ELECTROCHEMISTRY
[MH5; Chapter 18]
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Electrochemistry is the study of the relationship between chemical
reactions and electrical energy.
The transfer of electrons (what happens in an oxidation-reduction,
or redox reaction) constitutes electrical energy.
The reactions studied in electrochemistry are always redox
reactions.
Electrochemical reactions occur experimentally in an
ELECTROCHEMICAL CELL, of which there are two basic types.
Voltaic Cell
• Stored chemical energy is converted by a spontaneous reaction to
electrical energy; for example, a battery.
Electrolytic Cell
• A chemical reaction is made to occur by passage of an electric
current.
• Electrical energy is converted to chemical energy, for example,
electroplating.
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The reactions in an electrochemical cell take place on the surface of
the electrodes.
An electrode is an electronic conductor (usually a metal) in contact
with an ionically conducting medium (usually an electrolyte solution).
Recall when we learned to balance redox reactions; we used half
reactions.......
In an electrochemical cell, the Oxidation half reaction takes place
at one electrode, which we call the anode.
The Reduction half reaction takes place at the other electrode,
which we call the cathode.
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VOLTAIC or GALVANIC CELLS [MH5; 18.1]
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Recall that a voltaic cell is one in which the redox reaction occurs
spontaneously and electrical energy is produced.
Consider the following reaction:
Zn (s)
+
Cu
2+
(aq)
!
Zn
2+
(aq)
+
Cu (s)
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What happens if we combine just the reactants ?
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Electrical energy can only be produced if the two half reactions are
actually separated in space.....
What does this look like ?
What actually happens when the cell is “operating” ??
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The electrons produced in the oxidation half reaction at the anode
travel through a external circuit to the cathode where they are
consumed in the reduction half reaction.
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A simple voltaic cell, the DANIELL cell
[MH5, Figure 18.2]
Zn metal in
Zn2+(aq) solution
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Cu metal in
Cu (aq) solution
2+
A voltaic cell consists of two half cells; in this cell there is a zinc
half cell and a copper half cell
The Zn metal and the Cu metal are acting as the electrodes.
Recall that oxidation occurs at the anode, which has a negative
charge; reduction occurs at the cathode, which has a positive
charge.
The electrons must pass through an external circuit which connects
the two half cells (electrons flow from anode to cathode).
A “salt bridge” connects the solutions and prevents mixing; it allows
for ions to flow between the two solutions.
If the salt concentrations are 1 M, and a current is allowed to flow
by completing the external circuit, a potential difference of 1.10 V
(at 25EC) will be established; this is the Standard Cell Potential.
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The Redox reactions occurring at the electrodes:
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at Zn electrode:
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at Cu electrode:
Overall redox reaction:
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As this reaction occurs, there is a build-up in cation concentration at
the Zn anode, and a decrease in cation concentration at the Cu
cathode.
So anions migrate through the salt bridge from Cu side to Zn side
and cations pass in the opposite direction to preserve the charge
balance.
We abbreviate this cell as:
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The voltmeter reads the potential difference between the two
electrodes.
This potential difference (or voltage) is the “electromotive force”
(EMF) of the cell and is measured in Volts.
The potential of the cell is independent of the number of electrons
passing through the cell.
We cannot divide the 1.10V cell potential between Zn and Cu,
because we cannot study one electrode in isolation.
Electrode Potentials for a half-cell are measured relative to the
Standard Hydrogen Electrode reaction, on a Pt surface [catalyst]
with [ H+(aq) ] = 1.0 M, at 25EC
The reaction for this electrode is:
+ 2 e — !
2 H+ (aq)
H2 (g)
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The reduction potential for this cell has been defined as: 0.00 V
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Reduction Potentials for other half reactions can be measured by
setting up a cell using the half cell in question along with the
Standard Hydrogen Electrode............. [MH5; Figure 18.3]
This example uses the Zn half cell.
The measured cell potential is + 0.762 V; all of this voltage is
assigned to the Zn electrode.
For:
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EE, the STANDARD REDUCTION POTENTIAL
is - 0.762 V relative to the standard hydrogen electrode.
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For the cell:
EEcell is found to be + 0.339V
ˆ EE for the Cu2+/Cu reduction, must be + 0.339 V
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To combine these two half-reactions to make a Daniell cell, one must
be reversed, so its EE value changes sign.....
Oxidation Half Reaction:
Reduction Half Reaction:
Overall:
EEcell =
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When EE values for half-reactions are combined in this way, the
“unknown” contribution of H+/H2 (which we assumed was 0.00 V)
vanishes and the experimentally correct EEcell value is obtained.
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STANDARD VOLTAGES [MH5; 18.2]
• In tables of EE values, all half-reactions are given as reductions...
• So one reaction must be reversed (changing the sign of EE) when two
half reactions are combined to give an overall redox reaction.
Reaction will proceed spontaneously in
the direction for which EE cell is POSITIVE
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If we multiply equations for half-reactions to balance the overall
redox equation, the EE values do not change.
Half cell potentials are intensive properties, independent of the
amount of reagent.
EXAMPLE:
Given the cell:
AR (s) / AR
3+
(aq) // Fe
2+
(aq) , Fe
3+
(aq) / Pt (s)
Will AR metal reduce Fe3+ to Fe2+ ?
What is the anode half reaction ?
What is the cathode half reaction ?
Overall:
EE cell =
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What happens as reaction proceeds ?
• Ecell will change as reactant concentrations change.
• Eventually the battery or cell runs down as Ecell has fallen to zero.
• The reaction stops; the system has reached chemical equilibrium.
• In practice, most redox reactions go virtually to completion; for the
Daniell cell....
essentially no Cu2+ remains (not responsible for calculation).
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Once EE values have been measured relative to the H+/H2 halfreaction, a table of standard reduction potentials may be drawn up
(see tutorial problems, p. 17-18 or MH5; page 481) from which EEcell
for any overall redox reaction may be calculated........
A redox reaction and overall EEcell value can only be made up by
reversing one half-reaction (and its EE value) and adding
them................
Remember that EE values are NOT measured “per mole”!!
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How do we set up reactions which occur only in solution (no metals to
act as electrodes) ?
EXAMPLE:
a) Will the reaction Ce3+ (aq) + Fe3+ (aq) ! Ce 4+ (aq) + Fe2+ (aq)
proceed spontaneously ?
b) Write the cell diagram for this reaction.
Write the two half reactions... Look up Eo values
The Cell Diagram:
What Can We Determine from Eo values ?
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The relative strengths of oxidants (those species which are
reduced) and reductants (those species which are oxidized) may be
determined from their positions in a table of Standard Reduction
Potentials. [MH5; Table 18.1] or Electrochemistry Tutorial p. 17 &
18.
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Remember:
• Half-reactions with large +ve values occur readily.
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A strong oxidizing agent has a high +ve Eo value.
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A strong reducing agent has a high -ve Eo value.
EXAMPLE:
Which of the metals Ag, Cu, Sn and Fe will dissolve in aqueous acid ?
(Recall that acid means H+)....
The reduction half-reaction in each case is:
Write each metal half-reaction as an oxidation:
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Those metals appearing above H2 in the table of reduction EE values
(electro-chemical series) should dissolve in any aqueous acid.
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Reversibility of Cell Reactions
• Can the electrode reactions in a voltaic cell be reversed by applying
an external potential ? Sometimes !
EXAMPLE 1: Apply an external potential of > 1.10 V to the Daniell
cell.......
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At the anode:
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At the cathode:
# because H+ is reduced more easily than Zn2+
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This is an example of an electrolytic cell - electrical energy is being
used to drive a non-spontaneous chemical reaction.
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ELECTROLYSIS [MH5; 18.5]
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Electrolysis is the process by which a non spontaneous chemical
reaction is forced to occur by pumping electrical energy into the
system.
This process takes place in an electrolytic cell.
A battery provides a source of direct electrical current.
Two wires from the battery lead to the electrodes; the anode and
the cathode.
The battery is the electron “pump”; it pushes electrons into the
cathode and pulls them away from the anode.
If electrical neutrality is to be maintained, electrons must somehow
be consumed at the cathode and released at the anode.
A redox reaction will accomplish this; the oxidation reaction (which
occurs at the anode) releases electrons, while the reduction reaction
(which occurs at the cathode) consumes electrons.
This whole process is what is known as Electrolysis.
Quantitative Electrolysis
• Faraday’s Law defines the relationship between the amount of
electricity passed in an electrochemical cell and the amount of
reactants used or products formed in the corresponding redox
reaction.................
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EXAMPLES:
Half Reactions
Electrons
React / Prod
Ag+(aq) + e- ! Ag(s)
Fe(s) ! Fe2+(aq) + 2 eAR3+(aq) + 3 e- ! AR(s)
Cu2+(aq) + 2 e- ! Cu(s)
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Only chemists use moles for measuring electrons.........
The physicists and engineers use Coulombs for measuring charge and
Amperes for measuring current.
One Coulomb is the amount of charge produced when a current of
one Ampere flows for one second.........
So..........the amount of charge (denoted by Q) is equal to the amount
of current (in amperes, denoted by i) that flows multiplied by the
time it flows for in seconds (denoted by t).
This relationship is usually expressed by:
Q
=
i
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x
t
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The link between the charge and the number of moles of electrons is
known as the Faraday Constant.
This constant tells us that each mole of electrons that are passed
carries a charge of 96490 Coulombs.
EXAMPLE 1:
What mass, in g, of silver, Ag, can be plated out from a solution of
AgNO3 if a current of 0.750 amperes flows for 1 hour ?
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EXAMPLE 2:
We wish to electroplate 6.50 g of copper, Cu, from a solution of CuCR2.
How long will the electrolysis take if we use a current of 3.50 Amperes?
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EXAMPLE 3:
• Electroplating is the deposition of a thin layer of metal on a
conducting surface.
• The object to be electroplated is used as the cathode; the anode is
the electroplating metal.
• Chromium metal has been used for “chrome plating”; suppose we wish
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to chrome plate an object using a solution of CrCR3.
What happens at each electrode when a current is passed ?
At the anode
•
At the cathode
If a current of 1.65 amperes is passed for 3 hours, what mass, in
grams, of chromium metal would be plated ?
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EXAMPLE 4:
Two electrolysis cells were connected in series as follows:
Cu*Cu2+*CuCCCAg*Ag+*Ag
Cell 1
Cell 2
After a period of electrolysis, 0.295 g of Cu was deposited on the
cathode of the Cu cell (Cell 1) . What mass of Ag, in grams, was
deposited in the other cell (Cell 2)?
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Electrolysis of Water
• While we know that pure water does not conduct electricity very
well; it can participate in a redox reaction if there are some ions
present..............
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At the anode; the oxidation half reaction:
Eo = - 1.23 V
2 H2O(l) ! O2 (g) + 4 H+(aq) + 4 e—
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At the cathode; the reduction half reaction:
Eo = - 0.83 V
2 H2O(l) + 2 e— ! H2(g) + 2 OH—(aq)
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The overall reaction is:
6 H2O (l) ! 2 H2 (g) + 4 OH —(aq) + 4 H+(aq) + O2
OR:
2 H2O
(l)
! 2 H2
(g)
+ O2
(g)
(g)
Eocell =
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What does this mean in terms of performing electrolysis using
aqueous solutions ?
It depends on the Standard Reduction Potentials of the solutes
present in the solution...........
If quite a bit of energy is required to react the solute species, the
water will react first, using up the energy that was intended for the
solute ions.
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There are two possibilities for reduction at the cathode:
1) Cation to metal, characteristic of easily-reduced transition metal
cations and electroplating......
2) Water to hydrogen gas, occurring when the cation in solution is
difficult to reduce (such as Na + and K +).....
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There are also two possibilities for the reaction occurring at the
anode:
1) The anion may be oxidized to the corresponding non metal....
2) If the anion cannot be oxidized (for example, NO3—or SO42—, then
water will be oxidized to oxygen gas......
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EXAMPLE 1:
Can Sodium metal be produced by the electrolysis of
aqueous brine (a solution of NaCR) ?
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The desired reaction is:
Na+ (aq) + e — ! Na (s)
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But instead, at the cathode, the reaction is:
2 H2O + 2 e — ! H2 (g) + 2 OH — Eo = - 0.83 V
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At the anode:
2 CR
•
—
! CR2 (g)
+
2 e
E o = - 2.714 V
—
E o = - 1.36 V
Overall:
2 CR
—
+ 2 H2O ! H2 (g) + CR2 (g) + 2 OH
—
EXAMPLE 2: What reactions would occur at each electrode during
the electrolysis of aqueous CuSO4 ?
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COMMERCIAL CELLS [MH5; 18.6]
• Consider the electrolysis of aqueous NaCR ; (EXAMPLE 1, p. 107)
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At the cathode:
2 H2O + 2 e — ! H2 (g) + 2 OH
•
At the anode:
2 CR
—
! CR2 (g)
+
2 e
—
—
Eo = - 0.83 V
E o = - 1.36 V
• There are a variety of uses for the products formed in this
electrolysis reaction........
Batteries
• Batteries are voltaic, or galvanic cells; they undergo spontaneous
reactions.
• These reactions are exothermic (heat producing); but when reaction
happens in a cell, the energy is released in the form of electricity.
• There are two main types of batteries: dry cells (or primary
batteries) and rechargeable (secondary batteries).
•
A dry cell consists of a zinc container which acts as the anode, a
graphite (carbon) cathode and a paste of MnO2, NH4CR and ZnCR2
which acts as the electrolyte.....
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The half reactions are:
At the anode:
Zn (s) ! Zn2+ (aq) + 2 e
—
At the cathode:
2 MnO2 (aq) + 2 NH4+ (aq) + 2 e — ! Mn2O3 (s) + H2O (R) + 2NH3(aq)
Overall:
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E ocell = 1.5 V (not at standard conditions) and delivers ~ 0.5 A
current for about 6 hours.
The voltage decreases as Zn 2+ accumulates; eventually goes “dead”;
cannot be recharged.
In alkaline cells the NH4CR is replaced by KOH and the anode is a gel
made of Zn powder and KOH.
The voltage is also 1.5 V; but is delivered consistently and for a
longer period of time; these batteries are not rechargeable either.
The half reactions are:
At the anode:
Zn (s) + 2 OH
—
(aq) !
ZnO (s)
At the cathode:
2 MnO2 (s) + H2O (R) + 2 e —
Overall:
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!
+
H2O (R) + 2 e —
Mn2O3 (s)
+
2 OH
—
(aq)
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The mercury cell is a very small primary cell used in watches, hearing
aids, cameras and some calculators.
The anode is a zinc - mercury amalgam, with the reacting species
being zinc.
The cathode is mercury (II) oxide, HgO.
At the anode:
Zn (s) + 2 OH
—
(aq)
!
Zn(OH)2 (s)
+
2e
—
At the cathode:
HgO (s) + H2O + 2 e — ! Hg (R) + 2 OH — (aq)
Overall:
Storage (Rechargeable) Voltaic Cells
• Rechargeable batteries, or storage cells can be repeatedly
recharged..........
• The products of the reaction are deposited directly on the
electrodes when the battery is being used; the battery is acting as a
galvanic cell (or is said to be discharging).
• When recharging, a current is applied (battery is acting as an
electrolytic cell) and the discharge reaction is reversed.
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The lead storage battery is commonly used as a car battery; this is
a 12 V battery.
It consists of six 2 V cells, connected in series.
The anode consists of lead plates filled with “spongy” lead; the
cathode is lead plates filled with PbO2; these grills alternate
through the cell.
The large surface area supplied by these plates allows large
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currents to flow for short periods of time (like when starting the
car).
The electrolyte is sulfuric acid, H2SO4.
•
During the initial charge, the battery is acting as an electrolytic cell:
2 PbSO4 +
(anode)
•
2 H2O !
Pb (s) +
(cathode)
PbO2 (s) + 4 H+ + 2 SO42—
The reactions when battery is operating (starting the car, powering
the headlights etc) are:
Anode:
Pb (s) +
SO4 2— (aq)
!
PbSO4 (s)
+
2 e—
E o = + 0.36 V
Cathode:
PbO2 (s) + 4 H +(aq) + SO4 2— (aq) + 2 e — ! PbSO4 (s) + 2 H2O (R)
E o = + 1.68 V
Overall:
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When the cell is recharged, a current is applied to the cell which
provides the energy required for the non spontaneous reverse
reaction.
In the car, this happens when the car is running (this is what the
alternator does).
The extent of the charge of the battery is a function of the density
of the electrolyte; if the density falls too low, the battery cannot
usually be recharged.
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A discharged battery may also freeze, because the density of the
electrolyte is too low; if this happens the plates can warp, come in
contact with each other and short out the battery.........
These batteries cannot be recharged forever; too much repeated
“quick charging” (when battery goes dead) can result in Pb, PbO2 and
PbSO4 flaking off the electrodes.
These species collect as sludge at the bottom of the battery and
can cause short circuiting.
These batteries are heavy, the lead compounds can be toxic and the
sulfuric acid is corrosive.
NICAD batteries are also rechargeable and are used widely in
portable electronic equipment.
As the name suggests, these batteries use Nickel and Cadmium.....
The anode reaction is:
Cd (s) + 2 OH — (aq)
!
CdO (s)
The cathode reaction is:
NiO2 (s) + H2O (R) + 2 e —
!
+
H2O (R) + 2 e —
NiO (s)
+
2 OH
—
(aq)
Overall:
•
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The reaction is reversed when battery is being charged.
These batteries are lighter than the lead storage battery and have
a longer rechargeable life.
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EXAMPLE 1:
Consider the mercury cell described on p. 274. Suppose
this battery contains 0.030 g of Zn and 0.150 g of HgO. If the
battery can deliver a current of 0.00100 milliamps, how long, in hours,
will it take for the battery to completely discharge (go dead) ?
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EXAMPLE 2: Now consider the NICAD battery described on p. 276.
Initially, this battery contains 1.00 g of Cd and 0.900 g of NiO2.
When this battery “dies” it can be recharged. Assuming that a full
recharge is necessary, how long, in hours, would it take if a current of
1.50 amperes was used.
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Fuel Cells
• In a fuel cell, a fuel (usually hydrogen) is oxidized at the anode.
• In the reduction reaction at the cathode, oxygen is reduced.
• In fuel cells currently in use, the reaction takes place in alkaline
solution.
Anode:
Cathode:
Overall:
•
This reaction is very exothermic; the energy produced can be used
for power.
• A catalyst is necessary for this reaction to occur at low
temperatures (40 oC); platinum is usually used.
• Only very pure hydrogen can be used....if any carbon monoxide is
present it can poison the catalyst.
• Research is looking at replacing the internal combustion engine in
automobiles with fuel cells.
• The burning of hydrogen produces only water (not CO, CO2 and NO
such as the car engine does at present).........so this would be much
better for the environment.
•
Drawbacks to using fuel cells for car engines.....
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Redox Chemistry of some Transition Metals
• Transition metals often exhibit several different oxidation states.
• This makes their redox chemistry more complex and often colourful,
as some metal cations have different colours associated with their
different oxidation numbers.
• Transition metals can form cations by themselves at lower oxidation
numbers ( +1, +2, +3).
•
When these metals have higher oxidation numbers, (+4, +5, +6, +7),
they are only found in covalent compounds or complex ions.
•
A charge of +4 or greater on an ion would make it highly reactive, so
it would not be stable for long in an aqueous solution.
These different oxidation states are possible because the highest
energy electrons are easily lost to form cations.............
•
EXAMPLE:
•
The relative stabilities of these oxidation states are shown by Eo
values for interconversion.
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EXAMPLES:
[MH 5; table 20.3, p. 542]
Cr3+
- 0.41 V
!
Mn3+
1.56 V
!
3+
Fe
0.77 V
!
Cr2+
- 0.91 V
!
Cr(s)
Mn2+
- 1.18 V
!
Mn(s)
2+
Fe
- 0.41 V
!
1.95 V
Co3+
Cu
2+
Au
3+
!
0.16 V
!
1.40 V
!
Fe(s)
- 0.28 V
Co2+
Cu
!
+
Au
+
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Co(s)
0.52 V
!
Cu(s)
1.70 V
!
Au(s)
• Now let’s look at some questions that can be answered using the
information on the previous page.............
1)
Which is strongest oxidizing agent ?
2)
Which is the strongest reducing agent ?
3)
Which should disproportionate ? Look for an intermediate
oxidation state where.....
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4)
Given: O2 + 4 H+ + 4 e—
!
Which ions should oxidize water ?
5)
Which cations should reduce O2 to H2O ?
6)
Which cations should react with H+ ?
Recall: 2 H+ + 2 e— ! H2
Eo = 0.00V
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H2 O
EE = + 1.23 V