Transcript Electrochemistry

Electrochemistry
• Importance of Electrochemistry
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starting your car
use a calculator
listen to a radio
corrosion of iron
preparation of important industrial
materials
• Electrochemistry - the study of the
interchange of chemical and electrical
energy
Galvanic Cells
• Redox reactions
– electron transfer reactions
– oxidation - loss of electrons, i.e., an
increase in oxidation number
– reduction - gain electrons, i.e., a
decrease in oxidation number
Galvanic Cells
• Ex: 8 H+ + MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3 + 4 H2O
• If MnO4- and Fe+2 are present in the same
solution, the electrons are transferred directly
when the reactants collide.
– No useful work is done.
– Chemical energy is released as heat.
• How can the energy be harnessed?
– Separate the half reactions.
– Have the electron transfer go through a wire, which
can then be directed through a motor or other useful
device
Galvanic Cells
• Galvanic cell
– Uses a spontaneous redox reaction
to produce current to do work
– A device in which chemical energy
is changed to electrical energy
– Also known as a voltaic cell
Galvanic Cells
• Requirements for a Galvanic Cell
– Half reactions are in separate
compartments
– Compartments are connected by a salt
bridge
• a porous disk connection
• a salt bridge (U bridge) containing a strong
electrolyte held in a jello-like matrix
Galvanic Cells
• Why a salt bridge?
– Without a salt bridge, the reaction will
not go
– Current starts, but stops due to charge
buildup in both compartments
• one side becomes negatively charged as
electrons are added
• one side becomes positively charged as
electrons are lost
• Creating a charge separation requires a large
amount of energy
Galvanic Cells
• Reaction in a galvanic cell occurs at
each of the electrodes
• Be able to label the anode, cathode,
direction of electron flow, and
direction of ion flow
Galvanic Cells
• Cell Potential
– The oxidizing agent “pulls” electrons
towards itself from the reducing agent
– The “pull” or driving force on the
electrons is the cell potential, Ecell, aka,
the electromotive force (emf) of the
cell.
– Measured in volts, V
– 1 V = 1 Joule/Coulomb
Galvanic Cell
• Cell Potential
– Measure with a voltmeter
Standard Reduction Potentials
• Predict cell potentials if the half cell
potentials are known
– add half cell potentials to get cell
potential
– Eox + Ered = Ecell
– There is no way to measure half cell
potentials
Standard Reduction Potentials
• So how do we get these half cell
potentials if they can’t be measured
directly?
• Measure a cell potential, assigning one
reaction to be the “zero”
– 2H+ + 2 e- --> H2 Ered = 0.00 V
Standard Reduction Potentials
• So for 2H+ + Zn --> H2 + Zn+2 the
voltage is found to be 0.76 V
• If 2 H+ + 2 e- --> H2 Ered = 0.00 V
and Ecell = Eox + Ered,
then Eox = 0.76 - 0.00 V = 0.76 V
• So we can say
Zn --> Zn+2 + 2 e-
Eox = 0.76 V
Standard Reduction Potentials
• Combine other half reactions with
half reactions with known half cell
potentials to complete the table of
Standard Reduction Potentials
Standard Reduction Potential
• Things to know about standard
reduction potentials:
– Eo values correspond to solutes at 1 M
and all gases at 1 atm
– When a half reaction is reversed, the
sign of Eo is reversed
– When a half reaction is multiplied by an
integer, Eo remains the same!! Do not
multiply Eo by any number!
Standard Reduction Potentials
• The cell potential is always positive for a
galvanic cell
Given Fe+2 + 2e- --> Fe
Ered = -0.44 V
MnO4- + 5 e- + 8 H+--> Mn+2 + 4H2O Ered = 1.51 V
• Reverse the reaction that will result in an
overall cell potential that is positive, i.e.,
reverse the rxn involving iron
Standard Reduction Potentials
• Find the Eo for the galvanic cell
based on the reaction
Al+3 + Mg --> Al + Mg+2
• Sketch this cell and label the anode,
cathode, direction of electron and ion
flow, and indicate what reaction
occurs at each electrode
Standard Reduction Potentials
• Calculate Eo and sketch the galvanic
cell based on the (unbalanced)
reaction:
MnO4- + H+ + ClO3- --> ClO4- + Mn+2 + H2O
• Balance the reaction
Galvanic Cells
• Line Notation
– Used to describe electrochemical cells
– Anode components are listed on the left
– Cathode components are listed on the
right
– Separate half cells with double vertical
lines: ll
– Indicate a phase difference with a single
vertical line: l
Galvanic Cells
• Ex: Mg(s) l Mg+2(aq) ll Al+3(aq) l Al(s)
is the line notation for the example in slide
# 15
• For slide #16, the reactants and products
are all ions, and so cannot act as an
electrode
– an inert electrode is needed - use Pt
• Pt(s) l ClO3-(aq),ClO4- ll MnO4-(aq),Mn+2(aq) l Pt(s)
Cell Potential, Electrical Work, and Free Energy
• Relationship between free energy and cell potential
– after some serious derivations:
DGo = -nFE o
• n = number of moles of electrons
• F = the charge on 1 mole of electrons
= Faraday = 96,485 Coulombs/1 mole e-
– this equation provides an experimental means to
obtain DGo.
– Cell potential is directly related to DGo.
– Confirms that a galvanic cell runs in a direction
that results in a positive E o, because a positive E
o means a negative DGo, which means the reaction
is spontaneous
Cell Potential and Free Energy
• Calculate DGo for the reaction
Cu+2 + Fe --> Cu + Fe+2
Is this reaction spontaneous?
Cell Potential and Free Energy
• Predict whether 1 M HNO3 will
dissolve gold metal to form a 1 M Au+3
solution.
Dependence of Cell Potential on Concentration
• The standard reduction potentials can
be used to tell us the cell potential
for cells under standard conditions
(all concentrations = 1 M)
• What will happen to the cell potential
when the concentration is not 1 M?
Dependence of Cell Potential on Concentration
• Cu(s) + 2Ce+4(aq) --> Cu+2(aq) + 2 Ce+3(aq)
• Under standard conditions, Eo = 1.36 V.
What will the cell potential be if [Ce+4] is
greater than 1.0 M?
• Use Le Chatelier’s principle.
– Increasing the [Ce+4] (when the [Ce+4] > 1 M),
then the forward reaction is favored, so the
driving force on the electrons is greater, so the
cell potential is greater
Dependence of Cell Potential on Concentration
2 Al(s) + 3Mn+2(aq) --> 2 Al+3(aq) + 3 Mn(s) Eo = 0.48 V
• Predict whether Ecell is larger or smaller than Eocell
a. [Al+3] = 2.0 M, [Mn+2] = 1.0 M
b. [Al+3] = 1.0 M, [Mn+2] = 3.0 M
Concentration Cells
• Cell potential depends on concentration
– Make a cell (a concentration cell) in
which both compartments contain the
same components, but at different
concentrations
– Usually has a small cell potential
– The difference in concentration
produces the cell potential
Concentration Cells
• Given a cell in which both
compartments contain aquesous
AgNO3
• Ag+ + e- ---> Ag Eo = 0.80 V
• Left side: 0.10 M Ag+
• Right side: 1.0 M Ag+
• What will happen?
Concentration Cells
• There will be a positive cell potential due to
the difference in Ag+ concentrations
• To reach equilibrium, the driving force is to
equalize the Ag+ concentration.
– This can be done if the 1.0 M Ag+ could be
reduced and the 0.10 M Ag+ could be
increased.
– The Ag in the 0.10 M Ag+ compartment
will dissolve while the Ag in the 1.0 M Ag+
compartment will increase in mass.
The Nernst Equation
DG = DGo + RT ln Q
DGo= -nFEo
• -nFE = -nFE o + RT lnQ
• E = Eo - RT ln Q Nernst equation
nF
• The Nernst equation gives the relationship
between cell potential and the
concentrations of the cell components
The Nernst Equation
• @ 25oC, the Nernst equation can be
written as:
E = E
o
- 0.0592 log Q
n
• Use the Nernst equation to calculate
the cell potential when one or more of
the components are not in their
standard states
The Nernst Equation
• For 2 Al(s) + 3 Mn+2 --> 2 Al+3 + 3 Mn(s)
– Calculate E when [Mn+2] = 0.50 M and
[Al+3] = 1.50 M
– Applying Le Chatelier’s principle, we
would predict that the reverse reaction
would be favored, so E should be less
than Eo.
The Nernst Equation
• The cell potential calculated by the
Nernst equation is the maximum cell
potential before any current flows
• As an galvanic cell discharges, the
concentrations of the components will
change, so E will change over time
• A cell will spontaneously discharge
until equilibrium has been reached
The Nernst Equation
• Equilibrium has been reached when
Q = K, so Ecell = 0
• A dead battery is one in which the
cell reaction has reached equilibrium
– There is no driving force for electrons
to be pushed through the wire
DG = 0 at equilibrium, meaning the cell no
longer has the ability to do work
The Nernst Equation
• Describe the cell based on the following
half reactions and concentrations:
VO2+ + 2 H+ + e- --> VO+2 + H2O Eo = 1.00 V
Zn+2 + 2 e- --> Zn
Eo = -0.76 V
T = 25o C
[VO2+] = 2.0 M
[H+] = 0.50 M
[VO+2] = 1.0 x 10-2M
[Zn+2] = 1.0 x 10-1 M
Calculating the Keq for Redox Reactions
• For a cell at equilibrium,
Ecell = 0, and Q = K
Ecell = Eocell - 0.0592 log Q
0
n
= Eocell - 0.0592 log Keq
n
Eocell = 0.0592 log Keq
log Keq
n
= nEocell
0.0592
@ 25oC
Batteries
• A battery is
– a galvanic cell
– a group of galvanic cells connected in
series
• cell potentials of the individual cells add up
to give the total battery cell potential
– a source of direct current
Batteries
• Lead Storage Battery
– can function for several years under
temperature extremes from -30oF to
120oF
– 12 Volt storage battery made of six cells
– anode = Pb
– cathode = Pb coated with PbO2
– electrolyte solution = H2SO4 solution
Batteries
Pb + HSO4- --> PbSO4 + H+ + 2 ePbO2 + HSO4- + 3 H+ + 2 e- --> PbSO4 + 2 H2O
___________________________________________________________
Pb + PbO2 + 2HSO4- + 2 H+ --> 2 PbSO4 + 2 H2O
• PbSO4 adheres to the electrodes
• As the cell discharges, the sulfuric acid is
consumed. The condition of the battery can be
determined by measuring the density of the
solution. As the sulfuric acid concentration
decreases, the density decreases.
Batteries
• The lead storage battery can be recharged
because the products adhere to the
electrodes, so the alternator can force
current through the battery in the
opposite direction and reverse the reaction
• Even though the battery can be recharged,
physical damage from road shock and
chemical side reactions (e.g. electrolysis of
water)eventually cause battery failure
Batteries
• Dry Cell Battery
– invented more than 100 years ago by George
Leclanche
• Acid Version
– anode: Zn
– cathode: Carbon rod in contact with a moist
paste of solid MnO2, solid NH4Cl, and carbon
– 2NH4+ + 2 MnO2 + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O
– produces 1.5 V
Batteries
• Alkaline dry cell
– anode: Zn
– cathode: carbon rod in contact with a moist
paste of solid MnO2, KOH or NaOH, and carbon
– 2MnO2 + H2O + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O
– the alkaline dry cell lasts longer because the
zinc doesn’t corrode as fast under basic
conditions
Batteries
• Nickel-Cadmium battery
• Cd + NiO2 + 2 OH- + 2 H2O --> Cd(OH)2 + Ni(OH)2 + 2 OH-
• This battery can be recharged because,
like the lead storage battery, the products
adhere to the electrodes.
Fuel Cells
• Fuel Cell
– a galvanic cell for which the reactants
are continuously supplied
– used in the U.S. space program
– based on the reaction of hydrogen and
oxygen to form water:
• 2 H2(g) + O2(g) --> 2H2O(l)
– not exactly a portable power source as
this cell weighs about 500 pounds
Corrosion
• Corrosion is the return of metals to their natural
states, i.e., the ores from which they are obtained.
• Corrosion involves oxidation of the metal
• Corroded metal loses its structural integrity and
attractiveness
• Metals corrode because they oxidize easily
• Metals commonly used for decorative and
structural purposes have less positive reduction
potentials than oxygen gas
Corrosion
• Noble Metals
– copper, gold, silver, and platinum are
relatively difficult to oxidize, hence the
term noble metals
Corrosion
• Some metals form a protective oxide
coating, preventing the complete corrosion
of the metal
– Aluminum is the best example, forming Al2O3,
which adheres to, and protects the aluminum
– Copper forms an external layer of copper
carbonate, known as patina
– Silver forms silver tarnish which is silver
sulfide
– Gold does not corrode in air
Corrosion of Iron
• Important to control the corrosion of iron
because it is so important as a structural
material
• The corrosion of iron is not a direct
oxidation process (iron reacting with
oxygen), but is actually an electrochemical
reaction.
Corrosion of Iron
• Steel has a nonuniform surface
– These nonuniform areas are where iron can
be more easily oxidized (the anode regions)
than at other regions (the cathode regions)
– anode: Fe--> Fe+2 + 2 e- (the electrons flow
through the steel to the cathode)
– cathode: O2 + 2 H2O + 4 e- --> 4OH– The Fe+2 formed in the anodic regions travel
through the moisture to the cathodic regions.
There, the Fe+2 reacts with oxygen to form
rust, which is hydrated iron (III) oxide,
Fe2O3.nH2O
Corrosion of Iron
• For iron to corrode, moisture must be
present to act as a salt bridge between the
anode and cathode regions.
– Steel does not rust in dry air
• Salt accelerates the rusting process
– Cars rust faster where salt is used on roads to
melt ice and snow
– Salt on the moist surfaces increases the
conductivity of the aqueous solution formed
Corrosion Prevention
• Apply a protective coating
– apply paint or metal plating
• Chromium or tin are often used to plate steel
because they form protective oxides
• Zinc can be used to coat steel, a process
called galvanizing
– Fe--> Fe+2 + 2 eEo = 0.44 V
– Zn --> Zn+2 + 2 eEo = 0.76 V
– Zinc, then, is more likely to be oxidized than iron.
It acts as a protective coating because it will
react with the oxygen preferentially, so the
oxygen and the iron don’t come into contact. The
zinc is sacrificed.
Corrosion Prevention
• Alloying is used to prevent corrosion
– Stainless steel does not corrode like iron
• carbon, chromium and nickel have been added
to iron to form stainless steel, an alloy.
• These additions to iron form a protective
oxide coating that changes steel’s reduction
potential to that of a noble metal.
Electrolysis
• Electrolytic Cell
– Uses electrical energy to force a
nonspontaneous redox reaction to go
– Current is forced through a cell for
which the Eocell is negative
– Importance
• charging a battery
• producing aluminum metal
• chrome plating an object
Electrolysis
• Compare an electrolytic cell to a galvanic
cell. Be able to label the anode, cathode,
direction of ion and electron flow, which
half reactions take place at each
electrode, and where oxidation or
reduction are taking place:
• Zn+2 + Cu --> Cu+2 + Zn
vs.
• Zn + Cu+2 --> Zn+2 + Cu
Electrolysis
• Stoichiometry of electrolytic process
– 1 Ampere = 1 coulomb of charge /sec or
– 1 A.sec = 1 C
– 96485 C = the charge on 1 mole of electrons
• To plate something is to deposit the
neutral metal on the electrode by reducing
the metal ions in solution, e.g.
Cu+2 + 2 e---> Cu
Electrolysis
• How long must a currentof 5.00 A be
applied to a solution of Ag+ to produce 10.5
g of silver metal?
Electrolysis
• Electrolysis of a mixture of ions
– Which metal will plate out first?
– Look at the standard reduction
potentials
• The more positive the Eo, the more likely the
reaction will proceed
– Given a solution with Cu+2, Ag+, and Zn+2,
which metal will plate out first?
Electrolysis
• Given a solution containing Ce+4, VO2+,
and Fe+3, give the order of oxidizing
ability of these species and predict
which one will be reduced at the
cathode of an electrolytic cell at the
lowest voltage.
Electrolysis
• Overvoltage
– a complex phenomenon which results in more
voltage needing to be applied to cause a reaction
to occur than predicted from the Eo’s, i.e., there
will be exceptions!
• Ex: Electrolysis of a solution of NaCl
– Expect Na+, Cl-, and H2O to be the major species
2Cl- --> Cl2 + 2 eEo = -1.36 V
2H2O --> O2 + 4 H+ + 4 e- Eo = -1.23 V
It would appear that H2O would be easier to
oxidize because Eo is more positive, but in
reality, Cl2 is produced. No good explanation,
sorry!
Commercial Electrolytic Processes
• Metals are typically good reducing agents
– typically found combined with other substances
in ores, mixtures of ionic substances containing
oxides, sulfides, and silicate anions
– nobles metals, Cu, Ag, Pt, and Au, can be found
as pure metals
Commercial Electrolytic Processes
• Production of aluminum
– Al is the third most abundant element on earth
– Al is a very active metal found in nature as the
oxide in an ore called bauxite (for Les Baux,
France where it was discovered)
– Charles Hall (U.S.) and Paul Heroult (France)
discovered an electrolytic process to produce
pure aluminum almost simultaneously - Hall-
Heroult process
Commercial Electrolytic Processes
• Production of Aluminum
– Uses molten cryolite, Na3AlF6 as the solvent
for aluminum oxide
• water is more easily reduced than Al+3, so aluminum
oxide could not be dissolved in water to produce
aluminum
• melting an ionic substance allows for mobility of the
ions, but the m.p. of Al2O3 is too high at 2050oC to
make melting it practical
• Mixing Al2O3 and Na3AlF6 lowers the m.p. to a mere
1000oC, so electrolysis became economically feasible
• Price of Aluminum dropped from $100,000/lb to
$0.74 /lb!
Commercial Electrolytic Processes
• Aluminum produced in the Hall-Heroult
process is 99.5% pure
• To be useful as a structural material,
aluminum is alloyed with metals like zinc
and manganese.
• Production of aluminum uses about 5% of
all the electricity used in the U.S.
Electrolysis of Sodium Chloride
• Electrolysis of sodium chloride
– produces pure sodium metal
– melt solid sodium chloride (after mixing
with calcium chloride to lower the m.p.
from 800oC to 600oC), apply electricity,
and sodium is produced
Electrolysis of Sodium Chloride
• Electrolysis of aqueous sodium chloride
– aqueous sodium chloride, aka, brine
– process is the second largest consumer of
electricity in the U.S.
– produces chlorine gas and sodium hydroxide
– not a source of sodium metal because water is
more easily reduced than Na+
• Na+ + e- --> Na
Eo = -2.71 V
• 2 H2O + 2 e- --> H2 + 2 OH- Eo = -0.83V