Fundamentals of Electrochemistry - W.H. Freeman

1504T_ch14_270-297 1/23/06 11:35 Page 283
To find the concentration of Ag , we use the solubility product for AgCl. Because the cell
contains 0.033 4 M Cl and solid AgCl, we can say
Putting this value into the Nernst equation gives
which differs from the value calculated in Equation 14-18 because of the accuracy of and
the neglect of activity coefficients. Equations 14-17 and 14-19 give the same voltage because
they describe the same cell.
Advice for Finding Relevant Half-Reactions
When faced with a cell drawing or a line diagram, first write reduction reactions for each
half-cell. To do this, look in the cell for an element in two oxidation states. For the cell
we see Pb in two oxidation states, as Pb(s) and PbF 2 (s), and Cu in two oxidation states, as
and Cu(s). Thus, the half-reactions are
Cu 2
[Ag ] K sp(for AgCl)
[Cl ]
Right half-cell:
Left half-cell:
1.8 1010
0.033 4
Cu 2 2e T Cu(s)
5.4 10 9 M
1
E 0.799 0.059 16 log
5.4 10 0.309 9 9 V
Pb(s) 0 PbF 2 (s) 0 F (aq) Cu 2 (aq) 0 Cu(s)
PbF 2 (s) 2e T Pb(s) 2F
K sp
(14-20)
K sp [Ag ] [CI ]
The cell voltage cannot depend on how we
write the reaction!
How to figure out the half-cell reactions
You might have chosen to write the Pb half-reaction as
Left half-cell:
Pb 2 2e T Pb(s)
(14-21)
because you know that, if PbF is present, there must be some Pb 2
2 (s)
in the solution.
Reactions 14-20 and 14-21 are equally valid and should predict the same cell voltage.
Your choice of reactions depends on whether the F or Pb 2 concentration is easier to
figure out.
We described the left half-cell in terms of a redox reaction involving Pb because Pb is
the element that appears in two oxidation states. We would not write a reaction such as
F 2 (g) 2e T 2F , because F 2 (g) is not shown in the line diagram of the cell.
The Nernst Equation Is Used in
Measuring Standard Reduction Potentials
The standard reduction potential would be observed if the half-cell of interest (with unit
activities) were connected to a standard hydrogen electrode, as it is in Figure 14-7. It is
nearly impossible to construct such a cell, because we have no way to adjust concentrations
and ionic strength to give unit activities. In reality, activities less than unity are used in each
half-cell, and the Nernst equation is employed to extract the value of E° from the cell voltage.
12 In the hydrogen electrode, standard buffers with known pH (Table 15-3) are used to
obtain known activities of H .
Don’t invent species not shown in the cell. Use
what is shown in the line diagram to select the
half-reactions.
Problem 14-20 gives an example of the use of
the Nernst equation to find E°.
14-5 E and the Equilibrium Constant
A galvanic cell produces electricity because the cell reaction is not at equilibrium. The
potentiometer allows negligible current (Box 14-4), so concentrations in the cell remain
unchanged. If we replaced the potentiometer with a wire, there would be much more current
and concentrations would change until the cell reached equilibrium. At that point,
nothing would drive the reaction, and E would be 0. When a battery (which is a galvanic
cell) runs down to 0 V, the chemicals inside have reached equilibrium and the battery is
“dead.”
Now let’s relate E for a whole cell to the reaction quotient, Q, for the net cell reaction.
For the two half-reactions
Right electrode:
Left electrode:
aA ne T cC
dD ne T bB
E°
E°
At equilibrium, E (not E°) 0.
14-5 E° and the Equilibrium Constant 283