| Chem 110 |
|
Fall 2003 |
| Lecture Notes::Lec 13_24 September |
|
|
| PREVIOUS |
Last time we looked at the two half cells required to create a galvanic cell. We then connected them with a wire, discovering that there was only an instantaneous flow of currrent. So what can we do to allow the flow of electrons to continue? What we need to do is connect the solutions in the beakers so that the charges can be neutralized with a counter flow of ions. The connecting ionic fluid is referred to as a salt-bridge, as seen in the figure. Other arrangements are possible such as semi-permeable membranes etc. Such an arrangement is called a Galvanic cell.

Galvanic cells (or Voltaic cells) are cells in which the overall redox reactions occur spontaneously (equilibrium favors products) as written. They can serve as a source of electric power (as a battery).
Anodes vs. Cathodes.
Let's look again at a Galvanic Cell (note that Zumdahl's illustrations assumes a porous disk connection)
The Galvanic cell consists of two half cells. In each cell the reaction, by convention, is written as a half-reaction, which is in fact the chemistry taking place in the half cells.
Note that in tables of Reduction Potentials the reactions are written as reduction half reactions, with the potentials those which would occur if the half-cell were connected into a galvanic cell with a SHE.
For the Zn, Cu, system we then have
Zn half cell:
But of course in our cell, this goes backwards:
Cu Half Cell:
CELL :
But recall we also stated that these half-cell potentials are
always related to the SHE, which is defined as 0.00
V at all temperatures.
The voltages of half cells determined relative to SHE are the Standard Reduction Potentials for these half cells (1M [more properly, at an activity of 1], 1 ATM)
So how do we determine the V of a Galvanic Cell from half-cell voltages?
1. Electrons will flow from less positive to more positive cell.
Consider a Galvanic Cell of Ag+/Ag and Cu2+/Cu
What will the reaction be?
Ag+ + e - Æ Ag0 E° = +0.80 VCu2+ + 2e - Æ Cu0 E° = +0.34 V
Since e- flow toward the more positive half cell, then Ag+/Ag cell is cathode (reduction at cathode) and Cu2+/ Cu will be reversed.Now we need to balance electrons so
2 (Ag+ + e - Æ Ag0) E° = +0.80 V
Cu0 Æ Cu2+ + 2e - E° = -0.34 V
2 Ag+ + Cu0 Æ 2 Ag0 + Cu2+ E° = + 0.46 V
Instead of drawing cells we often draw a cell diagram using what your author refers to as "Line Notation." It is conventional to start with the anode on left. The cell diagram for the Zn/ Cu Galvanic Cell will then be represented as:
Both of these describe cells with a liquid junction, such as a porous disk. What do Lines represent? Changes in phase or boundaries. With a salt bridge we see:
Note the two lines in the center representing the salt bridge. We need two lines because there is a boundary between each end of the salt bridge and its respective half cell.
So the complete description of our cell includes:
If we think physics, then the work done by a system is:
Where E = potential difference and q = charge.
So for the system or reaction:
and
If we consider a system doing the maximum possible work (no currrent flow - so takes forever), then work equals free energy, or:
For standard conditions @ equilibrium:
Substituting -nFE for DG:
Dividing by -nF:
This is called the Nernst Equation, which relates the voltage produced by a cell to the concentrations of reactants and products in a system. Note that this is the electrochemical version of the equation for free energy in a non-equilibrium system.
| Syllabus/Schedule |
© R A Paselk
Last modified 25 September 2003