UNIT 8
How do we
harness chemical
energy?
The core concepts, ideas, and methodologies introduced
in previous Units of this course can be applied to design,
implement, and evaluate strategies to harness energy from
chemical reactions. The proper selection of materials and
conditions critically depends on the analysis of diverse factors that may influence the directionality, extent, and rate
of different chemical processes.
Modern devices used to harness chemical energy often rely
on reactions that result in the transfer of electrons from one
system to another. These batteries transform chemical energy into electrical energy and are used to power watches, cell
phones, tablet computers, and electric cars in modern times.
The current challenge is to produce devices that can store energy
in large amounts per unit mass and that can deliver that energy in
short periods of time.
The central goal of Unit 8 is to create opportunities for you to apply
the knowledge developed in this course to the analysis of electrochemical
processes and devices used to harness energy.
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436
Chemical Thinking
UNIT 8 MODULES
M1. Controlling Electron Transfer
Analyzing thermodynamic, kinetic,and practical
aspects of electrochemical systems.
437
438
Most of the energy used by modern societies results from the transformation of
chemical energy into thermal, electrical, or light energy. The major proportion of
that energy is generated by combustion of fossil fuels rich in
hydrocarbons that react with oxygen in the atmosphere
and produce carbon dioxide and water in a very exothermic process. Concerns with the role that CO2 has on
global warming and climate change on Earth has sparked
interest in the development of more durable and powerful batteries and fuel cells. These devices transform chemical
energy into electrical energy without generating greenhouse gases.
The design of electrical batteries and fuel cells demands a very good understanding of the thermodynamics and kinetics of the processes that take place inside
these devices. The challenges that scientists and engineers face to improve energy
storage and delivery are not small. Chemical reactions inside batteries and fuel
cells involve substances in different states of matter that should interact effectively
over time and under different conditions. Rechargeable batteries require chemical
reactions that can be reversed without major deterioration of components. Reactants and products should be contained within the device at all times and any heat
produced by internal processes should be quickly dissipated to avoid damaging the
battery and major hazards. In Unit 8 we will explore how chemical thinking can
be used to face some of these challenges.
THE CHALLENGE
Bio-Batteries
The understanding of how plants and animals store and use energy has guided
the development of bio-batteries that mimic biological systems.
•
•
Which biochemical processes for energy storage do you know?
What are some potential advantages and disadvantages of bio-batteries
compared to conventional batteries?
Share and discuss your ideas with one of your classmates.
This module will help you develop the type of chemical thinking that is used
to answer questions similar to those posed in the challenge. In particular, the
central goal of Module 1 is to help you apply your knowledge in the analysis of
electrochemical processes used to harness chemical energy.
By KyloDeel (Own work)
[Public Domain] via Wikimedia Commons
U8: MODULE 1
Controlling
Electron Transfer
Chemical Thinking
U8
How do we harness chemical energy?
Analyzing Electron Distribution
Chemical reactions result from interactions between regions with different electron density in the reacting particles. These interactions often lead to the breaking
of some chemical bonds and to the formation of new ones. In these processes,
valence electrons are redistributed between some atoms and their potential energy
changes. The chemical reactions commonly used to generate energy in modern
societies are exothermic processes in which energy is released due to the movement
of electrons from high to low potential energy states. Thus, identifying
chemical species in which electrons occupy states with different potential
energies can help us select combination of reactants that will react exothermically with each other.
Electrons are negatively charged particles that are attracted to centers
with a partial or net positive electric charge. Electrical forces thus induce
movement of electrons from regions with a partial or net negative charge
to areas with a partial or net positive charge. From an energetic perspective, the potential energy of electrons is higher in areas with a partial or net negative charge and lower in regions with a more positive partial or net charge (Figure
8.1). Transfer of electrons should then spontaneously occur from high potential
energy states to low potential energy states, but these processes often face kinetic
activation barriers that need to be overcome for the transfer to happen.
We can identify areas of partial positive or negative charge in molecules by
analyzing their composition and structure. This allows us to, for example, identify
electrophilic and nucleophilic centers as we did it in Unit 7. This approach, however, has limitations as it does not make explicit how many electrons are transferred
between reactants and how they are redistributed during a chemical reaction.
Electron Transfer and Redistribution
Consider the addition reaction
represented to the right:
•
•
•
•
Figure 8.1 In this reaction,
electrons in the double bond
(higher potential energy)
move towards the chlorine
atom where they occupy
lower potential energy states.
LET’S THINK
+ HBr
Assign partial charges to all
the atoms in reactants and
products.
Describe the major changes in electron density that you observe.
Evaluate whether you can determine how many electrons were transferred in this process
and how they were redistributed.
Highlight advantages and disadvantages to this approach to analyze electron transfer.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Analysis of the partial charges on different atoms in reactants and products
provides qualitative information about changes in electron density as a result of a
chemical reaction. However, chemists have developed other strategies that facilitate the identification of electron-rich and electron-poor sites in molecules and
provide more insights into how electrons are transferred or redistributed in chemical reactions.
439
440
MODULE 1
Controlling Electron Transfer
A useful strategy to analyze electron distribution in molecules and ions is to
calculate the oxidation state (or oxidation number) of each atom present in the
system. The oxidation state indicates the total number of valence electrons that
an atom would gain or lose when forming chemical bonds if we assumed that
bonding electrons belong to the most electronegative atom in each bond. Thus,
the oxidation state represents the charge of the atoms if we assumed that all bonds
are ionic. The oxidation state is often a fictitious quantity but it provides useful
information about electron distribution in chemical species.
To calculate the oxidation state (OS) of an atom in a system we take the difference between the number of valence electrons in the isolated atom and the
number of valence electrons in the bonded atom (assuming ionic bonding):
OS = Number of valence electrons in the isolated atom
– Number of valence electrons in the bonded atom
Let us apply this relationship to calculate the oxidation state of all atoms in a
molecule of ethanol CH3CH2OH. If we assign the different valence electrons to
the most electronegative atom in each bond and split bonding electrons equally
between atoms of the same type, as shown in Figure 8.2, the oxidation state for
each atom in the molecule can be calculated as:
OS(H) = 1 – 0 = +1
OS(O) = 6 – 8 = –2
1
2
OS(C1) = 4 – 7 = –3
OS(C2) = 4 – 5 = –1
Figure 8.2
Assignation
of valence electrons to
different atoms in the
ethanol molecule assuming ionic bonding.
The number of valence electrons in the isolated atoms can be inferred from their
position in the periodic table. The number of valence electrons in the bonded state
is determined by counting the electrons assigned to each atom based on relative
electronegativity values (Figure 8.2). Notice that the carbon atoms in ethanol do
not have the same oxidation states, making explicit the presence of an electronpoor site (C2) in this molecule. However, the oxidation states determined above do
not correspond to actual electric charges on any of the atoms.
LET’S THINK
Determine the oxidation state of each
atom in the species shown to the right:
•
What major patterns do you observe?
•
What is the relationship between the
oxidation states and the net charge of
a molecule or ion?
Clearly justify your ideas.
Oxidation State
Chemical Thinking
U8
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441
The determination of the oxidation states of many molecules and ions reveals
some basic patterns that simplify the calculation of this quantity for any atom in
any molecular species. These simplifying rules are listed below:
a) Elementary Substances: The oxidation state of atoms present in the particles of elementary substances, such as He(g), O2(g), Na(s) or He(g) is always
zero. For example, in the case of the molecule of Cl2 (Figure 8.3):
OS(Cl) = 7 – 7 = 0
Figure 8.3 Assignation of va-
b) Monoatomic Ions: The oxidation state of monoatomic ions present in
ionic compounds or as free ions in solution, such as Na+, Cl–, Mg2+, S2–, is the same
as the net charge of the ion. Thus:
lence electrons to the atoms
in the Cl2 molecule.
OS(Al3+) = +3
c) Reference Atoms: The oxidation state of some atoms is generally a fixed
number in many chemical compounds. These values can be used to infer the oxidation states for other atoms present in the same chemical species:
OS(Alkali Metals: Li, Na, K ..) = +1
OS(Alkaline Earth Metals: Mg, Ca, Sr ..) = +2
OS(H) = +1
OS(O) = –2
OS(F) = –1
d) Net Charge: The algebraic sum of the oxidation states of all the atoms present in a molecule or ion should be equal to the net charge of the chemical species.
If the molecule has no charge, the algebraic sum of the oxidation states should be
equal to zero. If the particle is a polyatomic ion, the algebraic sum of the oxidation states should be equal to the charge of the ion. Consider, for example, the
molecule of ethanol CH3CH2OH in Figure 8.2:
Net Charge = SOS = 6 x OS(H) + OS(C1)+ OS(C2)+ OS(O)
Net Charge = SOS = 6 x (+1) + (–3) + (–1) + (–2) = 0
Basic Rules
LET’S THINK
The following chemical species are commonly used in the synthesis or transformation of alcohols:
H2
H2CrO4
KMnO4
NaOCl
The following species are polyatomic ions present in different electric batteries:
OH–
SO42–
NO3–
NH4+
•
Determine the oxidation state of each atom in these different chemical species.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
442
MODULE 1
Controlling Electron Transfer
Oxidation-Reduction (REDOX) Reactions
When different chemical species interact with each other, electrons may be transferred from one atom to another. In this process, the oxidation state of some atoms may change as some bonds get broken and new bonds are formed. Chemical
reactions that result in changes in the oxidation state of participating atoms are
classified as oxidation-reduction reactions or redox reactions.
LET’S THINK
Electron Transfer
Consider the chemical reaction for the incomplete combustion of methanol (CH4O):
CH4O +
•
•
O2
CO + 2 H2O
Assign oxidation states to each atom in the reactants and products.
Use your results to identify which atoms lost electrons and which atoms gained electrons
during the reaction.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Redox reactions tend to be exothermic reactions due to the movement of electrons
from higher to lower potential energy states, and thus may be used to harness
chemical energy. The following two chemical reactions are typical examples of
these types of processes (oxidation states for each atom are shown):
-4 +1
C
Oxidized
CH4
CH4
Reducing
Agent
+
O2
O
2 O2 Oxidizing
Reduced
Agent
CO2
+
2 H2O
Figure 8.4 Oxidized and reduced atoms, and oxidizing
and reducing agents.
0
C H4 + 2 O 2
0
0
3 H2 + N2
+4 -2
+1 -2
C O2 + 2 H2 O
-3 +1
2 N H3
In the first of these reactions, the combustion of methane (CH4), the oxidation
state of carbon atoms increases from –4 to +4, while the oxidation state of oxygen atoms decreases from 0 to –2. These changes indicate a transfer of electrons
(or electron density) from carbon atoms to oxygen atoms during the combustion
process. In the synthesis of ammonia (NH3) represented by the second reaction,
electron transfer occurs from hydrogen atoms to nitrogen atoms.
During a redox reaction, the atoms that lose electrons or electron density are
said to be oxidized during the process. We can recognize oxidized atoms because
their oxidation state increases (it becomes less negative or more positive). In general, the term “oxidation” refers to the loss of electrons. On the other hand, the atoms
that gain electrons or electron density are said to be reduced. We can recognize
reduced atoms because their oxidation state decreases (it becomes less positive
or more negative). In general, the term “reduction” refers to the gain of electrons.
Consequently, during the combustion of CH4 carbon atoms are oxidized and oxygen atoms are reduced. The two reactants that exchange electrons during a redox
reactions are commonly named the oxidizing agent and the reducing agent. The
oxidizing agent holds the atoms that gain electrons during the process (i.e., the atoms that are reduced), while the reducing agent holds the atoms that lose electrons
as a result of the reaction (i.e., the atoms that are oxidized). In the combustion of
methane, O2 is the oxidizing agent and CH4 is the reducing agent (Figure 8.4).
Chemical Thinking
U8
How do we harness chemical energy?
Redox Reactions 443
LET’S THINK
Consider the following exothermic reactions :
CH4
•
H2 + 2 O2
HCl + NH3
+ 2 H2O
C2H4 + H2
2 H2O
NH4Cl
CO2 + 4 H2
C2H6
In each case, determine whether the process can be classified as a redox reaction. Identify
the oxidized and reduced atoms, and the oxidizing and reducing agents in redox reactions.
Many redox reactions used in electric batteries to generate energy involve metallic and ionic species that react readily when in contact with each other. For
example, when metallic zinc is added to an aqueous solution of a copper(II) salt,
the following exothermic and product-favored reaction takes place:
Zn(s) +
Cu2+(aq)
Zn2+ (aq) +
Cu(s)
Analysis of the oxidation states of reactants and products indicates that zinc atoms
are oxidized and copper ions are reduced in this process. Moreover, the change in
oxidation states reveals that every zinc atom loses two electrons and every copper
ions gains two electrons. It is common to represent the oxidation and reduction
processes as the following separate half reactions:
Zn
Cu2+
+
Zn2+ +
2 e–
2 e–
Cu
OXIDATION
REDUCTION
By Toby Hudson (Own work)
[Sahre Alike 3.0] via Wikimedia Commons
Each half-reaction represents the chemical species involved in electron transfer and
the number of electrons that are exchanged between them. By convention, electrons are represented in the product side in the oxidation half reaction and in the
reactant side in the reduction half reaction. The number of electrons exchanged
is always the same, as the electrons lost by the oxidized atoms are gained by the
reduced atoms. For example, consider the reaction that takes place when metallic
copper is in contact with an aqueous solution of a silver(I) salt (Figure 8.5):
Cu(s) + 2 Ag+(s)
Cu2+(s) + 2 Ag(s)
In this case, the half reactions can be expressed as:
Cu
2 Ag+
Cu
2+
+
2 e–
+
Figure 8.5 Oxidation of
2e
2 Ag
–
OXIDATION
REDUCTION
These half reactions reveal that each copper atoms provides the number of electrons needed to reduce two silver ions. Notice that in this case the copper atoms
are the oxidized species, which suggests that the direction of electron transfer depends on the nature of the chemical species in interaction.
metallic copper in a solution
of silver nitrate.
444
MODULE 1
Controlling Electron Transfer
LET’S THINK Half-Reactions
Consider the following unbalanced redox reactions:
Al(s) + Cu2+(aq)
Zn(s) + H+(aq)
Al(s) + Br2(l)
•
Al3+(aq) + Cu(s)
Zn2+(aq) + H2(g)
Al3+(aq) + Br–(aq)
Write the corresponding half reactions for each process. Identify the atoms that are oxidized
and reduced, and the minimum number of each species involved in electron exchange.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Electrochemical Devices
The electron transfer that takes place during some redox reactions can be used to
generate an electric current. Electrochemical devices are systems designed to control the transformation of chemical energy into electrical energy. These types of
devices often include three basic parts:
I. A chemical system that undergoes “oxidation”: A
An+ + nem+
II. A chemical system that undergoes “reduction”: B + meB
III. A mechanism to allow and control charge transfer.
These three parts must be connected to each other to allow a controlled transfer
of electrons from the system where the oxidation takes place to the system where
the reduction occurs.
LET’S THINK The image to the right depicts the basic components of a galvanic or voltaic cell, a prototypical electrochemical device. Although
the specific chemical substances used in
the cell may vary, as well as the manner in
which different components are linked to
each other, the underlying structure is present in most electrical batteries.
•
•
(–)
Anode
An Electrochemical Cell
Na2SO4 paste
Sodium Ions
Sulfate Ions
Identify the three basic components of
the electrochemical device.
Describe and represent in detail the
processes that are taking place in each
of the main components of the system.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
(+)
Cathode
Chemical Thinking
U8
How do we harness chemical energy?
Charge Transfer
High Electric Potential
++++++++++++
+
––––––––––––
Low Electric Potential
Figure 8.6 Movement of
charge due to a difference in
electric potential.
Figure 8.7 Major components
of an electrochemical cell
LET’S THINK
Consider the electrochemical cell depicted in Figure 8.7:
•
•
Carefully analyze the flow of charge in this electrochemical device. Discuss how electric
charge is transferred through the external circuit and through the salt bridge.
Analyze whether the salt bridge is necessary for the electrochemical cell to work. Think of
other arrangements that could be used to replace the salt bridge.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
–
By Shamsher Singh (Own work)
via UC Davis ChemWIki
In an electrochemical cell, the systems that undergo oxidation and reduction
are physically separated from each other to avoid the direct transfer of electrons
between them. Instead, they are connected via an external circuit that controls the
flow of electrons from the reducing agent to the oxidizing agent. Electrons spontaneously move from high to low potential energy states. The potential energy of
charged particles is determined by the electric potential (V) at the place where they
are located. The potential energy of positive charges decreases as they move from
regions where the electric potential is higher or more positive to regions where the
electric potential is lower or more negative. Conversely, the potential energy of
negatively charged particles, such as electrons, decreases as they move from areas
with a low or negative electric potential to regions with a higher or more positive
electric potential (Figure 8.6).
The two main systems that comprise an electrochemical cell, also called half
cells, have different electric potentials. The half cell that contains the oxidizing
agent has the higher or more positive electric potential and is associated with the
“cathode” of the cell (Figure 8.7). The half cell that contains the reducing agent has
the lower or more negative electric potential and is associated with the “anode.”
The difference in electric potential from the anode to the cathode is the electric
voltage of the cell (DV) and it drives the flow of electrons from one electrode to the
other. The electric voltage DV depends on the chemical nature and concentration
of the species in the cell, and on temperature. DV
DV
decreases as electrons transfer from anode to cathode and eventually becomes zero.
When electrons are allowed to move from the
anode to the cathode through the external circuit
of an electrochemical cell, a charge unbalance is
created. The anode becomes slightly more positive
and the cathode becomes slightly more negative.
This affects the distribution of ions in the aqueous solutions of the electrochemical cell. Negative
ions are repelled by the cathode and attracted by
the anode, while positive ions are pulled towards
the cathode and move away from the anode. This
flow of ions allows the oxidation and reduction processes to continue as the positive ions formed in the anode as a result of the oxidation (Cu2+ in Figure 8.7) are
continuously dispersed throughout the system, and the positive ions involved in
the reduction reaction (Ag+ in Figure 8.7) are pulled towards the reduction site.
445
446
MODULE 1
Controlling Electron Transfer
Electrochemical Cell Potentials
The voltage DV of an electrochemical cell determines the maximum amount of
electrical energy that the device may generate. DV is a measure of the chemical energy transformed into electrical energy per unit of electric charge that flows from
the anode to the cathode. The voltage DV is often expressed in volts (V) which
are a measure of the joules (J) of energy per coulomb of charge transferred. An
electrochemical cell with a DV = 2.5 V, for example, will produce 2.5 J of energy
when 1 C of charge flows through its external circuit.
Given that the magnitude of the electric charge of a single electron is very
small when expressed in coulombs |qe| = 1.602 x 10–19 C, it is common to express
and calculate the energy transformed by an electrochemical cell per mole of electrons transferred. The magnitude of this amount of charge is known as Faraday’s
constant F and it is equal to:
Ep
F = Na|qe| = 6.022 x 1023 x 1.602 x 10–19 = 9.647 x 104 C/mol
n eDG = – nFDV
Anode
Cathode
Figure 8.8 Change in
free energy due to electron transfer.
where Na is Avogadro’s number. If “n” moles of electrons are transferred from the
anode to the cathode in an electrochemical cell with a voltage DV, the amount
of energy involved is then given by the product nFDV. This quantity is equal in
magnitude to the decrease in potential energy of the electrons transferred from the
anode to the cathode (Figure 8.8). This change in potential energy is a measure
of the decrease in the Gibbs free energy DG of the electrochemical cell when “n”
moles of electrons flow from the anode to the cathode:
(8.1)DG = – n F DV
The negative sign in this expression accounts for the decrease in the potential energy of the system. This relationship highlights the importance of determining the
value of DV for any electrochemical cell of interest as this quantity can be used to
predict the maximum amount of electrical energy that device can produce.
LET’S THINK A Reference Cell
Consider the following information:
When an electrochemical device with the two half cells Ag(s)|Ag+(aq) (1 M) and H2(g)|H+(aq)
(1 M) is built, the following processes take place and a DV is established
Anode: H2
2 H+ + 2 e-
Cathode: Ag+ + e-
Ag
DV = 0.80 V
When an electrochemical device with the two half cells Zn(s)|Zn2+(aq) (1 M) and H2(g)|H+(aq)
(1 M) is built, the following processes take place and a DV is established:
Anode: Zn
•
Zn2+ + 2 e-
Cathode: 2 H+ + 2 e-
H2
DV = 0.76 V
What would you expect to happen when an electrochemical cell is built using the half cells
Ag(s)|Ag+(aq) (1 M) and Zn(s)|Zn2+(aq) (1 M). What would DV be? Justify your reasoning.
Chemical Thinking
U8
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447
The two electrodes in an electrochemical cell typically have different electric
potentials. The electric potential of the cathode is higher than the electric potential
of the anode and DV is a measure of such difference. If we build a set of electrochemical cells that have one half cell in common and measure their associated voltages, we can use these data to infer the DV for any electrochemical cell built with
any pair of half cells in the set. For example, consider these data:
Electrochemical Cell A: H2(g)|H+(aq) (1 M) || Cu(s)|Cu2+(aq) (1 M)
Anode:
H2
2 H+ + 2 eCathode: Cu2+ + 2 eCu
DV = 0.34 V
V
Cu|Cu2+
0.34 V
H2|H+
Electrochemical Cell B: Zn(s)|Zn2+(aq) (1 M) || H2(g)|H+(aq) (1 M)
Anode:
Zn
Zn2+ + 2 e- Cathode: 2 H+ + 2 eH2
DV = 0.76 V
0.76 V
Zn|Zn2+
These data suggest that the Cu(s)|Cu2+(aq) half cell has an electric potential that
is 0.34 V higher than that of the H2(g)|H+(aq) half cell, while the Zn(s)|Zn2+(aq)
half cell has an electric potential that is 0.76 V lower than that of the H2(g)|H+(aq)
half cell. Consequently, an electrochemical cell built with the copper and zinc half
cells should have these characteristics (Figure 8.9):
e-
Figure 8.9 Relative elec-
tric potential V of different half cells. When free,
electrons move from regions of low to high electric potential.
Electrochemical Cell C: Zn(s)|Zn2+(aq) (1 M) || Cu(s)|Cu2+(aq) (1 M)
Anode:
Zn
Zn2+ + 2 e- Cathode: Cu2+ + 2 eCu
DV = 0.34 + 0.76 = 1.1 V
Systems with a more positive value of Eored involve stronger oxidizing
agents than systems with the less positive or more negative standard
reduction potentials. The relative position of different redox pairs in the
scale can be used to predict which species undergo reduction (cathode)
and oxidation (anode) when connected in an electrochemical cell.
Zn2+ + 2 e-
Zn
–0.76
Al3+ + 3 e-
Al
–1.66
Mg + 2 e-
Mg
–2.37
2+
Li+ + e-
Li
–3.05
More Oxidizing
The DV measured using the H2(g)|H+(aq) half cell as a reference can be used
to define an standard reduction potential Eored for any given redox pair A|An+. The
magnitude of Eored is equal to the voltage DV of the electrochemical cell comprised
of the half cells A|An+ and H2(g)|H+(aq) under standard conditions: 25 oC of temperature, 1 M concentrations for all aqueous solutions, and 1 atm of pressure for
all gaseous substances. The sign of Eored depends on whether the half cell A|An+ acts
as the cathode (Eored > 0) or as the anode (Eored < 0) when paired with
Standard Reduction Potentials
the reference system. The standard reduction potentials for common
redox pairs are listed in the table to the right. By definition, the value of Reduction Half-Reaction Eored (V)
Eored for the reference H2(g)|H+(aq) system is equal to zero.
F2 + 2 e2 F–
+2.87
The standard reduction potentials can be used to calculate the volt–
Cl2 + 2 e2 Cl
+1.36
age, commonly called the standard cell potential Eocell of electrochemical cells formed using any combination of redox pairs. The value of Eocell
Ag+ + eAg
+0.80
o
is given by the difference between the E red for the half cell where reducI2 + 2 e2 I–
+0.54
tion takes place (cathode) and the Eored for the half cell where oxidation
Cu2+ + 2 eCu
+0.34
occurs (anode):
H+ + 2 eH2
0.0
o
o
o
(8.2)
DV = E cell = E red (cathode) – E red (anode)
2+
Pb + 2 ePb
–0.13
448
MODULE 1
Controlling Electron Transfer
LET’S THINK
Design your Own Devices
Based on the information presented in previous pages:
•
•
•
•
Design an electrochemical cell that will produce a standard cell potential close to 2.0 V.
Determine what half reactions will take place in the cathode and the anode of your cell.
Make a drawing of the electrochemical cell showing the chemical processes that are taking
place in each half cell and how charge is being transferred from one half cell to the other.
Determine the amount of electrical energy that your electrochemical cell will be produce per
mole of reducing agent consumed in the anode.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
The values of the reduction potentials Ered and the associated Ecell for an electrochemical device depend on the temperature and the concentration of all species
present in the system. For a reduction half reaction of the generic form:
Oxidizing Species + n e-
Reducing Species
the effects on the reduction potential Ered are determined by the following relationship, known as Nernst equation:
(8.3)
Ered = Eored –
RT
[Red]
ln
nF
[Ox]
In this equation, Eored is the standard reduction potential for the redox pair under
consideration, T is the temperature, and [Red] and [Ox] are the concentrations (or
pressures) of the reducing and oxidizing species, respectively.
Consider now the generic electrochemical cell: Red1|Ox1 || Red2|Ox2
Anode:
Red1
Ox1 + n e- Cathode: Ox2 + n eRed2
DV = Ecell = Ered(cat) – Ered(an)
Red1 +
with an overall redox reaction:
Ox2
Ox1 + Red2
If we apply Nernst equation to calculate the reduction potentials Ered for the anode
and the cathode, the cell potential Ecell is given by:
(8.4)
Ecell = Eocell –
RT
nF
ln
[Red2][Ox1]
[Ox2][Red1]
= Eocell –
RT
nF
ln Q
where Q is the reaction quotient for the overall reaction. As discussed in Unit 6,
Q = K at equilibrium, where K is the equilibrium constant for the process. When
chemical equilibrium is reached Ecell = 0 and no net charge is transferred between
the half cells.
Chemical Thinking
U8
449
How do we harness chemical energy?
What are the Effects? LET’S THINK
Analyze the implications of (8.4):
•
•
•
How does Ecell vary as the redox reaction proceeds and reactants are consumed?
How does Ecell change with increasing or decreasing temperature?
If Ecell = 0 and Q = K when the redox reaction reaches equilibrium, how is K related to Eocell?
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Electrochemical Cell Kinetics
The voltage of an electrochemical cell Ecell determines the amount of energy that
the device can produce per unit of charge transferred from the anode to the cathode. The electric power of the cell, however, depends on both Ecell and the rate at
which the charge is being transferred between electrodes as measured by the electric current “i”. The magnitude of the electric current i is determined by the rate
of the chemical reactions taking place in the half cells. Let us explore the effect of
different factors on such reaction rate.
LET’S THINK
Reaction Rate
The rate of reaction in an electrochemical cell depends on the various factors listed below:
a) Temperature of the system
c) The porosity of the electrodes
e) The concentration of ions at the interface
•
b) The activation energy of the reaction
d) The solvation of ions in solution
f ) The rate of ion diffusion in solution
Discuss and explain how each of these factors may affect reaction rate and, thus, the magnitude of the electric current generated by an electrochemical cell.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
++++++++++++
In the types of electrochemical cells analyzed in Unit 8, the chemical reactions taking place at the electrodes typically involve metal dissolution and
deposition, as well as gas evolution. When the device is set up, the reactants
can be expected to be plentiful and the electric current will be determined
by the rate at which electrons are transferred between metal atoms and ions
at the electrode/solution interface. As the reaction proceeds and reactants
are consumed, the availability of ions at the electrode/solution interface will
decrease, slowing down the overall process (Figure 8.10). The rate at which
ions diffuse from the bulk of the solution to the surface of the electrode will
affect the rate of the reaction. Increasing temperature will increase the diffusion rate and thus the electric current. The formation of films at the metal
surfaces, such as metal oxide films, often increases the resistance to charge
transfer and decreases electric current.
+
–
+
–
+
–
–
+
Figure 8.10 Submicroscopic representation of an
electrode/solution interface.
450
MODULE 1
Controlling Electron Transfer
Ecell
i
Cu
Cu2+ + 2e-
i1
0.34
–i1
Ecell
Figure 8.11
Relationship
between electric current i
and cell potential Ecell in a
Cu|Cu2+ || Ag|Ag+ cell.
Once an electric current is established in an electrochemical cell, conditions in
each half cell start to change. For example, positive ions produced by the oxidation
in the anode start accumulating near this electrode. In the other hand, the concentration of positive ions near the cathode decreases as the reduction half reaction
takes place. These changes affect the value of the voltage (Ecell) across the electrochemical cell. In particular, we can expect the electric potential of the anode to
increase due to the accumulation of positive charge in its vicinity and the electric
potential of the anode to decrease due to the depletion of positive ions near that
electrode. The faster the reaction, the greater the electric current but also the larger
the decrease of the cell potential Ecell = Ered(cathode) – Ered(anode).
The relationship between electric current i
at i = i1
and the reduction potential Ered of the electrodes
is represented in Figure 8.11 for a Cu|Cu2+ (1 M)
|| Ag|Ag+ (1 M) electrochemical cell. By convention, the anodic current is defined to be positive
(electron flow out of the anode) and the cathodic
current is taken to be negative (electron flow into
0.80
Ered (V) the electrode). Thus, on the current “i” axis in Figure 8.11 anodic current is plotted upward and cathodic current downward. The greater the electric
current, the larger the drop in the cell potential Ecell
+
Ag + e- Ag
and the lower the amount of energy that the device
generates per unit of charge transferred from the
at i = 0
anode to the cathode. Consequently, the design of
efficient electrochemical devices demands creating
conditions in which the decrease of the cell potential is minimal when an electric
current is actually circulating through the system. This requires lowering all types
of energetic barriers that hinder the transfer and movement of electrons and ions
in different parts of the electrochemical cell.
LET’S THINK
The graph shown to the right depicts the relationship between electric current i and cell potential
Ecell for an electrochemical device under two different conditions represented by the solid and dotted
lines, respectively.
i-Ecell Relationship
Ecell at i = i1
i
i1
Discuss and explain which set of i-Ecell data
Ered (V)
–i1
corresponds to:
a) A cell working at high versus low tempera
ture;
b) A cell with more porous versus less porous
Ecell at i = 0
electrodes;
c) A cell in which ions diffuse rapidly versus slowly.
• Discuss what other factors may affect the i-Ecell relationship and how they may influence it.
•
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
U8
FACING THE CHALLENGE
Bio-Batteries
A bio-battery is an electrochemical device that generates electricity from organic compounds, such as
sucrose, glucose, and fructose. The batteries rely
on biocatalysts to increase rates of oxidation and
facilitate electron transfer. The biocatalysts can be
enzymes or microorganisms that help process the
organic fuels. In passive bio-batteries, the reactive
substances are absorbed into the system by natural
diffusion. In active devices, the reactants are injected into the electrochemical cell. In the future,
miniature bio-batteries
could be used to power
Glucose
microsensors implanted
within a human patient
Anode
Enzymes
to monitor blood presAnode
Mediators
sure, temperature, and
the concentration of
different metabolites.
The chemical reace–
tions in a bio-battery
can proceed by direct
electron transfer beCathode
tween the biocatalysts
Mediators
(enzymes) and the elecCathode
Enzymes
trodes. However, it is
O2
common to use “mediators” that reduce the
kinetic barrier for electron transfer. Many of these
mediators are vitamin molecules, such as vitamin
C and vitamin K3. Direct transfer systems are
more desirable as they simplify the construction
of the batteries.
A bio-battery typically includes an anode
containing the carbohydrate to be oxidized, carbohydrate-digesting enzymes, and a mediator.
The cathode contains oxygen, oxygen-reducing
enzymes, and a mediator. The two half cells are
divided by a porous separator. When using glucose
as a fuel, the oxidation at the anode can be represented as follows:
Glucose
Glucanolactone + 2 H+ + 2 e–
How do we harness chemical energy?
Hydrogen ions H+ migrate to the cathode through
the porous separator, where they participate in the
reduction reaction
O2 + 4 H+ + 4 e–
2 H2O
Due to the selective reactivity of the enzymes at
each electrode, no cross reaction occurs between
the anode and the cathode.
Currently, the are no bio-battery templates
that allow the mass production of miniature devices with a size in the order of 1 cubic centimeter, as
needed for physiological sensors. Electron flow in
current devices is too slow, mostly due to the difficulty of establishing effective contact between
enzymes and electrodes.
Glucanolactone
Enzyme stability is also
a critical issue as these
Anode
bio-catalysts are very
sensitive to changes in
temperature and pH.
Electrolyte
Enzymes used in
H+
bio-batteries are exSeparator
tracted from microCathode
organisms an immobilized onto carbon
nanotubes that are used
as electrodes. NanoH2O
materials provide large
surface areas for the
attachment of enzymes, increasing the concentration of biocatalyst and improving battery power
density. Additionally, nanostructured materials exhibit great potential for stabilizing enzyme activity.
Carbon nanotubes, however, seem to induce the
denaturation of some proteins by dehydration.
An important advantage of bio-batteries over
other types of electrochemical devices is the possibility of instant recharge. If the organic fuel is
readily available in the surroundings, these batteries can continuously refuel. They are also a clean
and renewable energy source fueled by inexpensive materials, including different types of waste.
Unfortunately, existing prototypes are not yet
competitive in terms of energy storage and power
density.
451
452
MODULE 1
Controlling Electron Transfer
Let’s Apply
By Dervisoglu (Own work)
[Public Domain] via Wikimedia Commons
ASSESS WHAT YOU KNOW
Hydrogen Fuel Cell
Fuel cells are electrochemical devices in which reactants are continuously supplied to the system to sustain the redox reaction.
In a hydrogen fuel cell, for example, hydrogen gas and oxygen
gas from air are used to produce energy. The overall reaction
that takes place in the hydrogen fuel cell can be represented as:
H2(g) + 1/2 O2(g)
H2O(l).
Oxidation-Reduction
Consider the overall chemical reaction in a hydrogen fuel cell: H2(g) + 1/2 O2(g)
•
Determine the oxidation state of each atom in the reactants and products.
•
Identify the oxidized and reduced atoms in the process.
•
Identify the oxidizing agent and reducing agent in the reaction.
•
Determine the number of electrons exchanged per molecule of H2 consumed.
H2O(l)
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Thermodynamic Analysis
The table to the right lists relevant thermodynamic information for reactants and products in
a hydrogen fuel cell.
DHof (kJ/mol)
Sof (J/(mol K)
H2(g)
0
130.6
O2(g)
0
205.0
H2O(l)
–285.8
69.95
•
Qualitatively predict the signs of DHrxn and DSrxn for the overall reaction in a hydrogen
fuel cell. Predict the effect of changing temperature T on the extent of the reaction.
•
Verify your predictions using thermodynamic data to calculate DHorxn and DSorxn.
•
Calculate the DGorxn and 25 oC and 70 oC (typical temperature at which a hydrogen fuel
cell operates). Build a graph showing how DGorxn changes as a function of T.
According to Equation (8.1): DGorxn = – nFEocell, where Eocell is the cell potential generated by a
fuel cell at standard pressures (1 atm).
•
Estimate the value of Eocell at 25 oC and 70 oC. Build a plausible explanation for the
change of Eocell with temperature.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
U8
How do we harness chemical energy?
453
Cell Potential and Energy Output
The table to the right lists the standard reduction potentials for redox pairs relevant to the
operation of fuel cells under acidic or basic (alkaline) conditions.
•
Reduction Half-Reaction
Eored (V)
O2 + 4 H+ + 4 e-
2 H2O
+1.23
O2 + 2 H2O + 4 e-
4 OH–
+0.40
2 H+ + 2 e2 H2O + 2 e-
H2
0
H2 + 2 OH–
–0.83
Express the half reactions taking place
in the anode and the cathode of hydrogen fuel cell working under basic (alkaline)
conditions.
•
Determine Eocell for each of these fuel cells and compare your result with the value
obtained using the DGorxn from thermodynamic data.
•
Estimate the maximum amount of energy that each of these fuels cell produces per
mol of hydrogen consumed under standard conditions.
•
Express the Nernst equation for these fuel cells in terms of the temperature of the
system and the pressure of the gases used as reactants. Qualitatively analyze this relationship and determine the best conditions to maximize energy output.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Fuel Cell Kinetics
Given that reactants are in the gas phase in hydrogen fuel cells, inert metal electrodes are
used as surfaces where electron transfer takes place. The nature of the electrodes used affects the relationship between the electric current i established in the system and the cell
potential Ecell of the device.
The graph to the right represents i-Ecell
data for hydrogen fuel cells with two
different types of electrodes.
i
•
Which type of electrode would
be best to use?
i1
•
When a current is established,
the change in Ered is greater in
the cathode than in the anode.
Propose a plausible explanation
for this result.
–i1
•
Pt
Au
Ered (V)
0.0
1.23
Au
Pt
How would you expect temperature to affect the value of Ecell when an electric current flows through the fuel cell?
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
•
Express the half reactions taking place in
the anode and the cathode of hydrogen
fuel cell working under acidic conditions.
Standard Reduction Potentials
454
MODULE 1
Controlling Electron Transfer
Let’s Apply
Electric Batteries
Alkaline Battery
By Aney (Own work)
[Sahre Alike 3.0] via Wikimedia Commons
ASSESS WHAT YOU KNOW
Different types of batteries have been developed in the past 150 years. Some of them are commonly
used to power small electric devices, cell phones and laptop computers, and electric systems in
regular combustion cars. Let us explore the characteristics of the most used batteries.
Alkaline batteries are portable electrochemical systems
that can be rechargeable or not rechargeable. They account for over 80% of all manufactured batteries in the
US. The following chemical reactions take place in the
electrodes of this type of battery:
Zn(s) + 2 OH– (aq)
2 MnO2(s) + H2O(l)
ZnO(s) + H2O(l)
Mn2O3(s) + 2 OH–(aq)
•
Assign oxidation states to the atoms in these half reactions. Complete each of the half
reactions by representing the number of electrons lost or gained in each case.
•
Identify which of these reactions takes place in the anode of the alkaline cell and which
reaction occurs in the cathode.
•
Write the overall chemical reaction for the alkaline battery.
An alkaline battery generates a standard cell potential close to 1.5 V.
•
How much energy does an alkaline battery produce per gram of zinc consumed?
An aqueous solution of potassium hydroxide (KOH) is used as electrolyte in alkaline batteries.
This solution creates a basic environment.
•
Express the Nernst equation for the battery. Analyze
how the cell potential varies with pH, and discuss the
implications of your results for battery performance.
The graph shows how the cell potential (Ecell) of an alkaline
battery decreases as a function of time when the battery
produces the same current i at three different temperatures.
•
How would you justify these results?
20 oC
0 oC
–20 oC
Chemical Thinking
U8
How do we harness chemical energy?
455
Lead-Acid Battery
The lead–acid battery was developed more than 150 years ago. It
is a low cost rechargeable battery that provides the high current
required by automobile starter motors. Lead–acid battery sales account for over 40% of the value from batteries sold worldwide.
The following chemical reactions take place in the electrodes of this
type of battery:
PbSO4(s) + 2 H2O(l)
PbSO4(s) + H+(aq)
•
Assign oxidation states to each atom in these half reactions. Complete each of the
half reactions by representing the number of electrons lost or gained in each case.
•
Identify which of these reactions takes place in the anode of the lead–acid battery
and which reaction occurs in the cathode.
•
Draw a schematic diagram of a lead–acid battery showing the chemical composition
and structure of each half cell, as well as the flow of electric charge throughout the
system.
The standard reduction potentials in the cathode and anode of a lead–acid battery are
Eored = 1.685 V and Eored = –0.356 V, respectively.
•
Calculate the standard cell potential for this battery.
A typical car battery does not run under standard conditions. The concentration of sulfuric
acid (H2SO4; a strong acid that completely dissociates into HSO4– ions in aqueous solution) is close to 5 M and the pH is close to 0.
•
Write the overall chemical reaction for the alkaline battery.
•
Express the Nernst equation for a lead–acid battery and calculate its actual cell potential.
•
Express the Nernst equation for the battery as a function of pH. Make a graph of cell
potential versus pH and discuss its implications for cell voltage and power.
The graph sketches how the reduction potentials of the
cathode and the anode in a lead–acid battery change
when a given current i is established across the battery
at two different temperatures.
•
Explain why starting a car is more difficult under
cold weather conditions.
i
i1
–i1
10 oC
–10 oC
Ered (V)
ASSESS WHAT YOU KNOW
PbO2(s) + 3 H+(aq) + HSO4–(aq)
Pb(s) + HSO4–(aq)
456
MODULE 1
Controlling Electron Transfer
Let’s Apply
ASSESS WHAT YOU KNOW
Lithium-Ion Battery
Lithium-ion batteries are often used in consumer electronics. They are one of the most
popular types of rechargeable batteries for
portable devices, such as watches, cell phones,
and laptop computers. This type of battery
takes advantage of the ability of Li+ ions to
be reversibly inserted and removed from certain solids. The image represents the internal
structure of this type of battery.
Anode
Cathode
LixC6
Li1-xCoO2
Li+
The half cell reactions in the lithium-ion battery are:
ANODE: CATHODE: LixC6
x Li+ + x e- + C6
Li1-xCoO2 + x Li+ + x eLiCoO2
where LixC6 represents graphite (C6) intercalated with x moles of Li+ ions, and LiCoO2 represents lithium cobalt (III) oxide. When x moles of Co3+ ions are oxidized to Co4+, x moles of Li+
ions are released from the solid and Li1-xCoO2 is produced. This process is reversible for x < 0.5.
•
Analyze the information provided and generate a detailed explanation of how the lithium-ion battery works. Discuss what happens in each half cell as electrons are transferred
from the anode to the cathode, and Li+ diffuse between the two cells.
•
A lithium-ion battery can be recharged. Discuss why this is possible and how it can be
done. Analyze the distribution and movement of Li+ ions between the two half cells
when the battery is charging and when it is discharging.
A lithium-ion battery generates a standard cell potential close to 3.7 V. As shown in the graph, this voltage depends on temperature and the value of the electric current i (measure din milliampere, mA) flowing
through the system.
•
Analyze the data represented in the graph and
build a kinetic argument to justify the behavior
of the lithium-ion battery under different conditions.
0.2 mA
2.0 mA
20 mA
Chemical Thinking
U8
How do we harness chemical energy?
457
Unit 8: Image Attributions
Module 1
Most of the images in this module have been generated by the authors (Own work). Other attributions include, P436 & P437: Background
“Berlin Potdamer Platz” by Avda (Own work) [Share Alike 3.0] http://commons.wikimedia.org/wiki/File:Berlin_-_Potsdamer_Platz_-_E-Mobility-Charging.jpg; P438: Upper right “Lithium Battery CR2032” by KyloDee (Own work) [Public Domain] http://commons.wikimedia.
org/wiki/File:Battery-lithium-cr2032.jpg; P443: Middle right by Toby Hudson (Own work) [Share Alike 3.0] http://commons.wikimedia.org/
wiki/File:Precipitation_of_Silver_on_Copper_1.jpg; P445: Middle right by Shamsher Singh (Own wok) http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells; P452: Upper right by Dervisoglu (Own work) [Public domain] http://commons.wikimedia.org/
wiki/File:Solid_oxide_fuel_cell_protonic.svg; P454: Upper right “Alkaline batteries” by Aney (Own work) [Share Alike 3.0] http://commons.
wikimedia.org/wiki/File:Alkali_battery_5.jpg
General
Activity icons: Clip art from Microsoft Office; Molecular structures: Derived from online public software (Chemical Education Digital Library,
Jmol) and via Wikimedia Commons (Public domain images).