Electrochemistry: OxidationReduction Reactions

1
Electrochemistry Oxidation-Reduction Reactions
gaining to 2e-
Zn(s) 2H(aq) ? Zn2(aq) H2(aq)
loss of 2e-
oxidation involves the loss of electrons
reduction involves the gain of electrons
The transfer of electrons that occur in this
reaction produces heat energy or in many cases,
electrical energy, and proceeds spontaneously.
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Electrochemistry Oxidation-Reduction Reactions
By writing the oxidation number of each element
under the reaction equation, we can easily see
the oxidation state changes that occur
Zn(s) 2H(aq) ? Zn2(aq) H2(aq)
0 1 2
0
In any oxidation-reduction reaction (redox), both
oxidation and reduction must occur.
Zn(s) is oxidized and is the reducing agent
H is reduced and is the oxidizing agent
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Oxidation Number Method
3(4)12
4Al(l) 3MnO2 (s) ? 2Al2O3(s) 3Mn(l)
0 4 -2 3 -2
0
-4(3)-12
Write the unbalanced equation Assign oxidation
numbers and determine which elements undergo
changes in oxidation number during the
reaction Choose coefficients that make the total
increase in oxidation number equal the total
decrease in oxidation number Balance the
remaining elements by inspection
4
Electrochemistry Balancing Oxidation-Reduction
Reactions
Sample problem
2(5)
I2O5(s) 5CO(g) ? I2(s) 5CO2(g)
5 -2 2 -2 0 4 -4
-5(2)
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Half-Reaction Method
Sn2(aq) 2Fe3(aq) ? Sn4(aq) 2Fe2(aq)
Although oxidation and reduction occur
simultaneously, it is sometimes convenient to
think of the reaction as occurring in two
separate steps.
oxidation
Sn2(aq) ? Sn4(aq) 2e-
Half-reactions
reduction
2Fe3(aq) 2e- ? 2Fe2(aq)
Sn2(aq) 2Fe3(aq) ? Sn4(aq) 2Fe2(aq)
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Half-Reaction Method
MnO4-(aq) C2O42-(aq) ? Mn2(aq) CO2(g) (acid)
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Half-Reaction Method
Step 1 Write the unbalanced equation for the
reation
MnO4-(aq) C2O42-(aq) ? Mn2(aq) CO2(g)
Step 2 Divide the overall reaction into two
unbalanced half-reactions, one for oxidation and
the other for reduction
MnO4-(aq) ? Mn2(aq)
C2O42-(aq) ? CO2(g)
Step 3 Balance elements other than hydrogen and
oxygen
MnO4-(aq) ? Mn2(aq)
C2O42-(aq) ? 2CO2(g)
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Half-Reaction Method
Step 4 Balance the O atoms by adding H2O, and
then balance the H atoms by adding H
8H MnO4-(aq) ? Mn2(aq) 4H2O
C2O42-(aq) ? 2CO2(g)
Step 5 Balance charge on each side of the
half-reaction equations by adding e- to the
side with the greater positive charge
5e- 8H MnO4-(aq) ? Mn2(aq) 4H2O
C2O42-(aq) ? 2CO2(g) 2e-
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Half-Reaction Method
Step 6 Multiply the two half-reactions by
coefficients such that the overall electron loss
equals the overall electron gain.
Step 7 Add the two half-reactions simplify
where possible by eliminating terms appearing on
both sides of the equation
5e- 8H MnO4-(aq) ? Mn2(aq) 4H2O
2
And dont forget the trick we use when the
reaction is occurring in a basic solution
C2O42-(aq) ? 2CO2(g) 2e-
5
10e- 16H 2MnO4-(aq) ? 2Mn2(aq) 8H2O
5C2O42-(aq) ? 10CO2(g) 10e-
16H 2MnO4- 5C2O42- ? 2Mn2 10CO2 8H2O
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Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Balancing Oxidation-Reduction
Reactions
Electrochemistry Voltaic Cells
A voltaic cell is a device in which electron
transfer is forced to take place through an
external pathway rather than directly between
reactants
oxidation
reduction
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Electrochemistry Cell EMF
A voltaic cells work because it is favorable for
electrons to flow from the anode to the
cathode (electromotive force). This flow is
measured in terms of volts (1V 1J/C)
The potential difference between the two
electrodes of a voltaic cell may be viewed as a
driving force or electrical pressure that pushes
the electrons through the external circuit
Standard emf Standard potential Ecell
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Electrochemistry Cell EMF
The emf of any voltaic cell depends on the
chemical reactions taking place in the cell,
the concentrations of products and reactants and
the temperature of the system
Zn2(s) Cu2 (aq) ? Zn2(aq) Cu(s)
Ecell 1.10 V
Standard Electrode Potentials associated with
each of the two half-reactions determines the emf
of the system. The half-cell potential due to
the loss of electrons at the anode is called the
oxidation potential Eox, and the potential due to
electron gain at the cathode is called the
reduction potential, Ered
Ecell Eox Ered
Under standard conditions
Ecell E ox E red
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Electrochemistry Cell EMF
Half-cell potential cannot be directly measured
so the reduction of H to H2 is used as
a reference 2H (1 M) 2e- ? H2(1 atm, 25 C)
Ecell 0 V
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Electrochemistry Cell EMF
Half-cell potential cannot be directly measured
so the reduction of H to H2 is used as
a reference 2H (1 M) 2e- ? H2(1 atm, 25 C)
Ecell 0 V
Ecell 0.76 0 0.76 V
Note that the activity series is simply the
oxidation half-reactions of the metals ordered
from the highest to lowest
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Electrochemistry Cell EMF
Sample Problem. Balance the following redox
reaction and determine the standard emf for the
balanced system
Cr2O72-(aq) I-(aq) ? Cr3(aq) I2(s)
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Electrochemistry Cell EMF Oxidizing and
Reducing Agents
The more positive the E value for a half
reaction, the greater the tendency for that
reaction to occur as written.
A negative reduction potential indicates that the
species is more more difficult to reduce than H,
whereas a negative oxidation potential indicates
that the species is more difficult to oxidize
than H2.
Among the most common oxidizing agents are the
halogens, oxygen, and oxy- anions such as MnO4-,
Cr2O72-, and NO3-, whose central metal atoms have
high positive oxidation states.
Commonly used reducing agents include H2 and a
variety of metals having positive oxidation
potentials such as ZN and Fe, or Sn2 which can
be oxidized to Sn4.
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Electrochemistry Cell EMF Oxidizing and
Reducing Agents
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Electrochemistry Cell EMF Spontaneity and Extent
of Redox Reactions
A positive emf indicates a spontaneous process,
and a negative emf indicates a nonspontaneous one.
Sample Problem Using the standard electrode
potentials, determine whether the following
reaction are spontaneous under standard
conditions.
A. Cu(s) 2H(aq) ? Cu2(aq) H2
B. Cl2(g) 2I-(aq) ?2Cl-(aq) I2(s)
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Electrochemistry Cell EMF EMF and Free Energy
Change
Any redox reaction involves free energy change
(?G) which also may be used as a measure of
spontaneity
?G -n?E
Or under standard conditions
?G -n?E
In this equation, n is the number of moles of
electrons transferred in the reaction and ? is
Faradays constant . Faradays constant
describes the electrical charge on 1 mole of
electrons.
1 ? 96,500 C/mol e- 96,500 J/V-mol e-
Note that because n and ? are both positive
values, a positive value in E leads to a negative
value of ?G.
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Electrochemistry Cell EMF EMF and Free Energy
Change
Determining the free energy change associated
with a Redox reaction
Use the standard electrode potentials to
calculate the standard free energy change, ?G,
for the following reaction
2Br-(aq) F2(g)? Br2(l) 2F-(aq)
2Br-(aq) ? Br2(l) 2e- Eox
-1.06
F2(g) 2e- ? 2F-(aq) Ered
2.87
2Br-(aq) F2(g)? Br2(l) 2F-(aq) E
1.81 V
?G -n?E
?G -(2 mol e- )(96,500 J/volt-mol e-)( 1.81 V)
?G -3.49 x 105 J -349 kJ
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Electrochemistry Cell EMF EMF and Free Energy
Change
Determining the free energy change associated
with a Redox reaction
Balance the following redox reaction. Determine
the free-energy for the reaction and indicate
whether the reaction will occur spontaneously.
I2(s) Cu2(aq) ? IO3-(aq) Cu(s) (acidic
environment)
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Electrochemistry Cell EMF EMF and Equilibrium
Constant
Remember in Chapter 19, we related ?G to the
equilibrium constant, K?
?G -RT lnK
No I dont!
Where am I?
Nope!
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Electrochemistry Cell EMF EMF and Equilibrium
Constant
Well. Now the plot thickens
?G -RT lnK
?G -n?E
-n?E -RT lnK
Or if you like to use base-10 logarithms
-n?E -2.30 RT log K
E 2.30 RT log K
n?
Or if you just want to settle for describing the
reaction at 25 C (298K)
E 0.0591V logK
n
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Electrochemistry Cell EMF EMF and Equilibrium
Constant
HERE IT COMES!!!
Calculate the equilibrium constant for the
following reaction at 25 C
O2(g) 4H(aq) 4Fe2(aq) ? 4Fe3(aq) 2H2O(l)
O2(g) 4H(aq) 4e- ? 2H2O(l)
Ered 1.23 V
4Fe2(aq) ? 4Fe3(aq) 4e- Eox
-0.77 V
O2(g) 4H(aq) 4Fe2(aq) ? 4Fe3(aq) 2H2O(l)
E 0.46 V
E 0.0591V logK
n
n E
4(0.46V)

log K

31.1
0.0591V
0.0591V
K 1 x 1031
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Electrochemistry Cell EMF EMF and Equilibrium
Constant
Now Its Your Turn!!!
Calculate the equilibrium constant at 25 C for
the reaction
IO3-(aq) 5Cu(s) 12H(aq) ? I2(s) 5Cu2
(aq) 6H2O(l)
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Electrochemistry Cell EMF EMF and Concentration
The Problem is that many times voltaic cells
operate under nonstandard conditions
The emf generated under nonstandard conditions
can be calculated from E , temperature, and the
concentrations of products and reactants.
Remember Chapter 19?
?G -RT lnK
zzzzzzz
zzzzzzz
zzzzzzzzz
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Electrochemistry Cell EMF EMF and Concentration
Walther Hermann Nernst Gets to Have an Equation
named after him
?G -RT lnK
Standard Conditions
Non- Standard Conditions
?G ?G RT lnQ
Remember Ch 19?
?G ?G 2.30 RT log Q
-n?E -n?E 2.30 RT log Q
E E - 2.30 RT log Q
Nernst Equation
n?
When T298K
E E - 0.0591 log Q
n
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Electrochemistry Cell EMF EMF and Concentration
The Nernst Equation
Consider the following reaction
Zn(s) Cu2(aq) ? Zn2 (aq) Cu(s) E
1.10 V
Since the number of moles of electrons
transferred is 2. Then at 298 K the Nernst
equation would be constructed as follows
Zn2
E 1.10 - 0.0591 log
Cu2
2
Now. Lets say the the concentrations of these
ions in each of the half-cells is not the
standard 1.0 M and instead we construct a cell
where Zn20.05M, and Cu2 0.50 M
0.05
E 1.10 - 0.0591 log
1.16V
5.0
2
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Electrochemistry Cell EMF EMF and Concentration
The Nernst Equation
Calculate the emf generated by the the following
reaction
Cr2O72-(aq) 14 H 6I-(aq) ? 2Cr3(aq)
3I2(s) 7H2O(l)
Cr2O72- 2.0 M
H 1.0 M
I- 1.0 M
Cr3 1.0 x 10-5
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Electrochemistry Cell EMF Commercial Voltaic
Cells
Lead Storage Battery
anode Pb(s) SO42-(aq) ? PbSO4(s) 2e-
E 0.356 V
cathode PbO2(s) SO42-(aq) H(aq) 2e- ?
PbSO4(s)
2H2O(l) E 1.685 V
overall Pb(s) PbO2(s) 4H 2SO4(aq) ?
2PbSO4(s)
2H2O(l) E 2.041 V
Note that one advantage of the lead storage
battery is that it can be recharged because the
PbSO4 produced during discharge adheres to the
electrodes
recharge
2PbSO4(s) 2H2O(l) ? Pb(s)
PbO2(s) 4H 2SO4(aq)
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Electrochemistry Cell EMF Commercial Voltaic
Cells
Dry Cell
anode Zn(s) ? Zn 2(aq) 2e-
cathode 2NH4(aq) 2MnO2(s) 2e- ?
Mn2O3(s) 2NH3(aq)
H2O(l)
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Electrochemistry Cell EMF Commercial Voltaic
Cells
Fuel Cells
The direct production of electricity from the
conversion of H2 and O2
anode 2H2(g) 4OH-? 4H2O(l) 4e-
cathode 4e- O2(g) H2O(l) ? 4OH-(aq)
2H2(g) O2(g) ? 2H2O(l)
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Electrochemistry Cell EMF Electrolysis
Electrolytic Cells
voltage source acts like an electron pump
These electrodes are inert
Electrolysis is driven by an outside energy source
34
Electrochemistry Cell EMF Electrolysis
Electrolysis of Aqueous Solutions
Sodium cannot be prepared by electrolysis of
aqueous solutions of NaCl, because water is more
easily reduced than Na(aq)
2H2O(l) 2e- ? H2(g) 2OH-(aq) Ered
-0.83
Na(aq) e- ? Na(s)
Ered -2.71
Consequently, H2 is produced at the cathode
The possible anode reactions are
2Cl-(aq) ? Cl2(g)
Eox -1.36
2H2O(l) ? 4H(aq) O2(g) 4e- Eox
-1.23
Note that while the H2O appears to be easier to
oxidize the Cl- will oxidize instead because the
overvoltage for the formation of O2 is greater
than that for Cl2. (p.727). As a result
electrolysis of aqueous solutions known as
brines, produces H2 and Cl2.
2Cl-(aq) ? Cl2(g)
Eox -1.36
anode
2H2O(l) 2e- ? H2(g) 2OH-(aq)
Ered -0.83
cathode
Ecell gt -2.19
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Electrochemistry Cell EMF Electrolysis
Electrolysis of With Active Electrodes
When aqueous solutions are electrolyzed using
metal electrodes, an electrode will be oxidized
if the oxidation potential is greater than that
of water
Ni(s) ? Ni2(aq) 2e- Eox 0.28
2H2O(l) ? 4H(aq) O2(g) 4e- Eox -1.23
Ni(s) ? Ni2(aq) 2e-
anode
Ni2(aq) 2e- ? Ni(s)
cathode
Electroplating creates a silver lining
36
Electrochemistry Cell EMF Electrolysis
Quantitative Aspects of Electrolysis
For any half-reaction, the amount of a substance
that is reduced or oxidized in an electrolytic
cell is directly proportional to the number of
electrons passed into the cell
The quantity of change passing through an
electrical circuit is generally measured in
Coulombs. (1? 96,500 C charge of 1 mole of
electrons)
A coulomb may also be understood as the quantity
of electrical charge passing a point in a
circuit in one second when the current is one
ampere (A) (C amperes x seconds)
Calculate the mass of aluminum produced in 1.00
hr. by the electrolysis of molten AlCl3 if the
current is 10.0 A.
C amperes x seconds
(10.0A)(1.00 hr)
1C
3600 sec
3.6 x 104 C
1 A-s
1 hr
since the half-reaction for the reduction of Al3
is Al3 3e- Al
1 mol Al
27.0 g Al
1?
grams Al 3.60 x 10-4 C
3.36 g
3?
1 mol Al
96,500 C
37
Electrochemistry Cell EMF Electrolysis
Quantitative Aspects of Electrolysis
The half-reaction for the formation of magnesium
metal upon electrolysis of molten MgCl2 is
Mg2(aq) 2e- ? Mg. Calculate the mass of
magnesium formed upon passage of 60.0 A for a
period of 4.00 x 103 s
38
Electrochemistry Cell EMF Electrolysis
Electrical Work
since ?G wmax and ?G n?E then wmax n?E
Note that if the sign convention for work is
negative, then this would correspond to work
being done on the surroundings. This work
would have a positive E and indicate
a spontaneous process.
In an electrolytic cell, an external energy
source is required to initiate what would
ordinarily be a non- spontaneous process. In this
case ?G for the cell would be a positive value,
and the cell potential would be negative. The
minimum amount of work done on a system to cause
a non-spontaneous cell reaction to occur is
wmin n?E where w Joules or watt(W) or 1 J/s
The unit employed by electric utilities is
kilowatt-hour (kWh 3.6 x 106J)
39
Electrochemistry Cell EMF Electrolysis
Electrical Work
Calculate the minimum number of kilowatt-hours of
electricity required to produce 1000 kg of
aluminum by electrolysis of Al3 if the required
emf is 4.50 V.
wmin n?E
1 kWh
1000 g Al
1 mol Al
96,500 C
3 ?
1 J
4.5 V
1000 kg Al
1 kg Al
27g Al
1 mol Al
1 ?
3.6 x 106J
1 C-V
since (1V 1J/C), then
w 1.34 x 104 kWh
40
Electrochemistry Cell EMF Electrolysis
Electrical Work
Calculate the minimum number of kilowatt-hours of
electricity required to produce 1.00 kg of M from
electrolysis of MgCl2 is the applied emf is 5.00 v
41
Electrochemistry Cell EMF Corrosion
Corrosion reactions are redox reactions in which
a metal is attacked by some substance in its
environment and converted to an unwanted compound
All metals except gold and platinum are
thermodynamically capable of undergoing
oxidation in air at room temperature
42
Electrochemistry Cell EMF Corrosion
The corrosion of Iron
The rusting of iron is known to require oxygen
iron does not rust in water unless O2 is present.

Ered 1.23 V
Eox 0.44 V
Note that as the pH increases, the reduction of
O2 becomes less favorable
43
Electrochemistry Cell EMF Corrosion
The corrosion of IronProtecting the surface with
tin
It works OK if the tin surface remains intact
Fe(s) ? Fe2(aq) 2e- Eox 0.44 V
Sn(s) ? Sn2(aq) 2e- Eox 0.14 V
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Electrochemistry Cell EMF Corrosion
The corrosion of IronCathodic Protection
Fe(s) ? Fe2(aq) 2e- Eox 0.44 V
Zn(s) ? Zn2(aq) 2e- Eox 0.76 V
Galvanization
45
Electrochemistry Cell EMF Corrosion
The corrosion of IronCathodic Protection