Batteries: Producing Electricity via Chemical Reactions
• A battery is a cell or series of cells that generate an electrical
current.
The Leclanché (Dry) Cell
– Batteries are the means by which we harness the electrical work of a
galvanic cell and use it productively.
• Primary Cells.
– Single-use batteries that cannot be recharged
– Cell reaction is not reversible.
– The most prevalent type is the alkaline battery.
• Secondary Cells.
– Cell reaction can be reversed by passing electricity through the cell
(charging).
– Nickel-cadmium or “ni-cad” batteries are an example
• Flow Batteries and Fuel Cells.
– Materials pass through the battery which converts chemical energy to
electric energy.
Dry Cell
Alkaline Dry Cell
2MnO2(s) + Zn(s)  Mn2O3(s) + ZnO(s)
Zn(s) → Zn2+(aq) + 2 e-
Oxidation:
Reduction:
2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-
Acid-base reaction:
NH4+ + OH- → NH3(g) + H2O(l)
Precipitation reaction: NH3 + Zn2+(aq) + Cl- → [Zn(NH3)2]Cl2(s)
Alkaline Dry Cell
Reduction:
• The chemistry of
an alkaline dry
cell battery. The
net reaction is
shown above.
– The alkaline
battery is termed
a dry cell because
the KOH
electrolyte is in
the form of a
paste or gel.
The Silver-Zinc Cell: A Button Battery
2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-
Oxidation reaction can be thought of in two steps:
Zn(s) → Zn2+(aq) + 2 eZn2+(aq) + 2 OH- → ZnO + H2O
Zn (s) + 2 OH- → ZnO + H2O + 2 e-
Zn(s),ZnO(s)|KOH(sat’d)|Ag2O(s),Ag(s)
Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s)
Ecell = 1.8 V
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Zinc-Mercury Battery
• Mercury batteries are another type of primary cell and are
quite small. They are used for medical devices like
pacemakers.
– Zinc is the anode.
Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2e– Mercury(II) oxide is the cathode
cathode.
HgO(s) + H2O(l) + 2e-  Hg(l) +2OH-(aq)
• The mercury battery (also
called a zinc-mercuric
oxide cell) has a voltage
output that is extremely
stable over long times.
Zinc-Air Battery
• Zinc-air batteries are also primary cells.
– Zinc is the anode.
Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2e– Oxygen reacts at the cathode.
½ O2(g) + H2O(l) + 2e- 2OH-(aq)
• In a zinc-air battery,
one of the reactants is
oxygen from the
surrounding air. As a
result, these batteries
can offer a very
attractive energy
density.
Ni-Cd Battery
Ni-Cd Battery
• The anode for a ni-cad battery is cadmium.
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e-
• The complex cathode reaction can be represented
as
NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq)
• Important design features of a nickel-cadmium battery are shown
to the left.
Metal Hydride Battery
Nickel-MetalyHydride Battery
• Nickel-metal-hydride batteries are secondary
cells.
– The anode for a nickel-metal-hydride battery is M,
some metal or metal alloy.
MH(s) + OH-(aq)  M(s) + H2O(l) + e– The complex cathode reaction can be represented as
NiO(OH)(s) + H2O(l) + e-  Ni(OH)2(s) + OH-(aq)
• Nickel-metal-hydride batteries have become popular as rechargeable
cells. The design is quite similar to the Ni-Cd cell, but nickel-metalhydride cells are less prone to memory effects.
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Lead Acid Battery
Lead-Acid Battery
• The lead-acid storage battery found in cars is a secondary
cell.
Pb(s) + HSO4
-(aq)
 PbSO4(s) +
H+(aq)
Reduction:
PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- → PbSO4(s) + 2 H2O(l)
– The anode for a lead-acid battery is lead metal.
+2
e-
– The cathode for a lead-acid battery is lead oxide.
Oxidation:
Pb (s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2 e-
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e-  PbSO4(s) + 2 H2O(l)
• The lead-acid storage
battery consists of Pb
anodes alternating
with PbO2 cathodes,
all immersed in
sulfuric acid.
Fuel Cells
• A fuel cell is a voltaic cell in which the reactants
can be supplied continuously and the products of
the cell reaction are continuously removed.
• Most common type is based on the reaction of
hydrogen and oxygen to produce water.
2H2 + O2  2H2O
– Oxygen is reduced at the cathode.
O2 + 4H+ + 4e-  2H2O
– Hydrogen is oxidized at the anode.
H2  2H+ + 2e-
PbO2(s) + Pb(s) + 2 H+(aq) + 2HSO4-(aq) → 2 PbSO4(s) + 2 H2O(l)
E°cell = E°PbO2/PbSO4 - E°PbSO4/Pb = 1.74 V – (-0.28 V) = 2.02 V
Fuel Cells
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
2{H2(g) + 2 OH-(aq) → 2 H2O(l) + 2 e-}
2H2(g) + O2(g) → 2 H2O(l)
E°cell = E°O2/OH- - E°H2O/H2
= 0.401 V – (-0.828 V) = 1.229 V
Efficiency value
 = ΔG°/ ΔH° = 0.83
Electrolysis
Limitations of Batteries
• Corrosion is a major cause for the loss of
performance in batteries.
– Protective plating of materials used in batteries is an
attempt to limit the performance-diminishing effects of
corrosion on batteries.
• Electrolysis is the process of passing an electric
current through an ionic solution or molten salt to
produce a chemical reaction.
ect o yt c cells
ce s are
a e divided
d v ded into
to two categories
catego es
• Electrolytic
based on the nature of the electrodes used.
– Passive electrolysis: the electrodes are chemically inert
materials that simply provide a path for electrons.
– Active electrolysis: the electrodes are part of the
electrolytic reaction.
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Electrolysis:
Causing Non-spontaneous Reactions to Occur
Galvanic and Electrolytic Cells
Galvanic Cell:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
E°cell = 1.10 V
El
Electolytic
l
Cell:
C ll
Zn2+(aq) + Cu(s) → Zn(s) + Cu2+(aq)
E°cell = -1.10 V
Electrolysis is forcing a current through a cell to produce a
chemical change for which the cell potential is negative.
a) A standard galvanic cell based on the spontaneous reaction
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
b) A standard electrolytic cell. A power source forces the
opposite reaction: Zn2+(aq) + Cu(s)  Zn(s) + Cu2+(aq)
Passive Electrolysis in Refining Aluminum
• Electrolysis provides the means to overcome the
nonspontaneous reaction to separate aluminum
from its oxide.
– The Hall-Heroult refining process uses carbon
electrodes as inert sites for passive electrolysis.
– The Hall-Heroult process
involves the
electrolytic refining of
aluminum from Al2O3 to
produce aluminum metal
and oxygen gas.
Active Electrolysis and Electroplating
• The process of depositing a thin coat of metal on another
metal by using electrolysis is electroplating.
– In some cases, the thin coating is cosmetic, or to provide some vital
functionality for the coated piece, such as corrosion resistance or
desirable conductive properties.
• Silver is plated onto electrical devices because silver is a
good conductor and resistant to corrosion.
– The solution from which silver is plated contains CN–(aq) ions,
which form a complex with Ag+. The need for uniform coatings
makes this an important step.
Active Electrolysis and Electroplating
• The object being electroplated is the cathode.
Active Electrolysis and Electroplating
– Anode
Ag(s) + 2CN-(aq)  Ag(CN)2-(aq) + e– Cathode
A (CN)2-(aq)
Ag(CN)
( ) + e-  Ag(s)
A ( ) + 2CN-(aq)
( )
• Opposite reactions at the anode and cathode are
common for electroplating operations.
– Silver is transferred from the anode to the cathode,
coating the cathode in a thin layer of silver.
– The zero cell potential is not critical since an external
current drives electrolysis.
• Barrel plating is often used to apply coatings to small parts.
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Electroplating
Electrolysis and Stoichiometry
• For electroplating, it can be vitally important to
use carefully controlled amounts of materials.
– C
Controlling
t lli the
th flow
fl off electrons
l t
(current)
(
t) in
i an
electroplating operation provides a method to
accurately limit the amount of material deposited.
(a) A silver-plated teapot.
Net Equation: Ag(s)anode  Ag(s)cathode
(b) Schematic of the
electroplating of a spoon
– Electroplating is often used to prevent galvanic
corrosion in an electrical apparatus in places where
different metals come into contact with one another.
Transfer of material from the anode to the cathode
Quantitative Aspects of Electrolysis
How much chemical change occurs with the flow of a given
current for a specified time?
current and time  quantity of charge
Charge (C) = current (C/s) x time (s)
quantity of charge moles of electrons
1 mol e- = 96485 C
ne- = I x t
F
moles of electrons  moles of analyte  grams of analyte
Electrolysis Examples
1. A solution containing Cu2+(aq) is electrolyzed. The
cathode reduction half reaction is
Cu2+(aq) + 2e-  Cu(s).
What mass of copper will be deposited in 1.00hr by a
current of 1.5A?
2. A current of 25.5 A is used in the synthesis of MnO2(s) by
Mn2+(aq) + 2H2O  MnO2(s) +2H+(aq) + H2(g)
If the electrolysis is 85% efficient, how long would it take
to produce 1.000 kg MnO2(s)
Calculations in Electrolysis
• A knowledge of current, how long the current
flows, stoichiometry, and the number of electrons
required to reduce a metal cation are used to
answer the following questions.
questions
– How much material is plated given a specific current
for an allotted time or electrical energy expenditure?
– How long must a given current to pass through the cell
to yield a desired mass of plated material?
Corrosion
• Oxidation of metals
– Metals tend to be easily oxidized
– most metals have standard reduction potentials that are
less than oxygen
• 00.40
40 V in basic solutions
• 1.23 V in acidic solutions
– The oxidation of most metal by O2 will have a positive
Ecell, so they will spontaneously occur
– Some metals, such as copper, gold, silver and platinum,
are relatively difficult to oxidize. These are often called
noble metals.
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Corrosion: Unwanted Galvanic Cells
The Electrochemical Corrosion of Iron.
In neutral solution:
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
2 Fe(s) → 2 Fe2+(aq) + 4 e-
EO2/OH- = 0.40 V
EFe/Fe2+ = 0.44 V
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)
Ecell = 0.84 V
In acidic solution:
O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = 1.23 V
Corrosion Prevention
• The following observations can be used to reduce
corrosion in a number of ways.
Coatings
• Applying a protective coating to a material is the most
common way of protecting against corrosion.
– A coating can be applied with electroplating or painting.
– The coating protects the underlying material from exposure to water
and oxygen.
– Corrosion is a pervasive reaction, with a large, negative
free energy change.
– It is possible to predict what materials will corrode and
use this information to protect a material such as iron.
• Rust inhibitors can be added to paint to further inhibit
corrosion.
– Many of the common inhibitors contain the following ions:
phosphate, borosilicate, chromate, or phosphosilicate.
– All of these ions, as part of a paint coating, form compounds with
oxidized iron that inhibit further rust formation.
– This process is called passivation.
– Some materials, like aluminum, corrode readily, but the
product, in this case Al2O3, forms a protective layer that
eliminates further corrosion.
Cathodic Protection
Cathodic Protection
• Some materials are more easily oxidized than iron,
which provides a way to construct galvanic
corrosion conditions intended to protect the iron.
– Mg has a reduction potential more negative than Fe.
– When combined, Mg will oxidize and Fe will be
reduced.
• Magnesium can be used to prevent iron corrosion.
– The piece of magnesium is called a sacrificial anode.
– Connecting magnesium to iron forces iron to be the
cathode, preventing iron from oxidizing.
– This process is called cathodic protection.
– The sacrificial anode must be replaced periodically to
be effective.
•
Sacrificial anodes are one effective method of corrosion prevention.
The anode is preferentially oxidized relative to the protected metal.
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Corrosion Protection
Summary Examples
1.
Consider an aluminum-air battery
a) The Al3+ produced at the anode is obtained as the
complex ion [Al(OH)4]-. Write equations for the
oxidation and reduction half-reactions and the net
reaction that occur in the battery.
b) The E
E°cell of the aluminum-air
aluminum air cell is +2.73
+2 73 V.
V What
is the value of the standard reduction potential of the
reaction: [Al(OH)4] + 3e Al(s) + 4OH
c) Given that G°f[OH-]= -157kJ/mol and
G°f[H2O(l)]= -237.2 kJ/mol,
What is G°f [Al(OH)4]-?
d) What mass of Al is consumed if 10.0A of electric
current is drawn from the battery for 4.00 hr.
Al-Air Batteries
Summary Examples
2. Two electrochemical cells at 25°C are connected
as shown
Cell A
┌─Zn(s)|Zn2+(0.85 M)║Cu2+(1.10 M)|Cu(s) ┐
│
│
2 (1.05 M)║Cu
2+
2 (0.75 M)|Cu(s) ┘
└
└─Zn(s)|Zn
║ 2+
Cell B
4 Al(s) + 3 O2(g) + 6 H2O(l) + 4 OH- → 4 [Al(OH)4]-(aq)
a) Do the electrons flow in the clockwise or counter
clockwise direction?
b) What are the ion concentrations in the half-cells at the
point at which the current stops flowing.
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