Multiple Choice: 5 out of 75 questions
Free Response: Almost every year
 Anything
that uses batteries:
› Cell phones
› Game boys
› Flash lights
› Cars
 Jewelry—electroplating
The
study of the
interchange of chemical
and electrical energy
All electrochemistry
reactions are oxidationreduction reactions.
A
substance that is oxidized
loses electrons
 A substance that is reduced
gains electrons; a reducing
agent loses electrons.
 LEO says GER
 Processes must occur together.
 An
oxidizing agent causes
something else to be oxidized, so
it is reduced (gains electrons).
 A reducing agent causes
something else to be reduced, so
it is oxidized (loses electrons).
 Redox
reactions are often broken
into two half reactions, one
showing the oxidation and the
other showing the reduction.
 FeCl2 + Ca  Fe + CaCl2
› Fe2+ + Ca  Fe + Ca2+
Fe2+  Fe (reduction)
Ca  Ca2+ (oxidation)
 When
the two ions are in the
same solution, the electrons
are transferred directly from
one to the other in a collision.
 In order to harness the energy,
a flow of electrons (current)
must be created.
A
U-tube filled with an
electrolyte or a porous disk
that allows ions to flow
 The purpose of a salt bridge is
to prevent the build-up of
charge that would stop the
flow of electrons.
 Disk
with small openings that
allows ions to flow back and
forth
 Like a salt bridge—prevents
the build-up of ions on one
side of the cell
A
device in which chemical
energy is changed to electrical
energy
 Oxidation occurs at the anode;
reduction occurs at the
cathode.

An ox; red cat
 Cell
potential—the “pull” or
driving force on electrons
(emf)
 Unit = volt
 Potentials are calculated by
using a standard reference
electrode: Hydrogen
 Values
for emf at 25oC and 1M
concentration for solutions/1
atm pressure for gases
 Table—p. 833
 All
reactions are reduction
potentials.
 One reaction must be reversed to
show oxidation.
 The sign must change for the
reversed reaction.
 Coefficients do not matter.
 Cell runs spontaneously to produce
positive cell potential.
A
galvanic reaction is based
on the following reaction:
Al3+ + Mg  Al + Mg2+
 Give the balanced cell
reaction and find the cell
potential.
 Shorthand
to represent a
galvanic cell
 Mg(s)IMg2+(aq)IIAl3+(aq)IAl(s)
 Anode components
(oxidation) on left; cathode
components (reduction) on
right
1.
2.
3.
4.
Cell potential and balanced
cell reaction
Direction of electron flow
Designation of anode &
cathode
Nature of each electrode and
ions present in each
compartment
 Completely
describe the
galvanic cell based on the
following half reactions:
› Ag+ + e-  Ag
› Fe3+ + e-  Fe2+
E = 0.80 v
E = 0.77 V
w
w
= -qE
= work
q = quantity of charge
transferred (96 500
coulombs/mole electrons)
E = cell potential
A
cell has maximum potential
of 2.50 V. If the actual voltage is
2.10 V, how much work could
be done by the flow of 1.33 mol
of electrons?
 Think
of the free energy as the
energy that does the work.
 If w = -qE, then also DG = -qE
 For
q you may use nF (n =
number of moles of electrons x
Faraday's constant.)
 Is
this reaction spontaneous?
Cu2+ + Fe  Cu + Fe2+
DG = -qE = -nFE
 Gives
relationship between cell
potential and concentration of
cell components:
 DG = DGo + RTlnQ
 Remember DG = -nFE
 So, -nFE = -nFE* + RTlnQ
 At 25oC, E = E* (- 0.0591/n) log Q
 At
25o C: E = E* - {(.0591/n) log Q}
› E = Cell potential
› E* = Standard potential
› n = number of moles of electrons
› Q = reaction quotient
When equilibrium is reached, DG = 0
(Battery is dead.)
Opposite
of galvanic
Uses electrical energy to
produce chemical change
(electrolysis)
 Stoichiometry
problems
 1ampere (A) = 1 Coulomb/s
 1 mole e- = 96 485 Coulombs
 Current & time  quantity of
charge in coulombs moles of
electrons  moles of metal 
grams of metal
 How
long must a current of 5.00
A be applied to a solution of Ag+
to produce 10.5 g of sliver
metal?
 Using
a current of 1.00 x 106 A for
2.00 hours, what mass of aluminum
can be produced from aluminum
oxide?