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Readings for this section
Petrucci: Section 3-1, Chapter 20;
Introduction
Electrochemistry is the study of interchange of chemical and electrical energy.
Oxidation/Reduction involves the exchange of electrons from one chemical species to another.
Normally, this is done when the two chemicals contact each other in the activated complex (when two
species bump into each other in solution for example).
We are interested in separating the chemical species such that the electrons transfer via an external
circuit. That way, we can measure the electrochemical effects.
To properly understand the connection between the redox reaction and the electricity, we should
balance the overall redox reaction using a half-reaction method such as the one described in the
previous section of these notes. We can set up the physical reaction vessel such that the chemicals
from one half reaction are separated from those of the second half reaction. For reaction to occur, we
still need to connect the solutions to complete the circuit. This is done by attaching wires between
electrodes in the two half cells and by connecting the solutions of the two half cells via a salt bridge or
by some other device such as a semi-permeable membrane.
In general, such a cell is called an electrochemical cell. These cells could be used in one of two types
of situations:
1. The chemical reaction is spontaneous and produces electricity.
This is called a voltaic cell or a galvanic cell.
2. The chemical reaction is non-spontaneous and is forced by electricity from an external source.
This kind of cell is called an electrolysis cell.
We will look at the first situation first.
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Galvanic or Voltaic cells
Consider a piece of zinc foil placed in a beaker of copper sulphate. The copper sulphate solution is
blue because of the presence of the Cu2+ ions. When the zinc is added, the solution changes to
colourless and the zinc metal is dissolved and replaced by a reddish orange powder. The colourless
solution no longer has Cu2+ in it and the reddish orange powder is Cu(s). The Zn(s) is dissolved and is
now Zn2+. [I know this because of my vast knowledge of chemistry ;-)]
If we write an overall reaction for this process, we get:
Zn + Cu2+  Zn2+ + Cu. This doesn't help us in our quest for electrochemistry knowledge. Let's
rewrite this in half-reaction form.
Zn  Zn2+ + 2e–
Cu2+ + 2e–  Cu
Now, we can set up two half cells, one with a zinc electrode in a Zn2+ solution (say ZnSO4) and the
other with a copper electrode in a Cu2+ solution (say CuSO4) as follows.
The centre line in the diagram, recall, is either a semi-permeable membrane or a salt bridge.
Now, at the anode, we have the reaction
Zn(s)  Zn2+(aq) + 2e–(aq)
and at the cathode, we have
Cu2+(aq) + 2e–(aq)  Cu(s)
Electrons pass through the wires and SO42– pass through the membrane to keep the solutions neutral.
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Cell Potentials
Electrons from solution pushed onto the anode, around the external circuit and onto the cathode where
they are pulled out into the solution.
That's one way of thinking of the electrical circuit part of the electrochemical cell. The electrons
pushed around the external circuit can do work (run a motor, illuminate a light bulb, etc). The amount
of work possible is a function of both the voltage (potential) and of the current (number of electrons)
in the circuit.
Pushing one Coulomb of charge around a circuit at a potential of 1 Volt does One Joule of work. OR,
mathematically, 1J = 1V × 1C.
The cell potential (voltage of the cell) depends on the chemicals used. For example, the chemicals in
dry-cells (batteries) are such that the potential is always about 1.5 V. This has become a standard and
is now a limiting factor in deciding which chemicals can be used to create a battery.
The cell potential is given a symbol of Ecell. If all chemicals are at activity of 1 (conc. = 1 M, p = 1
bar) then the cell potential is the standard cell potential and is given as E°cell.
Any redox reaction has the potential (pun) to be used in an electrochemical cell. We merely need to be
able to divide the oxidizing and reducing agents into two half cells (half reactions).
Take for example, the reaction of zinc metal dissolving in hydrochloric acid. The reaction is:
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
We need to separate the zinc from the hydrogen. We can use zinc as an electrode but what about the
hydrogen.
In this case, we need to set up a special electrode, which allows H2 gas molecules to interact directly
with H+ dissolved in water and with the electrons from the external circuit simultaneously. Such a
system is pictured below.
A blow-up of the surface of the platinum electrode is shown below so the location of reaction can be
better understood.
The other half-cell would look much like that pictured in a previous diagram. The whole cell diagram,
of course would include the external part of the circuit and the salt bridge or membrane to complete
the circuit. We can abbreviate this diagram as follows:
Zn(s)/ZnSO4(aq)//H2SO4(aq)/H2(g),Pt(s)
[Anode
//
Cathode]
Where the single slash mark / represents the boundary between solution and electrode and the double
slash // represents the salt bridge or semi-permeable membrane. The external circuit, of course, joins
the two electrodes (solid) and is not explicitly shown here.
The overall cell voltage can be summed from the half-cell potentials of the oxidation and of the
reduction reactions.
Ecell = ERed + Eox.
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Standard Reduction Potentials
As before, we wish to tabulate thermodynamic data in some useable way. We define a standard set of
conditions under which all half-reactions can theoretically operate and we tabulate the half-cell
potentials for the reactions under standard conditions. Since the potential of an oxidation reaction
(loose electron) is merely the negative of the potential for a reduction reaction reaction (gain electron),
we choose to tabulate only the reduction half-reactions. Any oxidation half reactions are merely the
reverse of the tabulated reduction reaction and the oxidation potential is the negative of the tabulate
reduction potential.
Table D.4 from Petrucci lists many reduction half-reactions that you will need for problems, etc in the
text. Herein, I have duplicated only a few in the following table.
1
2
3
4
5
6
7
reduction
O2 + 4H+ + 4e–  2H2O
Ag+ + e–  Ag
Cu2+ + 2 e–  Cu
Fe2+ + 2 e–  Fe
Zn2+ + 2 e–  Zn
2H2O + 2e–  H2 + 2OH–
Na+ + e–  Na
Let's look at this small reduction potential table and see if we can use it
to explain certain experimentally observable properties of matter. First
off, I should point out that the more positive reduction potentials refer
to more spontaneous reduction reactions.
The first reaction in this mini-table shows the reduction of oxygen.
This is higher in the table than many other things. That makes sense
since we know that oxygen is a good oxidizing agent (it makes other
things oxidize while it itself is reduced).
Eº/volts
+1.229
+0.7996
+0.3419
–0.447
–0.7628
–0.83
–2.71
A few years ago, a city
experienced a number of very
serious water main breaks on
sections of pipe that had been
recently replaced. The City
engineer was fired. His only
defence was "How was I to
know that brass fittings (mostly
copper) on cast iron pipes would
corrode so fast." Was the city
justified in firing him? Can you
explain their reasoning for
calling him incompetent?
We can see that water can be both oxidized (-1) and reduced (2). It is
oxidized to oxygen and reduced to hydrogen. These two reactions, in
fact, serve as a set of boundaries for aqueous solutions. Any materials harder to oxidize than water
will not oxidize since the water will. Similarly, any materials harder to reduce than water will never
reduce (in water). For example, sodium ions will never reduce to sodium metal in the presence of
water since the water will reduce first. In fact, sodium metal in water will spontaneously
(explosively) oxidize to Na+ ions and the water will reduce to hydrogen gas and leave a basic solution
(equation 6). The overall reaction for this process is the sum (#6) – 2×(#7).
NOTE: when we add half-cells,
their potentials are strictly
additive. We don't multiply or
divide by any factors even
though we do so to the
stoichiometric coefficients of
the equations in order to
balance the overall
equation. Potentials don't
depend on the total amount or
size of the cell, just the
2Na + 2 H2O  2Na+ + H2 + 2OH–
Thus, the overall standard cell potential for this reaction is
Eºcell = Eºred + Eºox = –0.83 + 2.71 (reverse reaction => change sign) =
1.88 V.
concentrations. You already
know this. If you buy a AA cell
or a D cell, their voltages are
both 1.5V. However, their
capacity to do work is not the
same.
Positive cell potential means spontaneous reaction.
When pairing up metals, we know that zinc will dissolve in a copper (II) solution. Here we see that
the copper (II) (3) is easier to reduce than the zinc (II) (5).
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Gibbs' Free Energy and Cell potential
Let's now relate the electrochemistry ideas we've explored with the thermodynamic parameter called
Gibbs' Free Energy.
We saw,
w = –QE
(work = – charge times potential),
where Q is the charge and can be defined as Q = nF where F is the Faraday constant (96485 C/mol),
named after Michael Faraday.
In our case, we're interested in the maximum work since this can be related to the thermodynamic
parameter G.
Thus, for the case where the work is done infinitely slowly (chemical system is always at equilibrium
with the electrodes, electrical resistance is zero since current is essentially zero, etc…) we have
wmax = – QEmax where Emax for standard conditions is simply E° as calculated from the tabulated halfcell potentials.
Take for example, a cell with a maximum cell potential of 2.50 V. If 1.33 mol of e– passes through the
cell at an average potential E = 2.10 V. What is the efficiency?
w = – Q E = – nFE. = 1.33 mol × 96485 C/mol × 2.10 V
(V = J/C)
w = – 2.69 × 105 J = – 269 kJ
wmax = – nFEmax = 1.33 mol × 96485 C/mol × 2.50 V
= – 321 kJ
efficiency = w/wmax × 100% = –269/–321 × 100% = 83.8 %
Of course, since wmax is only achievable if the work is done reversibly (infinitely slowly), we can never reach 100%
efficiency in any system in the real world.
We already have seen that G is a measure of the maximum work obtainable from a system. Thus,
G = wmax
G = – Q E
G = – nFE. In this case, the potential is the cell potential Ecell.
G° = – RT lnK = – nFE°cell.
Thus, we now have a link between free-energy, equilibrium and electrochemical thermodynamic
parameters.
Example:
Is Fe2+ spontaneously oxidized by the oxygen of the air in acidic solution? Calculate G and K.
Two half-reactions can be determined by looking in the table of standard reduction potentials.
Oxidation of
iron:
[Fe2+  Fe3+ + e– ]
Reduction of
O2
O2 + 4H+ + 4e–  2 H2O
E°red = 1.23 V
4Fe2+ + O2 + 4H+  4Fe3+ + 2H2O
E°cell = 0.46 V
n = (4 mol electrons / mol equation)
(Positive means
spontaneous)
Overall:
–E°ox = –0.77 V
G = –nFE°cell = – 4 mol × 96485 C/mol × 0.46 V = – 1.8×105 J = –180 kJ (per mole of reaction)
K = e72.7 = 3.7×1031.
SUMMARY
Gibbs
Q vs. K
Cell potentials
Spontaneous
direction
G < 0
Q<K
Ecell > 0

G = 0
Q=K
Ecell = 0
Equilibrium
G > 0
Q>K
Ecell < 0

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Non-Standard Conditions: Nernst Equation
Consider the following set of half-reactions
Reduction
Fe2+ + 2e–
Oxidation
Zn
Overall
Zn + Fe2+
Fe
E°red = –0.44 V
Zn2+ + 2e–
E°ox = +0.76 V
Zn2+ + Fe
E°cell = 0.32 V
This is valid only if all species have concentrations of 1 M.
What if not all the species are at Standard Conditions?
Le Châtelier's principle gives us some ideas. Look at the overall reaction. Consider the reaction to be
at equilibrium (Ecell = 0 V). If we add more Fe2+ to the reaction mixture, The equilibrium will shift to
the right to use up the excess Fe2+. This will, of course cause a positive cell potential to be
measurable.
Conversely, if we removed Fe2+ from the reaction mixture, the equilibrium would shift to the left to
try to replace some of the missing Fe2+. This would result in a measurable negative potential (reaction
goes left).
If the reaction were not at equilibrium then these preceding two changes would me additive to the
measured voltage (say standard voltage). So, if we add Fe2+ to a mixture at Standard Conditions, then
the voltage would be more positive and if we removed some Fe2+ from a mixture at Standard
Conditions the measured voltage would be lower (less positive).
The reverse results would be found for addition or removal of Zn2+ (a product). More Zn2+ would give
a lower positive potential and vice versa.
Now we try to quantify the previous hand-waving discussion.
Recall the relation: G = G° + RT ln Q
(for non-standard conditions)
Substitute the new definition of G and we get
–nFE = –nFE° + RT ln Q
(This is called the Nernst Equation).
Thus, we can calculate the cell potential of any cell, given only the initial conditions (to calculate Q)
and the reduction potentials (to calculate E°).
Now, looking back at the reaction cell discussed at the beginning of this section we can calculate the
voltage for the situation where the concentrations are not standard. For example, consider
[Fe2+] = 0.1 M and [Zn2+] = 1.9 M.
For this reaction, we also know that there are two electrons transferred for each equation (n = 2).
Although the reaction is 90% complete (from Standard Conditions initial) the cell potential has only
dropped by a small amount (0.04 V). This is normal. For example, the batteries in your flashlight will
put out almost full voltage until the last traces of chemical are almost used up at which point the
voltage will drop off rather sharply. This is an especially useful characteristic of cells powering
electronic equipment, (like a calculator or CD player, etc.) which require a certain minimum (and
dependable) voltage to operate successfully.
We can also use the Nernst equation to calculate such things as equilibrium constants.
At equilibrium, Q = K and Ecell = 0 so the Nernst equation becomes
or
Example:
What is the equilibrium constant for the reaction of copper metal with bromine to form copper(II) ions
and bromide ions in aqueous solution at 25°C?
Cathode
Anode
Cell
Br2 + 2e–
Cu
Cu + Br2
2 Br–
Cu2+ + 2e–
Cu2+ + 2 Br–
E°red = 1.09 V
E°ox = –E°red = –0.34 V
E°cell = 0.75
Example:
What is the value of the solubility product constant for AgCl?
We could calculate this by breaking the overall Ksp reaction into a series of redox reactions as follows:
Anode
Cathode
Overall
( Ksp reaction)
Ag(s)  Ag+ + e–
E°ox = –0.80 V
AgCl + e–  Ag(s) + Cl–
E°red = 0.22 V
AgCl
Ag+ + Cl–
E°cell = –0.58 V (not very
spontaneous)
K = 6.3 × 109
Ion Concentrations can also be calculated...
Consider the Platinum-Hydrogen electrode coupled with a copper/copper(II) electrode.
The overall reaction is Cu2+ + H2  Cu + 2H+.
The Nernst equation is
If [Cu2+] = 1 M and P(H2) = 1 bar (both their activities are 1) then
E = E° – 0.0257 V ln [H+]
E = E° + 0.05917 V (–log [H+])
(Note: ln Y = 2.303 × log Y and .-257×2.303=.05917)
E = E° + 0.05917 V × pH.
We see here that the cell potential is a function of pH. The probes in pH meters are set up this way. A
complete electrochemical cell is contained within the probe casing. All chemicals are at standard
conditions and a porous glass membrane allows only H+ ions to pass through. The diagram of the setup is given here.
Concentration cells
A concentration cell is one in which the half-reactions in the two half cells are identical but the
concentration of the ions is different. In this case, the reduction and the oxidation half reaction are
identical. Take for example a concentration cell where both half cells have an iron electrode in a Fe2+
solution. We would have the following:
Consider the following set of half-reactions
Reduction
(half-cell 1)
Fe2+ + 2e–
Oxidation
(half-cell 2)
Fe
Overall
Fe
E°red = –0.44 V
Fe2+ + 2e–
E°ox = +0.44 V
Fe + Fe2+
Fe2+ + Fe
E°cell = 0.00 V
if all concentrations are identical (or if they are all standard) then obviously the cell voltage is zero. If,
however there is a different concentration on one half cell than on the other one, we could use the
nernst equation to calculate, for example the cell potential.
E = E° – RT/nF ln Q
NOTE: Q = a(Fe2+)2/a(Fe2+)1 Where the subscripts 1 and 2
refer to half-cells 1 and 2
NOTE: E° = 0.00
E = RT/nF ln {a(Fe2+)2/a(Fe2+)1}
Normally, we set the concentration of half-cell 1 to be fixed
(at say, activity=1) and vary the the activity of the Fe 2+ in
half-cell 2.
E = RT/nF ln a(Fe2+)(in half-cell 2)
if a (Fe2+)1 = 1
Half-cell calculations
The Nernst equation can also be used to calculate half-cell potentials. In this case, Q is simply the
ratio of ions of the half-cell equation rather than the full-cell and the value for E° is that of the halfcell. No other changes to the equation are made.
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Batteries
There are three classes of batteries:
Primary
Cells
Chemical reaction occurs irreversibly. The battery cannot be recharged. This
can happen for several reasons, including the chemical properties and the
design of the cell. Examples include standard Alkali batteries or their less
common precursors the carbon/zinc battery.
Leclanché cell (improperly called Zinc-Carbon dry cell)
This is the common dry-cell available in most stores that we use to power
toys and flashlights. These involve a zinc casing (anode) and a carbon rod in
the centre (cathode) with a paste of MnO2, NH4Cl, ZnCl2 and graphite.
The reactions involved are
electrode
anode
cathode
overall
reaction
Zn (s)  Zn2+ + 2 e–
2MnO2+2NH4++2e–  Mn2O3+2NH3+H2O
Zn+2MnO2+2NH4+Mn2O3+2NH3+H2O+Zn2+
~1.5 V
Notice that the products can slowly build up in this type of cell and cause the
overall cell potential to drop.
Alkaline cell (alkaline meaning basic)
In the alkaline cell, the product of the cathode reaction is used up again in the
anode reaction, no build up of products means the voltage drops (as water is
used up) more slowly than the previous cell. Since the cell voltage is also
about 1.5 V, this alkaline dry cell can be used in the same devices as the
normal dry cell. Notice, the only real difference between these is the fact that
the reaction in this case occurs in a basic medium.
electrode
anode
cathode
overall
reaction
Zn (s) + 2OH–  Zn(OH)2(s) + 2 e–
2MnO2+2H2O+2e–  2MnO(OH)(s)+2OH–
Zn+2MnO2+2H2O2MnO(OH)(s)+Zn(OH)2(s)
voltage
~1.5 V
Ruben-Mallory (mercury) cell
This type of cell is used in application where the cell potential must remain
very constant over the life of the battery (commonly computers and watches,
etc). The products and reactants are all pure substances (standard state). Since
the voltages of these cells are not the same as the previous two, this type of
battery is not interchangeable with the first two types.
electrode
anode
cathode
overall
reaction
Zn (s) + 2OH–  ZnO(s) 2H2O + 2 e–
HgO+H2O+2e–  Hg+2OH–
Zn + HgO  ZnO(s)+Hg
voltage
~1.35 V
Secondary Chemical reaction is reversible. The battery can be recharged. Examples
include NiCd batteries and Lead/acid batteries found in cars. These are also
Cells
called storage cells or rechargeable cells. In theory, alkaline batteries can also
be recharged, the process, however is dangerous and can lead to explosions if
not done with the proper equipment. Similarly, lead/acid car batteries can
explode if they are charged too fast but for a different reason.
Lead/acid battery has an overall reaction of
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–  2PbSO4(s) + 2H2O(l)
with a cell voltage of just over 2 V. Hence, six cells in series will make a
battery of cells with a combined voltage of 6×2V ~ 12 V. (actually closer to
13 V).
Nickel/Cadmium batteries have reactions
electrode
anode
cathode
overall
reaction
Cd (s) + 2OH–  Cd(OH)2(s) + 2 e–
NiO(OH)(s)+H2O+e–  Ni(OH)2(s)+OH–
Cd(s)+2NiO(OH)(s)2Ni(OH)2(s)+Cd(OH)2(s)
voltage
~1.35 V
.We see that none of the reactants or products are in solution so the voltage
will remain very constant during the discharge period. Unfortunately, the cell
voltage is lower than standard dry cells and are not always interchangeable.
Fuel Cells
Reactants are flowed through the cell. This is an irreversible reaction but by
refilling the reservoir of fuel the cell can be reused. The space shuttle uses
this type of battery to run much of its electrical equipment.
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Electrolysis and Electrolytic Cells
We've seen how spontaneous reactions can be harnessed to produce electricity. In electrolytic cells,
electricity is used to force non-spontaneous reactions to proceed. Electrolysis is used to prepare
certain compounds and metals. Products of electrolysis are found in every day life: Chrome plated
bumpers, Silver and gold plating on some inexpensive jewellery, among other things.
In setting up an electrolysis cell, we must take into account all possible reactions and make sure the
conditions are set so that the desired reaction occurs. We will do this at first by using only inert
electrodes and then by using pure electrolyte liquids rather than aqueous (or other) solutions.
We will use a non-aqueous system such as molten NaCl as our medium, and non reactive electrodes
like platinum. We can successfully separate out pure
Na. Consider the following electrolytic cell.
This is a very important method of producing
sodium metal. At the anode, Cl2 is evolved in the
reaction
2Cl–  Cl2(g)+ 2e–.
At the Cathode, the molten sodium ions are
converted to sodium metal (liquid at these
temperatures) in the reaction
Na+ + e–  Na(l).
Alkali earth metals which are also highly reactive can be prepared using electrolysis of molten
chloride salts as in the reaction MgCl(l)  Mg(l) + Cl2(g)
Other metals which are not quite as reactive as the above two families but are sufficiently reactive that
they cannot be produced in an aqueous cell can be produced this way. Aluminum is one such metal.
The Hall process invented before the turn of the century allowed for the production of aluminum on
an industrial scale.
A molten mixture of aluminum ore (Al2O3) and cryolite (Na+)3(AlF6–)(l). The cryolite is the solvent
and it is used because it has a lower melting point than pure Al2O3. A carbon electrode acts as the
anode and aluminum forms the cathode such that the following reactions occur:
Al3+ + 3e–  Al(l)
C(s) + 2O2–  CO2(g) + 4e–
This process needs lots of electricity per kg of aluminum so you will most commonly find aluminum
production plants very close to electric power stations to reduce the cost of electricity.
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Quantitative Aspects of Electrolysis:
Consider the reaction in the molten NaCl electrolysis cell.
2Cl–  Cl2(g) + 2e–.
2 moles of electrons per mole of Cl2(g)
Na+ + e–  Na(l).
1 mole of electrons per mole of Na metal
Recall that the charge on the electron is Q = nF where F is the Faraday constant (96485 C/mol),
named after Michael Faraday.
Thus, to produce one mole (23 g) of sodium, we need Q = 1×96485 C = 96500 C.
To produce one mole (70.9 g) of chlorine gas, we need 2×96500 = 1,930,000 C.
Current is defined as the amount of charge passing a point in a circuit in one second.
I = Q/s
The units are
1 Ampere = 1 Coulomb/second (1A = 1C/s).
Now, if we have a current of 50.0 A passing through an NaCl(l) electrolysis cell in 1 hour, how much
sodium and chlorine will we produce.?
50.0 A × 3600 s = 180,000 C
(NOTE: 1As = 1C)
n = Q/F = 180,000/96,485 = 1.87 mol e–
It takes lots of current to produce very little sodium and chlorine.
We can do other types of calculations similar to this.
What mass of aluminum will be produced in 1.00 h by electrolysis of Molten AlCl3 using a current of
10.0 A?
Q = I×t = 10.0 A × 3600 s = 3.60×104 C.
It takes three moles of electrons for every mole of Aluminum. So...
n = Q/F = 3.60×104 C / 96485C/mol e– ×1/3 × 27.0 g/mol = 3.36 g Al.
What volume of Cl2(g) at STP (0°C, 1 atm) will be produced by a current of 20.0 A in 2.00 h in the
same cell as used in the previous
example?
Note that in this problem, we used
R=0.08206 Latm/mol K.
This was to facilitate the conversion to volume.
You could have also used the more standard
value of R=8.314 J/mol K but conversion to
volume would have involved one more step of
calculations.
If we had tried to electrolyse sodium Chloride in water, we would need to consider not only the Na+
and Cl– as possible reactants but also the water. We can never produce sodium in aqueous solution,
because it will spontaneously react with the water to produce Na+(aq). Whether we tried to produce
sodium metal either chemically or electrolytically, the chance of success would be equally bad. Thus,
at the cathode of an aqueous NaCl cell, we would get the reduction of water happening
2H2O + 2e–  H2 + 2OH–
E°red = –0.83 V
rather than the reduction of the sodium
Na+ + e–  Na
E°red = –2.71 V
Notice that the reduction potential for the water reduction is a lot more positive than that for the
sodium. This is a way we can look up reduction potentials and tell what will happen in our solution
before hand.
Looking at the two possible reactions at the anode of the NaCl (aq) cell, we see that the Cl– and the
H2O are both candidates for oxidation. The two possible half-reactions are:
2H2O  O2 + 4H+ + 4e–
E°ox = –E°red = –1.23
2Cl–  Cl2 + 2e–
E°ox = –E°red = –1.36
Since the water has a more positive oxidation potential than the chlorine reaction, it should be
oxidized more readily. There is a complication, however. The water oxidation requires a considerable
over-voltage (extra voltage) to make the reaction rate appreciable. Since the two possible reactions are
quite close in potentials the extra voltage can quickly be sufficient to cause the chlorine oxidation to
occur. Since this latter reaction is quite rapid in comparison to the water  oxygen one, it will
dominate at only slightly elevated voltages.
If we had used Na2SO4 rather than NaCl as the electrolyte, then the anode reaction would have been
the water one since SO42– is very difficult to oxidize (very negative oxidation potential in comparison
to water).
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Electrolytic Purification of Metals
Consider the electrolysis of Copper(II)sulfate
solution.
We find O2 produced at the anode and copper
solid deposited at the cathode. This means, as we
saw before that the water is more easily oxidized
(to oxygen) than are the sulphate ions while the
copper(II) ions are more easily reduced (to copper
metal) than is water (to hydrogen). Hence, it is
quite easy to set up an electrolytic purification process for copper metal. Unlike aluminum, where we
needed molten baths of aluminum ore, we need only dissolve the copper in a water solution and then
electrolyse it out as copper metal. If we do a crude separation chemically of the copper from the ore,
we can use this impure copper as the anode and use pure copper as the cathode. Then, we will
electrolyse copper ions off of the impure copper anode and then reduce them onto the pure cathode in
the same solution. Impurities in the anode would merely settle to the bottom of the cell to be disposed
of. We need to take care to adjust the voltage so that only the copper is oxidized from the anode and
not some of the impurities. This done, we can successfully purify copper using electrolysis. The
slower we take the reaction (less over voltage) the more pure will be the copper.
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Products of Electrolysis
We've seen that the particular product of electrolysis we see at the electrodes depends on the ease of
production of that particular species. There are three factors that determine product that we will find
in any given electrolysis cell: the electrode material, the solvent and the electrolyte. Often, we use
inert electrodes and/or pure liquids to help alleviate some of the complications from competing
chemistry, other times, we must simply carefully adjust conditions (concentrations and voltages) to
ensure that the desired product is achieved.
@ Cathode
Reduction of substances most easily reduced will be the one observed.
For example, Na+, Ca+, Al+ all have more negative reduction potentials
than water so they will not be reduced in aqueous solutions
On the other hand, Cr3+, Cu2+, Ag+ have reduction potentials more
positive than that for water so they will be preferentially reduced at the
cathode.
@ Anode
Oxidation of substances most easily oxidized will be observed.
For example, Br–, I– will be oxidized but F– and Mn2+ will not. (check
reduction potentials)
Complication: Cl– we've seen to be more easily oxidized than water even though it has an oxidation
potential (negative of the reduction potential for Cl2) which is more negative than that for water. This
is due to kinetics. The water reaction is quite slow and requires a large voltage to force it to proceed
with any speed. The Cl– oxidizes to Cl2 much more rapidly and the over voltage we apply in an
attempt to force the water to oxidize is sufficient to produce Cl2 which occurs faster.
Summary:
Reaction involves:
Anode
Cathode
1
Electrolyte
Anions oxidized
Cations reduced
2
Solvent
Solvent oxidized
Solvent reduced
3
Electrode
Electrode oxidized
Electrode reduced
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Corrosion of metals
We've seen that the oxygen from the air can be reduced to water and that the reduction potential for
this half-reaction is very positive. More positive, in fact than that for most metal reduction process.
This means that any metal in contact with oxygen from the air and in the presence of an electrolyte
solution will oxidize. Since mining and production of metal is a fairly energy-intensive process, it is
very useful to be able to find ways to reduce or eliminate the oxidation of these metals if at all
possible. There are several ways to reduce oxidation of metals
Eliminate the oxygen/metal contact
This is achieved by coating the metal with some material that is impervious to either the
oxygen or to any aqueous salts. This prevents the electrical circuit from being complete in the
'electrochemical cell' that is spontaneously set up at the point of rusting.
To keep the O2 or the salt solution, we can use several methods
1)Insulate the metal
a) Paint the metal. This is a pure insulating technique and serves merely to keep the reacting
species separated. Unfortunately any small scratch will allow the reaction to start up and may
even accelerate the process.
b) Oil the metal. (various rust-proofing companies use this technique). The unfortunate
problem is that oil is a volatile liquid and eventually either runs off the surface or evaporates
away, leaving the metal unprotected. Special surfactants can be added to the oil to make it
cling to the metal surface and slow evaporation. You will find these surfactant agents in
treatments such as RustCheck and The new Canadian Tire rust-proofing treatments. No matter,
these treatments must be repeated regularly (yearly) to maintain protection.
c) anodize the metal: sometimes, the oxides of the metal are very stable and act to insulate the
metal from the air. This is most obvious in copper roofing where the (orange) copper metal is
quickly oxidized to the green oxide CuO. This oxide is quite stable and protects the underlying
metal from further corrosion. This is useful in that small scratches quickly 'seal themselves up'
with more oxide and the process is halted. Aluminum is often deliberately anodized in a
controlled electrochemical cell to form an aluminum oxide coating that is relatively stable (in
neutral solutions).
2) Coat the metal with a material that oxidizes easier than the metal you wish to protect. For
example, steel coated with zinc will not oxidize until first all the zinc is oxidized because the
zinc will change the electrical potential of the steel and prevent it from being oxidized even if
it is exposed to the air and water. This protection will last until the zinc is used up. at which
time, the steel will start to oxidize.
Cathodic protection
a) Use a sacrificial anode made of a material that oxidized easier than the metal you wish to
protect. This is similar to coating the metal with the material (as in galvanized steel) but is
simpler to do since it merely means attaching chunks of the anode material rather than going
through a coating process.
Pipelines are often protected this way, the steel pipes are attached by a wire to a large block of
zinc buried next to the pipeline in the ground. The zinc serves to electrically charge the steel
changing its potential and thus reducing the amount of oxidation.
b) Electrically change the metal's potential. Similar to the use of sacrificial anodes, this process
involves attaching the negative pole of an electrical system to the metal we need to protect. In
cars, for example, the negative pole of the battery is the ground. This serves to charge up the
car and change it's potential. Thus, the oxygen has to overcome both the iron's reduction
potential as well as the batteries artificial potential to force the reaction to happen. Cars would
rust out far quicker it they were grounded positive since that, in effect, would serve to increase
the rate of oxidation of the steel.
Back to Top
Prof. Michael J. Mombourquette.
Copyright (c) 1997
Revised:
Lecture Notes
Michael's Site
Chemistry 112
Computers in Chemistry
Prologue
1.Introduction
2.Nomenclature
3.Stoichiometry
4.Gas Laws
5.Thermochem. I
6.Solid Liquid Gas
7.Liquid Solutions
8.Colligative Prop.
9.Solids
10.Atomic Structure
11.Molecular Structure
12.Equilibrium
13.Acid/Base Chem.
14.Solubility
15.Thermochem. II
16.REDOX
17.Electrochemistry
18.Kinetics
19.Organic Chemistry
A.Review of Fundamentals
B.Dimension & Units
C.Scientific Notation
D.Special Prefixes
E.Significant Figures
F.Dimensional Analysis
G.Logs & exponentials
A.Valence Shell Electrons
B.Binary Compounds
C.IUPAC System
D.-OUS -IC SYSTEM
E.PREFIX SYSTEM
F.Compounds with Polyatomic Ions
A.Atomic Structure
B.Mass of Atoms
C.Conservation of Mass and Energy
D.The Mole
E.Formula/Molecular Mass
F.Calculations
A.Avagadro's Law
B.Pressures of Gases
C.Boyle's Law
D.Charles' Law
E.Idea Gas Law
F.Dalton's Law of Partial Pressures
G.Kinetic Molecular Theory
H.van der Waals' Equation
A.First Law of Thermodynamics
B.Enthalpy
C.Heat Capacity
D.Calorimetry
E.Standard Enthalpy Changes
F.Hess' Law
G.Standard Enthalpies of Formation
H.Temperature change and Enthalpy
I.Internal Energy
A.Introduction
B.Vapour Pressure
C.Boiling
D.Intermolecular Forces
E.Melting & Freezing
F.Phase Diagrams
A.Introduction
B.Composition of Solutions
C.Liquid vapour Equilibrium
D.Boiling of Solutions
E.Azeotropes
F.Freezing of Solutions
G.Solubility
H.Henry's Law
Molarity
molality
A.Introduction
B.Vapour Pressure Lowering
C.Boiling Point Elevation
D.Freezing Point Depression
E.Osmotic Pressure
F.Electrolyte Solutions
Types of Solids
Lattices
Cubic Lattices
Metals
Ionic Solids
Cell Calculations
A.Quantum Mechanics
B.Duality of Nature
C.Quantization Effects
D.Quantum Numbers
E.Spherical Harmonics
F.Electron Spin
G.Periodic Table
H.Trends
Hydrogen Atom
Rydberg Equation
Bohr Model
1-D
2-D
3-D
Pauli Exclusion Principle
Electronic Configurations
Atomic Radii
Ionization Energies
Electron Affinities
Reactivity
A.Lewis Dot Structures
B.VSEPR
D.Orbitals
E.Band Theory
F.Bond Energies
G.Electronegativity
H.Dipole Moments
Introduction
Ionic Bonds
Covalent Bonds
Hyper-Valences
Multiple Bonds
Formal Charge
Oxidation Numbers
Radicals
Resonance Structures
Hydrogen Bonding
Table of Hyper-Valences
Atomic Orbitals
Molecular Orbitals
LCAO
Valence Bond
A.General Discussion
B.Equilibrium Constant
C.Heterogeneous Equilibria
D.Applications
E.Le Châtelier's Principle
F.Real Gases
Relative Activities
Mathematical
Haber Process
Concentration Change
Pressure Change
Temperature Change
A.Introduction
B.Acid-base Neutralizations
C.Strengths of Acids and Basis
D.Equilibria in acid-base systems
E.Self-Ionization of water
F.pH Scale
G.Distribution Diagrams
H.Buffer Solutions
I.Summary
J.Salts
k.Polyprotic acids
l.Titrations
A.Introduction
B.Ksp
C.Common Ion
D.Effect of pH
E.Metal Sulfides
F.Complexation
A.Introduction
B.Spontaneous Processes
C.Entropy & 2nd Law
D.Std Molar Entropy
E.Gibbs Energy
F.Temp Dependance
G.Other Dependencies
H.Examples
A.Introduction
B.Galvanic or Voltaic Cells
C.Cell Potentials
D.Standard Reduction Potentials
E.Gibbs' Free Energy
F.Nernst Equation
G.Concentration Cells
H.Batteries
I.Electrolysis
J.Corrosion of Metals
A.Introduction
B.Reaction Rate
C.Rate Laws
D.Initial Rate Method
E.Integrated Rate Law Method
F.Half-life
G.Effect of Temperature
H.Mechanisms
I.Catalysis
A.Alkanes
B.Structural Isomers
C.Conformers
D.Cycloalkanes
E.Naming Alkanes
F.Branched Alkyls
G.Geometric Isomers
H.Functional Groups
I.Reactions