3 CN− + 2 MnO4−  3 CNO‐ + 2 MnO2 + H2O
Add 2 OH‐ to both sides to cancel out the H+
2 OH‐ +2 H+ + 3 CN− + 2 MnO4−  3 CNO‐ + 2 MnO2 + H2O + 2 OH‐
(Remember that H+ + OH‐  H2O)
2 H2O + 3 CN− + 2 MnO4−  3 CNO‐ + 2 MnO2 + H2O + 2 OH‐
The balanced equation is
3 CN− + 2 MnO4− + H2O  3 CNO‐ + 2 MnO2 + 2 OH‐
A History of Electricity/Electrochemistry
•
•
•
Thales of Miletus (640‐546 B.C.) is
credited with the discovery that
amber when rubbed with cloth or
fur acquired the property of
attracting light objects.
The word electricity comes from
"elektron" the Greek word for
amber.
Otto von Guericke (1602‐1686)
invented the first electrostatic
generator in 1675. It was made of
a sulphur ball which rotated in a
wooden cradle. The ball itself was
rubbed by hand and the charged
sulphur ball had to be
transported to the place where
the electric experiment was
carried out.
Thales of Miletus
Otto von Guericke
• Eventually, a glass globe replaced the sulfur sphere used by Guericke • Later, large disks were used
• Ewald Jürgen von Kleist (1700‐1748), invented the Leyden Jar in 1745 to store electric energy. The Leyden Jar contained water or mercury and was placed onto a metal surface with ground connection. • In 1746, the Leyden jar was independently invented by physicist Pieter van Musschenbroek (1692‐
1761) and/or his lawyer friend Andreas Cunnaeus in Leyden/the Netherlands
• Leyden jars could be joined together to store large electrical charges
• In 1752, Benjamin Franklin (1706‐1790) demonstrated that lightning was electricity in his famous kite experiment
• In 1780, Italian physician and physicist Luigi Aloisio Galvani (1737‐1798) discovered that muscle and nerve cells produce electricity. Whilst dissecting a frog on a table where he had been conducting experiments with static electricity, Galvani touched the exposed sciatic nerve with his scalpel, which had picked up an electric charge. He noticed that the frog’s leg jumped.
Count Alessandro Giuseppe Antonio Anastasio Volta (1745 – 1827) developed the first electric cell, called a Voltaic Pile, in 1800.
A voltaic pile consist of alternating layers of two dissimilar metals, separated by pieces of cardboard soaked in a sodium chloride solution or sulfuric acid.
Volta determined that the best combination of metals was zinc and silver
Volta’s electric pile (right)
A Voltaic pile at the Smithsonian Institution, (far right)
•
In 1800, English chemist William Nicholson (1753–1815) and surgeon Anthony Carlisle (1768‐1840) separated water into hydrogen and oxygen by electrolysis.
•
Johann Wilhelm Ritter (1776‐1810) repeated Nicholson’s separation of water into hydrogen and oxygen by electrolysis. Soon thereafter, Ritter discovered the process of electroplating. He also observed that the amount of metal deposited and the amount of oxygen produced during an electrolytic process depended on the distance between the electrodes
•
Humphrey Davy (1778‐1829) utilized the voltaic pile, in 1807, to isolate elemental potassium by electrolysis which was soon followed by sodium, barium, calcium, strontium, magnesium. William Nicholson
Johann Wilhelm Ritter
Humphrey Davy
•
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•
•
Michael Faraday (1791‐1867) began his career in 1813 as Davy's Laboratory Assistant.
In 1834, Faraday developed the two laws of electrochemistry: The First Law of Electrochemistry
The amount of a substance deposited on each electrode of an electrolytic cell is directly proportional to the amount of electricity passing through the cell. The Second Law of Electrochemistry
The quantities of different elements deposited by a given amount of electricity are in the ratio of their chemical equivalent weights.
•
Faraday also defined a number of terms:
The anode is therefore that surface at which the electric current, according to our present expression, enters: it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, etc., are evolved; and is against or opposite the positive electrode. The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity; the combustible bodies, metals, alkalies, and bases are evolved there, and it is in contact with the negative electrode. Many bodies are decomposed directly by the electric current, their elements being set free; these I propose to call electrolytes.... Finally, I require a term to express those bodies which can pass to the
electrodes, or, as they are usually called, the poles. Substances are frequently spoken of as being electro‐negative or electro‐positive, according as they go under the supposed influence of a direct attraction to the positive or negative pole...I propose to distinguish such bodies by calling those anions which go to the anode of the decomposing body; and those passing to the cathode, cations; and when I have occasion to speak of these together, I shall call them ions. …the chloride of lead is an electrolyte, and when electrolyzed evolves the two ions, chlorine and lead, the former being an anion, and the latter a cation.
•
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John Frederic Daniell (1790‐1845), professor of chemistry at King's College, London. Daniell's research into development of constant current cells took place at the same time (late 1830s) that commercial telegraph systems began to appear. Daniell's copper battery (1836) became the standard for British and American telegraph systems. In 1839, Daniell experimented on the fusion of metals with a 70‐
cell battery. He produced an electric arc so rich in ultraviolet rays that it resulted in an instant, artificial sunburn. These experiments caused serious injury to Daniell's eyes as well as the eyes of spectators.
Ultimately, Daniell showed that the ion of the metal, rather than its oxide, carries an electric charge when a metal‐salt solution is electrolyzed. Left: An early Daniell Cell
Right:Daniell cells used by Sir William Robert Grove, 1839.
Voltaic Cells
In spontaneous oxidation‐
reduction (redox) reactions, electrons are transferred and energy is released.
Voltaic Cells
• If the reaction is separated into two parts, we can use that energy to do work by making the electrons flow through an external device.
• This type of setup is called a voltaic cell.
Voltaic Cells
• This is a typical voltaic cell
• A strip of zinc metal is immersed in a solution of Zn(NO3)2
• A strip of copper metal is immersed in a solution of Cu(NO3)2
• The two solutions are connected by a salt bridge containing NaNO3
• The oxidation occurs at the anode (Zn)
• The reduction occurs at the cathode (Cu)
Voltaic Cells
•
•
•
The salt bridge is used to prevent electrons flowing directly from the zinc to the copper
The salt bridge consists of a U‐
shaped tube that contains a salt solution, sealed with porous plugs, or an agar gel solution of the salt
The salt bridge keeps the charges balanced and forces the electron to move through the wire – Cations move toward the cathode.
– Anions move toward the anode.
Voltaic Cells
• In the cell, electrons leave the anode and flow through the wire to the cathode.
• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.
• Eventually, if the cell is used for a long time, the anode (zinc) will dissolve
Voltaic Cells
• As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.
• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.
• Eventually, if the cell is used for a long time, all the copper ions will plate onto the copper cathode
Electromotive Force (emf)
• Water only spontaneously flows one way in a waterfall.
• Likewise, electrons only spontaneously flow one way in a redox reaction—
from higher to lower potential energy.
Electromotive Force (emf)
• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
• It is also called the cell potential, and is designated Ecell.
Cell Potential
Cell potential is measured in volts (V).
J
1 V = 1 C
Where J = Joules
C = Coulombs
Recall that 1 electron has a charge of 1.6 x 10‐19 C
Standard Reduction Potentials
The cell potential is the difference between two electrode potentials.
By convention, electrode potentials are written as reductions
Reduction potentials for most common electrodes are tabulated as standard reduction potentials.
Standard Hydrogen Electrode
• Electrode potentials are referenced to a standard hydrogen electrode (SHE).
• By definition, the reduction potential for hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−  H2 (g, 1 atm)
Standard Cell Potentials
The cell potential at standard conditions is calculated
 = Ered (cathode) − Ered
Ecell
 (anode)
Substance reduced Substance oxidized
Because cell potential is based on the potential energy per unit of charge, it is an intensive property.
Cell Potentials
Oxidation: E°red = -0.76 V
Reduction: E°red = +0.34 V
Cell Potentials
 = Ered
 (cathode) − Ered
 (anode)
Ecell
= +0.34 V − (−0.76 V)
= +1.10 V
Generally, most of the common cells used, on the average,
generate approximately 1.5 V
Oxidizing and Reducing Agents
• The strongest oxidizers have the most positive reduction potentials.
• The strongest reducers have the most negative reduction potentials.
Oxidizing and Reducing Agents
The greater the difference between the two, the greater the voltage of the cell.
Free Energy
G for a redox reaction can be found by using the equation
G = −nFE°
where: n is the number of moles of electrons transferred
F is a constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V‐mol
E° = The standard cell potential in V
Under standard conditions,
G = −nFE
Nernst Equation
• Remember that
G = G + RT ln Q
• This means
−nFE = −nFE + RT ln Q
• Dividing both sides by −nF, we get the Nernst equation:
RT
ln Q
E = E −
nF
or, using base‐10 logarithms,
2.303 RT
log Q
E = E −
nF
Nernst Equation
At room temperature (298 K), and R = 8.314 J/mol K
F = 96,485 J/V‐mol
2.303 RT
F
= 0.0592 V
The final form of the Nernst Equation becomes
0.0592
E = E −
n
log Q
Walther Hermann Nernst (1864 ‐1941) •
•
•
•
Nernst's early studies in electrochemistry were inspired by Arrhenius' dissociation theory of ions in solution. In 1889 he elucidated the theory of galvanic cells by assuming an "electrolytic pressure of dissolution" which forces ions from electrodes into solution and which was opposed to the osmotic pressure of the dissolved ions. Also, in 1889, he showed how the characteristics of the current produced could be used to calculate the free energy change in the chemical reaction producing the current. This equation, known as the Nernst Equation, relates the voltage of a cell to its properties. Independently of Thomson, he explained why compounds ionize easily in water. The explanation, called the Nernst‐Thomson rule, holds that it is difficult for charged ions to attract each other through insulating water molecules, so they dissociate.
Concentration Cells
• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
• For such a cell, E°cell would be 0, but Q would not.
• Therefore, as long as the concentrations are different, E will not be 0.
Applications of Oxidation‐
Reduction Reactions
Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial production
of chemicals such
as Cl2, NaOH, F2
and Al
• Biological redox
reactions
The heme group
BATTERIES
Primary, Secondary, and Fuel Cells
Batteries
Most commercial batteries produce 1.5 V. To get a higher
voltage, batteries are joined together.
Dry Cell Battery
Primary battery — uses redox reactions that cannot be restored by recharge.
Anode (-)
Zn  Zn2+ + 2eCathode (+)
2 NH4+ + 2e-2 NH3 + H2
Alkaline Battery
Nearly same reactions as in common dry
cell, but under basic conditions.
Anode (‐): Zn + 2 OH‐  ZnO + H2O + 2e‐
Cathode (+): 2 MnO2 + H2O + 2e‐  Mn2O3 + 2 OH‐
Alkaline Batteries
Lead Storage Battery
• Secondary battery
• Uses redox
reactions that can be reversed.
• Can be restored by recharging
Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4-  PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e-  PbSO4 + 2 H2O
Ni-Cad Battery
Anode (-)
Cd + 2 OH-  Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e-  Ni(OH)2 + OH-
Fuel Cells: H2 as a Fuel
•Fuel cell ‐ reactants are supplied continuously from an external source.
•Cars can use electricity generated by H2/O2 fuel cells.
•H2 carried in tanks or generated from hydrocarbons.
Hydrogen—Air Fuel Cell
Hydrogen Fuel Cells
H2 as a Fuel
Comparison of the volumes of substances required to store 4 kg of hydrogen relative to car size. Storing H2 as a Fuel
One way to store H2 is to adsorb the gas onto a metal or metal alloy. Electrolysis
Using electrical energy to produce chemical change.
Sn2+(aq) + 2 Cl‐(aq)  Sn(s) + Cl2(g)
Electrolysis of Aqueous NaOH
Electric Energy f Chemical Change
Anode (+)
4 OH-  O2(g) + 2 H2O + 4e-
Cathode (-)
4 H2O + 4e-  2 H2 + 4 OH-
Eo for cell = -1.23 V
Anode
Cathode
Electrolysis
Electric Energy f Chemical Change
• Electrolysis of molten NaCl.
• Here a battery +
“pumps” electrons Anode
from Cl‐ to Na+.
• NOTE: Polarity of electrodes is reversed from batteries.
electrons
BATTERY
Cathode
Cl-
Na+
Electrolysis of Molten NaCl
See Figure 20.18
Electrolysis of Molten NaCl
electrons
Anode (+)
BATTERY
2 Cl-  Cl2(g) + 2e-
+
Anode
Cathode
Cl-
Na+
Cathode (-)
Na+ + e-  Na
Eo for cell (in water) = E˚c ‐ E˚a
= ‐ 2.71 V – (+1.36 V)
= ‐ 4.07 V (in water)
External energy needed because Eo is (‐). Electrolysis of Aqueous NaCl
electrons
Anode (+)
2 Cl-  Cl2(g) + 2e-
BATTERY
Cathode (-)
2 H2O + 2e-  H2 + 2
Eo for cell = -2.19 V
OH-
Note that H2O is more
easily reduced than
Na+.
+
Anode
Cathode
Cl- Na +
H 2O
Also, Cl‐ is oxidized in preference to H2O because of kinetics.
Electrolysis of Aqueous NaCl
Cells like these are the source of NaOH and Cl2.
In 1995: 25.1 x 109 lb Cl2 and 26.1 x 109 lb NaOH
Also the source of NaOCl for use in bleach.
Electrolysis of Aqueous NaI
Anode (+):
2 I-  I2(g) + 2eCathode (-): 2 H2O + 2e-  H2 + 2 OHEo for cell = -1.36 V
Electrolysis of Aqueous CuCl2
Anode (+)
electrons
2 Cl-  Cl2(g) + 2e-
BATTERY
Cathode (-)
Cu2+ + 2e-  Cu
+
Anode
Cathode
Eo for cell = -1.02 V
Note that Cu is more
easily reduced than
either H2O or Na+.
Cl- Cu2+
H2O
Electrolytic Refining of Copper
Impure copper is oxidized to Cu2+ at the anode. The aqueous Cu2+ ions are reduced to Cu metal at the cathode.
The copper at the cathode is over 99% pure
Producing Aluminum
2 Al2O3 + 3 C f 4 Al + 3 CO2
Charles Hall (1863‐1914) developed electrolysis process. Founded Alcoa.
Corrosion and…
…Corrosion Prevention