Galvanic Cell Concept
• Separating the oxidation and reduction half-reactions
will make it so the energy in the electron transfer can
be harnessed.
• A salt bridge or porous disk is needed to connect the
half-reactions so ions can flow and electrons don’t
build up on one side of the reaction (one beaker)
• **VIDEO?
Galvanic Cell Definition
• Device which chemical energy
is changed to electrical energy.
• Oxidation occurs at the
ANODE
• Reduction occurs at the
CATHODE (cat gets fat =
cathode gains electrons a.k.a.
reduction)
– An ox and a red cat
(anode/oxidation,
reduction/cathode)
Electrodes
• If there is an element (not ion) in either
half-reaction, it is what that particular
electrode is made of ***comes up later!
• When all reactants/products are in
solution (aq) Pt or graphite can be used
Galvanic Cell Picture
(Parts)
Cell Potential/Electromotive Force
• (EMF) Represented by E°cell
• Unit = volt (V) = 1 joule/coulomb
• Measured with a voltmeter (not completely
accurate b/c of heat). A potentiometer is used
instead where the maximum cell potential can
be measured.
Standard Reduction Potentials
• If we can find the potential for each halfreaction (Table 18.1 pg. 829), we can determine
the cell potential (E°cell)
• Half-reaction manipulations (DO NOT
MANIPULATE VOLTAGE):
– One must be reversed (oxidation)…can reverse E
so you have -E = -voltage…
– Electrons lost must = electrons gained, so
multiplication of reaction may be needed (DO NOT
MULTIPLY VOLTAGE BY THIS NUMBER!)
• EQUATION (MEMORIZE)
E°cell = E°(cathode) - E°(anode)
Table 18.1
• Better oxidizing agents: easily reduced, LEFT
side rxn. = largest, most positive standard
reduction potential
• Better reducing agents: easily oxidized, RIGHT
side = most negative standard reduction
potential (aka most positive standard oxidizing
potential)
• Example: Which is best reducing agent Cu+, F-,
H-, H2O, I2, K (find in reactants)
• Answer: K (-2.92) > H- (-2.23) > Cu+ (0.16) > I2
(1.20) > H2O (1.23) > F- (2.92)
Galvanic Cell Example
• Calculate the emf values (E°cell) for the
following
Mg(s) + 2H+(aq) -> Mg2+(aq) +H2(g)
• Answer: E°cell = +2.37V
In Galvanic Cells…
• When E°cell is positive, the reaction will run
spontaneously. (last slide)
• If negative, it will run in the opposite
direction (will NOT run spontaneously as
written).
Cell Potential, Electrical Work,
and Free Energy
• Spontaneous IF:
– Positive cell potential
– Negative Gibbs Free Energy (we will learn
more about Gibbs Free Energy in a later
chapter)
Line Notation
• Not required for AP Exam
• A double vertical line separates the anode on
left and cathode on right
– Represents a salt bridge or porous disk
• A single vertical line separates different phases
• Ex:
anode Cd -> Cd2+ + 2ecathode Hg2+ + 2e- -> Hg
Line notation: Cd(s) I Cd2+(aq) II Hg2+(aq) I Hg
Galvanic Cell: Complete Description
In half-reaction descriptions…FOUR items needed:
1. The cell potential - positive when E°cell = E°(cathode)
- E°(anode) and the balanced cell reaction
2. The direction of electron flow, obtained by inspecting
the half-reactions and using the direction that gives a
positive E°cell.
3. Designation of the anode and cathode.
4. The nature of each electrode and the ions present in
each compartment. A chemically inert conductor is
required if none of the substances participating in
the half-reaction is a conducting solid.
Example: Complete Description
•
Describe a galvanic cell based on the two halfreactions below.
Cu2+ + 2e- -> Cu E° = 0.34 V
Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O E° = 1.33V
1. Balanced cell rxn: 3Cu(s) + Cr2O72-(aq) + 14H+(aq) ->
3Cu2+(aq) + 2Cr3+(aq) + 7H2O(l)
E°cell = 0.99V (needs to be positive)
2. 1.33 (cathode) -0.34 (anode) means Cu needs to be
reversed. Cu will be giving off e- which will travel
from Cu (anode) to cathode (platinum electrode).
Continued…
3. Anode (copper metal electrode), cathode
(platinum electrode)
4. The copper metal electrode (anode) will be
in the Cu/Cu2+ compartment while the
platinum electrode (cathode) will be in the
Cr2O72-/Cr3+ compartment