Download Lecture 11: Cell Potentials

Lecture 11: Cell Potentials
•  Reading: Zumdahl 11.2
•  Recommended Problems:11.15,11.17,11.19
11.21
•  Outline
–  Redox Review
–  What is a cell potential?
–  SHE, the electrochemical zero.
–  Using standard reduction potentials.
Galvanic Cells
•  In redox reaction, electrons are transferred from
the oxidized species to the reduced species.
•  Imagine separating the two 1/2 cells physically,
then providing a conduit through which the
electrons travel from one cell to the other.
Porous Cap
Galvanic Cells (cont.)
e-
e-
Cu
Ag
Cu2+ (aq)
Ag+ (aq)
Cu
Ag+ + e-
Cu2+ + 2e-
Ag x 2
Galvanic Cells (cont.)
Anode: Electrons are lost
Oxidation Cathode: Electrons are gained
Reduction
Cell Potentials
•  In a galvanic cell, a species is oxidized at the
anode, a another species is reduced at the cathode,
with electrons flowing from anode to cathode.
This electron current can do work: w=-qEcell where
q is the charge transferred from anode to cathode.
•  The force on the electrons causing them to full is
referred to as the electromotive force Ecell (EMF).
The unit used to quantify this force is the volt (V)
•  1 volt = 1 Joule/Coulomb of charge
V = J/C
Cell Potentials (cont.)
•  We can measure the magnitude of the EMF
causing electron (i.e., current) flow by measuring
the voltage.
e-
Anode
Cathode
Cell Potentials (cont.)
•  Imagine an experiment in which we simply
measure the voltage for a variety of species.
e-
-
Anode
Anode
?
+
Cathode
V
Cu/Cu2+ Ag/Ag+
0.46
Cu/Cu2+ Au/Au3+
1.16
Al/Al3+
2.00
Cu/Cu2+
Cathode
Can we use this information to predict what the cell potential will be?
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
Zn+2 + 2e-
2H+ + 2eZn + 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 as reductions.
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.
The Lead Storage Battery
•  The lead storage battery commonly used in
automobiles consists of 6 identical cells joined in
series. Each cell has a lead anode and a lead dioxide
cathode. Both the cathode and anode are immersed in
an aqueous solution of sulfuric acid. Under normal
operating conditions, each cell produces 2V.
The Lead Storage Battery (discharge mode)
•  The Reduction (Cathode): PbO2 + 4H+(aq) + SO42-(aq) +2e- 2PbSO4 + 2 H2O
Eo=1.69V (reduction potential)
•  The Oxidation (Anode)
Pb + SO42-(aq)  PbSO4 + 2e-
Eo=-0.35V (reduction potential)
•  Overall:
Pb + PbO2 + 4H+(aq) + 2 SO42-(aq)  2 PbSO4 + 2 H2O
Ecell=1.69V-(-0.35V)=2.04V
The Lead Storage Battery (recharge mode)
•  In recharge mode a voltage is applied to the battery by the alternator.
Ideally the battery runs in reverse
2 PbSO4 + 2 H2O  Pb + PbO2 + 4H+(aq) + 2 SO42-(aq) Ecell=-2.04V
•  Gassing: When excessive voltage is applied to the battery
causing hydrogen production
Oxidation at the anode:
•  H2SO4(aq) + 2e-  H2(g) + SO42-(aq)
Reduction at the cathode:
•  Pb(s) + 2 H2O(l) + 2 SO42-(aq)  PbO2(s) + 2 H2SO4(aq) + 4e-