Transcript of Electrochemistry Importance of Electrochemistry –starting your car –use a calculator –listen...
- Slide 1
- Electrochemistry Importance of Electrochemistry 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
- Slide 2
- 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
- Slide 3
- Galvanic Cells Ex: 8 H + + MnO 4 - + 5 Fe +2 --> Mn +2 + 5
Fe +3 + 4 H 2 O If MnO 4 - 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
- Slide 4
- 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
- Slide 5
- 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
- Slide 6
- 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
- Slide 7
- 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
- Slide 8
- 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, E cell, aka,
the electromotive force (emf) of the cell. Measured in volts, V 1 V
= 1 Joule/Coulomb
- Slide 9
- Galvanic Cell Cell Potential Measure with a voltmeter
- Slide 10
- Standard Reduction Potentials Predict cell potentials if the
half cell potentials are known add half cell potentials to get cell
potential E ox + E red = E cell There is no way to measure half
cell potentials
- Slide 11
- Standard Reduction Potentials So how do we get these half cell
potentials if they cant be measured directly? Measure a cell
potential, assigning one reaction to be the zero 2H + + 2 e -
--> H 2 E red = 0.00 V
- Slide 12
- Standard Reduction Potentials So for 2H + + Zn --> H 2 + Zn
+2 the voltage is found to be 0.76 V If 2 H + + 2 e - --> H 2 E
red = 0.00 V and E cell = E ox + E red, then E ox = 0.76 - 0.00 V =
0.76 V So we can say Zn --> Zn +2 + 2 e - E ox = 0.76 V
- Slide 13
- Standard Reduction Potentials Combine other half reactions with
half reactions with known half cell potentials to complete the
table of Standard Reduction Potentials
- Slide 14
- Standard Reduction Potential Things to know about standard
reduction potentials: E o values correspond to solutes at 1 M and
all gases at 1 atm When a half reaction is reversed, the sign of E
o is reversed When a half reaction is multiplied by an integer, E o
remains the same!! Do not multiply E o by any number!
- Slide 15
- Standard Reduction Potentials The cell potential is always
positive for a galvanic cell Given Fe +2 + 2e - --> Fe E red =
-0.44 V MnO 4 - + 5 e - + 8 H + --> Mn +2 + 4H 2 O E red = 1.51
V Reverse the reaction that will result in an overall cell
potential that is positive, i.e., reverse the rxn involving
iron
- Slide 16
- Standard Reduction Potentials Find the E o 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
- Slide 17
- Standard Reduction Potentials Calculate E o and sketch the
galvanic cell based on the (unbalanced) reaction: MnO 4 - + H + +
ClO 3 - --> ClO 4 - + Mn +2 + H 2 O Balance the reaction
- Slide 18
- 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
- Slide 19
- 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 ClO 3 -
(aq),ClO 4 - ll MnO 4 - (aq),Mn +2 (aq) l Pt(s)
- Slide 20
- Cell Potential, Electrical Work, and Free Energy Relationship
between free energy and cell potential after some serious
derivations: G o = -nF E 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 G o. Cell
potential is directly related to G o. Confirms that a galvanic cell
runs in a direction that results in a positive E o, because a
positive E o means a negative G o, which means the reaction is
spontaneous
- Slide 21
- Cell Potential and Free Energy Calculate G o for the reaction
Cu +2 + Fe --> Cu + Fe +2 Is this reaction spontaneous?
- Slide 22
- Cell Potential and Free Energy Predict whether 1 M HNO 3 will
dissolve gold metal to form a 1 M Au +3 solution.
- Slide 23
- 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?
- Slide 24
- Dependence of Cell Potential on Concentration Cu(s) + 2Ce +4
(aq) --> Cu +2 (aq) + 2 Ce +3 (aq) Under standard conditions, E
o = 1.36 V. What will the cell potential be if [Ce +4 ] is greater
than 1.0 M? Use Le Chateliers 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
- Slide 25
- Dependence of Cell Potential on Concentration 2 Al(s) + 3Mn +2
(aq) --> 2 Al +3 (aq) + 3 Mn(s) E o = 0.48 V Predict whether E
cell is larger or smaller than E o cell a. [Al +3 ] = 2.0 M, [Mn +2
] = 1.0 M b. [Al +3 ] = 1.0 M, [Mn +2 ] = 3.0 M
- Slide 26
- 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
- Slide 27
- Concentration Cells Given a cell in which both compartments
contain aquesous AgNO 3 Ag + + e - ---> Ag E o = 0.80 V Left
side: 0.10 M Ag + Right side: 1.0 M Ag + What will happen?
- Slide 28
- 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.
- Slide 29
- The Nernst Equation G = G o + RT ln Q G o = -nF E o -nF E = -nF
E o + RT lnQ E = E o - RT ln QNernst equation nF The Nernst
equation gives the relationship between cell potential and the
concentrations of the cell components
- Slide 30
- The Nernst Equation @ 25 o C, 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
- Slide 31
- 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 Chateliers principle, we would predict that the reverse
reaction would be favored, so E should be less than E o.
- Slide 32
- 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
- Slide 33
- The Nernst Equation Equilibrium has been reached when Q = K, so
E cell = 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 G = 0 at equilibrium, meaning the cell no
longer has the ability to do work
- Slide 34
- The Nernst Equation Describe the cell based on the following
half reactions and concentrations: VO 2 + + 2 H + + e - --> VO
+2 + H 2 O E o = 1.00 V Zn +2 + 2 e - --> Zn E o = -0.76 V T =
25 o C [VO 2 + ] = 2.0 M [H + ] = 0.50 M [VO +2 ] = 1.0 x 10 -2 M
[Zn +2 ] = 1.0 x 10 -1 M
- Slide 35
- Calculating the K eq for Redox Reactions For a cell at
equilibrium, E cell = 0, and Q = K E cell = E o cell - 0.0592 log Q
n 0 = E o cell - 0.0592 log K eq n E o cell = 0.0592 log K eq n log
K eq = n E o cell @ 25 o C 0.0592
- Slide 36
- 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
- Slide 37
- Batteries Lead Storage Battery can function for several years
under temperature extremes from -30 o F to 120 o F 12 Volt storage
battery made of six cells anode = Pb cathode = Pb coated with PbO 2
electrolyte solution = H 2 SO 4 solution
- Slide 38
- Batteries Pb + HSO 4 - --> PbSO 4 + H + + 2 e - PbO 2 + HSO
4 - + 3 H + + 2 e - --> PbSO 4 + 2 H 2 O
___________________________________________________________ Pb +
PbO 2 + 2HSO 4 - + 2 H + --> 2 PbSO 4 + 2 H 2 O PbSO 4 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.
- Slide 39
- 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
- Slide 40
- 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 MnO 2, solid NH 4 Cl, and
carbon 2NH 4 + + 2 MnO 2 + Zn --> Zn +2 + Mn 2 O 3 + 2 NH 3 + H
2 O produces 1.5 V
- Slide 41
- Batteries Alkaline dry cell anode: Zn cathode: carbon rod in
contact with a moist paste of solid MnO 2, KOH or NaOH, and carbon
2MnO 2 + H 2 O + Zn --> Zn +2 + Mn 2 O 3 + 2 NH 3 + H 2 O the
alkaline dry cell lasts longer because the zinc doesnt corrode as
fast under basic conditions
- Slide 42
- Batteries Nickel-Cadmium battery Cd + NiO 2 + 2 OH - + 2 H 2 O
--> 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.
- Slide 43
- 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 H 2 (g) + O 2
(g) --> 2H 2 O(l) not exactly a portable power source as this
cell weighs about 500 pounds
- Slide 44
- 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
- Slide 45
- Corrosion Noble Metals copper, gold, silver, and platinum are
relatively difficult to oxidize, hence the term noble metals
- Slide 46
- Corrosion Some metals form a protective oxide coating,
preventing the complete corrosion of the metal Aluminum is the best
example, forming Al 2 O 3, 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
- Slide 47
- 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.
- Slide 48
- 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: O 2 + 2 H 2 O + 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, Fe 2 O 3. nH 2 O
- Slide 49
- 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
- Slide 50
- 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 e - E o = 0.44 V Zn
--> Zn +2 + 2 e - E o = 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
dont come into contact. The zinc is sacrificed.
- Slide 51
- 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 steels reduction potential to that of a noble metal.
- Slide 52
- Electrolysis Electrolytic Cell Uses electrical energy to force
a nonspontaneous redox reaction to go Current is forced through a
cell for which the E o cell is negative Importance charging a
battery producing aluminum metal chrome plating an object
- Slide 53
- 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
- Slide 54
- 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
- Slide 55
- Electrolysis How long must a currentof 5.00 A be applied to a
solution of Ag + to produce 10.5 g of silver metal?
- Slide 56
- Electrolysis Electrolysis of a mixture of ions Which metal will
plate out first? Look at the standard reduction potentials The more
positive the E o, the more likely the reaction will proceed Given a
solution with Cu +2, Ag +, and Zn +2, which metal will plate out
first?
- Slide 57
- Electrolysis Given a solution containing Ce +4, VO 2 +, 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.
- Slide 58
- Electrolysis Overvoltage a complex phenomenon which results in
more voltage needing to be applied to cause a reaction to occur
than predicted from the E o s, i.e., there will be exceptions! Ex:
Electrolysis of a solution of NaCl Expect Na +, Cl -, and H 2 O to
be the major species 2Cl - --> Cl 2 + 2 e - E o = -1.36 V 2H 2 O
--> O 2 + 4 H + + 4 e - E o = -1.23 V It would appear that H 2 O
would be easier to oxidize because E o is more positive, but in
reality, Cl 2 is produced. No good explanation, sorry!
- Slide 59
- 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
- Slide 60
- 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
- Slide 61
- Commercial Electrolytic Processes Production of Aluminum Uses
molten cryolite, Na 3 AlF 6 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 Al 2 O 3 is too
high at 2050 o C to make melting it practical Mixing Al 2 O 3 and
Na 3 AlF 6 lowers the m.p. to a mere 1000 o C, so electrolysis
became economically feasible Price of Aluminum dropped from
$100,000/lb to $0.74 /lb!
- Slide 62
- 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.
- Slide 63
- 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 800 o C to 600 o C),
apply electricity, and sodium is produced
- Slide 64
- 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 E o = -2.71 V 2
H 2 O + 2 e - --> H 2 + 2 OH - E o = -0.83V