Lecture 11: Cell Potentials
• Reading: Zumdahl 11.2
• Outline
– What is a cell potential?
– SHE, the electrochemical zero.
– Using standard reduction potentials.
Reminder
• “Redox” Chemistry: Reduction and Oxidation
• Oxidation: Loss of electrons
(increase in oxidation number)
• Reduction: Gain of electrons
(a reduction in oxidation number)
• Electrons are transferred from the reducing
agent to the oxidizing agent
• Electrons are transferred from the species being
oxidized to that being reduced.
Galvanic Cells (cont.)
8H+ + MnO4- + 5eFe2+
Mn2+ + 4H2O
Fe3+ + e-
x5
Galvanic Cells (cont.)
• Galvanic Cell: Electrochemical cell in which chemical
reactions are used to create spontaneous current (electron)
flow
Galvanic Cells (cont.)
Anode: Electrons are lost
Cathode: Electrons are gained
Oxidation
Reduction
Cell Potentials
• In a galvanic cell, we had a species being oxidized
at the anode, a species being reduced at the
cathode, and electrons flowing from anode to
cathode.
• The force on the electrons causing them to full is
referred to as the electromotive force (EMF). The
unit used to quantify this force is the volt (V)
• 1 volt = 1 Joule/Coulomb of charge
V = J/C
Cell potential and work
• From the definition of electromotive force (emf):
Volt = work (J)/charge (C)
• 1 J of work is done when
1 C of charge is transferred between a potential
difference of 1 V.
Cell Potentials (cont.)
• We can measure the magnitude of the EMF
causing electron (i.e., current) flow by measuring
the voltage.
e-
Anode
Cathode
1/2 Cell Potentials
• What we seek is a way to predict what the voltage
will be between two 1/2 cells without having to
measure every possible combination.
• To accomplish this, what we need to is to know
what the inherent potential for each 1/2 cell is.
• The above statement requires that we have a
reference to use in comparing 1/2 cells. That
reference is the standard hydrogen electrode
(SHE)
1/2 Cell Potentials (cont.)
• Consider the following galvanic cell
• Electrons are spontaneously flowing from the
Zn/Zn+2 half cell (anode) to the H2/H+ half cell
(cathode)
1/2 Cell Potentials (cont.)
• We define the 1/2 cell potential of the hydrogen
1/2 cell as zero.
SHE
P(H2) = 1 atm
[H+] = 1 M
2H+ + 2e-
H2
E°1/2(SHE) = 0 V
1/2 Cell Potentials (cont.)
• With our “zero” we can then measure the voltages
of other 1/2 cells.
• In our example, Zn/Zn+2 is the anode: oxidation
Zn+2 + 2e-
Zn
2H+ + 2e-
Zn + 2H+
H2
E°Zn/Zn+2 = 0.76 V
E° SHE = 0 V
Zn+2 + H2
E°cell = E°SHE + E°Zn/Zn+2 = 0.76 V
0
Standard Reduction Potentials
• Standard Reduction Potentials: The 1/2 cell
potentials that are determined by reference to the
SHE.
• These potentials are always defined with respect
to reduction.
Zn+2 + 2e-
Zn
E° = -0.76 V
Cu+2 + 2e-
Cu
E° = +0.34 V
Fe+3 + e-
Fe+2
E° = 0.77 V
Standard Potentials (cont.)
• If in constructing an electrochemical cell, you
need to write the reaction as a oxidation instead of
a reduction, the sign of the 1/2 cell potential
changes.
Zn+2 + 2eZn
Zn
E° = -0.76 V
Zn+2 + 2e-
E° = +0.76 V
• 1/2 cell potentials are intensive variables. As
such, you do NOT multiply them by any
coefficients when balancing reactions.
Writing Galvanic Cells
For galvanic cells, Ecell > 0
In this example:
Zn/Zn+2 is the anode
Zn+2 + 2e- E° = +0.76 V
Zn
Cu/Cu+2 is the cathode
Cu+2 + 2e-
Cu E° = 0.34 V
Writing Galvanic Cells (cont.)
Zn+2 + 2e- E° = +0.76 V
Zn
Cu+2 + 2e-
Cu E° = 0.34 V
Cu+2 + Zn
Cu + Zn+2
E°cell = 1.10 V
Notice, we “reverse” the potential
for the anode.
E°cell = E°cathode - E°anode
Writing Galvanic Cells (cont.)
Shorthand Notation
Zn|Zn+2||Cu+2|Cu
Anode
Cathode
Salt bridge
Predicting Galvanic Cells
• Given two 1/2 cell reactions, how can one
construct a galvanic cell?
• Need to compare the reduction potentials of the
two half cells.
• Turn the reaction for the weaker reduction
(smaller E°1/2) and turn it into an oxidation. This
reaction will be the anode, the other the cathode.
Predicting Galvanic Cells (cont.)
• Example. Describe a galvanic cell based on the
following:
Ag+ + e-
Ag
E°1/2 = 0.80 V
Fe+3 + e-
Fe+2
E°1/2 = 0.77 V
Weaker reducing agent – turn it around
Ag+ + Fe+2
Ag + Fe+3
E°cell = 0.03 V
E°cell > 0….cell is galvanic
Another Example
• For the following reaction, identify the two half
cells, and use these half cells to construct a
galvanic cell
3Fe+2(aq)
+2
Fe(s) + 2Fe+3(aq)
reduction
0
+3
oxidation
Fe+2(aq) + 2e-
Fe(s)
E° = -0.44 V
Fe+3(aq) + e-
Fe+2(aq)
E° = +0.77 V
Another Example (cont.)
weaker reduction – turn it around
Fe+2(aq) + 2e2 x Fe+3(aq) + e-
2Fe+3(aq) + Fe(s)
Fe(s)
Fe(s)
E° = -0.44 V
Fe+2(aq)
E° = +0.77 V
3Fe+2(aq)
Fe+2(aq) + 2e-
E°cell = 1.21 V
E° = +0.44 V