Electrochemistry
AP Chem/Mrs. Molchany
(0808)
Drill
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Use AP Review Drill #75-77
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Objectives
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SWBAT
Determine which electrode supports
oxidation and which supports reduction.
Write a half reaction for the anode and
cathode of a galvanic cell.
Explain the purpose of a salt bridge in a
galvanic cell.
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Voltaic Cell (or Galvanic Cell)
A voltaic cell is a device in which the transfer
of electrons takes place through an external
pathway rather than directly between
reactants.
 In a voltaic cell, chemical energy is changed
to electrical energy.
 The energy released in a spontaneous redox
reaction can be used to perform electrical
work.
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Half Cells
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The voltaic cell is thought of as being
comprised of two "half-cells."
One cell is where oxidation occurs and the
other is reduction.
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Half Cells
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Each half cell is made of a metal strip in
contact with a solution of its ions.
In the oxidation half cell, the metal electrode
loses electrons.
In the reduction half cell, the ions in solution
gain electrons and create metal atoms which
deposit onto the electrode.
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http://tutors4you.com/electrochemicalcell.jpg
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Anode/Cathode
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The two solid metals that are connected by
the external circuit are called electrodes.
The electrode at which oxidation occurs is
called the anode. This electrode strongly
attracts negative ions in the solution, these
ions are then called anions.
The electrode at which reduction occurs is
called the cathode. Cations will be deposited
onto this electrode as a metal.
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Half Reactions
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Anode (oxidation half-reaction):
Zn (s)
→
Zn+2 (aq) + 2e-

Cathode (reduction half-reaction):
Cu+2 (aq) +2e- → Cu (s)
Determining half reactions will be shown later.
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OIL – RIG
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Remember the acronym OIL – RIG
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Oxidation Is Loss of electrons
Reduction Is Gain of electrons
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Red-Ox Reaction
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In a red-ox reaction, one substance must be
oxidized and another substance must be
reduced.
The substance that is “oxidized” is the
“reducing agent”.
The substance that is “reduced” is the
“oxidizing agent”.
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Cell Operation (using the Zn/Cu cell)
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Electrons become available when the zinc metal is
oxidized at the anode.
The electrons flow through the external circuit to the
cathode, where they are consumed as Cu+2 is
reduced.
Because zinc is oxidized in the cell, the zinc
electrode loses mass, and the concentration of the
Zn+2 solution increases as the cell operates.
At the same time, the Cu electrode gains mass, and
the Cu+2 solution becomes less concentrated as the
Cu+2 is reduced to Cu (s).
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Cell Operation
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As the voltaic cell operates, oxidation of Zn
introduces additional Zn+2 ions into the anode
compartment. Unless a means is provided to
neutralize this positive charge, no further
oxidation can take place.
At the same time, the reduction of Cu+2 at the
cathode leaves an excess of negative charge
in solution in that compartment.
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Salt Bridge
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Electrical neutrality of the system is
maintained by a migration of ions through the
porous glass disc (salt bridge) that separates
the two compartments.
A salt bridge consists of a U-shaped tube that
contains an electrolyte solution whose ions
will not react with other ions in the cell or with
the electrode materials.
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Salt Bridge
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As oxidation and reduction proceed at the
electrodes, ions from the salt bridge migrate
to neutralize charge in the cell
compartments.
The cell diagram slide shows the Zn+2 ions
migrating into the salt bridge from the left cell
and the SO4-2 ions migrating into the salt
bridge from the right cell.
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Salt Bridge
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The salt bridge is required for completing the
circuit. It allows the movement of ions from
one solution to the other so that a charge
does not build up in either of the cells.
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Ion Migration
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Anions migrate toward the anode and cations
toward the cathode.
No measurable electron flow will occur
through the external circuit unless a pathway
is provided for ions to migrate through the
solution from one electrode compartment to
another, completing the circuit.
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Electron Flow
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In any voltaic cell the electrons flow from
the anode through the external circuit to
the cathode.
Because the negatively charged electrons
flow from the anode to the cathode, the
anode in a voltaic cell is labeled with a
negative sign and the cathode with a positive
sign.
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Oxidation Numbers
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Remind yourself how to determine oxidation
numbers.
Oxidation numbers show what the charge of
each atom would be (in a molecule or ion), if
each atom were an ion.
Go to the text and revisit the Rules for
Assigning Oxidation Numbers section.
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Balancing Red-Ox Equations
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Obey the Law of Conservation of Mass
Gain and loss of electrons must be balanced
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Half Reactions
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Cu + 2Ag+ →

Determine the oxidation number for each
substance in the reaction
Write two half reactions based on one
substance being oxidized and one being
reduced.
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2 Ag + Cu+2
Half Reactions
Cu
→
Cu+2
2e- + 2Ag+ →


+ 2e-
2 Ag
oxidation
reduction
Half reactions show the number of electrons gained
or lost by the substance, standard redox equations
do not.
Go to ChemReview packet to practice half reaction
writing.
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Review Sample Ex: 20.4
Try Practice Exercises
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Cell EMF
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What is the driving force that pushes the electrons
through an external circuit in a voltaic cell?
An oxidizing agent in one compartment pulls
electrons through a wire from a reducing agent in the
other compartment.
The pull (or driving force) on the electrons is called
the cell potential ( Ecell ) or electromotive force (emf)
of the cell.
Or, electrons flow from the anode to the cathode (in
a voltaic cell) because of a difference in potential
energy.
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Potential Energy of Electrons
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The potential energy of electrons is higher in
the anode than in the cathode. Therefore,
electrons spontaneously flow through an
external circuit from the anode to the
cathode.
For any cell reaction that proceeds
spontaneously (i.e. voltaic cell), the cell
potential will be positive.
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Standard Conditions
Standard Conditions:
* 1M concentrations for reactants and
products in solution
* 1 atm pressure (for gases)
* 25 ˚C
Under Standard Conditions the EMF is called the
“standard emf” or the standard cell potential (E ˚cell )
Keep in mind that the superscript ˚ denotes
standard-state conditions.

Note: Read the symbol “ ˚ ” as the word not
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Cell Potential
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The cell potential of a voltaic cell depends on the
particular cathode and anode half-cells involved.
Standard potentials have been assigned to each
individual half-cell. You can look them up in your AP
Packet.
Then use the half-cell potentials to determine E ˚cell .
The cell potential is the difference between two
electrode potentials (anode and cathode).
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Cell Potential
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The potential associated with each electrode
is chosen to be the potential for reduction to
occur at that electrode.
Therefore, your AP Packet only shows
Standard Reduction Potential reactions and
EMF values.
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Standard Reduction Potentials
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Standard electrode potentials are tabulated for
reduction reactions, so they are called standard
reduction potentials ( E ˚red ).

E ˚cell = E ˚red (cathode) - E ˚red (anode)
Or
E ˚cell = E ˚red (cathode) + E ˚ox (anode)
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Standard Reduction Potentials
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I will use the following equation:
E ˚cell = E ˚red (cathode) + E ˚ox (anode)
When using this equation:
* Determine which half reaction is the oxidation
reaction.
* “Flip” the reaction by writing it backward.
* Remember to change the sign of the EMF
value.
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Standard Reduction Potentials
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The hydrogen half-reaction was used as a
reference for all of the other half-reactions.
The standard reduction potential for the hydrogen
half-reaction is assigned a standard reduction
potential of exactly 0 V.
It is also called the Standard Hydrogen Electrode.
(SHE)
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Calculating EMF
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We will use standard reduction potentials to
calculate the EMF of a voltaic cell.
Because electrical potential measures potential
energy per electrical charge, standard reduction
potentials are intensive properties.
Changing the stoichiometric coefficient in a halfreaction does not affect the value of the standard
reduction potential.
Do Not multiply the Standard Reduction Potential
value by the number of moles !!!
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FYI
 In
general, the more active the
metal, the lower its potential.
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Metal Activity
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The most active metal, Li, has the lowest
tendency to gain electrons (to be reduced).
You will see on the chart that it’s reduction
potential value is extremely low.
Li has the strongest tendency of the metals
listed to lose electrons (to be oxidized).
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Metal Activity
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The least active metal, Au, is at the bottom of
the table. It has a greater tendency to be
reduced.
Gold also has a lower tendency to lose
electrons (to be oxidized).
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In the cell diagram:
Zinc replaces copper

Zn + CuSO4 → ZnSO4 + Cu
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Calculating EMF
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Try sample exercise 20.5
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E ˚red and spontaneity
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The more positive the E ˚red value for a halfreaction, the greater the tendency for the
reactant of the half-reaction to be reduced,
and, therefore, to oxidize another species.

Try sample exercise 20.8
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EMF and Free Energy Change
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ΔG = - nFE
“n” is a positive # that represents the # of
electrons transferred in the reaction
“F” is Faraday’s constant
(this constant is the quantity of electrical
charge on 1 mole of electrons)
1F = 96,500 C/mol = 96,500 J / V mole
The units for ΔG are J/mole
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EMF and Free Energy Change
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ΔG = - nFE
If “n” and “F” are positive values,
a positive value of E leads to a negative ΔG.
Remember that a negative ΔG indicates a
spontaneous reaction.
The equation can be altered slightly if the
reactants and products are in their standard
states: ΔG˚ = - nFE˚
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Cell EMF Under Nonstandard
Conditions
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The next set of slides explains how to
calculate the cell emf while working under
non-standard conditions, specifically
concentrations other than 1M.
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Concentration and Cell EMF
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Remember ΔG = ΔG˚ + RT ln Q
Combining
ΔG = ΔG˚ + RT ln Q and ΔG = - nFE
You have the Nernst equation:
E = E˚ – RT ln Q
nF
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Nernst Equation
The Nernst equation can be expressed two
ways:
E = E˚ – RT ln Q
nF
E = E˚ – 2.303RT log Q
nF
Base 10 logs are related to natural logarithms
by a factor of 2.303.
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Nernst Equation Variation
The equation can be simplified if the cell is run
under standard conditions:
E = E˚ – 0.0592 V log Q
n
* Temp. is 298 K
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Remember when we talked about equilbrium
situations,
E = E˚ – 0.0592 V log [products ]
n
[reactants ]

Remember: this form of the Nernst equation
can only be used is the temp. is 298K.
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Use the Nernst Equation to work through the
example on the bottom of page 773
Try the sample exercise 20.11 and practice
exercise
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Electrolysis
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Electrolysis reactions are driven by an
outside source of electrical energy and take
place in electrolytic cells.
This process is the opposite of a voltaic cell.
Voltaic cells are spontaneous processes.
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Electrolysis
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An electrolytic cell consists of two electrodes
in a molten salt or a solution.
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Ch 20 Problems
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6, 7, 9, 14, 24, 25 a,b, 27, 29, 31, 36, 43, 44,
46, 49, 51, 52, 56, 59
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