Transcript Chapter 20 Electrochemistry

Lecture Presentation
Chapter 20
Electrochemistry
Dr. Subhash Goel
South GA State College
Douglas, GA
© 2012 Pearson Education, Inc.
Oxidation Numbers
• An oxidation occurs when an atom or ion loses
electrons.
• A reduction occurs when an atom or ion gains
electrons.
• One cannot occur without the other.
• In electrochemical reactions, electrons are
transferred from one species to another.
• In order to keep track of what loses electrons
and what gains them, we assign oxidation
numbers.
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Oxidation and Reduction
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron, and they
combine to form H2.
Electrochemistry
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Oxidation and Reduction
• What is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons
from it, and hence it is oxidizing
agent.
• What is oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons,
and hence it is reducing agent.
Electrochemistry
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Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents
The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity:
Cd(s) + NiO2(s) + 2 H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)
Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and
which is the reducing agent.
Solution
Solve
The oxidation state of Cd increases from 0 to +2, and that of Ni decreases
from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the
reducing agent. The oxidation state of Ni decreases as NiO2 is converted into
Ni(OH)2. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent.
Electrochemistry
Oxidation Numbers
To determine if an oxidation–reduction
reaction has occurred, we assign an
oxidation number to each element in a
neutral compound or charged entity.
Electrochemistry
Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.
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Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Oxygen has an oxidation number of 2,
except in the peroxide ion, which has an
oxidation number of 1.
– Hydrogen is 1 when bonded to a metal,
and +1 when bonded to a nonmetal.
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Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Fluorine always has an oxidation number
of 1.
– The other halogens have an oxidation
number of 1 when they are negative.
They can have positive oxidation
numbers, however; most notably in
oxyanions.
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Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
5. The sum of the oxidation numbers in
a polyatomic ion is the charge on
the ion.
Electrochemistry
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Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the
equation of an oxidation-reduction
reaction is via the half-reaction method.
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Balancing Oxidation-Reduction
Equations
This method involves treating (on paper
only) the oxidation and reduction as two
separate processes, balancing these
half-reactions, and then combining them
to attain the balanced equation for the
overall reaction.
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The Half-Reaction Method
1. Assign oxidation numbers to
determine what is oxidized and what
is reduced.
2. Write the oxidation and reduction
half-reactions.
Electrochemistry
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The Half-Reaction Method
3. Balance each halfreaction.
a. Balance elements other
than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding
electrons.
Electrochemistry
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The Half-Reaction Method
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.
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.
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The Half-Reaction Method
Consider the reaction between MnO4 and C2O42:
MnO4(aq) + C2O42(aq)  Mn2+(aq) + CO2(aq)
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The Half-Reaction Method
First, we assign oxidation numbers:
+7
+3
+2
+4
MnO4 + C2O42  Mn2+ + CO2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
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Oxidation Half-Reaction
C2O42  CO2
To balance the carbon, we add a
coefficient of 2:
C2O42  2CO2
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Oxidation Half-Reaction
C2O42  2CO2
The oxygen is now balanced as well.
To balance the charge, we must add
2 electrons to the right side:
C2O42  2CO2 + 2e
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Reduction Half-Reaction
MnO4  Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to
the right side:
MnO4  Mn2+ + 4H2O
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Reduction Half-Reaction
MnO4  Mn2+ + 4H2O
To balance the hydrogen, we add
8H+ to the left side:
8H+ + MnO4  Mn2+ + 4H2O
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Reduction Half-Reaction
8H+ + MnO4  Mn2+ + 4H2O
To balance the charge, we add 5e to
the left side:
5e + 8H+ + MnO4  Mn2+ + 4H2O
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Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42  2CO2 + 2e
5e + 8H+ + MnO4  Mn2+ + 4H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2:
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Combining the Half-Reactions
5C2O42  10CO2 + 10e
10e + 16H+ + 2MnO4  2Mn2+ + 8H2O
When we add these together, we get:
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
Electrochemistry
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Combining the Half-Reactions
10e + 16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2 +10e
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16H+ + 2MnO4 + 5C2O42 
2Mn2+ + 8H2O + 10CO2
Electrochemistry
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Sample Exercise 20.2 Identifying Oxidizing and Reducing Agents
Complete and balance this equation by the method of half-reactions:
Cr2O72(aq) + Cl(aq)  Cr3+(aq) + Cl2(g)
(acidic solution)
Solution
Analyze We are given an incomplete, unbalanced (skeleton) equation for a redox reaction occurring in acidic
solution and asked to complete and balance it.
Plan We use the half-reaction procedure we just learned.
Solve
Step 1: We divide the equation into two halfCr2O72(aq)  Cr3+(aq)
reactions:
Cl(aq)  Cl2(g)
Step 2: We balance each half-reaction. In the first
half-reaction the presence of one Cr2O72 among
the reactants requires two Cr3+ among the
products. The seven oxygen atoms in Cr2O72 are
balanced by adding seven H2O to the products.
The 14 hydrogen atoms in 7 H2O are then
balanced by adding 14 H+ to the reactants:
We then balance the charge by adding electrons to
the left side of the equation so that the total charge
is the same on the two sides:
14 H+(aq) + Cr2O72(aq)  2 Cr3+(aq) + 7 H2O(l)
6 e + 14 H+(aq) + Cr2O72(aq)  2 Cr3+(aq) + 7
H2O(l)
Electrochemistry
Balancing in Basic Solution
• If a reaction occurs in a basic solution,
one can balance it as if it occurred
in acid.
• Once the equation is balanced, add OH
to each side to “neutralize” the H+ in the
equation and create water in its place.
• If this produces water on both sides,
you might have to subtract water from
each side.
Electrochemistry
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Sample Exercise 20.3 Balancing Redox Equations in Basic Solution
Complete and balance this equation for a redox reaction that takes place in basic solution:
CN(aq) + MnO4(aq)  CNO(aq) + MnO2(s) (basic solution)
Solution
Analyze We are given an incomplete equation for a basic redox reaction and asked to balance it.
Plan We go through the first steps of our procedure as if the reaction were occurring in acidic solution. We
then add the appropriate number of OH ions to each side of the equation, combining H+ and OH to form
H2O.We complete the process by simplifying the equation.
Solve
Step 1: We write the
incomplete, unbalanced
half-reactions:
Step 2: We balance each
half-reaction as if it took
place in acidic solution:
CN(aq)  CNO(aq)
MnO4(aq)  MnO2(s)
CN(aq) + H2O(l)  CNO(aq) + 2 H+(aq) + 2 e
3 e + 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
Electrochemistry
Voltaic Cells
In spontaneous oxidation-reduction (redox)
reactions, electrons are transferred and energy is
released.
• We can use that energy to do work
if we make the electrons flow through an external
device.
• We call such a setup a voltaic cell.
Electrochemistry
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Voltaic Cells
• A voltaic cell consist of two half cells that are
electrically connected. Each half cell is the portion of
an electrochemical cell in which half-reaction takes
place.
• A simple half cell can be made from a metal strip
that dips into a solution of its metal ions.
• Example zinc in zinc salt solution (zinc electrode) or
copper in copper salt solution (copper electrode).
• The oxidation occurs at the anode.
• The reduction occurs at the cathode.
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Electrochemistry
Voltaic Cells
Once even one electron flows from the anode to
the cathode, the charges in each beaker would not
be balanced and the flow of electrons would stop.
• Therefore, we use a salt bridge, usually a Ushaped tube that contains a salt solution, to keep
the charges balanced.
– Cations move toward the cathode.
– Anions move toward the anode.
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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.
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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.
Electrochemistry
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Sample Exercise 20.4 Describing a Voltaic Cell
The oxidation-reduction reaction
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured
into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution
(such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a
voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current.
Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the
direction of ion migration, and the signs of the electrodes.
Solve In one half-reaction, Cr2O72 (aq) is converted into Cr3+ (aq). Starting with these ions and then completing
and balancing the half-reaction, we have
Cr2O72(aq) + 14 H+(aq) + 6 e  2 Cr3+(aq) + 7 H2O(l)
In the other half-reaction, I (aq) is converted to I2(s):
6 I(aq)  3 I2(s) + 6 e
Now we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first half-reaction is the
reduction process (electrons on the reactant side of the equation). By definition, the reduction process occurs at the
cathode. The second half-reaction is the oxidation process (electrons on the product side
of the equation), which occurs at the anode.
The I ions are the source of electrons, and the Cr2O72 ions accept the electrons. Hence, the electrons flow
through the external circuit from the electrode immersed in the KI solution (the anode) to the electrode immersed
in the Cr2O72H2SO4 solution (the cathode). The electrodes themselves do not react in any way; they merely
provide a means of transferring electrons from or to the solutions. The cations move through the solutions toward
the cathode, and the anions move toward the anode. The anode (from which the electrons move) is the negative
electrode, and the cathode (toward which the electrons move) is the
Electrochemistry
positive electrode.
Electromotive Force (emf)
• Water only
spontaneously flows
one way in a
waterfall.
• Likewise, electrons
only spontaneously
flow one way in a
redox reaction—from
higher to lower
potential energy.
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Electromotive Force (emf)
• The potential difference between the
anode and cathode in a cell is called the
electromotive force (emf).
• It is also called the cell potential and is
designated Ecell.
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Cell Potential
Cell potential is measured in volts (V).
J
1V=1
C
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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) + 2e  H2(g, 1 atm)
Electrochemistry
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Standard Cell Potentials
The cell potential at standard conditions
can be found through this equation:
 = Ered
 (cathode)  Ered
 (anode)
Ecell
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property
(do not depend on mass).
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Electrochemistry
Standard Reduction Potentials
Reduction
potentials for many
electrodes have
been measured
and tabulated.
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Cell Potentials
• For the oxidation in this cell,
 = 0.76 V
Ered
• For the reduction,
 = +0.34 V
Ered
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Cell Potentials
 = Ered
 (cathode)  Ered
 (anode)
Ecell
= +0.34 V  (0.76 V)
= +1.10 V
Electrochemistry
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Sample Exercise 20.5 Calculating
from
For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have
Given that the standard reduction potential of Zn2+ to Zn(s) is 0.76 V,
calculate the
for the reduction of Cu2+ to Cu:
Cu2+(aq, 1 M) + 2 e  Cu(s)
Solution
Analyze We are given
and
for Zn2+ and asked to calculate
for Cu2+.
Plan In the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given
for Zn 2+ is
(anode). Because Cu2+ is reduced, it is in the cathode half-cell. Thus, the unknown reduction
potential for Cu2+ is
(cathode). Knowing
and
(anode), we can use Equation 20.8 to
solve for
(cathode).
Solve
Electrochemistry
Sample Exercise 20.6 Calculating
from
Use Table 20.1 to calculate for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
Solution
Analyze We are given the equation for a redox reaction and asked to use data in Table 20.1 to calculate
the standard cell potential for the associated voltaic cell.
Plan Our first step is to identify the half-reactions that occur at the cathode and anode, which we did in
Sample Exercise 20.4. Then we use Table 20.1 and Equation 20.8 to calculate the standard cell
potential.
Solve The half-reactions are
Cathode: Cr2O72(aq) + 14 H+(aq) + 6 e  2 Cr3+(aq) + 7 H2O(l)
Anode:
6 I(aq)  3 I2(s) + 6 e
According to Table 20.1, the standard reduction potential for the reduction of Cr2O72 to Cr3+ is +1.33 V
and the standard reduction potential for the reduction of I2 to I (the reverse of the oxidation halfreaction) is +0.54 V.We use these values in Equation 20.8:
Electrochemistry
Sample Exercise 20.7 Determining Half-Reactions at Electrodes and
Calculating Cell Potentials
A voltaic cell is based on the two standard half-reactions
Cd2+(aq) + 2 e  Cd(s)
Sn2+(aq) + 2 e  Sn(s)
Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at the anode
and (b) the standard cell potential.
Solution
Analyze We have to look up
for two half-reactions. We then use these values first to determine the
cathode and the anode and then to calculate the standard cell potential,
.
Plan The cathode will have the reduction with the more positive
value, and the anode will have the less
positive
. To write the half-reaction at the anode, we reverse the half-reaction written for the reduction, so
that the half-reaction is written as an oxidation.
Solve
(a) According to Appendix E,
(Cd2+/Cd) = 0.403 V and
(Sn2+/Sn) = 0.136 V. The standard
reduction potential for Sn2+ is more positive (less negative) than that for Cd2+. Hence, the reduction of
Sn2+ is the reaction that occurs at the cathode:
The anode reaction, therefore, is the loss of electrons by Cd:
Electrochemistry
Sample Exercise 20.7 Determining Half-Reactions at Electrodes and
Calculating Cell Potentials
Continued
(b) The cell potential is given by the difference in the standard reduction potentials at the cathode and anode
(Equation 20.8):
Notice that it is unimportant that the
values of both half-reactions are negative; the negative values merely
indicate how these reductions compare to the reference reaction, the reduction of H + (aq).
Electrochemistry
Oxidizing and Reducing Agents
• The strongest oxidizers
have the most positive
reduction potentials.
• The strongest reducers
have the most negative
reduction potentials.
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Oxidizing and Reducing Agents
The greater the
difference between
the two, the greater
the voltage of the cell.
Electrochemistry
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Sample Exercise 20.8 Determining Relative Strengths of Oxidizing
Agents
Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents:
NO3 (aq), Ag+ (aq), Cr2O72 (aq).
Solution
Plan The more readily an ion is reduced (the more positive its
value), the stronger it is as an oxidizing agent.
Solve From Table 20.1, we have
Because the standard reduction potential of Cr2O72 is the most positive, Cr2O72 is
the strongest oxidizing agent of the three. The rank order is Ag+ < NO3 < Cr2O72.
Electrochemistry
Free Energy
• Eo = Eo reduction process) - Eo(oxidation process)
• This equation can be used to determine
spontaneity of the reaction.
+ ve Eo indicates spontaneous reaction
- ve Eo indicates a non-spontaneous rx.
(
Electrochemistry
Sample Exercise 20.9 Determining Spontaneity
Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions.
(a) Cu(s) + 2 H+(aq)  Cu2+(aq) + H2(g)
(b) Cl2(g) + 2 I(aq)  2 Cl(aq) + I2(s)
Solution
Electrochemistry
Sample Exercise 20.9 Determining Spontaneity
Continued
Solve
(a) For oxidation of Cu to Cu2+ and
reduction of H+ to H2, the half-reactions
and standard reduction potentials are
Notice that for the oxidation, we use the
standard reduction potential from Table
20.1 for the reduction of Cu2+ to Cu. We
now calculate E by using Equation
20.10:
Because E is negative, the reaction is
not spontaneous in the direction written.
Copper metal does not react with acids
in this fashion. The reverse reaction,
however, is spontaneous and has a
positive E value:
Thus, Cu2+ can be reduced by H2.
Cu2+(aq) + H2(g)  Cu(s) + 2 H+(aq)
E = +0.34 V
Electrochemistry
Sample Exercise 20.9 Determining Spontaneity
Continued
(b) We follow a procedure analogous to
that in (a):
In this case
E = (1.36 V)  (0.54 V) = +0.82 V
Because the value of E is positive, this reaction is spontaneous and could be used to build a voltaic cell.
Electrochemistry
Free Energy
G for a redox reaction can be found by
using the equation
G = nFE
where n is the number of moles of
electrons transferred, E is cell energy
and F is a constant, the Faraday:
1 F = 96,485 C/mol = 96,485 J/V-mol
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Free Energy
Under standard conditions,
Electrochemistry
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Sample Exercise 20.10 Using Standard Reduction Potentials to
Calculate G and K
(a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change,
G, and the equilibrium constant, K, at 298 K for the reaction
(b) Suppose the reaction in part (a) is written
What are the values of E, G, and K when the reaction is
written in this way?
Electrochemistry
Sample Exercise 20.10 Using Standard Reduction Potentials to
Calculate G and K
Continued
Solve
(a) We first calculate E by breaking
the equation into two half-reactions
and obtaining
values from Table
20.1 (or Appendix E):
Even though the second half-reaction
has 4 Ag, we use the
value
directly from Table 20.1 because emf
is an intensive property.
Using Equation 20.10, we have
E = (1.23 V)  (0.80 V) = 0.43 V
The half-reactions show the transfer
of four electrons. Thus, for this
reaction n = 4. We now use Equation
20.12 to calculate G:
The positive value of E leads to a negative value of G. The per mol part of the unit relates to the
balanced equation, 4 Ag(s) + O2(g) + 4 H+(aq)  4 Ag+(aq) + 2 H2O(l). Thus, 170 kJ is associated with 4
mol Ag, 1 mol O2 and 4 mol H+, and so forth, corresponding to the coefficients in the balanced equation.
Electrochemistry
Sample Exercise 20.10 Using Standard Reduction Potentials to
Calculate G and K
Continued
Now we need to calculate the
equilibrium constant, K, using G =
RT. Because G is a large negative
number, which means the reaction is
thermodynamically very favorable, we
expect K to be large.
K is indeed very large! This means that we expect silver metal to oxidize in acidic aqueous environments,
in air, to Ag+.
Notice that the emf calculated for the reaction was E = 0.43 V, which is easy to measure. Directly
measuring such a large equilibrium constant by measuring reactant and product concentrations at
equilibrium, on the other hand, would be very difficult.
(b) The overall equation is the same as
that in part (a), multiplied by . The
half-reactions are
The values of
are the same as they
were in part (a); they are not changed
by multiplying the half-reactions by .
Thus, E has the same value as in part
a):
E = +0.43 V
Electrochemistry
Sample Exercise 20.10 Using Standard Reduction Potentials to
Calculate G and K
Continued
Notice, though, that the value of n has
changed to n = 2, which is one-half the
value in part (a). Thus, G is half as
G = (2)(96,485 J/V-mol)(+0.43 V) = 83 kJ/mol
large as in part (a):
The value of G is half that in part (a)
because the coefficients in the
chemical equation are half those in (a).
Now we can calculate K as before:
Electrochemistry
Nernst Equation
• Remember that
G = G + RT ln Q
• This means
nFE = nFE + RT ln Q
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Nernst Equation
Dividing both sides by nF, we get the
Nernst equation:
RT
ln Q
E = E 
nF
or, using base-10 logarithms,
2.303RT
log Q
E = E 
nF
Electrochemistry
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Nernst Equation
At room temperature (298 K),
2.303RT
= 0.0592 V
F
Thus, the equation becomes
0.0592
log Q
E = E 
n
We can use this equation to find emf E produced by a cell
under nonstandard conditions or to determine
concentration of reactant or product by measuring E for the
Electrochemistry
cell.
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Nernst Equation
• Consider following example:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this case n = 2 and standard emf is
+1.10 V. Hence at 298 K
E= 1.10 V – 0.0592V/2.log [Zn2+/C2+]
For [Cu2+] = 5.0 M and [Zn2+] = 0.05M
E = 1.10V – 0.0592V/2.log[0.050/5.0]
E = 1.16 V
Electrochemistry
Concentration Cells
• Notice that the Nernst equation implies that a cell
could be created that has the same substance at
both electrodes.
 would be 0, but Q would not.
• For such a cell, Ecell
• Therefore, as long as the concentrations
Electrochemistry
are different, E will not be 0.
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Sample Exercise 20.11 Cell Potential under Nonstandard Conditions
Calculate the emf at 298 K generated by a voltaic cell in which the reaction is
Cr2O72(aq) + 14 H+(aq) + 6 I(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
when [Cr2O72] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0  105 M.
Solution
Solve We calculate E for the cell from standard reduction potentials (Table 20.1 or Appendix E). The
standard emf for this reaction was calculated in Sample Exercise 20.6: E = 0.79 V. As that exercise shows,
six electrons are transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is
Electrochemistry
Sample Exercise 20.11 Cell Potential under Nonstandard Conditions
Continued
Using Equation 20.18, we have
Electrochemistry
Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell
If the potential of a Zn-H+ cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn2+] = 1.0 M and
PH2 = 1.0 atm, what is the H+ concentration?
Solution
Plan We write the equation for the cell reaction and use standard reduction potentials to calculate E for
the reaction. After determining the value of n from our reaction equation, we solve the Nernst equation,
Equation 20.18, for Q. Finally, we use the equation for the cell reaction to write an expression for Q that
contains [H+] to determine [H+].
Solve The cell reaction is
The standard emf is
Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g)
Because each Zn atom loses
two electrons,
n=2
Electrochemistry
Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell
Continued
Using Equation 20.18, we can
solve for Q:
Q has the form of the
equilibrium constant for the
reaction:
Solving for [H+], we have
Comment A voltaic cell whose cell reaction involves can be used to measure or pH. A pH meter is a
specially designed voltaic cell with a voltmeter calibrated to read pH directly. (Section 16.4)
Electrochemistry
Sample Exercise 20.13 Determining pH Using a Concentration Cell
A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown
concentration of H+(aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 atm, [H+] = 1.00 M). At 298 K
the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the
external circuit to electrode 2. Calculate for the solution at electrode 1.What is the pH of the solution?
Plan We can use the Nernst equation to determine Q and then use Q to calculate the unknown concentration.
Because this is a concentration cell,
.
Solve Using the Nernst equation, we
have
Because electrons flow from electrode 1 to
electrode 2, electrode 1 is the anode of
the cell and electrode 2 is the cathode.
The electrode reactions are therefore as
follows, with the concentration of H+ (aq)
in electrode 1 represented with the
unknown x:
Electrochemistry
Sample Exercise 20.13 Determining pH Using a Concentration Cell
Continued
Thus,
At electrode 1, therefore, the pH of the
solution is
pH = log[H+] = log(2.7  104) = 3.57
Comment The concentration of H+ at electrode 1 is lower than that in electrode 2, which is why
electrode 1 is the anode of the cell: The oxidation of H2 to H+(aq) increases [H+] at electrode 1.
Electrochemistry
Applications of
Oxidation-Reduction
Reactions
Electrochemistry
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Batteries
Electrochemistry
© 2012 Pearson Education, Inc.
Lead Storage Battery
• six cells in series
• electrolyte = 30% H2SO4
• anode = Pb
Pb(s) + SO42−(aq)  PbSO4(s) + 2 e−
• cathode = Pb coated with PbO2
• PbO2 is reduced
PbO2(s) + 4 H+(aq) + SO42−(aq) + 2 e−
 PbSO4(s) + 2 H2O(l)
• cell voltage = 2.09 V
• rechargeable, heavy
74
Electrochemistry
Alkaline Dry Cell
• same basic cell as acidic dry cell,
except electrolyte is alkaline KOH
paste
• anode = Zn (or Mg)
Zn(s)  Zn2+(aq) + 2 e−
• cathode = brass rod
• MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e−
 2 NH4OH(aq) + 2 Mn(O)OH(s)
• cell voltage = 1.54 V
• longer shelf life than acidic dry cells
and rechargeable; little corrosion of
75
zinc
Electrochemistry
NiCad Battery
• electrolyte is concentrated KOH solution
• anode = Cd
Cd(s) + 2 OH−(aq)  Cd(OH)2(s) + 2 e− E0 = 0.81 V
• cathode = Ni coated with NiO2
• NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e−  Ni(OH)2(s) + 2OH− E0 = 0.49 V
• cell voltage = 1.30 V
• rechargeable, long life, light—however,
recharging incorrectly can lead to battery
breakdown
Electrochemistry
76
NiMH Battery
• electrolyte is concentrated KOH solution
• anode = metal alloy with dissolved hydrogen
– oxidation of H from H0 to H+
M∙H(s) + OH−(aq)  M(s) + H2O(l) + e− E° = 0.89 V
• cathode = Ni coated with NiO2
• NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e−  Ni(OH)2(s) + 2 OH− E0 = 0.49 V
• cell voltage = 1.30 V
• rechargeable, long life, light, more environmentally
friendly than NiCad, greater energy density than NiCad
Electrochemistry
77
Lithium Ion Battery
• electrolyte is concentrated KOH
solution
• anode = graphite impregnated with
Li ions
• cathode = Li ─ transition metal
oxide
– reduction of transition metal
• work on Li ion migration from
anode to cathode causing a
corresponding migration of
electrons from anode to cathode
• rechargeable, long life, very light,
more environmentally friendly,
greater energy density
78
Electrochemistry
Hydrogen Fuel Cells
Electrochemistry
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Electrolytic Cell
• uses electrical energy to overcome the energy
barrier and cause a nonspontaneous reaction
– must be DC source
• the + terminal of the battery = anode
• the − terminal of the battery = cathode
• cations attracted to the cathode, anions to the
anode
• Cations pick up electrons from the cathode and
are reduced; anions release electrons to the
anode and are oxidized.
• Some electrolysis reactions require more voltage
Electrochemistry
than Etot, called the overvoltage.
80
Electrochemistry
81
Electroplating
In electroplating, the
work piece is the
cathode.
Cations are reduced at
the cathode and plate
to the surface of the
work piece.
The anode is made of
the plate metal. The
anode oxidizes and
replaces the metal
cations
in the solution.
82
Electrochemistry
Faraday’s Law
• The amount of metal deposited during
electrolysis is directly proportional to the
charge on the cation, the current, and the
length of time the cell runs.
– charge that flows through the cell = current x time
Electrochemistry
83
Sample Exercise 20.14 Relating Electrical Charge and Quantity
of Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl 3 if the
electrical current is 10.0 A.
Solution
Analyze We are told that AlCl3 is electrolyzed to form Al and asked to calculate the number of
grams of Al produced in 1.00 h with 10.0 A.
Plan Figure 20.27 provides a road map for this problem. First, the product of the amperage and the
time in seconds gives the number of coulombs of electrical charge being used (Equation 20.21).
Second, the coulombs can be converted with Faraday’s constant (F = 96,485 C/mol electrons) to
tell us the number of moles of electrons being supplied. Third, reduction of 1 mol of to Al requires
3 mol of electrons. Hence, we can use the number of moles of electrons to calculate the number of
moles of Al metal it produces. Finally, we convert moles of Al into grams.
Solve First, we calculate the
coulombs of electrical charge
passed into the electrolytic
cell:
Second, we calculate the
number of moles of electrons
that pass into the cell:
Electrochemistry
Sample Exercise 20.14 Relating Electrical Charge and Quantity
of Electrolysis
Continued
Third, we relate number of
moles of electrons to number of
moles of aluminum formed,
using the half-reaction for the
reduction of Al3+:
Al3+ + 3 e  Al
Thus, 3 mol of electrons are
required to form 1 mol of Al:
Finally, we convert moles to
grams:
Because each step involves
multiplication by a new factor,
we can combine all the steps:
Electrochemistry
Electrolysis
• Electrolysis is the process of
using electricity to break
apart a compound.
• Electrolysis is done in an
electrolytic cell.
• Electrolytic cells can be used
to separate elements from
their compounds.
– generate H2 from water for fuel
cells
– recover metals from their ores
86
Electrochemistry
Electrolysis of Pure Compounds
• must be in molten (liquid) state
• electrodes normally graphite
• cations reduced at cathode to metal
element
• anions oxidized at anode to nonmetal
element
Electrochemistry
87
Corrosion
• Corrosion is the spontaneous oxidation
of a metal by chemicals in the
environment.
• Since many materials we use are active
metals, corrosion can be a very big
problem.
Electrochemistry
88
Rusting
• Rust is hydrated iron(III) oxide.
• Moisture must be present.
– Water is a reactant.
– required for flow between cathode and anode
• Electrolytes promote rusting.
– enhance current flow
• Acids promote rusting.
– lower pH = lower E°red
Electrochemistry
89
Corrosion and…
Electrochemistry
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