Transcript No Slide Title

Homework Problems
Chapter 18 Homework Problems: 4, 7, 16, 40, 42, 46, 50, 53, 63, 64,
66, 68, 74, 99, 102, 118, 122
CHAPTER 18
Electrochemistry
Oxidation Number
An oxidation number is a number assigned to an atom in a
molecule or ion indicating whether that atom is electron rich (negative
oxidation number), electron poor (positive oxidation number), or neutral
(oxidation number of zero).
Oxidation numbers do not indicate charges of atoms, but changes
in oxidation number indicate whether atoms have gained or lost electron
density.
A reaction where some of the oxidation numbers change in going
from reactants to products is called an oxidation-reduction (redox)
reaction.
Rules For Assigning Oxidation Numbers
1) The sum of the oxidation numbers of a species is equal to the
charge of the species.
2) Atoms in elemental forms have an oxidation number of 0.
3) Hydrogen has an oxidation number of +1 when bonded with
nonmetals (molecular compounds) and -1 when combined with metals
(ionic compounds).
4) Oxygen usually has an oxidation number of -2. Exceptions:
Elemental forms, peroxides (H2O2), OF2.
H2O2 H +1, O -1
OF2
F -1, O +2
5) Halogens. F in compounds is always -1. Other halogens are
often -1, but in compounds with oxygen or other halogens the oxidation
number may take on different values.
6) For assigning oxidation numbers in ionic compounds, it is
useful to break the compounds up into ions.
Examples:
Assign oxidation numbers for each atom in the following
substances.
H2O
Fe(NO3)2
SO42O3
MnO4I3-
H2O
H +1; O -2
Fe(NO3)2
= Fe2+ and NO3so Fe +2, O -2, and so N +5
SO42-
O -2, and so S +6
O3
elemental form, so O = 0
MnO4-
O -2, and so Mn +7
I3-
I = - 1/3 (unusual, but not impossible)
Oxidation-Reduction Reactions
In an oxidation-reduction (redox) reaction the oxidation numbers
of some of the atoms involved in the reaction change in going from
reactants to products.
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
+3 -2
0
0
+4 -2
We use the following terms in reference to redox reactions.
Oxidation - An increase in the value for the oxidation number.
Reduction - A decrease in the value for the oxidation number.
There will always be one oxidation and one reduction process per
reaction.
C
0 to +4, and so is oxidized.
Fe
+3 to 0, and so is reduced.
Oxidizing and Reducing Agents
We call the species that is oxidized in a redox reaction the
reducing agent, and the species that is reduced in a redox reaction an
oxidizing agent. This seems confusing, but is based on the idea that
there must be both an oxidation and a reduction taking place in a redox
reaction.
SO2(g) + 2 Fe3+(aq) + 2 H2O()  2 Fe2+(aq) + SO42-(aq) + 4 H+(aq)
+4 -2
+3
+1 -2
+2
+6 -2
+1
In the above reaction S is oxidized (from +4 to +6) and so SO2 is a
reducing agent, and Fe is reduced (from +3 to +2) and so Fe3+ is an
oxidizing agent.
Balancing Oxidation-Reduction Reactions
(acid or base conditions)
1) Write the unbalanced net ionic equation.
2) Assign oxidation numbers; identify the substance being
oxidized and the substance being reduced.
3) Write balanced half reactions for oxidation and reduction.
Include electrons as products (oxidation) or reactants (reduction) to
change oxidation numbers. You may use H2O and H+ in balancing the
half reactions.
4) If necessary, multiply one or both of the half reactions by
whole numbers to get the electrons to cancel.
5) Combine the half reactions to get the net ionic equation.
6) If the reaction is carried out in base conditions, add OH- to
both sides of the equation to cancel any H+ ions present.
Example: Balance the following reaction for acid conditions.
Ag+(aq) + Cu(s)  Ag(s) + Cu2+(aq)
Example: Balance the following reaction for acid conditions.
Ag+(aq) + Cu(s)  Ag(s) + Cu2+(aq)
+1
0
0
ox
Cu(s)  Cu2+ + 2 e-
red
Ag+(aq) + e-  Ag(s)
+2
Need to multiply the reduction reaction by 2 to get electrons to cancel.
ox
Cu(s)  Cu2+ + 2 e-
red
2 Ag+(aq) + 2 e-  2 Ag(s)
net
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Example: Balance the following reaction for acid conditions.
PbO2(s) + Mn2+(aq)  Pb2+(aq) + MnO4-(aq)
Example: Balance the following reaction for acid conditions.
PbO2(s) + Mn2+(aq)  Pb2+(aq) + MnO4-(aq)
+4 -2
ox
+2
+2
+7 -2
Mn2+(aq)  MnO4-(aq) + 5 eMn2+(aq) + 4 H2O()  MnO4-(aq) + 5 e- + 8 H+(aq)
red
PbO2(s) + 2 e-  Pb2+(aq)
PbO2(s) + 2 e- + 4 H+(aq)  Pb2+(aq) + 2 H2O()
Now, must multiply the oxidation reaction by 2 and the reduction
reaction by 5 to get electrons to cancel, giving us
ox 2 Mn2+(aq) + 8 H2O()  2 MnO4-(aq) + 10 e- + 16 H+(aq)
red 5 PbO2(s) + 10 e- + 20 H+(aq)  5 Pb2+(aq) + 10 H2O()
net 2 Mn2+(aq) + 5 PbO2(s) + 4 H+(aq)
 2 MnO4-(aq) + 5 Pb2+(aq) + 2 H2O()
Example: Balance the following reaction for base conditions.
SeO32-(aq) + Cl2(g)  SeO42-(aq) + Cl-(aq)
Example: Balance the following reaction for base conditions.
SeO32-(aq) + Cl2(g)  SeO42-(aq) + Cl-(aq)
+4 -2
ox
0
+6 -2
-1
SeO32-(aq)  SeO42-(aq) + 2 eSeO32-(aq) + H2O()  SeO42-(aq) + 2 e- + 2 H+(aq)
red
Cl2(g) + 2 e-  2 Cl-(aq)
ox
SeO32-(aq) + H2O()  SeO42-(aq) + 2 e- + 2 H+(aq)
red
Cl2(g) + 2 e-  2 Cl-(aq)
net SeO32-(aq) + Cl2(g) + H2O()  SeO42-(aq) + 2 Cl-(aq) + 2 H+(aq)
2 OH-(aq)  2 OH-(aq)
base SeO32-(aq) + Cl2(g) + 2 OH-(aq) SeO42-(aq) + 2 Cl-(aq) + H2O()
Notice that to balance for base conditions we add OH-(aq) to
convert all H+(aq) ions into H2O().
Electrochemical Cells
An electrochemical cell is a device that can interconvert chemical
and electrical energy. There are two general types of cells:
Galvanic cell (battery) - In a galvanic cell a chemical reaction is
used to generate a voltage.
Electrolytic cell - In an electrolytic cell an external voltage is
used to drive a chemical reaction in a particular direction.
Galvanic Cells
In a galvanic cell a chemical reaction is used to generate a
voltage. There will be a half-cell oxidation reaction and a half-cell
reduction reaction, which combine to give the net cell reaction.
The sites where the oxidation and reduction reactions take place
are called electrodes. By convention
anode - electrode where oxidation reaction occurs
cathode - electrode where reduction reaction occurs
In a galvanic cell the anode is negative and the cathode is
positive. Current flow is from the anode to the cathode.
Salt bridge - A gel containing ions that can transfer charge and
therefore complete the electrical circuit.
Cell convention:
Anode - Electrode where oxidation occurs.
Cathode - Electrode where reduction occurs.
Cell Notation
It is convenient to have a method for indicating the elements of a
galvanic cell. This is done as follows:
1) List the elements of the cell, in order, from the anode (left) to
the cathode (right).
2) Use a single vertical line to indicate a change in phase.
3) Use a double vertical line to indicate a salt bridge.
We can also indicate the concentration or partial pressure of
substances in the galvanic cell for cases where that information is useful.
Cu(s) | Cu2+(aq)
||
Ag+(aq) | Ag(s)
ox
Cu(s)  Cu+2(aq) + 2 e-
red
2 ( Ag+(aq) + e-  Ag(s) )
cell
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Example: Give the half-cell oxidation reaction, the half cell
reduction reaction, and the net cell reaction for the galvanic cells with the
following shorthand notation.
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu
Pt(s) | H2(g) | H+(aq) || Cl-(aq) | AgCl(s) | Ag(s)
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu
anode
cathode
ox
Zn(s)  Zn2+(aq) + 2 e-
red
Cu2+(aq) + 2 e-  Cu(s)
cell
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Pt(s) | H2(g) | H+(aq) || Cl-(aq) | AgCl(s) | Ag(s)
anode
cathode
ox
H2(g)  2 H+(aq) + 2 e-
red
2 ( AgCl(s) + e-  Ag(s) + Cl-(aq) )
cell
H2(g) + 2 AgCl(s)  2 Ag(s) + 2 H+(aq) + 2 Cl-(aq)
Standard hydrogen electrode (SHE) Electrode where the following process
takes place (either as an oxidation or as
a reduction half reaction).
ox
H2(g)  2 H+(aq) + 2 e-
red
2 H+(aq) + 2 e-  H2(g)
Platinum is used as the electrode since
it is chemically inert.
Standard Cell Potential
The standard half-cell potential (E) is the voltage generated by a
half-cell reaction. We can talk about either a standard half-cell reduction
potential (Ered) or a standard half-cell oxidation potential (Eox) which
correspond to the potential generated by these processes when all
reactants and products are present at standard concentration.
Since galvanic cells must have both an oxidation and a reduction
half-cell reaction, the standard half-cell reduction potential is defined
relative to the potential generated by the standard hydrogen electrode
(SHE).
2 H+(aq) + 2 e-  H2(g)
Ered = 0.00 v (by definition)
H2(g)  2 H+(aq) + 2 e-
Eox = 0.00 v
Based on this definition experimental values for other half-cell
reduction potentials can be found.
Example: Consider the following galvanic cell.
Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)
ox
H2(g)  2 H+(aq) + 2 e-
Eox
red
Cu2+(aq) + 2 e-  Cu(s)
Ered
cell
H2(g) + Cu2+(aq)  2 H+(aq) + Cu(s)
Ecell
The experimental value for the cell potential is Ecell = 0.34 v.
Since Ecell = Eox + Ered
Ered = Ecell - Eox = 0.34 v - 0.00 v = 0.34 v
All half-cell reduction potentials can be found by the use of this
method.
A table of half-cell reduction potentials is found in Table 18.1, or
Appendix II-D of the text.
Properties of Half-Cell Potentials
1) Reversing the direction of the reaction (changing the reaction
from a reduction to an oxidation) changes the sign of the potential.
Cu2+(aq) + 2 e-  Cu(s)
Ered = 0.34 v
Cu(s)  Cu2+(aq) + 2 e-
Eox = - 0.34 v
2) Multiplying a reaction by a positive constant has no effect on
the value of the half-cell potential. This is because the potential is
defined as that found for standard conditions.
Cu2+(aq) + 2 e-  Cu(s)
Ered = 0.34 v
2 Cu2+(aq) + 4 e-  2 Cu(s) Ered = 0.34 v
Finding Cell Potentials For Standard Conditions
The standard cell potential for a galvanic cell can be found by
adding together the half-cell oxidation potential and the half-cell
reduction potential.
Example:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu
ox Zn(s)  Zn2+(aq) + 2 e-
Eox = 0.76 v
red Cu2+(aq) + 2 e-  Cu(s)
Ered = 0.34 v
cell Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Ecell = 1.10 v
Because half-cell oxidation potentials can be found from half-cell
reduction potentials (by reversing the direction of the reaction and the
sign of the potential) we typically only keep tables of the half-cell
reduction potentials.
Free Energy and Cell Potential
The connection between galvanic cells and thermodynamics is
through the following relationship (which can be derived).
Grxn = - n F Ecell
Grxn = - n F Ecell
In the above expressions n is the number of electrons transferred
per mole of reaction (which can be found from either the oxidation or the
reduction half-cell reaction) and F is the Faraday constant, a conversion
factor between Coulombs (MKS unit of charge) and moles of charge.
F = 96485 C/mol
Note (1 volt) (1 Coulomb) = 1 Joule
In the previous galvanic cell n = 2 and Ecell = 1.10 v, so
Grxn = - (2) (96485 C/mol) (1.10 v) = - 212.3 kJ/mol
(spontaneous)
Spontaneous Cell Reactions
Based on the above results we can determine when a cell reaction
will occur spontaneously. Since
Grxn = - n F Ecell
we have the following possibilities
Ecell > 0 means Grxn < 0 and the reaction is spontaneous
Ecell = 0 means Grxn = 0 and the system is at equilibrium
Ecell < 0 means Grxn > 0 and the reaction is not spontaneous
Note that in the last case the reverse reaction will be spontaneous
(a consequence of free energy being a state function).
Since the above is true in general it is also true for cells where
standard conditions (so true if we add “  ” to Grxn and Ecell in the above
expressions.
Electrochemical Series
The electrochemical series is a list of half-cell reduction
potentials, in order, from most positive to most negative voltage.
Properties of the Electrochemical Series
Since reactions where Ecell is positive are spontaneous, we may
say the more positive a half-cell potential the more likely it will lead to a
spontaneous cell reaction.
That means that reactions at the top of the series are most likely
to occur as reduction reactions, and reactions at the bottom of the series
are least likely to occur as reduction reactions. Because changing from a
reduction to an oxidation reaction changes the sign of the half-cell
potential, it also means that reactions at the bottom of the series are most
likely to occur as oxidation reactions, and reactions at the top of the
series are least likely to occur as oxidation reactions.
Oxidizing and Reducing Agents
Oxidizing agent – Is reduced in a chemical reaction, and so
oxidizes another substance.
Reducing agent – Is oxidized in a chemical reaction, and so
reduces another substance.
In the electrochemical series substances at the top left of the
series are the best oxidizing agents (since they are most likely to be
reduced) and substances at the lower right of the series are the best
reducing agents (since they are most likely to be oxidized).
Example: Which of the following reactions will occur
spontaneously for standard conditions?
Ni2+(aq) + Zn(s)  Ni(s) + Zn2+(aq)
Zn2+(aq) + Ni(s)  Zn(s) + Ni2+(aq)
In the first reaction Ni2+ is an oxidizing agent, while in the second
reaction Zn2+ is an oxidizing agent.
Based on the above portion of the electrochemical series Ni2+ is
more likely to be reduced than Zn2+, and so it is the first reaction that will
be spontaneous (in fact Ecell = + 0.50 v for the first process and – 0.50 v
for the second process).
The Nernst Equation
The central equation for galvanic cells is the Nernst equation. It
may be derived in the following manner.
Recall the following thermochemical relationship
Grxn = Grxn + RT ln Q
However, we have a connection between free energy and cell
potential
Grxn = - n F Ecell
If we use the above to substitute, we get
- n F Ecell = - n F Ecell + RT ln Q
Ecell = Ecell – RT ln Q
nF
the Nernst equation.
Uses of the Nernst Equation
The Nernst equation can be used to predict the value for the cell
potential for cases where reactants and products are not at their standard
concentrations.
Example: Consider the following galvanic cell
Zn(s) | Zn2+(aq, 0.010 M) || Cu2+(aq, 1.00 M) | Cu
What is Ecell for the above galvanic cell?
Zn(s) | Zn+2(aq, 0.010 M) || Cu2+(aq, 1.00 M) | Cu
What is Ecell for the above galvanic cell?
From the Nernst equation
Ecell = Ecell – RT ln Q
nF
ox Zn(s)  Zn2+(aq) + 2 ered Cu2+(aq) + 2 e-  Cu(s)
cell Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Eox = 0.76 v
Ered = 0.34 v
Ecell = 1.10 v
n = 2 Q = [Zn2+] = (0.010) = 0.010
[Cu2+] (1.00)
So Ecell = 1.10 v – (8.314 J/mol.K) (298.2 K) ln(0.010)
(2) (96485 C/mol)
= 1.10 v + 0.059 v = 1.16 v
The Nernst Equation and Equilibrium
The Nernst equation says
Ecell = Ecell – RT ln Q
nF
Consider a system at equilibrium. In this case Grxn = 0, which
means Ecell = 0, and Q = K, the equilibrium constant. If we substitute
this into the Nernst equation we get
0 = Ecell – RT ln K
nF
Ecell = RT ln K
nF
ln K = n F Ecell
RT
Using this expression we can obtain values for the equilibrium
constant for any reaction that can be represented by a galvanic cell.
Example: The solubility product for silver bromide (AgBr) is Ksp
= [Ag+][Br-]. Use data from Appendix II-D for half-cell reduction
potentials to find a numerical value for Ksp.
Example: The solubility product for silver bromide (AgBr) is Ksp
= [Ag+] [Br-]. Use data from Appendix II-D for half-cell reduction
potentials to find a numerical value for Ksp.
The solubility product reaction is
AgBr(s)  Ag+(aq) + Br-(aq)
ox
Ag(s)  Ag+(aq) + e-
Eox = - 0.80 v
red
AgBr(s) + e-  Ag(s) + Br-
Ered = + 0.07 v
cell
AgBr(s)  Ag+(aq) + Br-(aq)
Ecell = - 0.73 v
So ln K = n F Ecell = (1) (96485 C/mol) (- 0.73 v) = - 28.41
RT
(8.314 J/mol.K) (298.2 K)
Ksp = e-28.41 = 4.6 x 10-13
Electrochemical Determination of pH
We may construct a galvanic cell that allows us to measure pH
based on an electrochemical measurement.
H2(g) | H+(aq, MH+) || “reference electrode”
where “reference electrode” refers to a cathode that will give a stable
value for Ered, and MH+ is the concentration of hydrogen ion for the
anode solution (the solution whose pH we wish to determine). Note that
our anode is the standard hydrogen electrode (SHE) except that the 1.0
M H+ solution has been replaced by the solution whose pH we wish to
determine.
ox
H2(g)  2 H+(aq) + 2 e-
Eox = 0.00 v
red
“reference electrode”
Ered
cell
Ecell = Ered
Ecell = Ecell - RT ln Q
n = 2 ; Q = [H+]2/(pH2)
nF
= Ered - (8.314 J/mol.K) (298.2 K) ln{([H+]2/(pH2)}
(2) (96485 C/mol)
For pH2 = 1.00 atm, this becomes
Ecell = Ered - (0.0256 v) ln[H+]
= Ered + (0.0256 v) { - ln[H+]}
= Ered + (0.0592 v) { - log10[H+]}
= Ered + (0.0592 v) pH
If we solve this for pH, we get
Ecell = Ered + (0.0592 v) pH
(Ecell - Ered) = (0.0592 v) pH
pH = (Ecell - Ered)/(0.0592 v)
What this means is that by
measuring a cell potential for an
appropriate galvanic cell we can find
the pH of a solution. To measure pH
to a precision of  0.03 pH units
requires a precision in measuring
voltage of (0.03) (0.0592 v)   0.002
v, not a difficult thing to do.
In practice, because of the
complexity of the standard hydrogen
electrode we usually use other types
of electrodes in pH measurements,
such as the glass-calomel electrode.
Lead Storage Battery
The lead storage battery is typically used as a car battery because
it is rugged, reliable, rechargable, and capable of storing a large amount
of electrical energy. The cell reactions are as follows
ox Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2 e-
0.296 v
red PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2e PbSO4(s) + 2 H2O()
1.628 v
cell Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq)
 2 PbSO4(s) + 2 H2O() 1.924 v
A typical 12 v lead car battery consists of six galvanic cells in
series, each generating a potential of approximately 2 v.
cell Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq)
 2 PbSO4(s) + 2 H2O() 1.924 v
Corrosion
Corrosion is the general term given for the deterioration of metals
by oxidation.
The reduction of oxygen in corrosion is the process
O2(g) + 4 H+(aq) + 4 e-  2 H2O()
Because the reduction potential for O2(g) is large and positive,
oxygen is expected to oxidize most metals, including copper, lead,
nickel, iron, zinc, and aluminum. Some metals, like zinc and aluminum,
develop a protective oxide coat (ZnO; Al2O3) that prevents further oxidation from taking place. The oxides for other metals, such as iron, are
porous, and so the metals can oxidize until completely rusted through.
The economic cost of corrosion in metals is substantial. In the
United States, corrosion of metals is estimated to cost $400 -$500 billion
dollars per year (about 3% of the US-GNP).
Electrolysis
In electrolysis an external voltage is used to force a chemical
reaction to take place in a particular direction.
Anode - oxidation reaction (+)
Cathode - reduction reaction (-)
As in galvanic cells, we can write down the half-cell oxidation
and reduction reactions and the net cell reactions for an electrolytic cell.
ox
2 Cl-()  Cl2(g) + 2 e-
red
2 ( Na+() + e-  Na() )
cell
2 Na+() + 2 Cl-()  2 Na() + Cl2(g)
Notice that in electrolysis the reactions that take place are
opposite of those that would normally occur.
The Nernst equation does not apply for electrolysis, since we are
not generating a voltage in an electrolytic cell. All that is required is that
the external voltage being used is large enough to drive the reaction in
the desired direction.
Identifying the Species Being Oxidized or Reduced
In an electrolytic cell the species involved in the reactions are
usually the ones that require the most positive (least negative) voltage to
be oxidized or reduced. We can use the electrolytic series to identify
which substances this will be.
Example: Consider the electrolysis of aqueous sodium chloride
(NaCl(aq)). What oxidation and reduction reactions will occur?
Example: Consider the electrolysis of aqueous sodium chloride
(NaCl(aq)). What oxidation and reduction reactions will occur?
red
ox
Na+(aq) + e-  Na(s)
E = - 2.71 v
2 H2O() + 2 e-  H2(g) + 2 OH-(aq)
E = - 0.83 v
2 Cl-(aq)  Cl2(g) + 2 e-
E = - 1.36 v
2 H2O()  O2(g) + 4 H+(aq) + 4 e-
E = - 1.23 v
Based on the above we would expect production of H2(g) at the
cathode, and production of O2(g) at the anode.
In fact, while H2(g) is produced at the cathode, it is Cl2(g), and
not O2(g) that forms at the anode. This is because for cases where the
potentials for different possible reactions are close (within ~ 0.5 v) there
are other factors that come into play that may determine the reaction that
occurs.
The electrolysis of aqueous sodium chloride is a major industrial
reaction, used to produce chlorine and hydrogen gas, and sodium
hydroxide.
Calculations Involving Electrolytic Cells
The main type of calculation done in connection with electrolytic
cells is finding the mass of a particular substance that can be produced in
a particular cell, given the operating conditions.
To do these calculations requires:
1) The appropriate half cell reaction.
2) The number of moles of electrical charge transferred.
3) The molecular mass of the product.
Example: Electrolysis is carried out for molten sodium chloride.
The cell is operated at a current i = 20.0 amp (1 amp = 1 C/s) for a period
of 1.00 hour. How many grams of Cl2 (MW = 70.90 g/mol) will be
produced?
Electrolysis is carried out for molten sodium chloride. The cell is
operated at a current i = 20.0 amp (1 amp = 1 C/s) for a period of 1.00
hour. How many grams of Cl2will be produced?
1) The appropriate half cell reaction.
2 Cl-()  Cl2(g) + 2 e2) The number of moles of electrical charge transferred.
mol charge = 1.00 hr 3600 s 20.0 C
1 hr
s
1 mol
= 0.746 mol
96485 C
3) The molecular mass of the product. M(Cl2) = 70.90 g/mol.
g Cl2 = 0.746 mol charge 1 mol Cl2
2 mol charge
70.90 g Cl2 = 26.5 g
mol Cl2
Electrolytic Production of Aluminum Metal
Naturally occurring aluminum occurs in the +3 oxidation state
(often in the form of aluminum oxide, Al2O3). Production of aluminum
metal requires reduction of Al3+ to the zero oxidation state.
The production of aluminum metal is carried out using the HallHeroult process, discovered in 1886.
How many moles and coulombs of electrons are required to
produce 1.00 kg of aluminum metal? (MW(Al) = 27.0 g/mol)
How many moles and coulombs of electrons are required to
produce 1.00 kg of aluminum metal? (MW(Al) = 27.0 g/mol)
Al3+ + 3 e-  Al
mol e- = 1000. g Al 1 mol Al
3 mol e- = 111.1 mol e-
27.0 g Al 1 mol Al
coulombs e- = 111.1 mol e- 96485 C = 1.07 x 107 C
mol
It takes a substantial amount of electrical power to produce
aluminum metal. In fact, 3% of the electrical power generated in the
United States goes into the production of aluminum (an average of 15
kilowatt hours per kilogram of aluminum produced). This is one reason
why recycling of aluminum products makes economic sense (production
of recycled aluminum uses only 5% of the electricity used to produce
aluminum from ore).
End of Chapter 18
“Anode comes from the Greek  (upward) and  (a way),
and therefore suggests the rising of the sun in the east. Cathode comes
from  (downward), and is related to the setting of the sun, in the
west.” - Keith Laidler, in The World of Physical Chemistry, explaining
how Michael Faraday came up with the terms “anode” and “cathode”.
“…the total annual estimated direct cost of corrosion in the U.S.
(in 1998) is a staggering $276 billion - approximately 3.1% of the
nation’s Gross Domestic Product (GDP).”
- Corrosion Costs and Preventive Strategies in the United States
“Rust never sleeps.” - Neil Young (borrowed from Devo)