AP Chemistry: Electrochemistry Notes
p.1
Chapter 20: Electrochemistry
Galvanic Cells
redox reactions involve a transfer of electrons. Oxidation is loss; reduction is gain.
We can use a redox reaction to create a current! The reaction between MnO4- and Fe2+ is
an example: 8H+ + MnO4- + 5Fe2+  Mn2+ + 5Fe3+ + 4H2O
 This can be broken into half reactions:
o reduction is 8H+ + MnO4- + 5e-  Mn2+ + 4H2O
o oxidation is 5(Fe2+  Fe3+ + e-)
 when MnO4- and Fe2+ are in the same solution, e- are transferred directly when
reactants collide. No useful work is achieved from the chemical energy of the
reaction. But if we separate the reactants and connect them with a wire, e - should
flow through the wire!
 but in reality, the current flows for just a millisecond and then stops, because
charges build up in the compartments. The left compartment (receiving e-) becomes
negative, and the right becomes positive. Electrons won’t flow into a negative
charge.
 we can solve this problem by connecting the containers so ions can flow and keep the
overall charge equal to zero.
o salt bridge: U-shaped tube filled with electrolyte
o porous disk
 the connections complete the circuit, allowing e- to flow from reducing agent to
oxidizing agent. Other ions flow the opposite direction in order to keep charge
balanced.
 galvanic cell: a device in which chemical energy is changed into electrical energy
(named after Luigi Galvani (1737-1798), who discovered electricity!). This is also
called a voltaic cell after Alessandro Volta (1745-1827) who built the first one in
1800.
o anode: where oxidation occurs (remember because oxidation and anode both
start with a vowel)
o cathode: where reduction occurs (both reduction and cathode start with
consonants)
AP Chemistry: Electrochemistry Notes
p.2

cell potential: Ecell, the “pull” of e- by the oxidizing agent from the reducing agent.
This is also called the electromotive force (emf) of a cell.
o the unit is the volt (V) which is 1 J of work/Coulomb of charge (1 V = 1J/C)
o cell potential is measured by a voltmeter, which draws current through a
known resistance (and V = IR). But often, frictional heating through the
wire loses some of the cell’s energy, and a lower potential is actually
determined.
o we can also use a potentiometer, which is a variable-voltage device in
opposition to the cell potential. Voltage is adjusted until no current flows in
the cell; the cell voltage is then equal and opposite of that to the
potentiometer.
17.2 Standard Reduction Potentials
 a galvanic cell always have a redox reaction. It would be nice if we could assign a
potential to each half reaction so that we can build a cell our of any two halfreactions and calculate the overall cell potential!
o 2H+ + Zn  Zn2+ + H2
o the anode has a zinc metal electrode with 1M Zn2+ and SO42- ions in solution.
The reaction is Zn  Zn2+ + 2e-. The Zn(s) is giving up e-, which flow through
the wire.
o the cathode does the reduction: 2H+ + 2e-  H2(g) This contains a platinum
electrode (chemically inert conductor) in contact with 1M H+ ions and bathed
in H2(g). This is called the standard hydrogen electrode.
o the hydrogen electrode is arbitrarily assigned a value of 0 V. This means
that if the total cell potential is 0.76 V, that must be the potential of the
zinc half-reaction, because Ecell = EH+ + EZn
 if we do a zinc and copper redox reaction:
Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s)
we get half-reactions: Zn  Zn2+ + 2eand
Cu2+ + 2e-  Cu
If the cell potential is 1.10 V, then we can calculate the potential of the
copper half-reaction: E°cell = E°Zn + E°Cu2+
1.10 V = 0.76 V + E°Cu2+ so E°Cu2+ = 0.34 V
 there are tables of half-reaction potentials based on the standard hydrogen
electrode being zero (including on your AP formula sheet!!)
 the half-reactions are always written as reductions
 E values assume all solutes are at 1M and all gases are at 1 atm, and are called
standard reduction potentials.
o if a reduction is reverse, the sign of the potential is also reversed
o if half-reactions are multiplied by an integer to balance an equation, the
potential is NOT multiplied (it is an intensive property that does not depend
on how many times a reaction has occurred).
 for a galvanic cell:
Fe3+ + Cu  Cu2+ + Fe2+
we get half-reactions: Fe3+ + 2e-  Fe2+ E° = 0.77 V
Cu2+ + e-  Cu
E° = 0.34 V
2+
We must reverse the second reaction: Cu  Cu + 2e-E° = -0.34 V
We must double the first reaction: 2Fe3+ + 2e-  2Fe2+
E° = 0.77 V
AP Chemistry: Electrochemistry Notes
p.3
Now we can get the overall cell reaction: 2Fe3+ + Cu  Cu2+ + 2Fe2+
and E°cell = 0.77 V - 0.34 V = 0.43 V
 Sample Exercise
A cell has the reaction MnO4- + H+ + ClO3-  ClO4- + Mn2+ + H2O
the half reactions are: MnO4- + 5e- + 8H+  Mn2+ + 4H2O E° = 1.51 V
ClO4- + 2H+ + 2e-  ClO3- + H2O
E° = 1.19 V
Give the balanced cell reaction and calculate E°cell
Reverse the second reaction and multiply both by an integer:
2(MnO4- + 5e- + 8H+  Mn2+ + 4H2O) E° = 1.51 V
5(ClO4- + 2H+ + 2e-  ClO3- + H2O)
-E° = -1.19 V
2MnO4- + 6H+ + 5ClO3-  5ClO4- + 2Mn2+ + 3H2O
E°cell = 1.51 – 1.19 = 0.32V
 line notation: is shortcut method for describing electrochemical cells
o the anode is listed on the left; cathode on the right (oxidation, then
reduction), separated by a double vertical line to indicate the salt bridge
o for the cell reaction Al3+ + Mg  Mg2+ + Al, the shortcut notation is
o

Mg(s)|Mg2+(aq)║Al3+(aq)|Al(s)
 the single vertical line is a phase boundary
 the actual materials that make up the anode and the cathode are
listed on the far left and the far right, respectively.
for the cell reaction in Sample Exercise, we get
Pt(s)|ClO3-(aq),ClO4-(aq), H+(aq)║H+(aq), MnO4-(aq), Mn2+(aq)|Pt(s)
 Pt is used as an electrode because all of the cell components are ions
complete description of a galvanic cell
o A cell has the following half-reactions:
 Fe2+ + 2e-  Fe(s)
E° = -0.44 V
+
2+
 MnO4 + 5e + 8H  Mn + 4H2O
E° = 1.51 V
o a cell always runs spontaneously in the direction that produces a positive
potential. So we must reverse the first half-reaction and multiply by the
correct integers:
 5( Fe  Fe2+ + 2e-)
-E° = 0.44 V
 2(MnO4- + 5e- + 8H+  Mn2+ + 4H2O)
E° = 1.51 V
+
2+
 overall: 2MnO4 + 5Fe + 16H  2Mn + 5Fe2+ + 8H2O
E°cell = 1.95 V
o physical setup:
Left compartment is the anode, which
has Fe(s) and 1.0M Fe2+. Oxidation
takes place here. Electrons flow from
the anode through a wire.
*the negative ions in sol’n depend on
what iron salt we used for the sol’n
Right compartment is the cathode,
which has Pt as the nonreactive solid
electrode. Reduction takes place
here. Electrons are received here
from the anode.
AP Chemistry: Electrochemistry Notes
p.4
Fe(s)|Fe2+(aq)║MnO4-, Mn2+(aq)|Pt(s)
o the complete description of a galvanic cell includes:
 cell potential and balanced reaction
 direction of e- flow
 designation of anode and cathode
 the nature of each electrode and ions present in each
compartment
 Sample Exercise
Describe completely the galvanic cell with the following half-reactions:
Ag+ + e-  Ag(s) E° = 0.80 V
Fe3+ + e-  Fe2+
E° = 0.77 V
In order for E°cell to be positive, we must reverse the second reaction:
Ag+ + e-  Ag(s) E° = 0.80 V
Fe2+  Fe3+ + e-E° = -0.77 V
+
Ag + Fe2+  Ag + Fe3+
E°cell = 0.03 V
Electrons will flow from the compartment with Fe2+ to the compartment with Ag+. The
anode (where oxidation occurs) will have a Pt electrode and contains 1M Fe2+ and Fe3+
ions. The cathode (where reduction occurs) will have a silver electrode and contains 1M
Ag+.
The line notation is: Pt(s)|Fe2+, Fe3+(aq)║Ag+(aq)|Ag(s)
17.3 Cell Potential, Electrical Work, and Free Energy
The work done when e- travel through a wire depends on the “push” behind the
electrons. The emf is defined in terms of a potential difference (V) between two
points in a circuit.
work (J)
charge (C)

emf = potential difference (V) =

Work is viewed from the point of view of the system. So work flowing out of the
system is negative. E = -w/q, where w=work and q=charge
When work flows out of the system, the cell produces a current, so the cell
potential is positive. So –w = qE and –wmax = qEmax
q is the quantity of charge in Coulombs. The charge on 1 mole of e- is a constant
called a faraday (F), which is 96,485 coulombs/mole e-. So q = nF where n is the
number of moles of electrons.
Example: A galvanic cell has a maximum potential of 2.50 V. In an experiment, 1.33
moles e- were passed through the cell at an average actual potential of 2.10 V. How
much work is done?
w = -qE = -nFE = (1.33 mole e-)(96,485 C/mole e-)(2.10 J/C)
= -2.69 x 105 J
The maximum work done is wmax = -qEmax = -nFEmax
= (1.33 mole e-)(96,485 C/mole e-)(2.50 J/C)
= -3.21 x 105 J



w
 2.69x10 5
x100% 
x100% = 83.8%
So the efficiency of the cell is
w max
 3.21x10 5

How do we relate the potential of a cell to free energy? Remember that the
maximum useful work is equal to ΔG! So wmax = ΔG = -qEmax = -nFEmax
AP Chemistry: Electrochemistry Notes
p.5
Assume that from now on, when we write E, we mean Emax, so ΔG = -nFE, and under
standard conditions, ΔG° = -nFE°
o this means that the maximum cell potential is directly related to the free
energy difference between reactants and products, and
o a galvanic cell will run in the direction that gives a positive value for Emax,
which corresponds to a negative value for ΔG!
 Sample Exercise
Calculate ΔG° for Cu2+ + Fe  Cu + Fe2+
Using half-reactions, we get
Cu2+ + 2e-  Cu
E° = 0.34 V
2+
Fe  Fe + 2e -E° = 0.44 V
So for Cu2+ + Fe  Cu + Fe2+ E°cell = 0.78 V
ΔG° = -nFE°
There are 2 moles of e- involved in the reaction, so
ΔG° = -2(96,485 C/mole e-)(0.78 J/C) = -1.5 x 105 J
ΔG° < 0 so the reaction is spontaneous (but we knew that because E°cell is > 0)
 Sample Exercise
Predict whether 1M HNO3 will dissolve gold metal to form 1M Au3+ solution.
The half-reactions are: NO3- + 4H+ + 3e-  NO + 2H2O E° = 0.96 V
Au  Au3+ + 3e-E° = -1.50 V
Total reaction: Au + NO3- + 4H+  Au3+ + NO + 2H2O
E°cell = -0.54 V
The reaction will not occur because E° cell is negative.

17.4 Dependence of Cell Potential on Concentration
 If the concentration of reactants is > 1M, LeChatelier’s Principle predicts that the
forward reaction will be favored, and so Ecell > E°cell
 If the concentration of products is > 1M, the opposite will be true and Ecell < E°cell
concentration cells
 We can make a galvanic cell where both compartments have the same components
but at different concentrations!
Both compartments have AgNO3(aq)
but at different molarities. The half-reaction
is Ag+ + e-  Ag(s) E° = 0.80 V
 One side is 1M, the other 0.1M
 The cell will try to equalize
concentrations of Ag+. this can’t be done
by transferring Ag+, but it can be done by
transferring e- to reduce Ag+ in the 1M
side!
 So the silver electrode on the 0.1M side
will oxidize and lose e- and transfer them
to the electrode on the 1M side. This has
two effects: it creates more Ag+ ions on
the 0.1M side and reduces Ag+ to Ag on
the 1M side!
AP Chemistry: Electrochemistry Notes
p.6
Concentration cell is one in which both compartments have the same component but
at different concentrations. Voltages are typically small.
 Sample Exercise
Determine the direction of e- flow and designate the anode and cathode for the cell
with 0.01M Fe2+ and an Fe electrode on the left compartment and 0.1M Fe2+ and an
Fe electrode in the right compartment.
The cell will attempt to equalize the concentrations of Fe2+:
 e- will flow from the 0.01M side (where Fe will oxidize to Fe2+) to the
0.1M side (where Fe2+ will be reduced to solid Fe)
 the left side (0.01M) is the anode (oxidation occurs); the right side
(0.1M) is the cathode (reduction occurs)
The Nernst Equation
 Remember that ΔG depends on concentration through the equation ΔG = ΔG° + RTlnQ
 But ΔG = -nFE and ΔG° = -nFE°, so -nFE = -nFE° + RTlnQ




RT
lnQ This is the Nernst Equation
nF
0.0591
Assuming 25°C and converting natural log to base 10 log, we get E = E° logQ
n
Dividing through by nF, we get E = E° -
Example: E°cell = 0.48 V for 2Al + 3Mn2+  2Al3+ +3Mn
What is the potential if [Mn2+] = 0.50M and [Al3+] = 1.50 M?
E = E° -
0.0591
0.0591
(1.50) 2
logQ = 0.48 log
= 0.47 V
n
6
(0.50) 3
There is very little change in potential, and it has been decreased because reactants
are < 1M and products are > 1M, so the concentrations work toward the reverse
reaction.
 As the cell discharges and current flows from the anode to the cathode,
concentrations change, and so Ecell will also change. The cell will spontaneously
discharge until it reaches equilibrium, at which point Q = K and E°cell = 0!
 A “dead” battery is one in which the cell reaction has reached equilibrium!! At
equilibrium, the components in the two cell compartments have the same free energy
and ΔG= 0. The cell does not have the ability to do work!
 Sample Exercise
Describe the cell: VO2+ + 2H+ + e-  VO2+ + H2O
E° = 1.00 V
2+
Zn + 2e  Zn
E° = -.76 V
T = 25°C [VO2+] = 2.0M, [H+] = 0.50M, [VO2+] = 1.0 x 10-2M, [Zn2+] = 1.0 x 10-1M
to get a balanced cell reaction, multiply the first reaction x 2 and reverse the second:
2VO2+ + 4H+ + 2e-  2VO2+ + 2H2O
E° = 1.00 V
2+
Zn  Zn + 2e
-E° = 0.76 V
2VO2+ + 4H+ + Zn  2VO2+ + 2H2O + Zn2+ E°cell = 1.76 V
oxidation occurs at the anode with a Zn electrode
reduction occurs at the cathode with a Pt electrode
0.0591
E = E° logQ
n
[VO 2 ] 2 [Zn 2 ]
n = 2 and Q =
[VO 2 ] 2 [H  ] 4
AP Chemistry: Electrochemistry Notes
p.7
E = 1.76 -
0.0591
(.01) 2 (.1)
log
= 1.89 V
n
(2) 2 (.5) 4
Ion-selective electrodes
 Because cell potential is sensitive to concentrations of reactants and products,
measured potentials can be used to determine ion concentrations!
 A pH meter has three components: a standard electrode of known potential, a glass
electrode that changes potential depending on [H +], and a potentiometer that
measures the potential between the electrodes. The potentiometer reading is
converted to a direct reading of pH.
 Ion-sensitive electrodes are sensitive to concentrations of a particular ion
calculation of equilibrium constants


0.0591
logQ and at equilibrium, Q = K and E°cell = 0
n
0.0591
nE 0
So at that point, E° =
logK, so logK =
(at 25°C)
n
.0591
We know that E = E° -
 Sample Exercise
S4O62- + Cr2+  Cr3+ + S2O32find E° and K at 25°C
22S4O6 + 2e  S2O3
E° = 0.17 V
2(Cr2+  Cr3+ + e-)
-E° = 0.5 V
S4O62- + 2Cr2+ 2 Cr3+ + S2O32E°cell = 0.67 V
There are 2 moles of e- so logK =
2(.67)
= 22.6 so K = 1022.6 = 4 x 1022
.0591
Batteries
Battery: a galvanic cell (or group of cells connected in series) where the potentials of
individual cells are added to give total battery potential
Lead storage battery: lead is the anode, and lead coated with PbO2 is the cathode. Both
electrodes dip into an electrolyte solution of H2SO4
anode: Pb + HSO4-  PbSO4 + H+ +2ecathode: PbO2 + HSO4- + 3H+ + 2e-  PbSO4 + 2H2O
cell reaction: Pb + PbO2 + 2H+ + 2HSO4-  2PbSO4 + 2H2O
 Typical car lead batteries have 6 cells connected in series. each gives 2V for a total
of 12V. Sulfuric acid is consumed as the battery discharges, changing the density
of the electrolyte solution (that’s how the detectors know when the battery is
almost “out”)
 The battery is charged by forcing a current through the battery in the opposite
direction
 Lead storage batteries have been around without much improvement for 85 years!
They last 3-5 years.
Other batteries
 dry cell batteries: were invented over 100 years ago!
o In the acid version, a zinc inner case acts as the anode and a carbon rod in
contact with a moist paste of solid MnO2, solid NH4Cl, and carbon acts as
the cathode. Cell potential is about 1.5 V.
AP Chemistry: Electrochemistry Notes
p.8
o
o
o
An approximation of the half-reactions has the anode with Zn  Zn2+ + 2eand the cathode with 2NH4+ + 2MnO2 + 2e-  Mn2O3 + 2NH3 + H2O
In the alkaline version, the solid NH4Cl is replaced with KOH or NaOH
Nickel-cadmium batteries have these reactions:
anode
Cd + 2OH-  Cd(OH)2 + 2ecathode
NiO2 + 2H2O + 2e-  Ni(OH) + 2OH-
Fuel cells
 These are galvanic cells for which the reactants are continuously supplied. An
example reaction is CH4 + 2O2  CO2 + 2H2O + energy
 In a fuel cell, electrons flow from the reducing agent, CH 4, to the oxidizing agent,
O2, through a conductor
 The space program uses a fuel cell based on the production of water (and I have car
kits that do the same thing!): anode 2H2 + 4OH-  4H2O + 4ecathode 4e- + O2 + 2H2O  4OHoverall: 2H2 + O2  2H2O
Corrosion
 Involves the oxidation of metal.
 Most metals oxidize spontaneously. Many of them actually develop a thin coating of
metal oxide, which inhibits further corrosion.
 Gold has a much higher reduction potential than O2 and shows no corrosion in air!
Corrosion of iron
 Steel has areas of nonuniformity where iron is more easily oxidized (anodic regions)
than at others (cathodic regions). In the anodic region, Fe  Fe2+ + 2e- Electrons
flow through the steel to cathodic regions, where O 2 + 2H2O + 4e-  4OHo The Fe2+ ions formed in anodic regions travel to cathodic regions through
moisture on the surface, and form rust: 4Fe2+ + O2 + (4+2n)H2O 
2Fe2O3.nH2O + 8H+
o Moisture acts as a salt bridge, allowing ions to travel. Steel won’t rust in
dry air! But salt increases rust formation, because salt dissolved in water
increases its conductivity.
Prevention of Corrosion
 Coatings are used to protect metal from oxygen and water.
 Chromium and tin are used to plate steel, as is zinc (in a process called galvanizing).
The zinc is more active than iron and “sacrifices” itself to form a zinc oxide coating
which protects the steel underneath.
 Alloying (forming a mixture of two or more metals) also protects. Stainless steel
contains chromium and nickel, which form oxide coatings that change the reduction
potential of steel.
 Cathodic protection: uses an active metal, like magnesium, connected by a wire to an
underground pipeline or tank. Because Mg is a better reducing agent than iron,
electrons come from Mg rather than iron, keeping iron from being oxidized. The Mg
anode must be replaced periodically.
AP Chemistry: Electrochemistry Notes
p.9
Electrolysis
 electrolytic cell: uses electrical energy to produce chemical change
 electrolysis: forcing a current through a cell to produce a chemical change for which
the cell potential is negative.
 In an electrolytic cell, the anode and the cathode are reversed from what they
would spontaneously be.
 Example: How much copper is plated out when a current of 10.0 amps (an ampere, A,
is the unit for current and = 1 C/sec) is passed for 30.0 min through a solution
containing Cu2+?
10 amps = 10.0 C/sec x 30.0 min x 60 sec/min = 1.8 x 104 C
1.8 x 104 C x 1 mol e-/96,485 C = 0.187 mol e0.187 mol e- x 1 mol Cu/2 mol e- xx 63.546 g/mol Cu = 5.94 g Cu
 Sample Exercise
How long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g
Ag(s)?
10.5 g Ag x 1 mol/107.868 g x 1mol e-/1 mol Ag = 9.37 x 10-2 mol e9.37 x 10-2 mol e- x 96,485 C/mol e- = 9.39 x 103 C of charge
time =
9.39 x 10 3 C
= 1.88 x 103 sec = 31.3 min
5.00 C/sec
Electrolysis of Water
We can force the splitting of water by running a current through it:
anode: 2H2O  O2 + 4H+ + 4e-E° = -1.23 V
cathode: 4H2O + 4e-  2H2 + 4OHE° = -0.83 V
+
6H2O  2H2 + O2 + 4(H + OH )
2H2O  2H2 + O2
E°cell = -2.06 V
Pure water has so few ions that almost no reaction is observed when you pass a current
through it. But if a salt is added, bubbles are immediately visible.
Electrolysis of Mixtures of Ions
In which order will different metals be plated out in a mixture of metal ions? Whichever
has the most positive potential will plate out first. So, for:
Ag+ + e-  Ag
E° = 0.80 V
Cu2+ + 2e-  Cu
E° = 0.34 V
2+
Zn + 2e  Zn
E° = -0.76 V
Silver is the most easily reduced. Ag+ > Cu2+ > Zn2+
Commercial Electrolytic Processes
 Production of aluminum: aluminum is the third most abundant element on earth,
after oxygen and silicon. It exists as an ore called bauxite. It was very difficult to
purify until the Hall-Heroult process was discovered in 1886. It is a very
complicated process, but it produces 99.5% pure Al. This process is the number one
use of electricity in the US, counting for 5% of total electricity!
 Electrorefining: impure metals can be purified by reduction. Copper is purified this
way to 99.95% purity.
 Metal plating: we can coat certain easily oxidizable metals with a thin layer of a
corrosion-resistant metal:
AP Chemistry: Electrochemistry Notes
p.10
“Tin Cans” are actually steel cans coated with tin.
Steel bumpers in cars are coated with chrome.
The object to be coated is made the cathode (where reduction occurs). The
cell has ions of the metal that will do the coating. As electrons flow into the
cathode, the metal ions are reduced onto it, and it becomes coated.
Electrolysis of sodium chloride: Aqueous sodium chloride (brine) is electrolyzed to
form Cl2(g) and NaOH. This process is the second largest use of electricity in the
US!
o
o
o

A summary of terminology of POSITIVE and NEGATIVE:
The cathode is the electrode where reduction takes place, period. The anode is
where oxidation takes place. So that means that electrons enter the cathode and leave the
anode. This is true for BOTH galvanic (spontaneous) and electrolytic (nonspontaneous) cells.
In a galvanic cell, a spontaneous chemical reaction generates electricity. At the
anode, electrons are given off, so the electrode is negatively charged and is labeled
NEGATIVE. Electrons are moving toward the cathode however, so that is labeled
POSITIVE.
In an electrolytic cell, electricity is causing a non-spontaneous reaction to occur. A
power source is pumping electrons through the wire. The negative side of the power source
is pumping electrons into the electrode where electrons are gained and something is
reduced—the cathode. Since it is attached to the negative side of the power source, it is
labeled NEGATIVE. The positive side of the power source is pumping electrons toward it,
drawing them away from the anode where they have been generated. Since the anode is
attached to the positive side of the power source, it is labeled as POSITIVE.