Download Commercial Voltaic Cells

Batteries: Commercial Voltaic Cells
• The first known battery, a Zn/Cu battery was built in 1836 by English chemist
John Frederick Daniell.
Batteries consist of a series of self-contained voltaic cells
connected anode to cathode. When connected in series, anode
to cathode, the voltage of the battery is the sum of the individual
voltaic cells. Following voltaic cell labeling conventions, the
electrodes of a battery are labeled (+) for the cathode and (–) for
the anode.
! Desired qualities: compact, high current, constant voltage,
long life, low cost, limit hazards
! Two major types of batteries:
Primary:
Secondary:
Electrochemistry
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Alkaline Batteries: Primary Batteries
Anode is reaction:
Zn(s) + 2 OH-(aq) —> Zn(OH)2(s) + 2 eCathode reaction:
2 MnO2(s) + 2 H2O(l) + 2 e- —> 2 Mn(OH)2(s) + 2 OH-(aq)
Single cell emf = 1.5 V at room temperature. How many voltaic cells are connected in series
in a 9 V battery?
Advantages:
Low cost, constant voltage, relatively long shelf life
Disadvantages: Disposal of KOH and MnO2
Electrochemistry
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Lead Storage (Car) Battery: Secondary Battery
Anode Reaction:
Pb(s) + HSO4–(aq) —> PbSO4(s) + H+(aq) + 2e–
Eox = +0.356 V
Cathode Reaction:
PbO2(s) + HSO4–(aq) + 3H+(aq) + 2e– —> PbSO4(s) + 2 H2O(l)
Ered = +1.685 V
Upon discharge, the battery generates electrical energy. Six
voltaic cells are connected in series. What is the total volatge?
Rechargeable by reversing the cell reaction since the product,
PbSO4(s), remains attached to each Pb electrode. What is the
minimum voltage needed to recharge?
Advantages:
Low cost, produces high current needed to start the engine
(100 to 200 amps) in hot and cold weather.
Voltage relatively constant since emf does not depend on
concentrations of PbO2, Pb and PbSO4 (Why is this?).
Disadvantages, can you list some?
Spongy lead: metallic lead
brought to a spongy form by
reduction of lead salts, or by
compressing finely divided
lead; -- used in secondary
batteries and otherwise.
[Websters].
Electrochemistry
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Lead Storage Battery Problem
What mass of Pb is consumed upon start up of a car engine if start up requires 150
amps of current for 2.0 seconds? What is ∆G for this amount of Pb consumed?
Electrochemistry
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Nickel-Cadmium (“ni-cad”) Rechargeable
Anode Reaction:
Cd(s) + 2 OH-(aq) —> Cd(OH)2(s) + 2 eCathode reaction:
NiO(OH)(s) + H2O(l) + e- —> Ni(OH)2(s) + OH-(aq)
Advantages: light weight, constant voltage
Disadvantages: high cost, short life, disposal of toxic Cd
(Has been replaced by nickel-metal hydride battery)
Questions:
a) Given the following reduction potentials, calculate the equilibrium constant at 25°C for
the overall ni-cad battery reaction:
Cd(OH)2(s) + 2 e-—> Cd(s) + 2 OH-(aq) E°red = –0.76 V
NiO(OH)(s) + H2O(l) + e —> Ni(OH)2(s) + OH (aq)
E°red = +0.49 V
Electrochemistry
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Nickel-Cadmium (“ni-cad”) Rechargeable
b) What would happen to the cell voltage if the concentration of hydroxide ion in the anode
was increased? What effect does an increase in [OH–] have on the equilibrium constant?
c) Write the overall balanced reaction for the recharge of a nickel-cadmium rechargeable
battery.
Electrochemistry
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PEM (proton-exchange membrane) Fuel Cells
Electrochemical cells that
operate with an external
reservoir of reactants. A
combustion reaction
separated into two half-cells
is used to produce electricity.
Anode:
The fuel does not ‘burn’ in
the traditional sense.
2 H2(g) —> 4 H+(aq) + 4 eUsed by NASA as a source
of energy and water on
spacecraft.
Disadvantages: Not selfcontained (source of fuel
continually needed).
Electrodes are short-lived
and expensive.
Questions:
What’s the main difference between a fuel cell
and a battery?
What is the overall reaction for the hydrogen
fuel cell shown?
Cathode:
O2(g) + 4 H+(aq) + 4 e- —> 2 H2O(l)
Ecell = 1.2 V
Electrochemistry
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Fuel Cells-Example Applications
• Spacecraft use fuel cells to provide electricity. One cell that has been proposed is based on
hydrazine (N2H4) and O2. The standard reduction potentials for the two half-cell reactions are the
following:
N2(g) + 4 H2O(l) + 4e- → N2H4(aq) + 4 OH–(aq)
E°red = –1.16 V
O2(g) + 2H2O(l) + 4e- → 4OH–(aq) E°red = +0.40
Use the standard reduction potentials provided above to determine and then write the balanced
chemical equation for the reaction that occurs at the anode of this fuel cell.
What current is produced by this fuel cell if it takes 50.0 hours for 7.5 g of hydrazine to react?
(The molar mass of hydrazine is 32.05 g/mol.)
For you Bio-Nerds look into Microbial Fuel Cells
Electrochemistry
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Corrosion: Rust Formation on Iron Surfaces
The rusting of steel is of major economic concern. The iron in steel is spontaneously
oxidized to Fe2O3(s) in the presence of water and dissolved oxygen. The structural integrity
of the steel is compromised. The corrosion of iron, Fe(s), is a complex electrochemical
process. The process can be broken down into four steps. Several features of the corrosion
should be noted:
1.
Moisture, H2O(l), must be present.
2.
O2 must be present in the moisture.
3.
Iron rusts faster in ionic solutions and low pH (acidic) solutions.
4.
The loss of Fe(s) and the deposition of rust (Fe2O3) can occur at different places.
5.
Iron rusts slower in the presence of active metals (Zn, Mg for example).
Electrochemistry
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Corrosion: Rust Formation on Iron Surfaces
4
3
2
1
Step 1:
Oxidation of solid Fe at the anodic region to Fe2+:
Fe(s) → Fe2+(aq) + 2 e–
Why must water be present?
Step 3:
Reduction of O2 at the cathodic region:
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) (faster) or
O2(g) + 2 H2O(l) + 4 e– → 4 OH-(aq) (slower)
Why acidic conditions favored?
The damage is done at this point. A pit is
formed where the Fe(s) was oxidized.
Step 2:
Electrons migrate to cathodic
region
Step 4: a second redox reaction with oxygen
Oxidation of the Fe2+ from step 1 by dissolved O2:
4 Fe2+(aq) + O2(g) + 4 H2O(l) + 2 xH2O(l) → 2Fe2O3•xH2O(s) + 8H+(aq)
Rust is a hydrated iron(III) oxide.
Electrochemistry
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Corrosion: Rust Formation on Iron Surfaces
The overall redox reaction for rust formation is the sum of the reactions shown in steps 1, 3, and 4:
4 Fe(s) + 3 O2(g) + 2 xH2O(l) → 2 Fe2O3•xH2O(s)
What role does the H+(aq) play in this reaction sequence?
Why can the rust form at a different place than where the Fe(s) is oxidized?
Why do steel bridges form rust mostly at the water line, but not above or far below the water line?
Preventing Corrosion
There are several ways to stop or slow down corrosion:
1.
Wash off road salt from cars!
2.
Paint the surface to prevent contact with water and oxygen. Watch out for scratches!
3.
Seal the surface with a passive coating such as Al2O3. Watch out for scratches!
Passivation: Some active metals form oxides on their surfaces that are impermeable to O2 and H2O. The oxide
provides a coating that protects the underlying metal from further corrosion.
Examples: Mg, Al
NOTE: The tendency of these metals to “passivate” makes them ineffective as electrodes in voltaic cells. You
probably noticed that the voltage was unstable when using Mg and Al electodes in our voltaic cell experiment.
4.
Provide a SACRIFICIAL ANODE such as Zn or Mg. This CATHODIC PROTECTION provides a more favorable
oxidation than that of Fe.
Electrochemistry
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Cathodic Protection of Iron by “Active” Metals
When a sacrificial anode is used, instead of being oxidized the iron in the steel simply serves as a
pathway (inactive cathode) to conduct electrons to oxygen that is reduced.
Mg(s) —> Mg2+(aq) + 2 e– Zn(s) —> Zn2+(aq) + 2 e– Fe(s) —> Fe2+(aq) + 2 e– E°ox = +2.37 V
E°ox = +0.76 V
E°ox = +0.44 V
Galvanized: Steel objects such as nails or pipes coated with zinc. Objects can be galvanized by
electroplating zinc onto the surface or by dipping the object in molten zinc.
Questions:
Explain why zinc or magnesium are preferentially oxidized instead of the iron.
Explain why galvanizing is preferable to painting.
Which of the following metals would be a good choice to use for a sacrificial anode to protect iron
from corrosion? Ag, Na, Cr
Electrochemistry
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Speeding Up Corrosion-NOT GOOD!
When the Fe(s) is in contact with some inactive metals such as Cu. The presence of the
less active affects the kinetics of the reaction, but not the chemistry.
1)
The Cu acts as a conductor for the electrons produced in step 1.
2)
These electrons are used to reduce the O2(g) in step 3. STEP 3 is the slow or rate limiting
step in the reaction sequence. If step 3 takes place at the surface of Cu(s) the rate of
reduction increases, therefore the overall corrosion rate also increases.
3)
A copper water pipe should NEVER be directly attached to a steel pipe! An insulating
spacer must be used between them.
3
1
2
Electrochemistry
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