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)
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
as the electrode.
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
half-reactions 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
Changing Concentration
In chemical reactions…
Standard conditions = 1M for all
When reactants are >1M, it will increase
product concentration and will increase the cell
potential
When products are <1M, it will decrease the
product concentration (oppose forward
reaction), decreases cell potential
Mg(s) + 2H+(aq) -> Mg2+(aq) +H2(g)
Concentration Cell
A galvanic cell that has the same component on
each side but at different concentrations
Causes a cell potential
Voltages are typically small
Ex: a cell has on its left side a 0.20 M Cu2+
solution and a 0.050 M Cu2+ solution on the
right side