How does a battery work?
A chemical Redox reaction
involves the transfer of
electrons from one species
to another (usually metals).
Without the battery setup,
the reaction would just
proceed and produce no
electrical work. By running
the electrons through an
external circuit the reaction
can produce electrical (I.e.
non-PdV) work.
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The chemical reaction:
Zn → Zn +2 + 2e −
Cu +2 + 2e − → Cu
Net Reaction
Zn ( s ) + Cu +2 ( aq ) → Zn +2 ( aq ) + Cu ( s )
This net reaction has a K and a Gibbs Energy:
o
∆Grxn
= ∆G of ( Zn +2 ( aq ) ) − ∆G of ( Cu +2 ( aq ) ) = −147.1 − 65.5 = −212.6kJ / rxn
∆Go 197
ln K = −
=
= 85.5
RTo 2.48
What do the electrons see? They move from a region of relatively
negative potential to one of more positive potential because they
moved from the Zn (metal) to the Cu ion.
The change in potential must be such that the change in energy of
the electrons is identical to the change in energy of the system. It is
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just a different perspective on the same reaction.
Electrical Energy
• The potential (electrical) energy for a charge, q, in a
potential is:
E = qΦ or ∆E = q∆Φ
• If we know the charge on an electron, which is negative,
then
q = −nF
• F is the Faraday, which, in SI units, is the number of
coulombs per mole of electrons, and n is the number of
moles of electrons. F = 96,500Coul / mole
• The potential is in volts, and the charge is in coulombs and
the energy is in Joules. A Volt is one Joule per Coulomb.
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Energy Change
• The first line is the change in overall Gibbs energy due to a
small amount of reaction using the net reaction as written.
• The second equation is the same overall change but from the
electrons’ view, of a potential change and some electrons
moving. As the reaction advancement is tied to the number of
electrons moving, by the stoichiometric coefficient, ν e , of the
half-reactions, which give the s.c. for the electrons.
• The minus sign is needed because if the potential is positive and
the electrons move, the energy must drop (the electrons are
negative and are attracted to positive regions of space).
dG = ∆Grxn dX
dEe = −∆ΦFdne = −∆Φν e FdX
Equating these two
energies:
∆Grxn = −ν e F ∆Φ
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Battery Potentials and the Nernst Equation
• Now that we have the relation between the reaction Gibbs
Energy and the potential of the battery, we can use all the
machinery of thermodynamics to tell how much potential
we have and how much work we will get from a battery.
o
∆Grxn
∆Grxn
o
o
∆Φ = E =
and @ Standard Conditions ∆Φ = E =
−ν e F
−ν e F
The max electrical work done is:
welect = dG = ∆Grxn dX
 RT 
∆Grxn
o
E=
=E − 
 ln Qa ( X )
−ν e F
νeF 
o
∆Grxn
o νeF 
=E 
ln K = −

RT
RT


Nernst Equation
RTo
= 26mV
F
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Battery Terminal-ology
•
•
•
•
•
The chemical reactions are redox, so we have reduction and oxidation at the
different battery poles.
Faraday named the poles the cathode, because it attracted cations, and the
anode, because anions migrated to it.
To my mind these terms have not helped me understand a battery, but we need
them anyway.
Consider the electrochemical cell running as a battery: Begin with the pole
where oxidation occurs. A good reducing agent gives up electrons, that stay
on the pole; so a good reducing agent is oxidized (LeoGrr). So the pole is
negative and cations left the pole, so it must be the anode. The anode is where
oxidation occurs and is negative. So the cathode must be where reduction
occurs and is positive (it is accepting the electrons). (RedCat)
The reaction is written as the two half reactions, the oxidation is on the left and
reduction on the right. This is done so that the metal electrodes are on the
outside of the notation, and electrons flow from left to right. The bar
represents an interface or phase separation.
M M + n M ′+ n M ′
oxidation reduction
negative positive
Caveat: You don’t know
which way the reaction will
run until you measure
(calculate) it.
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Half Cells
•
•
•
You can’t have a half cell do a reaction. However, because the
potential is proportional to the Gibbs energy it is a state function.
Therefore we can have any reference reaction we want.
The reference reaction is the Standard Hydrogen Electrode (SHE). All
other half cell energies are just the full cell energy using the SHE as
the other half of the battery.
When computing other half cell potentials, or even whole cell
potentials, the safest course it to convert to the reaction Gibbs energies,
then divide by the electron S.C. to go back to half cell potentials.
Zn ( s ) + Cu +2 ( aq ) → Zn +2 ( aq ) + Cu ( s )
Eo
Zn + 2 H + + 2e − → Zn +2 + H 2 ( g ) + 2e −
Eoxo ( Zn )
Cu +2 + H 2 ( g ) + 2e − → Cu + 2 H + + 2e −
o
Ered
( Cu )
o
E o = Eoxo ( Zn ) + Ered
( Cu ) = 0.76 + 0.34 = 1.10V
o
∆Grxn
212.6
=
= 1.10V
−ν e F 2 ⋅ 96.5
o
Eoxo ( M ) = − Ered
(M )
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From the two half cell reactions for Cu(I) compute the half
cell reaction for Cu(II) (Notice, it is not the sum)
Cu +1 + e − → Cu
o
Ered
(1) = 0.521
Cu +2 + e − → Cu +1
o
Ered
( 2 ) = 0.153
Cu +2 + 2e − → Cu
o
Ered
( 3) = ?
Rxn3 = Rxn1 + Rxn 2
o
o
o
ν 3 Ered
( 3) = ν1Ered
(1) + ν 2 Ered
( 2)
o
red
E
0.521 + 0.153
= 0.337V
( 3) =
2
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Charging up a battery
•
•
•
•
How does a battery change in description when you are charging it up.
The car battery: A lead/acid battery, that will start your car over 2,000
times in its life time, but is only good for about 6-10 tries when your
car won’t start.
To recharge you must run the battery backward, so you put electrons
into the pole where they came out. Therefore the negative terminal
stays negative but the reaction is opposite, i.e. it was an oxidation
reaction, it is now a reduction reaction, and before it was the cathode it
is now the anode. Reduction always takes place at the cathode
whether the system is run as a battery or being recharged.
Jumping a car that won’t start: Do you hook the negative terminal to
the negative terminal or to the positive terminal? Why?
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The lead acid battery
•
•
The car battery is very ingenious. It can deliver 40-70 amps of current at a
shot. It is a 12V battery.
The chemical reaction:
Pb ( s ) ! Pb +2 + 2e −
PbO2 ( s ) + 4 H + + 2e − ! Pb +2 + 2 H 2O
•
•
•
Add the the two reactions to get the net reaction. The acid is needed to
provide the protons to react with the oxygen. The addition of sulfuric acid is
particularly good because lead sulfate is not soluble, so it precipitates and
allows for more lead to come off the plates. The redox is rather interesting
because the product is only lead II. But the lead oxide is lead IV so you have a
redox, of lead going to +2, (oxidation) and lead oxide going from IV to +2 (or
reduction).
The reaction is exothermic (as you would guess). And the entropy change is
positive, which happens because molecules move off the solid surfaces and go
into water as ions. So this is entropically positive, and not much entropy is
lost in precipitating the lead sulfate.
The reaction is 2Volts, and the entropy contributes about 40% to the energy of
the reaction. Don’t every say entropy is wasted energy.
o
o
o
−377 kJ / rxn = ∆Grxn
= ∆H rxn
− To ∆Srxn
= −227 − 298 ⋅ 0.501
o
∆Grxn
1.95V = Eo =
−ν e F
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The Cu-Cu Battery
•
•
•
Using the Nernst Equation, compute the voltage you get from a battery
where both electrodes or plates are copper metal, but one beaker (or
half cell) has a copper sulfate solution that is 1 M, and the other half
cell has a very dilute solution, as 0.0001M at the start.
If they both have the same volume of solutions, what will the
concentration be in each beaker when the battery is dead (at
equilibrium).
How does the concentration compare to just mixing the two solutions?
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The Acid Base Battery
•
Water tends to ionize, but not by much.
H 2O " H + + OH −
•
•
•
•
•
•
− log10 K w = pK w = 14
Use Table 4.1 in the back to determine that
Convert this K or Gibbs reaction energy to volts.
Compare this voltage with a reaction you can find in table 11.2
Now make a battery as follows: In one beaker 1M HCl and in another
beaker, 1M NaOH. Insert inert, Pt, electrodes in each beaker, and a salt
bridge, and bubble 1Atm H2 gas through both beakers and over the electrodes
(for about an hour to get equilibrium). Now measure the voltage. What do
you expect for the voltage?
The punch line: You can measure the energy of an acid base reaction without
reacting anything, or any heat change, and all substances can be in standard
state. Moreover, the voltage is enormous but the Kw is so terribly small, it
would be impossible to detect it directly by examining pure water.
An alternative way to compute Kw or the E: Use the Nernst equation and
assume that the NaOH is a beaker that contains a small amount of H+ that is
way off from its standard concentration; and that for this system Eo would in
fact be zero, when viewed this way.
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