AP Chemistry Ch 18 Electrochemistry I

Balancing oxidation-reduction reactions
o Half-reactions
 It is useful to separate redox reactions into two half-reactions: one involving oxidation
and the other involving reduction.
 Start by identifying and writing the equation for each half-reaction. Then balance
each half reaction and equalize the electron transfer between them. Lastly, add the
half-reactions. Check that the elements and charges are balanced.
Exercise 1
Half Reactions (Acidic)
Balance the following reaction.
MnO4-(aq) + Fe2+(aq)  Fe3+(aq) + Mn2+(aq)
How many electrons are transferred in this reaction?
o We generally balance redox reactions as if they took place in acidic solution—if water
molecules are added to one side, the hydrogen is balanced on the other side using hydrogen
ions. The reaction is Exercise 1 is balanced as if it took place in an acidic solution.
o Oxidation-reduction reactions can also occur in basic solution. Only a small change is
made to these balanced equations—after the equation is balanced “normally”, as it would be
acidic solution, hydroxide ions are added to both side of the equation to “cancel out” the
hydrogen ions, resulting in the formation of water to one side and the net addition of
hydroxide ions to the other.
Exercise 2
Half Reactions (Basic)
Balance the following reaction.
Ag(s) + CN-(aq) + O2(g)  Ag(CN2)-(aq)
How many electrons are transferred in this reaction?

Galvanic cells—commonly found in FRQs
o In redox reactions, electrons are transferred from one
species to another. When these species are in the same
solution, no useful work can be obtained from the chemical
energy involved in the reaction (this energy is instead
released as heat). This energy can be harnessed,
however, if we separate the oxidizing agent from the
reducing agent, thus requiring the electron transfer to occur
through a wire. However, a wire isn’t enough—we also
need to link the two half cells with a salt bridge or a porous
disk to complete the circuit. This is called a galvanic cell.
 The reaction in an electrochemical cell occurs at the
interface between the electrode and the solution
where the electron transfer occurs. The electrode
compartment in which oxidation occurs is called the
anode; the electrode compartment in which
reduction occurs is called the cathode.


The “pull”, or driving force, on the electrodes in a galvanic cell is called the cell potential (ξcell), or
the electromotive force (emf) of the cell. The unit of electrical potential is the volt, which is defined
as one joule of work per coulomb of charge transferred.
Standard reduction potentials
o A standard hydrogen electrode consists of a platinum electrode in contact with 1 M H+ ions
and bathed by hydrogen gas at 1 atm. The half-reaction potential of the reaction at this
electrode has been “set” to 0 V, and all other half-reaction potentials have been determined
using this as a standard.
2 H+ + 2 e-  H2
half-reaction potential = 0 V
o We can measure the total potential of a cell, there is no way to measure the potentials of the
individual electrode processes. To get around this problem, we base half-reaction potentials
(called standard reduction potentials) on hydrogen.
o The accepted convention is to give the potentials of half-reactions as reduction processes.
o Combining two half-reactions to obtain a balanced oxidation-reduction reaction often requires
two manipulations:
 One of the reduction half-reactions must be reversed (since redox reactions must
involve a substance being oxidized and a substance being reduced). The halfreaction with the largest positive potential will run as written (as a reduction) and the
other half-reaction will be forced to run in reverse (will be the oxidation reaction). The
net potential of the cell will be the difference between the two. Since the reduction
process occurs at the cathode and the oxidation process occurs at the anode, we can
write
ξ°cell = ξ°(cathode) - ξ°(anode)
Because subtraction means “change the sign and add”, an equivalent operation is
changing the sign of the oxidation (anode) reaction when we reverse it and adding it
to the reduction (cathode) reaction.
 Since the number of electrons lost must equal the number gained, the half-reactions
must be multiplied by integers as necessary to achieve the balanced equation.
However, the value of ξ° is not changed when a half-reaction is multiplied by an
integer—this is because a standard reduction potential is an intensive property.
Exercise 3
Galvanic Cell I
Consider a galvanic cell based on the redox reaction Fe3+(aq) + Cu(s)  Cu2+(aq) + Fe2+(aq).
The half-reactions are
Fe3+ + e-  Fe2+
ξ° = 0.77 V
2+
Cu + 2 e  Cu
ξ° = 0.34 V
Give the balanced cell reaction, and calculate ξ° for the cell.
+0.43 V
Exercise 4
Galvanic Cell II
Consider a galvanic cell based on the redox reaction Al3+(aq) + Mg(s)  Al(s) + Mg2+(aq)
The half-reactions are
Al3+ + 3e-  Al
ξ° = -1.66 V
Mg2+ + 2 e-  Mg
ξ° = -2.37 V
Give the balanced cell reaction, and calculate ξ° for the cell.
+0.32 V

Line Notation
o Electrochemical cells can also be described using line notation, in which
 The anode components are listed on the left
 The cathode components are listed on the right
 The anode and cathode are separated by double vertical lines
 A phase boundary is indicated by a single vertical line
o
Some electrochemical cells require the use of an inert conductor, such as platinum, as an
electrode (ex. all of the components are ions, one of the components is a gas).
Ex. Pt(s) | ClO3-(aq), ClO4-(aq), H+(aq) || H+(aq), MnO4-(aq), Mn2+(aq) | Pt (s)
Exercise 5
Line Notation
Describe the cells in Exercises 3 and 4 using line notation.
Exercise 3:
Exercise 4:

Complete description of a galvanic cell
o Let’s consider a galvanic cell based on the following half-reactions:
Fe2+ + 2 e-  Fe
ξ° = -0.44 V
+
2+
MnO4 + 5 e + 8 H  Mn + 4 H2O
ξ° = +1.51 V
In a working galvanic cell, one of these reactions must run in reverse. A cell will always run
spontaneously in the direction that produces a positive cell potential. Reversing the first
equation would give us a positive cell potential, so we’ll do that.
Fe  Fe2+ + 2 e-ξ° = +0.44 V
+
2+
MnO4 + 5 e + 8 H  Mn + 4 H2O
ξ° = +1.51 V
In the first reaction, the iron is oxidized, so this reaction happens at the anode. In the second
reaction, the manganese is reduced, so this reaction happens at the cathode.
This gives ξ°cell = ξ°(cathode) - ξ°(anode) = 1.51 + 0.44 V = 1.95 V.
Now we’ll talk about the physical setup of this cell.
In the left compartment the active components in
their standard states are pure metallic iron (Fe)
and 1.0 M Fe2+. The anion present depends on
the iron salt used. In this compartment the anion
does not participate in the reaction but simply
balances the charge. The half-reaction that takes
place at the electrode is Fe  Fe2+ + 2 e- which is
an oxidation reaction, so this compartment is the
anode. The electrode consists of pure iron metal.
In the right compartment the active components in
their standard states are 1.0 M MnO4-, 1.0 M H+,
and 1.0 M Mn2+, with appropriate unreacting ions (counterions) to balance the charge. The
half-reaction in this compartment is MnO4- + 5 e- + 8 H+  Mn2+ + 4 H2O which is a
reduction reaction, so this compartment is the cathode. Since neither MnO4- nor Mn2+ ions
can serve as the electrode, a nonreacting conductor such as platinum must be used.
The next step is to determine the direction of electron flow. Iron is oxidized in the left
compartment; permanganate is reduced in the right compartment. Thus the electrons flow
from Fe to MnO4- in this cell, or from the anode to the cathode, as is always the case. The
line notation for this cell is
Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq) | Pt(s)
Exercise 6
Description of a Galvanic Cell
Completely describe the galvanic cell based on the following half-reactions under standard conditions.
Include ξ°cell, direction of electron flow, identification and description of the anode and cathode, and
descriptions of the electrodes in each compartment. Also fill out the diagram and write the line notation
for this cell.
Ag+ + e-  Ag
Fe3+ + e-  Fe2+

ξ° = 0.80 V
ξ° = 0.77 V
Cell potential, electrical work, and free energy
o The driving force behind electron flow, the emf, is defined in terms of a potential difference
(in volts) between two points in a circuit. A volt represents a joule of work per coulomb of
charge transferred:
emf = potential difference (V) = _work (J)_
charge(C)
Thus 1 joule of work is produced or required (depending on the direction) when 1 coulomb of
charge is transferred between two points in the circuit that differ by a potential of 1 volt.
Here, work is viewed from the point of view of the system—work flowing out of the system
is indicated by a negative sign.
o
When a cell produces a current, the cell potential is positive, and the current can be used to
do work (ex. to run a motor). The cell potential ξ and the work w have opposite signs:
ξ = -w/q
or
-w = qξ
From this explanation it can be seen that the maximum work in a cell would be obtained at
the maximum cell potential:
-wmax = qξmax
or
wmax = -qξmax
However, there is a problem. To obtain electrical work, current must flow. When current
flows, some energy is inevitably wasted through frictional heating. In any real, spontaneous
process some energy is always wasted—the actual work realized is always less than the
calculated maximum—consequence of entropy.
Although we can never achieve maximum work through the actual discharge of a galvanic
cell, we can measure the potential using a potentiometer (no current = no waste of energy).
The actual work done by a galvanic cell is
w = -qξ
Where ξ represents the actual potential difference at which the current flows and q is the
quantity of charge in coulombs transferred. The charge on one mole of electrons is a
constant called the farady (F), which has the value 96,485 coulombs of charge per mole of
electrons. In other words,
q = nF = moles * 96,485 coulombs/mole of electrons
The efficiency of a cell can be determined by calculating its w and wmax (if given ξ and ξmax).
% efficiency = (w/wmax) * 100%
Next we want to relate the potential of a galvanic cell to free energy. For a process carried
out at constant temperature and pressure, the change in free energy equals the maximum
useful work obtainable from that process.
wmax = ΔG
Since wmax = -qξmax and q = nF we get ΔG = -nFξmax
or, more simply (and at standard conditions),
ΔG° = -nFξ°
This equation states that the maximum cell potential is directly related to the free energy
difference between the reactants and the products in the cell.
Exercise 7
Calculating ΔG° for a Cell Reaction
Using the table of Standard Reduction Potentials at the end of these notes, calculate the ΔG° for the
reaction Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq). Is this reaction spontaneous?
ΔG° = -1.5 x 105 J, yes
Exercise 8
Predicting Spontaneity
Using the table of Standard Reduction Potentials at the end of these notes, predict whether 1 M HNO 3
will dissolve gold metal to form a 1 M Au3+ solution.
ξ°cell = -0.54, no
AP Chemistry Schedule: Ch 18 Electrochemistry and 19 Nuclear Chemistry
4/7
Review
4/8
Review
4/9
Ch 9/10 Exam
4/10
Start Ch 18
4/11
Go over HW
HW: Read 18.1, do
pg. 877-878 #
17ab, 21, 29ab,
33bc, 33
Continue Ch 18
4/18
Good Friday
4/14
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4/15
Go over HW
4/16
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4/17
Go over HW
Continue Ch 18
Continue Ch 18
Continue Ch 18
Finish Ch 18
HW: Read 18.4, do
pg. 880-881 # 55,
59-65 odd
HW: Read 18.5, do
pg. 881 # 67-71
odd, 77
HW: do pg. 882 #
79, 81, 85, 87
HW: read 18.8, do
pg. 883 # 93a,
95ab, 99, 103-109
odd
HW: Read 18.218.3, do pg. 879880 # 35-39 odd,
41-45 a only, 47
5th Six Weeks Report Cards
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4/22
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4/23
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4/24
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4/25
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Ch 18 Quiz
Start Ch 19
Continue Ch 19
Finish Ch 19
Review
HW: Read 19.119.2, do pg. 919923 # 11, 15ab, 17,
27-33 odd, 59
HW: Read 19.319.5, do pg. 921922 # 37, 41-45
odd
HW: Read 19.619.7, do pg. 922923 # 53-57 odd,
67
4/28
Ch 18/19 Exam
4/29
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4/30
Review
5/1
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5/2
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5/5
AP Chemistry
Exam 8:00 am
5/6
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5/9
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