Download Batteries: 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 are made from a voltaic cell or a series of 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.
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Desired qualities:
compact, high current, constant voltage, long life, low cost, limit hazards
Three major types of batteries:
Primary, secondary and fuel cell
Electrochemistry
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Alkaline Batteries: primary batteries (not rechargeable; throw away)
Anode is zinc:
Zn(s) + 2 OH-(aq) —> Zn(OH)2(s) + 2 eCathode is MnO2:
2 MnO2(s) + 2 H2O(l) + 2 e- —> 2 MnO(OH)(s) + 2 OH-(aq)
Single cell emf = 1.55 V at room temperature
Advantages:
Low cost, constant voltage, longer shelf life than older dry cell, no
gases produced.
Disadvantages: Disposal of KOH and MnO2
Nickel-Metal Hydride (“NiMH”): secondary battery (rechargeable)
Used in hybrid gas-electric cars; considered “mature technology”
Cathode is NiO(OH): (Nickel (III) oxyhydroxide) The “Ni” in NiMH
NiO(OH)(s) + H2O(l) + e- —> Ni(OH)2(s) + OH-(aq)
The anode is commonly a ZrNi2 alloy embedded with hydrogen atoms, represented as
“MH” in NiMH. The hydrogen (0 oxidation state) is oxidized to H+ ions. The alloy material
does not react.
MH(s) + OH– —> M + H2O(l) + e(ZrNi2)H + OH– —> (ZrNi2) + H2O(l) + eAdvantages over ni-cad rechargeable batteries:
Longer life (Up to 8 years), no Cd to dispose of.
A disadvantage: poor performance in cold weather
Electrochemistry
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Lead Storage Battery (used in automobiles, rechargeable)
6 Lead plates immersed in H2SO4(aq):
3 Anode Plates:
Pb(s) + HSO4–(aq) —> PbSO4(s) + H+(aq) + 2e– (+0.356 V)
3 Cathode Plates:
PbO2(s) + HSO4–(aq) + 2H+(aq) + 2e– —> PbSO4(s) + 2 H2O(l)(+1.685 V)
Write the overall cell reaction on discharge. What is Ecell?
Series of 6 cells:
Total Voltage = ?
Rechargeable by reversing each half-cell reaction since the product,
PbSO4(s), remains attached to each Pb electrode.
Advantages:
Low cost, high current.
Voltage relatively constant since emf does not depend on
concentrations of PbO2, Pb and PbSO4 (Why is this?).
Disadvantages:
Heavy, Pb disposal. Fails over time due to mechanical dislodging (loss)
of the PbSO4 from the electrodes.
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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Nickel-Cadmium (“ni-cad”) Rechargeable
Anode is Cd:
Cd(s) + 2 OH-(aq) —> Cd(OH)2(s) + 2 e-
Cathode is NiO(OH):
NiO(OH)(s) + H2O(l) + e- —> Ni(OH)2(s) + OH-(aq)
Advantages: light weight, constant voltage
Disadvantages: high cost, short life, disposal of Cd.
a) Write the overall balanced reactions for the discharge and charge of a nickel-cadmium rechargeable battery.
b) 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
c) 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?
Electrochemistry
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PEM Fuel Cells (proton-exchange membrane)
Electrochemical cells
operatewith an external
reservoir of reactants. A
combustion reaction
separated into two half-cells
is used to produce electricity.
The fuel does not ‘burn’ in
Anode is H2(g)
the traditional sense.
Used by NASA as a fuel
2 H2(g) —> 4 H+(aq) + 4 esource for spacecraft.
Disadvantages: Not selfcontained (source of fuel
continually needed).
Electrodes are short-lived
Question:
What is the overall reaction?
Cathode is O2
O2(g) + 4
H+(aq)
Electrochemistry
+4
e-
5
—> 2 H2O(l)
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 three chemical 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
Step 1:
Oxidation of solid Fe at the anodic region to Fe2+:
Fe(s) → Fe2+(aq) + 2 e–
Why must water be present?
The damage is done at this point. A pit is left
where the Fe(s) was oxidized.
Step 2:
Electrons migrate to cathode.
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?
Step 4: a second oxidation 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 rust formation:
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 is 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, the iron in the steel simply serves as a pathway to conduct electrons to oxygen
that is reduced, instead of being oxidized. The Fe becomes an “inert” electrode for the cathode reduction.
Mg(s) —> Mg2+(aq) + 2 e– Zn(s) —> Zn2+(aq) + 2 e– Fe(s) —> Fe2+(aq) + 2 e– E° = +2.37 V
E° = +0.76 V
E° = +0.44 V
Galvanized Nails: Steel nails coated with zinc.
Objects can be galvanized by electroplating zinc onto the surface or by
dipping the object in molten zinc.
Explain why zinc or magnesium is 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
Cu(s) 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.
Electrochemistry
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Combustion, Fuel Cells and Thermodynamics
•
Hydrogen gas has the potential as a clean fuel in reaction with oxygen. The relevant reaction is
2H2(g) + O2(g) → 2H2O(l)
Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases
are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii)
hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at 85 C.
a) Using thermodynamic data, calculate ΔH° and ΔS° for the above reaction. We will assume that these values do not
change appreciably with temperature.
•
Based on the values from part (a), what trend would you expect for the magnitude of ΔG for the above reaction as
the temperature increases?
•
What is the significance of the change in the magnitude of ΔG with temperature with respect to the utility of
hydrogen as a fuel?
a) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to
generate energy from hydrogen?
Electrochemistry
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