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Practical Examples of Galvanic Cells
There are many practical examples of galvanic cells in use in our everyday lives. We are
familiar with batteries of all types. One of the most common is the lead-acid battery, which has
the capability of delivering enough energy (measured in total charge or ampere-hours) to start an
automobile.
The electrodes are Pb(s), and PbO2(s). The supporting electrolyte is sulfuric acid.
Here are the important reactions:
Anode:
Pb(s) + HSO4- → PbSO4(s) + H+ + 2e-
lead is oxidized from the 0 to the +2 state
Cathode:
PbO2(s) + HSO4- + 3H+ + 2e- → PbSO4(s) +2H2O lead is reduced from the +4 to the +2 state
Net reaction:
Pb(s) + PbO2(s) + 2H+ + 2HSO4- → 2PbSO4(s) + 2H2O
The cell voltage for a lead acid battery is 2 volts, so six cells must be connected in parallel to
achieve the 12v operation required in automobiles.
Under ordinary conditions, current flows in the external
circuit from the anode (negative electrode) to the cathode
(positive electrode). To charge the battery, a potential of
opposite polarity is applied acrosse the terminals, and
reactions at each electrode are reversed. This battery can be
recharged because the reaction products adhere to the
electrodes.
Note that as the battery discharges, the concentration of the
sulfuric acid electrolyte decreases. The density of
concentrated H2SO4 is 1.28 g/cm3, and the density of H2O is
1.00 g/cm3, so a measurement of the density of the
electrolyte gives a measure of how charged the battery is.
Over the long haul, the electrodes degrade – sulfate
“whiskers” eventually short out the terminals, and the
battery is “sulfated”, and cannot be recharged.
Why does battery performance degrade with
temperature? Contrary to intuition, the temperature
dependence of the cell potential is very small. The voltage does not decrease significantly. The
real reason for the poor behavior of the battery at low temperature is the following: when the
battery discharges, H+ ions must move toward the cathode, SO42- ions move toward both
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electrodes. The viscosity of the H2SO4 electrolyte increases enough to reduce the mobilities of
the ions in solution. The effect is much larger for HSO4-. A much smaller current can flow
through the battery because of this reduction in mobility.
Dry cells
One of the most common examples of a battery
is the alkaline dry cell, sometimes called a
Leclanché cell
Anode:
Zn → Zn2+ + 2eCathode:
2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O
The cell potential is 1.5 volts.
Here’s an old exam question:
In a typical Leclanché cell, 30 g of MnO2 is combined with an excess of Zn, NH4+, graphite, and
NH4Cl electrolyte. In driving an iPod®, the cell was operated at 100 ma until all of the MnO2 in
the cell was consumed. Calculate how long current would flow. Assume the cell potential
remains constant.
Solution:
What determines the battery lifetime? The MnO2 provides charge. 30 grams of MnO2 is (30/55)
moles. Each mole of MnO2 gives one mole of electrons. So, the MnO2 is capable of providing
(30/55)moles × 96485 Coulombs per mole of electrons = 5.3 × 104 Coulombs
The iPod is using current at the rate of 100 ma × 10-3 amps/ma = 0.100 amps = 0.100 Coulombs
per second.
Q = it . Q, the total charge, is 5.3 × 104 Coulombs. Time = Q/i = 5.3 × 104 Coulombs/0.100
Coulombs sec-1 = 5.3 × 105 sec. There are 3600 seconds in an hour, so the battery will last
(5.3 × 105 )/(3.6 × 103) = 147 hours, a little less than a week of continuous play.
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A couple of other examples:
Calculator batteries:
Zn + HgO → ZnO + Hg and Zn + Ag2O → ZnO + 2Ag
High energy density rechargeable batteries:
The lithium ion battery is one of the most popular batteries for computer
operations:
Li + 2MnO2 → Li2O + Mn2O3
Hybrid vehicles like the Toyota Prius use a rechargeable battery known as a nickel metal
hydride (NiMH) battery.
Anode reaction: MH + OH- → M + H2O + eThe typical metal hydride compound is LaNi5H6
Cathode reaction: NiOOH + H2O + e- → Ni(OH)2 + OH-
-E°anode = +0.83 v
E°cathode= 0.52
The electrolyte in such a cell is 6M KOH.
Fuel Cells
One final example of galvanic cell operation is that of the fuel cell. Consider the following
reactions:
Cathode reaction:
Anode reaction:
½O2 (g) + 2H+ + 2e- → H2O
H2 (g)
→ 2H+ +2e-
Net:
H2 (g)+ ½O2 (g)
→ H2O
E°cathode = +1.23 v
E°anode = 0.00 v
E°cell = +1.23 volts
Note that this is a combustion reaction. That is why we call the
electrochemical representation of this reaction a fuel cell, as we
noted a couple of lectures ago.
The diagram on the left shows how an H2/O2 fuel cell operates.
The electrolyte is molten KOH, so operating conditions are
severe.
Efficiency =
work − ∆G 237
=
=
= 0.83
heat − ∆H 286
Note that this efficiency is much larger than Carnot
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Electrolysis:
The process of electrolysis is the opposite of galvanic cell operation. In such a system, we
attempt to achieve conditions that are essentially the reverse of galvanic cell operation. We have
to supply an external voltage to make this energetically unfavorable process occur.
Why would we want to carry out electrolysis? The principal reason for carrying out electrolysis
is to produce desirable materials. In the laboratory, electrolysis of water produces lab scale
amounts of hydrogen and oxygen. Industrially, electrolysis can be used to produce high purity
copper. Electrolysis of brine (concentrated alkali halide solutions like NaCl) can be used to
produce alkali metals and halogens. This last process is called the “Chloralkali” process.
Let’s consider electrolysis of water. An apparatus like the following diagram is often used.
First, let’s consider electrolysis in water.
Here are the possible reduction reactions:
½O2 (g) + 2H+ + 2e- → H2O E° = +1.23
2H+ + 2e- → H2(g)
E° = +0.00
In electrolysis, we take the reaction with
the lower reduction potential and let that
reaction as written be the cathode reaction.
The reaction with the higher reduction
potential will be forced to run in the opposite
direction by the application of an external
voltage.
So, in electrolysis in water, we have the following reactions:
Cathode reaction:
2H+ + 2e- → H2(g)
E°cathode = +0.00 v
Anode reaction:
H2O → ½O2 (g) + 2H+ + 2e-
-E°anode = -1.23 v
So, E°cell = E°cathode -E°anode = -1.23 volts. The only time a negative cell potential makes sense is
in electrolysis, in which an external potential at least as large as the negative of the cell
potential is applied to overcome the negative cell potential. The way we make these reactions
run is to apply an external voltage of magnitude +1.23 volts or higher.
Think of electrolysis this way: the electrons given off at the anode (where O2 is made) are
pushed uphill in energy to the cathode, where H+ accepts them and makes H2 gas.
This is all pretty simple, and is summarized by the following energy diagram.
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Cathode:
2H+ + 2e- → H2(g)
0.00 v
Applied voltage ≥ +1.23 v
Anode:
H2O → ½O2(g) + 2H+ + 2e-1.23 v
Net: H2O → H2(g) + ½ O2 (g)
Reduction still takes place at the cathode and oxidation still takes place at the anode. The
power supply provides the driving force to push the electrons formed at the anode uphill
in energy to the cathode. This electron flow occurs through the solution.
It is pretty simple to calculate the amount of material produced by electrolysis from the current
and the time.
Suppose you run a current of 1 ampere (1 Coulomb per sec) for 30 minutes.
Total charge Q = (1 Coulomb sec-1)(30)(60)sec = 1800 Coulombs
Moles of electrons passed = 1800 Coulomb /96485 Coulombs mol-1) = 1.87 × 10-2 moles
Moles of H2 = 1.87 × 10-2 moles electrons × 1 mole H2/ 2 moles electrons = 9.33 × 10-3 moles
Mass of H2 = 9.33 × 10-3 moles × 2 grams /mol = 0.0187 grams.
Volume of H2 at STP = 9.33 × 10-3 moles × 22400 cm3/mol = 209 cm3
Moles of O2 = 1.87 × 10-2 moles electrons × 1 mole O2/ 4 moles electrons = 4.68 × 10-3 moles
Mass of O2 = 4.68 × 10-3 moles × 32 grams /mol = 0.150 grams.
Volume of O2 at STP = 4.68 × 10-3 moles × 22400 cm3/mol = 105 cm3
So, the preceding experiment was done with a normal H2 electrode, in which [H+] = 1.00 M., i.e.,
pH = 0.00.