Chapter 19: Electrochemistry: The Quest for Clean Energy
Problems: 19.2-19.3, 19.13-19.58, 19.69-19.72, 19.75-19.78
19.1 REDOX CHEMISTRY REVISITED
electrochemistry: the study of the conversion between chemical and electrical energy
Almost everything we do involves the use of a battery, which is a product of chemical energy.
– To understand how batteries work, we need to understand electrochemistry and
electrochemical cells.
REVIEWING OXIDATION NUMBERS (Ox. #’s) and REDOX REACTIONS
The reactions that occur in these electrochemical cells are all oxidation-reduction (“redox”)
reactions where electrons are transferred from one reactant to another.
Oxidation numbers:
actual or hypothetical charge of element or an atom in a compound if it
existed as a monatomic ion
Knowing the oxidation number of each element in a reaction allows us to track the electrons
being transferred to determine what was oxidized and what was reduced in a redox reaction.
Guidelines for Assigning Oxidation Numbers
1. The oxidation number of an element in its natural form is 0.
– e.g. the oxidation number is zero for each element in H2, O2, Cl2, P4, Na, etc.
2. The oxidation number of a monatomic ion is the charge on the ion in a binary compound.
– e.g. In CaCl2, the ions are Ca+2 and Cl–, so oxidation #’s: Ca=+2 and Cl=-1
In Al2O3, the ions are Al+3 and O2–, so oxidation #’s: Al=+3 and O=-2
3. In a compound or polyatomic ion,
– Group I elements are always +1.
– Group II elements are always +2.
– Fluorine is always –1.
– Oxygen is usually –2 (except in the peroxide ion, O22–, when O is –1)
– Hydrogen is usually +1
(except when it is with a metal, like NaH or CaH2, then it is –1)
– The oxidation number for all other elements depend on the other elements.
4. In a compound, the sum of all oxidation numbers must equal 0.
In a polyatomic ion, the sum of all oxidation numbers must equal charge.
Note: The oxidation numbers are constant for atoms in polyatomic ions.
– e.g., in CO32-, the oxidation numbers are always C=+4 and O=-2.
→ Breaking up ionic compounds into their constituent ions simplifies the process of
determining oxidation numbers for each atom in the compound.
CHEM162: Gilbert Chapter 19 page 1 Example: Determine the oxidation number for each atom in the following:
a. SO42–:
S: ____, O: ____
e. CO: C: ____, O: ____
b. Fe2(CO3)3: Fe: _____, C: ____, O: ____
d. KNO3: K: ____, N: ____, O: ____
c. H3PO4:
H: ____, P: ____, O: ____
g. ClO3–: Cl: ____, O: ____
d. CaCr2O7:
Ca: ____, Cr: ____, O: ____
h. C3H8: C: ____, H: ____
Consider the following redox (electron-transfer) reactions:
Oxidation half-reaction:
Zn(s) → Zn2+(aq) + 2 e
Reduction half-reaction: Cu2+(aq) + 2 e
−
−
→ Cu(s)
In this redox reaction, the zinc metal strip, Zn(s), loses 2 electrons to form zinc ion, Zn2+(aq)
while the copper(II) ion, Cu2+(aq), gains the 2 electrons from the zinc to form copper granules,
Cu(s), the fuzzy black solid forming on the zinc strip.
Notice that when Zn(s) loses electrons, its oxidation number increases, and
when Cu2+(aq) gains electrons, its oxidation number decreases.
Oxidation: process of losing electrons (oxidation number ↑)
Reduction: process of gaining electrons (oxidation number ↓)
CHEM162: Gilbert Chapter 19 page 2 In a redox reaction
– One reactant Loses Electrons/is Oxidized (LEO)
– Another reactant Gains Electrons/is Reduced (GER)
An easy way to remember is “LEO the lion goes GER!”
Oxidizing Agents (Oxidizers) and Reducing Agents (Reducers)
In a redox reaction,
– The element/reactant oxidized reduced the other reactant, so it is the reducing agent.
– The element/reactant reduced oxidized the other reactant, so it is oxidizing agent.
Ex. 1: For each of the following reactions,
i. Balance the equation (to balance the electrons transferred).
ii. Assign oxidation numbers to each atom in each reactant and product.
iii. Write the half reactions for the atoms that are oxidized and reduced.
iv. Identify the element or atom with its oxidation number in a compound oxidized
and the element or atom with its oxidation number in a compound reduced.
v. Indicate which reactant is the oxidizing agent and which is the reducing agent.
vi. Indicate the number of electrons transferred.
a.
Zn(s) +
AgNO3(aq)
→
Zn(NO3)2(aq) +
Ag(s)
What is oxidized? ____________________ What is reduced? ____________________
The oxidizing agent is ________________, and the reducing agent is __________________.
The number of electrons transferred is ______.
b.
MnO2(s) +
HCl(aq) →
MnCl2(s)
+
Cl2(g)
+
H2O(l)
What is oxidized? ____________________ What is reduced? ____________________
The oxidizing agent is ________________, and the reducing agent is __________________.
The number of electrons transferred is ______.
CHEM162: Gilbert Chapter 19 page 3 c.
NaClO(aq) +
H2O2(aq)
→
H2O(l) +
NaCl(aq) +
O2(g)
What is oxidized? ____________________ What is reduced? ____________________
The oxidizing agent is ________________, and the reducing agent is __________________.
The number of electrons transferred is ______.
d.
Ca(s) +
H2O(l)
Ca(OH)2(aq) +
→
H2(g)
What is oxidized? ____________________ What is reduced? ____________________
The oxidizing agent is ________________, and the reducing agent is __________________.
The number of electrons transferred is ______.
Note: In this reaction, only half of the hydrogen atoms were reduced; thus, not every atom of an
element may be oxidized or reduced in a redox reaction.
e.
C3H8(g) +
O2(g)
→
CO2(g) +
H2O(g)
What is oxidized? ____________________ What is reduced? ____________________
The oxidizing agent is ________________, and the reducing agent is __________________.
The number of electrons transferred is ______.
CHEM162: Gilbert Chapter 19 page 4 19.2 ELECTROCHEMICAL CELLS
When redox reactions occur in solutions, the electron transfer occurs in the solution, so no work
can be done with the electron flow.
– For example, when Zn metal is placed in a solution with Cu2+ ions, the electrons transfer from
the Zn atoms to the Cu2+ ions, causing the plating of Cu atoms onto the Zn metal.
However, if we can separate the oxidizing agent from the reducing agent and force the electron
transfer to occur through a wire, the current produced from the reaction can be directed to do
useful work → electrochemical cell.
Electrochemical cells can convert the chemical energy of a spontaneous redox reaction into
electrical energy (e.g. batteries).
There are two types of electrochemical cells:
1. galvanic cells (also called voltaic cells): cells in which a spontaneous chemical reaction
generates an electrical current
– named after Italian scientists Luigi Galvani (1737-1798), Alessandro Volta (1745-1827)
2. electrolytic cells: cells in which an electrical current runs a nonspontaneous reaction
GALVANIC (or VOLTAIC) CELLS
Let's design a galvanic (voltaic) cell:
1. Choose a system, so electrons flow spontaneously from the anode to the cathode
Oxidation half-reaction occurs at anode:
Reduction half-reaction occurs at cathode:
Zn(s) → Zn2+(aq) + 2 e
Cu2+(aq) + 2 e → Cu(s)
−
−
2. Prepare the two half cells by dipping electrodes of the solid metal anode and the solid metal
cathode into their corresponding solutions.
CHEM162: Gilbert Chapter 19 page 5 3. Prepare the external circuit with a digital voltmeter (drawing a negligible amount of current, so
it effectively measures the current flowing through) with leads to the cathode and anode.
Ex. 1: a. As the oxidation and reduction half reactions occur, the concentration of which ions
increases in the Zn half cell? Indicate if the solution contains more cations or anions.
b. As the oxidation and reduction half reactions occurs, the concentration of which ions
decreases in the Cu half cell? Indicate if the solution contains more cations or anions.
– When the cation concentration increases too much in the Zn half cell, the solution
becomes too positive.
→ The oxidation of more Zn atoms to Zn2+ ions is inhibited.
– When the anion concentration increases too much in the Cu half cell, the solution
becomes too negative and attracts the Cu2+ ions.
→ The reduction of Cu2+ ions to Cu metal is inhibited.
– Thus, a salt bridge or porous separator is added to prevent a build up of charges in both
cells.
4. A salt bridge allows cations and anions to flow to each half cell, preventing a charge buildup.
– As more Zn2+ ions are produced, the overall charge in the left cell becomes too positive.
– As more Cu2+ ions are reduced to Cu atom, the overall negative charge increases due
to the increasing concentration of SO42- ions relative to the number of Cu2+.
– These charge buildups impede the flow of electrons.
– A salt bridge is usually a U-tube filled with solution or a saturated piece of filter paper that
allows ions to flow to/from each compartment to maintain the charge balance.
We can represent this electrochemical cell by writing a cell diagram (or line notation):
Zn | Zn2+ || Cu2+ | Cu
using the following internationally accepted conventions:
1. The anode reaction (oxidation) is shown at the left.
– Remember oxidation at the anode both begin with vowels.
2. The cathode reaction (reduction) is shown at the right.
– Remember reduction at the cathode both begin with consonants.
3. The single vertical line "|" indicates a phase boundary (e.g. a solid electrode in a solution).
4. The separation of the two half cells is indicated by the symbol ||
An easy way to remember the anode goes before the cathode is "a before c."
CHEM162: Gilbert Chapter 19 page 6 Cells That Involve Gas Production
– Consider a different redox reaction: Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
– As with the previous cell, the Zn | Zn2+ half-cell and salt bridge are the same.
– However, since hydrogen exists naturally as a gas, an inert electrode is required.
– Frequently, the cathode is made out of platinum, a relatively unreactive metal.
– The cell diagram for this galvanic cell is: Zn | Zn2+ || H+ | H2 | Pt
and variations of these type of cells are commonly used as pH meters.
For any voltaic or galvanic cell, be able to diagram the cell, and be able to identify:
1. The reaction at the cathode
2. The reaction at the anode
3. The electron flow in the cells and through the external circuit
4. The ion flow in both cells
19.3
STANDARD POTENTIALS
A galvanic cell consists of an oxidizing agent in one compartment that “pulls” electrons through a
wire from the reducing agent in the other compartment. The “pull” (or driving force) on the
electrons is called the electromotive force (emf) of the cell.
– This is measured as a voltage called the cell potential (Ecell), which is direction proportional
to the chemical (potential) energy released when Zn atoms are oxidized and Cu2+ atoms are
reduced simultaneously.
– The unit of electrical potential is the volt (abbreviated V),
J
which is defined as 1 joule of work per coulomb of charge: 1 V = 1
C
CHEM162: Gilbert Chapter 19 page 7 19.5 A REFERENCE POINT: THE STANDARD HYDROGEN ELECTRODE
Standard Reduction Potential (Ered°):
– The cell potential for a reduction reaction at an electrode when all the solute concentrations
are 1M, and all gases are at 1 atm.
– Appendix 6 (pp. A-26 and A-77at the back of the Gilbert text) gives the Standard Reduction
Potentials for various species.
– These standard reduction potentials are based on hydrogen's Ered°, which was set as
2 H+(aq) + 2 e– → H2(g)
Ered°=0.000 V
→ Why a platinum electrode in a 1M H+ solution and bathed in hydrogen gas at 1 atm is
called the standard hydrogen electrode (or SHE).
The following table includes the most commonly used standard reduction potentials:
CHEM162: Gilbert Chapter 19 page 8 Standard Oxidation Potential (Eox°):
– Since oxidation is the opposite of reduction,
Eox° = − Ered°
Thus, if
Zn2+(aq) + 2 e– → Zn(s)
Ered°=−0.76 V
then
Zn(s) → Zn2+(aq) + 2 e–
Eox°=+0.76 V
Note that standard cell potentials are intensive properties (i.e. independent of amount, or in this
case, the number of electrons moving through the cell).
→ They only depend on the redox reaction occurring if the concentration of the species is 1M.
→ When calculating the cell potential for a galvanic (or voltaic) cell, the number of
electrons must be balanced to show the correct number of electrons being transferred, but
each standard reduction or oxidation potential is not multiplied by any integer used
to balanced the half reactions.
Calculating Ecell° from Eox° and Ered°:
– The cell potential (Ecell°) is the sum of the two half reactions:
Thus,
Eox°= 0.76 V
Ered°= 0.34 V
Zn(s) → Zn2+(aq) + 2 e–
2+
Ecell° = Eox° + Ered°
–
Cu (aq) + 2 e → Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Ecell°=1.10 V
General Guidelines for Voltaic Cells
1. Calculated cell potentials, Ecell°, must be positive for reactions in a galvanic/voltaic cell.
2. Because cell potentials are intensive properties (independent of amount), never multiply by
the coefficients in a half reaction.
Example: Calculate the cell potential for the following galvanic cells, then write the cell diagram
(or line notation) for each cell:
a. A galvanic cell consists of a magnesium electrode in a 1.0M Mg(NO3)2 solution and a silver
electrode in a 1.0M AgNO3 solution. Calculate the standard potential for the cell.
CHEM162: Gilbert Chapter 19 page 9 b. A galvanic cell consists of an iron electrode in a 1.0M Fe(NO3)3 solution and a copper
electrode in a 1.0M copper(II) nitrate solution. Calculate the standard potential for the cell.
c. A galvanic cell consists of a magnesium electrode in a 1.0M Mg(NO3)2 solution and an
aluminum electrode in a 1.0M aluminum nitrate solution. Calculate the standard potential for
the cell.
d. A galvanic cell consists of a copper electrode in a 1.0M Cu(NO3)2 solution and a platinum
electrode in 1.0M nitric acid. Calculate the standard potential for the cell. If the copper and
nitric acid were placed in the same container, the reaction would produce brown NO2 gas
(the primary component of smog).
Consider the following reduction potential:
NO3 (aq) + 2 H+(aq) + e → NO2(g) + H2O(l)
−
−
CHEM162: Gilbert Chapter 19 Ered°=0.78 V
page 10 Strength of Oxidizing and Reducing Agents
A strong oxidizing agent has a strong attraction for electrons.
– The strength of an oxidizing agent is directly related to its standard reduction potential, Ered°.
→ The more positive the Ered°
→ the stronger the oxidizing agent (or tendency to be reduced).
– e.g. F2 is the strongest oxidizing agent—this makes sense since F has a strong
electron affinity, so it tends to gain electrons.
→ Generally, F2 cannot be used in water solutions because it’s such a strong
oxidizing agent, it would cause explosive reactions.
A strong reducing agent has a strong tendency to lose electrons.
→ The more negative Ered°
→ the stronger the reducing agent (or tendency to be oxidized).
– e.g. Li is the strongest reducing agent.
→ Li is the highest metal on the Activity Series because of its tendency to lose e−s.
Ex. 1: Using the table of standard reduction potentials to arrange the following species in order of
increasing strength as oxidizing agents: MnO4–, Sn2+, Al3+, NO3–, and Ag+.
_____________ < _____________ < _____________ < _____________ < _____________
weakest oxidizing agent
strongest oxidizing agent
Ex. 2: Consider the following reduction half-reactions and the corresponding standard reduction
potentials, Ered°:
I2(aq) + 2 e– → 2 I (aq)
−
2+
–
Pb (aq) + 2 e → Pb(s)
2 H (aq) + 2 e– → H2(g)
+
Cr3+(aq) + 3 e– → Cr(s)
Ered°= 0.54 V
Ered°= −0.13 V
Ered°= 0.00 V
Ered°= −0.73 V
Note that an oxidizing agent must be able to gain electrons to oxidize another substance, and a
reducing agent must be able to lose electrons to reduce another substance.
Rank the oxidizing agents in terms of increasing strength:
_____________ < _____________ < _____________ < _____________
weakest oxidizing agent
strongest oxidizing agent
Rank the reducing agents in terms of increasing strength:
_____________ < _____________ < _____________ < _____________
weakest reducing agent
CHEM162: Gilbert Chapter 19 strongest reducing agent
page 11 19.4 CHEMICAL ENERGY AND ELECTRICAL WORK
The unit of electrical potential is the volt (abbreviated V),
which is defined as 1 joule of work per coulomb of charge:
1V=1
J
C
The work that can be done when electrons are transferred through a wire from a galvanic/voltaic
cell can be shown as follows:
w = n F Ecell
where w=work,
n=# of moles of electrons transferred,
J
C
and F=faraday =96,485
=96,485
mol ⋅ V
mol
Because the maximum amount of useful work that a system can do is –ΔG, the maximum
amount of electrical work that can be done, wmax, can be shown as wmax = –ΔG, so
ΔG = – n F Ecell
This relationship allows experimental determinations for ΔG for any electrochemical reaction.
Note that this expression is valid at nonstandard and standard conditions, ΔG° = – n F
Thus, because reactions are spontaneous when
→ reactions are spontaneous when
Ecell°.
ΔG is negative (–),
Ecell is positive (+).
Spontaneity of Redox Reactions
Up to now, we've predicted single-replacement redox reactions based on the Activity Series:
Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn >
Fe > Cd > Co > Ni > Sn > Pb > (H) > Cu > Ag > Au
The Activity Series is actually based on standard reduction potentials, Ered°, which indicate the
relative strength of oxidizing and reducing agents.
We know a redox reaction will occur only if Ecell° is positive.
→ If Ecell° is negative, the reverse reaction is spontaneous.
Example:
If alkaline batteries produce a cell potential of 1.5 V, calculate ΔG° for an alkaline
battery.
CHEM162: Gilbert Chapter 19 page 12 19.6
THE EFFECT OF CONCENTRATION ON Ecell
So far we have considered galvanic cells at standard conditions, but what happens when the
concentrations of the solutions used are not 1M?
For example, consider silver electrodes dipped into a half-cell with 1M AgNO3 and another halfcell with 0.1M AgNO3:
The tendency is for nature to equalize the concentrations, and the only way to do this is through a
flow of electrons from the less concentrated solution to the more concentrated solution.
→ As electrons flow to the more concentration solution, more Ag+ ions can be reduced to Ag
atoms, effectively lowering the concentration of the 1M AgNO3.
→ Thus, one can prepare a concentration cell in which both compartments contain the
same components but at different concentrations.
– If the concentration difference is large enough, a cell potential may be produced and
measured.
The cell potential for a concentration cell can be calculated by substituting the following:
ΔG = – n F Ecell
and
ΔG° = – n F Ecell°
into
ΔG = ΔG° + RT ln Q
– n F Ecell = – n F Ecell° + RT ln Q
to get
and dividing both sides by – nF yields
the Nernst equation:
CHEM162: Gilbert Chapter 19 page 13 Ecell = Ecell° – RT ln Q
nF
Since most reactions occur at room temperature (25°C) and R and F are constants, we can
simplify the Nernst equation for reactions at 25°C:
Ecell = Ecell
J ) (298.15K)
(
8.3145 mol
⋅K
°–
ln Q =
J )
n (96,485 mol
⋅V
Ecell° – 0.025693 V ln Q
n
and to convert ln Q to log Q = (ln 10)(ln Q) = (2.302585…) ln Q
Ecell =
Ecell° – 0.025693 V (2.302585…) ln Q = Ecell° – 0.059159.. . V log Q
n
n
which is usually rounded to
Ecell = Ecell° – 0.0592 V log Q.
n
Thus, the two most commonly used forms of the Nernst equation are
Ecell = Ecell° – 0.025693 V ln Q
n
and
Ecell = Ecell° – 0.0592 V log Q
n
These will be provided on Exams.
Ex. 1: Calculate the potential for the cell prepared by dipping silver electrodes into a half-cell with
1M AgNO3 and another half-cell with 0.1M AgNO3, which has the following cell diagram:
Ag(s) | Ag+ (0.10M) || Ag+ (1.0M) | Ag(s)
Ex. 2: Calculate the potential for the cell prepared by dipping silver electrodes into a half-cell with
1M Zn(NO3)2 and another half-cell with 0.0010M Zn(NO3)2, which has the following cell
diagram:
Zn(s) | Zn+2 (0.0010M) || Zn+2 (1.0M) | Zn(s)
CHEM162: Gilbert Chapter 19 page 14 Ex. 3: Calculate [H+] for a voltaic cell for the reaction, Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g),
where the [Zn2+]=1.0 M, PH2 =1.0 atm, and Ecell=0.54 V at 25°C.
Thus, we can calculate [H+] for any solution which can be used to determine the pH,
so using a voltaic cell to get [H+] and pH → a pH meter!
Calculation of Equilibrium Constants from Redox Reactions
At equilibrium, Q=K and ΔG=0
so combining these yields
At equilibrium,
Ecell° =
0=
→ since ΔG = – n F Ecell then Ecell=0.
Ecell° – RT ln K
nF
0.025693 V
ln K
n
and
Ecell° = 0.0592 V log K
n
Example: Use the table of standard reduction potentials to calculate the equilibrium constant, K,
for the reaction at 25°C:
Cd(s) + Pb2+(aq) → Cd2+(aq) + Pb(s)
CHEM162: Gilbert Chapter 19 page 15 DRY-CELL BATTERIES
– Flashlight, AA, and AAA batteries are examples of dry-cell batteries that do not contain large
amounts of water.
Alkaline dry-cell batteries consist of a Zn case with
OH , a base (hence the name alkaline) that serves as
the anode and a carbon rod and MnO2 paste that
serves as the cathode:
−
Anode half-reaction (oxidation):
Zn(s) + 2 OH (aq) → Zn(OH)2(s) + 2 e
−
−
Cathode half-reaction (reduction):
2 MnO2(s) + 2 H2O(l) + 2 e →
2 MnO(OH)(s) + 2 OH (aq)
−
−
The driving force causing electrons to flow is called the voltage (or potential difference).
– For these two half-reactions product above, the voltage is produced is 1.5 V,
so higher voltages would require two or more batteries to be connected in a series.
LEAD-ACID STORAGE BATTERIES
– Common automobile batteries are examples of lead-acid storage batteries.
– They consist of six electrochemical cells wired in series, with each cell producing 2 V for a
total of 12 V.
– The anode (negative electrode) consists of a grid made of porous Pb while the cathode
(positive electrode) consists of PbO2, with both anode and cathode immersed in H2SO4(aq)
Anode half-reaction (oxidation):
Pb(s) + HSO4−(aq) →
PbSO4(s) + H+(aq) + 2 e−
E°= 0.356V
Cathode half-reaction (reduction):
PbO2(s) + 3H+(aq) + HSO4−(aq) + 2e− →
PbSO4(s) + 2 H2O(l)
E°= 1.685V
Ex. 1a:
Write the overall reaction, and determine the E° for the overall reaction.
CHEM162: Gilbert Chapter 19 page 16 Ex. 1b:
Write the reaction quotient for the overall reaction.
Ex. 1c:
Use the Nernst equation to calculate the cell potential when the sulfuric acid
concentration drops from 1.00M to 0.100M.
Ex. 1d:
What does your answer to Ex. 1c indicate about the nature of batteries when the
concentrations of reactants decrease by an order of magnitude?
Note that the product for both half-reactions is PbSO4, so if the battery runs for a long time
without recharging, the PbSO4 builds up, and the battery goes dead.
→ However, the alternator in a car produces an electrical current that reverses the half-reactions
shown above, converting the PbSO4 back into Pb and PbO2, and thus, recharging the battery.
– Thus, electrical currents can be used to reverse spontaneous redox reactions.
→ This process is called electrolysis.
19.7 RELATING BATTERY CAPACITY TO QUANTITIES OF REACTANTS
One important characteristic of a battery is its capacity to do work—i.e., its ability to delivery
electrical charge at the designed cell potential.
The amount of electrical work a battery can do is as follows:
where C=quantity of electrical charge in coulombs
and a coulomb (C) is defined as follows:
w elec = − C
1 C = 1 A⋅s
where an ampere (A) is the SI base unit of electrical current.
These units can be related to joules as follows:
CHEM162: Gilbert Chapter 19 page 17 1 J = 1 C⋅V = 1 A⋅s⋅V
E cell
However, joules are such small energy units that larger units are need to express power and
energy capacity, which leads us to watts (W) and kilowatt-hours (kWh) as follows:
1 W = 1 J/s
1 kilowatt = 1 kW = 1000 W
1 kW⋅h = (1000 W)(1 h)
Thus, a 100W incandescent light bulb uses 100 J every second.
Most power companies report and bill electricity usage in kilowatt-hours (kWh)—e.g. an Energy
Star rated 22 ft3 refrigerator made after 2001 uses between 500-600 kWh per year while a
refrigerator made before 1990 uses over 1100 kWh.
Example: A 100 W light bulb running for 10.0 hours requires 1 kWh of energy. If the average
Seattle home uses 25 kWh/day, this is equal to how many 100 W incandescent light
bulbs running nonstop for one day?
Nickel-Metal Hydride Batteries
Many rechargeable batteries are made of nickel-metal hydride (NiMH) cells, in which the anode
contains hydrogen as a hydride with a transition metal or metal alloy (generally shown as MH).
The cathode and anode half reactions
for a NiMH cell are as follows:
cathode:
NiO(OH)(s) + H2O(l) + e– →
Ni(OH)2(s) + OH–(aq) E°=1.32V
anode:
MH(s) + OH–(aq) →
M(s) + H2O(l) + e–
overall reaction:
MH(s) + NiO(OH)(s) →
M(s) + Ni(OH)2(s)
The overall E !cell can’t be
determined without knowing the
specific metal hydride involved, but in
general the standard oxidation
potential is similar to that for the SHE,
which is 0.0 V.
→ Most NiMH are rated for 1.2 V.
CHEM162: Gilbert Chapter 19 page 18 Quantitative Relationships
– Michael Faraday developed the quantitative treatment for electrochemical cells
– The mass of product formed (or reactant consumed) at an electrode is directly proportional to:
1. the amount of electricity (or electrons) transferred at the electrode
2. the molar mass of the substance
Units to know:
faraday = 1 mole of electrons = 96,485 coulombs
1 coulomb = amperes · seconds or 1 C = 1 A·s
1 joule = volts · coulombs or
1 J = 1 V·C
Ex. 1
One type of NiMH rechargeable battery has a rated voltage of 1.2V and a rated capacity
of 2700 mAh (milliampere-hours). Use the cathode half-reaction to calculate the mass of
NiO(OH) consumed to meet the rated capacity.
Lithium-Ion Batteries
Li-ion batteries are used
to power laptops, cell
phones, and many digital
cameras. They’re also
used in hybrid and
electric cars since they
have much higher energy
capacity relative to size
compared to NiMH
batteries.
In a Li-ion battery, Li+
ions are stored in a
graphite anode, and
during discharge the Li+
ions migrate through a
nonaqueous electrolyte to
a porous cathode made
of transition metal oxides
or phosphates.
CHEM162: Gilbert Chapter 19 page 19 At the cathode, the Li+ ions form stable complexes with the cathode materials—e.g. one popular
cathode material is CoO2.
The cell potential is about 3.6 V and the cell reaction for a battery with this cathode is as follows:
Li1-xCoO2(s) + LixC6(s) →
C(s) +
LiCoO2(s)
For a fully charged cell, x=1, making the cathode lithium-free CoO2. As the cell discharges, Li+
ions migrate to the cathode, and x falls towards zero.
– To balance the flow of positive ions, electrons flow from the anode to the cathode through an
external circuit.
– Since the electrodes in a Li-ion battery may react with water and oxygen, the background
electrolytes (e.g. LiPF6) are dissolved in polar organic solvents (e.g. tetrahydrofuran, ethylene
carbonate, and propylene carbonate).
19.9 ELECTROLYTIC CELLS AND RECHARGEABLE BATTERIES
While a spontaneous reaction occurs in a voltaic cell, a nonspontaneous reaction can be made to
occur if a power supply is used → electrolysis.
Electrolysis: the process of using electrical energy to cause a nonspontaneous redox reaction
– This is the process used to plate objects with metals (e.g. silver-plated or gold-plated jewelry)
or to obtain a pure element from a compound.
CHEM162: Gilbert Chapter 19 page 20 electrolytic cell: the type of cell where electrolysis occurs
– Consider the electrolytic cell shown above.
– The power supply provides direct electrical current.
– Two wires lead from the terminals of the power source to the electrolytic cell.
– Again, oxidation occurs at the anode and reduction occurs at the cathode.
– The external power source acts as an electron pump.
– The power source draws electrons from the anode, which must be connected to the
positive terminal (usually shown in red) on the power source.
– The power source forces electrons to the cathode, which must be connected to the
negative terminal (usually shown in black) on the power source.
→ The external power source (usually a battery or electrical source) allows the
nonspontaneous reaction to occur.
An example of an electrolytic cell
– Consider a Cu-Zn electrolytic cell.
– The storage battery provides direct
electric current.
– Two wires lead from the terminals of
the battery to the electrolytic cell.
– The cell consists of two electrodes
dipped into a solution with its
corresponding ions, Cu into Cu(NO3)2
and Zn into Zn(NO3)2.
– The battery acts as an electron pump,
pushing electrons to the cathode and
removing them from the anode.
– The external power source (the
battery) allows the nonspontaneous
reaction to occur.
Ex. 1: Write the half reactions and overall reaction for the electrolytic cell shown above:
anode:
cathode:
overall:
CHEM162: Gilbert Chapter 19 page 21 Cell Reactions in Water Solutions
– Consistent with other redox reactions in galvanic cells, reduction occurs at the cathode and
oxidation occurs at the anode for electrolytic cells.
The following reductions can occur:
1. The reduction of cations to its corresponding metal
– characteristic of electroplating processes
Ag+(aq) + e–
→
Cu+2(aq) + 2 e–
Ag(s)
→ Cu(s)
2. The reduction of water molecules to hydrogen gas
– usually occurs when the cation in solution is such a strong reducing agent that it’s
difficult to reduce (e.g. K+, Na+)
– i.e., rather than the K+ being reduced to K or Na+ being reduced to Na, the H2O
molecules are reduced instead.
2 H2O(l) + 2 e– → H2(g) + 2 OH (aq)
−
E red°=-0.83 V
E red (actual) =-0.41 V
since [OH-]=10-7M in H2O
The following oxidations can occur:
1. The oxidation of anions to its corresponding nonmetal, like 2 Cl (aq) →
−
Cl2(g) + 2 e–.
2. The oxidation of water molecules to oxygen gas
– occurs when anion present is such a strong oxidizing agent that is almost impossible
to oxidize (e.g. F–, SO42–)
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e–
E ox°=-1.23 V
E ox (actual) =-0.82 V
since [H+]=10-7M in H2O
Ex. 1 a. Calculate the actual E cell for the electrolysis of water: 2 H2O(l) → 2 H2(g) + O2(g).
(When writing the overall reaction, consider what happens to the ions produced in the
half reactions when they’re combined.)
b. Is the reaction spontaneous?
CHEM162: Gilbert Chapter 19 Yes
page 22 No
Units to know:
faraday = 1 mole of electrons = 96,485 coulombs
1 coulomb = amperes · seconds or 1 C = 1 A·s
1 joule = volts · coulombs or
1 J = 1 V·C
Ex. 1: Salt water chlorination systems for pools are becoming more common. These systems
often use sodium chloride which can be converted into chlorine gas by electrolysis.
a. Write the two half reactions that occur at the two electrodes. (Note that the reduction
half-reaction is NOT the reduction of sodium!)
b. If a current of 25 amps passes through the electroytic cell in a pool’s chlorinator,
calculate the mass of chlorine gas produced in 1.0 h.
Ex. 2: An aqueous solution containing Pt4+ ions is electrolyzed by passing a current of 2.50 A.
How many hours must the current run to get 10.0 g of metallic platinum to form at the
cathode?
CHEM162: Gilbert Chapter 19 page 23 Consider again the electrolysis of water:
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e–
E ox=-0.82 V
2 H2O(l) + 2 e– → H2(g) + 2 OH (aq)
E red=-0.41 V
−
a. Write the overall reaction above (with the electrons
balanced and the coefficients simplified).
b. How would the volume of H2 gas produced differ from the
volume of O2 gas? Explain why.
FUEL CELLS
fuel cell: a voltaic cell that relies
on oxidizing a fuel (e.g. H2) to
generate electrical energy
Fuel cells used in electric vehicles use
acidic electrolytes and a proton exchange
member (PEM).
The half reaction at the anode results from
H2 fuel diffusing (moving) into a porous
graphite electrode, where it’s oxidized to
+
H ions:
+
anode: H2(g) → 2 H (aq) + 4 e
–
Electrons are pushed through an external
circuit by a cell potential of about 0.7 V,
about half the cell potential for the
reaction,
H2(g) + O2(g) → 2 H2O(l)
As electrons get pumped out of the cell
through the anode, H+ ions migrate to
through the PEM to the cathode, where
they help in the reduction of O2 from air:
+
–
cathode: O2(g)+ 4H (aq) + 4 e → 2H2O(l)
An individual PEM fuel cell produces a
potential of about 1.0 V, less than a 1.5V
AA battery, but hundreds of these cells can be assembled into fuel cell stacks that are capable of producing 100 kW
of electrical power, enough to power a car to run about 100 mph.
CHEM162: Gilbert Chapter 19 page 24