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Transcript
Chapter 21
Electrochemistry
AP Chemistry
Milam
Overview
• In physics electricity typically deals with
electrons flowing through metals, the flow
of charge is called current
• In chemistry we look at what causes
electrons to flow, construction of batteries
and which chemicals will work best for a
specific function such as a battery
• Some concepts from both will be present
and some unique to chemistry will be there
as well
Overview
• An electrolytic cell is one where external
energy is used to cause a nonspontaneous
chemical reaction to occur
• A voltaic cell is where a spontaneous
chemical reaction (usually redox) produces
electricity and can supply it to something
external
21-1 Electrical Conduction
• As mentioned earlier, charge is the key
thing that must move for electrical currents.
In physics charge can be an electron, and in
chemistry we will see electron flow, as well
as ion flow.
• Both of these occur in typical cells
(batteries)
21-2 Electrodes
• Electrodes are surfaces where reduction or
oxidation reactions occur
• In a battery, a redox reaction occurs, but
typically the reduction/cathode is separated
from the oxidation/anode by a wire and a
salt solution called a salt bridge
• The separation allows for electron and ion
flow between the two ½ reactions giving us
current
21-2
• The cathode has the reduction occur, so it
gains electrons and likewise the anode
(oxidation) loses electrons. Therefore it
only makes sense that electrons will flow
from the anode to the cathode.
• The directionality of electron flow is
typically 1 point on an electrochemistry
question on the AP exams.
21-3 The Electrolysis of Molten
Sodium Chloride (The Downs Cell)
• This is a good example to show the basics
of redox reactions as well as provide a good
visual example
• Here we have molten NaCl and by running
electrical current through it, we can produce
sodium metal and chlorine gas
• Na+ + e-  Na (l)
21-3
• Since the sodium ion picks up an electrons
or Gains an Electron it is Reduced (GER)
• 2Cl-  2e- + Cl2(g)
• Since the chloride Loses Electrons it is
Oxidized (LEO)
• Always remember that LEO the lion goes
GER
21-3
• To combine the two ½ reactions, you need
to have the same number of electrons so
that they cancel, thus we multiply the Na ½
reaction by 2
• 2x [Na+ + e-  Na (l)]
• + 2Cl-  2e- + Cl2(g)
• 2Na+ + 2Cl-  2Na(l) + Cl2(g)
• Since Na is reduced, it is formed at the
cathode and Chlorine is formed at the anode
21-4 The Electrolysis of Aqueous
Sodium Chloride
• This reaction is similar, but occurs in
solution rather than molten sodium chloride
• The redox reaction is the same, but after
Na(s) is produced, it immediately begins to
react with water to make Na+ and H2 and
OH-
21-5 The Electrolysis of Aqueous
Sodium Sulfate
• This is actually just an electrolysis of water
to produce hydrogen and oxygen gas
• While it appears to be a simple reaction, it
might be a good example to analyze by
splitting up the net reaction into ½ reactions
• H2O(l)  H2(g) + O2(g)
• The hydrogen in water is a +1 oxidation
state, it goes to 0 so it is reduced and
oxygen is therefore oxidized
21-5
• 2H2O  O2 + 4H+ + 4e• We can easily now see the loss of electrons for this
oxidation ½ reaction
• 2H2O + 2e-  H2 + 2OH• Additionally we can see the hydrogen is reduced
and gains electrons
• To combine these reactions, we would multiply
the 2nd reaction by 2, and the OH- and H+ would
combine to form H2O and cancel with 4 waters
from the reactant side
21-6 Counting Electrons:
Coulometry and Faraday’s Law
of Electrolysis
• In physics, current is measured in Amps, or
a Couloumb/second, a Couloumb (C) is a
measure of charge
• In chemistry we can use Amps to determine
how much electrolysis will occur, similar to
stoichiometry, using ½ reactions with the #
of electrons as mol ratios
21-6
• A mol of electrons is called a Faraday and
has 96, 485 C of charge.
• Therefore we can convert between time and
amount (mass/mol/volume/particles) or
material by using time, amps, Faraday’s
constant, # of electrons and the ratios
between # of electrons and amount of
substance
21-7 Commercial Applications of
Electrolytic Cells
• Plating of metals is very common, but also
expensive so typically it is only done as a
surface and sometimes very thin surfaces.
The chromium surface of bumpers is only
0.0002 mm thick.
• How many atoms thick is that?
21-8 The Construction of Simple
Voltaic Cells
• Voltaic (galvanic) cells have spontaneous
redox reactions occuring, separated into two
halves. The half-cells are separated and
connected by a wire and a salt bridge to
maintain neutrality of charge and allow
charge to flow and be extracted.
• A standard cell has all solutions as 1M and
all gases at 1atm
21-9 The Zinc-Copper Cell
• In this cell we have a zinc electrode in
contact with 1M Zinc Sulfate solution and
the other ½ has a copper electrode in 1M
copper sulfate solution connected to the
zinc by a wire and a salt bridge
• The reaction is:
• Cu+2(aq) + Zn(s)  Cu(s) + Zn+2 (aq)
21-9
• If a statement like the electrode loses mass,
or gains mass is said, this tells you whether
reduction or oxidation is occurring at that
electrode. If metal is plating the electrode
to make it gain mass, then it is more than
likely reduction and the cathode
• In the zinc copper cell the copper electrode
will gain mass and the current flows from
the zinc electrode to the copper electrode
via the wire
21-9
• The equation for this cell can be simplified
to Zn|Zn+2 (1M) ||Cu+2 (1M)|Cu
• The Double line in the middle indicates a
salt bridge, the single lines separate the
electrodes and the ions being reduced or
oxidized on those electrodes as well as the
concentrations.
• The anode is written on the left
21-9
• If you drop a piece of metal zinc into copper
sulfate solution, the exact same reaction will
occur. The point of the salt bridge and the
separate ½ reactions is to cause a flow of
electrons from one electrode to another. It
is the separation that allows us to capture
the electrical potential energy in these
chemicals.
21-10 The Copper-Silver Cell
• This is another specific example of a cell,
we won’t get into details on this until after
we go through some thermodynamics and
how to find cell potentials (voltages)
• There is a reference here back to activity
series of metals for single displacement
reactions way back in chapter 4, this is how
that list is developed.
21-11 The Standard Hydrogen
Electrode
• We cannot determine the potential of a
single electrode, both reduction and
oxidation occur at the same time so you
cannot just measure one, you can only
measure one relative to another
• For this reason, we set the standard
hydrogen electrode (SHE) to be 0 Volts
• From there we can compare electrodes to
this one and have a set of values to use
21-11
• The SHE converts protons (H+) into
hydrogen gas (H2) and vice versa. Both
reactions are assigned a potential of 0 V
21-12 The Zinc-SHE Cell
• This means that we can compare a zinc
electrode (Zn|Zn+2 (1M)) in conjunction
with the SHE
• This produces a cell with a potential
(voltage) of 0.763 V and we can therefore
assign this voltage to zinc
• The voltage of a cell is determined by
adding the voltage of the anode and cathode
21-12
• Since the SHE is 0V and the total was
0.763V, the zinc electrode must also be
0.763V.
• Zn+2 (aq)+ 2e-  Zn(s) Eored = -0.763V
• Zn(s)  Zn+2 (aq)+ 2e- Eoox = +0.763V
• A spontaneous cell will have a + Ecell and so
in this case the zinc metal will oxidize and
the hydrogen will be reduced to produce
hydrogen gas
21-13 The Copper-SHE Cell
• Likewise with copper, we can arrange a cell
and measure the potential difference
• Cu+2(aq) + 2e-  Cu(s) Eored = +.337 V
• Cu(s)  Cu+2(aq) + 2e- Eoox = -.337 V
• The reverse will always produce the same
potential with the opposite sign
21-14 Standard Electrode
Potentials
• Now we can use the SHE against all kinds
of electrode and end up with a list of
potentials of all electrodes, in addition we
can now compare 2 electrodes where
neither is a SHE
• If the electrode has a + Eored then it will
undergo reduction with the SHE and if it is
– then it will undergo oxidation with the
SHE
21-14
• The more positive the value of Eored, the
more likely that substance is to undergo
reduction at the cathode in a galvanic cell or
the better an oxidizing agent it will be
• The reverse is true with large negative
values for Eored
21-15 Uses of Standard Electrode
Potentials
• The uses of these potentials includes
determining which direction in a cell will
occur spontaneously as well as determining
the potential of that given cell.
• The more positive Eored will be the reaction
at the cathode being reduced in the
spontaneous reaction, with zinc and copper
we will see copper reduced
21-15
•
•
•
•
•
•
•
Cu+2(aq) + Zn(s)  Cu(s) + Zn+2 (aq)
Cu+2(aq) + 2e-  Cu(s) Eored = +.337 V
Zn(s)  Zn+2 (aq)+ 2e- Eoox = +0.763V
Then we can also calculate the Eocell
Eocell = Eored + Eoox
Eocell = +.337V + (+0.763V) = 1.100 V
So we will get 1.100 V and electrons will
flow from the Zn electrode to the Cu
electrode
21-16 Standard Electrode
Potentials for Other HalfReactions
• A summary of everything you should know
by now:
• Labeling oxidation, reduction, oxidizing
agents, reducing agents, cathode, anode,
direction of flow of electrons
• Calculating the potential of a cell given the
net reaction
• Going from ½ reactions to a full balanced
reaction and vice versa
21-16
• Making calculations involving Faraday’s Constant
96485 C and amount of substance deposited
during electrolysis
• Predicting which metal will be reduced in a
galvanic cell based on potentials
• Determine spontaneity based on overall cell
potential
• Describe metals as good oxidizing or reducing
agents based on reduction potentials
21-16
• Standard conditions (1M, 1atm, 25 °C)
• Describe how cell potentials can be
determined using a SHE and also know that
potentials are relative and arbitrarily
assigned
• Be familiar with common examples such as
decomposition of water, Zn/Cu cells and so
forth.
21-17 Corrosion
• Corrosion occurs when metals react with
oxygen, typically with moisture present
• Corrosion is a big problem because a metal
surface becomes ionic where it can crumble
and not have the protection that metals offer
• For iron the reaction is 4Fe(s) + 3O2 (g) +
xH2O (l)  Fe2O3*xH2O (s)
21-18 Corrosion Protection
• To protect from corrosion there are several
things you can do, interestingly you can add
a layer of metal on the surface of the iron,
and the metal can either be a more reactive
or less reactive metal.
• The less reactive metal will do nothing, but
if it breaks, the iron will corrode very
quickly because of the subsequent reaction
available
21-18
• Zinc is more reactive than iron, so
galvanized steel is often used. The zinc will
not rust, and if the iron is exposed, the iron
will not react with oxygen until there is no
longer any zinc metal in contact.
21-19 The Nernst Equation
• The Nernst Equation is used for cells that
do not have standard amounts, for example
if I create a battery where the concentrations
are much higher than 1M
• E = Eo – 0.0592/n * logQ
• E is the new potential, n is the number of
electrons transferred, Q is the reaction
quotient (from equilibrium), Eo is the
standard reduction potential
21-20 Using Electrochemical Cells
to Determine Concentrations
• The Nernst Equation can be used to calculate
potentials using concentrations, or if the
potentials are measured, we can measure
concentration.
• Common examples of this include pH meters
which use voltage to determine pH in a
solution.
• You do not need to know about saturated
calomel electrodes not glass electrodes for pH
meters
21-20
• The only thing you need to be able to do
with the Nernst Equation is the calculations
of concentration and potential
21-21 The Relationship of Eocell
to ΔGo and K
• Eocell = RTlnK/(nF)
• lnK = nFEocell/(RT)
• R is 8.314 J/(mol*K), n is the # of electrons
transferred, F is Faraday’s constant
(9.65x104 J)
• If you know the cell potential you can
calculate the equilibrium constant and vice
versa
21-21
• It’s important to be able to calculate the cell
potential and equilibrium constant, but it’s
also important to be able to analyze
relationships between Eocell, ΔGo, and K
• For example, if ΔGo is -, which means the
reaction is spontaneous in the forward
direction, what type of ΔEo would also
provide a spontaneous forward reaction?
21-22 Dry Cells – 21-25 The
Hydrogen-Oxygen Fuel Cell
• These sections go into specific types of
batteries and show how some can be
recharged and others cannot. If you’re
curious about what chemicals are in car
batteries or Duracell’s read on. It won’t be
on the AP test and we won’t have the time
to cover it.