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Electrochemical Reactions
Chapter 20
Electrochemistry
In electrochemical reactions, electrons
are transferred from one species to
another.
Oxidation Numbers
What are the oxidation numbers for…
Review Section 4.4
• An atom in its elemental form = 0
• monatomic ions = charge
• oxygen = -2
 major exception! Peroxide O22-, oxidation # = -1
• hydrogen = +1 with a nonmetal
• hydrogen = -1 with a metal
• oxidation #s in a neutral compound = 0
• P2O5
• NaH
• Cr2O72-
• P+5, O-2
• Na+, H• Cr-6, O-2
• charge #+ or #• oxidation number +# or -#
Redox Reaction
Oxidation and Reduction
• oxidation – losing electrons
• reduction – gaining electrons
• OIL RIG or LEO the lion says GER
• oxidation and reduction occur simultaneously
• # of electrons gained MUST equal # of electrons lost
• Zinc is oxidized – it is the reducing agent
• H+ is reduced – it is the oxidizing agent
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Half-Reaction Method
Balancing Redox Reactions
Half-Reaction Method
(Acidic Solution)
1. Assign oxidation numbers to
determine what is oxidized and what is
reduced.
2. Separate the oxidation and reduction
half-reactions.
Half-Reaction Method
(Acidic Solution)
5. Add the half-reactions, subtracting
things that appear on both sides.
6. Make sure the equation is balanced
according to mass.
7. Make sure the equation is balanced
according to charge.
Redox Rxns in Basic Solution
• Balance the reaction like it is in an acidic
soltuion
• Add OH− to each side to “neutralize” the
H+ in the equation and create water
• Cancel out any excess water that shows
up on both sides of the equation
(Acidic Solution)
3. Balance each half-reaction.
a.
b.
c.
d.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.
Half-Reaction Method
(Acidic Solution)
MnO4−(aq) + C2O42−(aq)
Mn2+(aq) + CO2(aq)
16 H+ + 2 MnO4− + 5 C2O42−
2 Mn2+ + 8 H2O + 10 CO2
Balancing Redox (Basic Solution)
CN- + MnO4-  CNO- + MnO2
3 CN- + 2 MnO4- + H2O 
3 CNO- + 2 MnO2 + 2 OH-
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Voltaic (galvanic) Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.
Voltaic Cells
• Electrons transfer
through an external
pathway
• Energy created from
the flow of electrons
is used to do work
Voltaic Cells
• reaction would stop if
solutions are not
neutral
• salt bridge – allows for
the migration of
cations and anions so
that the solutions
remain neutral
• half cell – each
compartment where
a half reaction occurs
• anode – where
oxidation occurs
• cathode – where
reduction occurs
 Cations move toward
the cathode.
 Anions move toward
the anode.
Voltaic Cells
• In the cell, then,
electrons leave the
anode and flow
through the wire to
the cathode.
• As the electrons
leave the anode, the
cations formed
dissolve into the
solution in the
anode compartment.
Voltaic Cells
• As the electrons
reach the cathode,
cations in the
cathode are
attracted to the now
negative cathode.
• The electrons are
taken by the cation,
and the neutral
metal is deposited
on the cathode.
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Electromotive Force (emf)
• Electrons flow from a
high potential energy
to a lower potential
energy
Standard Reduction Potentials
Electromotive Force (emf)
• electromotive force (emf ) - potential
difference between the anode and
cathode (height of waterfall)
• also called the cell potential (Ecell)
• measured in Volts (1 V = 1 J/C)
Standard Hydrogen Electrode
• Their values are referenced to a standard
hydrogen electrode (SHE).
• By definition, the reduction potential for
hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−
H2 (g, 1 atm)
Standard Cell Potentials
Cell Potentials
• oxidation of Zn,
Ecell = Ered (cathode) + Eox (anode)
Ered = −0.76 V
Ered (cathode) = – Eox (anode)
Eox = +0.76 V
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property.
• reduction of Cu,
Ered = +0.34 V
Ecell = Ered (cathode) + Eox (anode)
= +0.34 V + 0.76 V = +1.10 V
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Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.
Oxidizing and Reducing Agents
The greater the
difference between
the reduction
potentials for the
half cells, the
greater the voltage
of the cell.
Calculate E
Free Energy
G = −nFE
2 Al (s) + 3 I2 (s)  2 Al3+ (aq) + 6 I- (aq)
2.20 V
Is the reaction reactant or product
favored?
Product (+E)
Is the following reaction
spontaneous?
I2 (s) + 5 Cu2+ (aq) + 6 H2O 
2 IO3- + 5 Cu + 12 H+
n = of moles of electrons transferred
F (Faraday)
1 F = 96,485 C/mol = 96,485 J/V-mol
E = emf
+E (- G) indicates a spontaneous reaction
-E (+ G) indicates a nonspontaneous reaction
Nernst Equation
At non-standard conditions
E=E −
RT
nF
ln Q
No, G = 828 kJ
At room temperature (298 K),
0.0592
log Q
E=E −
n
Practice Exercise 20.11
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At Equilibrium…
Fe (s) + Cd2+ (aq)  Fe2+ (aq) + Cd (s) Ecell = + 0.04 V
• What is the equilibrium constant at room
temperature?
• K = 20
• Q=K
• E=0
• The Nernst Equation becomes
log K =
nE
0.0592
Electrolysis
• What are the concentrations of the ions
at equilibrium if the initial concentration of
each was 1.0 M?
• [Fe2+] = 1.9 M
[Cd2+] = 0.10 M
Electrolytic Cell
• lysis – to break
• anode – oxidation
• cathode – reduction
• applying a potential to force a
nonspontaneous reaction
• charges are opposite
of a voltaic cell
Watch out for water!
• Molten salts are easily electrolyzed…
• For aqueous solutions, you must be
sure that the water is not more easily
reduced or oxidized than the salt
• Nicely…water usually requires an
overvoltage
• What is the minimum emf required for the
electrolysis of a 1.0 M CuCl2 solution?
• E = -1.02 V
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G = maximum amount of useful work
• G = -nFE
• W max = -nFE on a galvanic cell
• W min = -nFE on an electrolytic cell
• Units for work are Joules
• 1 kWh = 3.6 x 106 J
• Calculate the number of kilowatt-hours
of electricity to produce 1.00 kg of Mg
from the electrolysis of molten MgCl2 if
the applied emf is 5.00 V. (Assume the
process is 100% efficient.)
• 11.0 kWh
Current
•
•
•
•
Q = It
Q = charge (coulombs)
I = current (amperes)
t = time (seconds)
• What is the mass of Mg formed upon
the passage of current of 60.0 A for
4.00 x 103 s?
• 30.2 g Mg
• 1 F (faraday) = 96500 C = charge on 1
mole of electrons
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