Lead Acid Car Batteries. In class exercise/ Discussion/Semester review
Key
Created by Sheila Smith, University of Michigan- Dearborn (sheilars@umd.umich.edu) and posted on VIPEr
(www.ionicviper.org) on May 26, 2010. Copyright Sheila Smith 2010. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
The lead acid battery can be represented as
Pb(s),PbSO4(s)|HSO4-(aq),H+(aq)||HSO4-(aq) ,H+(aq) |PbO2(s),PbSO4(s)
Write the balanced chemical equation represented by this notation.
Pb(s) + HSO4-(aq)  PbSO4(s) + 2e- + H+(aq) (anodic half reaction)
3H+(aq) + 2e- + PbO2(s) + HSO4-(aq)  PbSO4(s) + 2 H2O(l) (cathodic half reaction)
2H+(aq) + 2HSO4-(aq) + Pb(s) + PbO2(s) 2PbSO4(s) + 2H2O(l)
E°cell=2.04V (12 V is achieved by connecting 6 of these in series)
Calculate Ecell at 25°C for this battery when [H2SO4] = 4.5 M).
°
E cell = E cell
−
Q=
RT
lnQ
nF
1
2
[H ] [HSO ]
+
−
4
2
=
1
( 4.5)
4
= 2.4 x10 −3
J
)(298K)
Kmol
= 2.04V −
ln(2.4 x10 −3 )
(4mol)(96500C /mol)
= 2.12V
(8.314
E cell
E cell
(6 of these in series is actually a 12.7 V battery. )
€
Sheila Smith 5/25/10 2:11 PM
Comment: The lead acid battery, so named
because it contains lead species in a bath of
sulfuric acid, is a battery with which many
students are familiar (a car battery). I usually
introduce the exercise by asking what they
know about this battery. Often they talk about
the voltage or the need to recycle, or the acid
leakage that can occur onto the poles of the
battery.
Sheila Smith 5/25/10 1:59 PM
Comment: The shorthand notation for
Voltaic cells (and we know this battery is
voltaic because we use this to start our
engines) includes the anode first separated
from the cathode by the salt bridge (denoted by
the double vertical line). Inside each half cell,
phases are separated by phase boundaries
represented by single vertical lines. Thus the
anode contains Pb and PbSO4 in a bath of
sulfuric acid (shown as H+ and HSO4- since
H2SO4 is a strong acid that dissociates 100% in
water)
Sheila Smith 5/25/10 2:01 PM
Comment: Since the anode is where
oxidation occurs, the Lead must be going from
Pb(0) to Pb(2+) in the anode, losing 2 electrons.
Balancing in acid gives us the anodic half
reaction.
Sheila Smith 5/25/10 2:02 PM
Comment: Similarly, we can arrive at the
reduction reaction that is occurring at the
cathode.
Sheila Smith 5/26/10 10:39 AM
Comment: Finally, by adding the two halfreactions together, we arrive at the balanced
redox reaction for a lead acid battery. Be sure
to make sure electrons lost = electrons gained
before adding the two half reactions!
Sheila Smith 5/26/10 11:23 AM
Comment: We use the Nernst equation
because the conditions specified are not
standard (1 M H2SO4). We go back to the
now- balanced redox equation to get the form
of our reaction quotient. Good time to remind
students that solids and liquids don’t factor in
the equilibrium expression. I usually ask
students to predict which direction
LeChatelier’s principle predicts this change
will move the Voltage before doing the
calculation. Since we are effectively
increasing the concentration of two reactants
(compared to standard conditions of 1 M), the
reaction should become more spontaneous
... [1]
(shift toward products) and Ecell should
increase, which it does.
Lead Acid Car Batteries. In class exercise/ Discussion/Semester review
Key
Created by Sheila Smith, University of Michigan- Dearborn (sheilars@umd.umich.edu) and posted on VIPEr
(www.ionicviper.org) on May 26, 2010. Copyright Sheila Smith 2010. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
Using the Thermodynamic data given below, calculate ΔH°,ΔG° and ΔS° for the reaction.
Substance
Pb(s)
PbO2(s)
PbSO4(s)
H+(aq)
HSO4-(aq)
H2O(l)
ΔHf° (kJ/mol)
0
-276.65
-918.4
0
-885.75
-258.8
ΔGf° (kJ/mol)
0
-218.99
-811.2
0
-752.87
-237.2
Sf° (kJ/mol)
64.89
76.57
147.28
0
126.86
69.9
°
ΔH rxn
= ∑ ΔH °f Pr oducts − ∑ ΔH °f Re ac tan ts
Sheila Smith 5/26/10 11:07 AM
Comment: We use Hess’ Law to calculate
the energies of reaction from the heats of
formations. The enthalpy of reaction is
negative, this is an exothermic reaction.
°
ΔH rxn
= 2(−918.4) + 2(−258.8) − 2(0) − 2(−885.75) − (0) − (−276.65)
°
ΔH rxn
= −306.3kJ
°
ΔGrxn
= −372.1kJ
J
°
ΔSrxn
= +39.1
K
€
€
Sheila Smith 5/26/10 11:43 AM
Comment: Similarly, we an calculate
°
.
ΔGrxn
Since we know this is a battery and
°
is spontaneous, we predict that ΔGrxn will be
negative (exergonic), which it is.
€
€
°
is a little more complicated. I usually
ΔSrxn
remind students that
€ if° no gases are involved
(as in this case), ΔSrxn may be positive or
negative but it will be relatively small. If you
have curious students, you can talk about the
order associated with hydrated ions in solution,
since the loss of this order apparently
€
dominates the
€
°
term.
ΔSrxn
Lead Acid Car Batteries. In class exercise/ Discussion/Semester review
Key
Created by Sheila Smith, University of Michigan- Dearborn (sheilars@umd.umich.edu) and posted on VIPEr
(www.ionicviper.org) on May 26, 2010. Copyright Sheila Smith 2010. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
Battery acid (in a fresh battery) is 37% by mass H2SO4. Calculate the molality and the
molarity. The density of the solution is 1.29g/mL.
m=
37gH 2 SO4 10 3 gH 2O 1molH 2 SO4
= 6.0m
63gH 2O 1kgH 2O 98.04gH 2 SO4
M=
37gH 2 SO4 1.29gsolution 10 3 mL 1molH 2 SO4
= 4.9M
100gsolution 1mLsolution 1L 98.04gH 2 SO4
Sheila Smith 5/26/10 12:06 PM
Comment: This is a good opportunity to
review the need for different concentration
units. Also, I usually remind students here that
we only expect M to be equal to m in dilute
solution, and this is not dilute.
€ point I ask students if any of them have had the experience of one of these
At this
batteries failing. You go out on a cold February morning and get nothing… Why might
this happen? Invariably, someone will suggest that the battery is too cold/ frozen.
“Too cold” brings us back to our answer for the heat of reaction. Earlier, we calculated
that the reaction is exothermic. In this case, heat is a product, and removing a product
actually favors the reaction, so this battery should work better at cold temps.
Calculate the freezing point of this solution. Assume HSO4- dissociation is negligible. (kf
for water is 1.86°C/m)
ΔTf = ik f m
ΔTf = (2)(1.86°
C
)(4.9m)
m
ΔTf = 22.3°C
So the freezing point of€this solution (4.5 M H2SO4) is -22.3 °C. A quick conversion to
Fahrenheit gives us ~-8°F, not unheard of here in MI, but also not common enough to
think that this is the principal cause of battery failure. However, it’s fun to discuss the
fact that this an be a real issue in some places, such as Minnesota and North Dakota,
where garages and carports come equipped with engine block heaters to prevent
batteries from freezing.
What happens to the freezing point as the battery discharges?
Going back to the balanced equation, we see that as the battery discharges the reactants
are converted into a solid and water. So as the battery gets older, the freezing point of
the solution rises, causing freezing of the battery to be a more likely issue. Depending on
Sheila Smith 5/26/10 12:51 PM
Comment: This assumption is reasonable
because in a concentrated solution of H+, as in
a 4.5 M solution of H2SO4, a strong acid) the
weak acid HSO4- is not going to be able to
contribute further to the acidity of the solution.
Sheila Smith 5/26/10 12:52 PM
Comment: Colligative properties like
freezing point depression do not depend on the
nature of the solute, so we need the kf for the
solvent, H2O.
Sheila Smith 5/26/10 12:59 PM
Comment: The Van’t Hoff factor i=2 for
H2SO4 if we assume HSO4- ionization is
negligible. Sulfuric acid is a strong acid that
dissociates 100% into H+ and HSO4-.
We use the molality unit in this calculation,
because it wouldn’t make sense to use a
temperature dependent concentration unit to
calculate a temperature change. Again, this
emphasizes the need for different
concentration units.
Lead Acid Car Batteries. In class exercise/ Discussion/Semester review
Key
Created by Sheila Smith, University of Michigan- Dearborn (sheilars@umd.umich.edu) and posted on VIPEr
(www.ionicviper.org) on May 26, 2010. Copyright Sheila Smith 2010. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
your audience, you might choose to add a discussion of the alternator of a car, which
uses some of the energy from the combustion of fuel to push this spontaneous reaction
back uphill, recharging your battery.
Discuss any possible reasons why/circumstances where this battery might not be useful.
In summary, if your car battery is old and partially discharged, and it’s really cold
outside, and maybe your alternator isn’t working great, you could get stuck.