lecture 11: cell potentials
DESCRIPTION
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). - PowerPoint PPT PresentationTRANSCRIPT
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- + 5e- Mn2+ + 4H2O
Fe2+ Fe3+ + e- x 5
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 Oxidation
Cathode: Electrons are gained 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):
• 1 J of work is done when 1 C of charge is transferred between a potential difference of 1 V.
Volt = work (J)/charge (C)
Cell Potentials (cont.)• We can measure the magnitude of the EMF
causing electron (i.e., current) flow by measuring the voltage.
Anode Cathode
e-
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.
Zn Zn+2 + 2e-
E° SHE = 0 V
• In our example, Zn/Zn+2 is the anode: oxidation
2H+ + 2e- H2
Zn + 2H+ Zn+2 + H2
E°cell = E°SHE + E°Zn/Zn+2 = 0.76 V
0
E°Zn/Zn+2 = 0.76 V
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 + 2e- Zn E° = -0.76 V
Zn 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
Cu/Cu+2 is the cathode
Zn Zn+2 + 2e- E° = +0.76 V
Cu+2 + 2e- Cu E° = 0.34 V
Writing Galvanic Cells (cont.)
Zn Zn+2 + 2e- E° = +0.76 V
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) Fe(s) + 2Fe+3(aq)+2 0 +3 oxidation
reduction
Fe+2(aq) + 2e- Fe(s) E° = -0.44 V
Fe+3(aq) + e- Fe+2(aq) E° = +0.77 V
Another Example (cont.)
Fe+2(aq) + 2e- Fe(s) E° = -0.44 V
Fe+3(aq) + e- Fe+2(aq) E° = +0.77 V
weaker reduction – turn it around
Fe(s) Fe+2(aq) + 2e- E° = +0.44 V
2 x
2Fe+3(aq) + Fe(s) 3Fe+2(aq) E°cell = 1.21 V