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>>
View the other Key Equations and Concepts in this
chapter
Concentration Dependence of Cell Potential
>> Parts of this equation/concept include:
Standard potentials assume that the concentration of every substance
in the reaction is 1 M. At other concentrations, the Nernst
equation is used to correct for the difference.
| E |
= |
E° |
– |
|
 |
log Q |
This equation assumes that the temperature is 298 K (room
temperature). Q is the reaction quotient from Chapter
15 (the form of the equilibrium constant, but not at equilibrium).
In the Nernst equation, the concentration is in the form of partial
pressures with units of atmospheres for gases and molarity for aqueous
solutes, even if the units are mixed.
Keeping track of significant figures is difficult in problems like
this. As the math in all four of the following examples was worked
out, the appropriate number of significant figures was written.
In the calculations extra significant figures were used. All the
types of rules were required.
| +/– |
Answer keeps fewest decimal places |
/ |
Answer keeps fewest significant
figures
(Note: 0.0592 is a calculated number with three
significant figures.) |
| log |
For y = log x, the
number of decimal places in y is the same as the number
of significant figures in x. |
>> Example 1
What is the potential of an electrochemical cell with the reaction
O2 + 2 H2CO 
2 HCO2H
if PO2 = 0.200 atm, [H2CO] = 0.0137
M and [H2CO2H] = 0.124 M.
Solution:
The half–reactions are
O2 + 4 H+ + 4 e–
2 H2O E°
= +1.2291 V
H2CO + H2O HCO2H + 2 H+
+ 2 e– E°
= +0.237 V
| E |
= |
E° |
– |
|
|
log |
| [HCO2H]2 |
|
| PO2
[HCO2H]2 |
|
The value of E° = 1.2291 – 0.237 = 0.922 V.
The number of electrons transferred, (n) is 4.
| E |
= |
0.922 |
– |
|
|
log |
| [0.124)2 |
|
| (0.200)(0.0137)2 |
|
| E |
= |
0.922 |
– |
|
|
log |
(410) |
| E |
= |
0.922 |
– |
|
|
2.613 |
E = 0.922 – 0.0385
E = +0.883 V
>> Example 2
What is the potential of an electrochemical cell with the reaction?
2 H+ + H2O2 + Sn Sn2+
+ 2 H2O
if [Sn2+] = 0.100 M, [H2O2]
= 0.759 M, and pH = 3.49.
Solution:
The half–reactions are
Sn0 Sn2+ + 2 e– E°
= –0.141 V
2 e– + 2 H+ + H2O2
2 H2O E°
= +1.763 V
| E |
= |
E° |
– |
|
|
log |
|
The value of E° = +1.763 – (–0.141) = +1.904
V.
Solids, like Sn0, and solvents, like H2O,
are not included in Q.
The value of n = 2.
If the pH = 3.49, then [H+] = 10–3.49
= 3.2 x 10–4 M.
| E |
= |
1.904 |
– |
|
|
log |
| [0.10] |
|
| [0.759][3.2 X 10–4]2 |
|
| E |
= |
1.904 |
– |
|
|
log |
(6.1) |
| E |
= |
1.904 |
– |
|
|
0.79 |
E = 1.904 – 0.18
E = 1.72
>> Example 3
What is the potential of an electrochemical cell with the reaction?
3 Ni + 4 H2O + SO42–
3 Ni(OH)2 + S + 2 OH–
if [SO42–] = 0.17 M and pH = 12.66.
Solution:
| E |
= |
E° |
– |
|
|
log |
|
Ni, Ni(OH)2 and S are solids. H2O is the
solvent.
The half–reactions are
Ni + 2 OH– Ni(OH)2 + 2 e– E°
= –0.714 V
SO42– + 4 H2O + 6 e–
S + 8 OH– E°
= –0.751 V
The value of E° = –0.751 – (–0.714)
= –0.010 V. The value of n = 6. If pH = 12.66, pOH = 14 – 12.66 = 1.34, so [OH–]
= 0.046 M.
| E |
= |
–0.010 |
– |
|
|
log |
|
| E |
= |
–0.010 |
– |
|
|
log |
(0.012) |
| E |
= |
–0.010 |
– |
|
|
–1.90 |
E = –0.010 + 0.0187
E = +0.009 V
| A. Relative Concentrations at Equilibrium |
At equilibrium, the cell potential is zero. The ratio of various
components at equilibrium can then be determined by the standard
cell potential and reaction conditions. The Nernst equation is as
valid for half–reactions as for total cell reactions.
>> Example 4
At pH = 5.50 and at equilibrium, what is the ratio of [ClO3–]
to [ClO4–]? The standard potential
for the reduction of perchlorate ion to chlorate ion in acidic
solution is +1.266 V.
Solution:
First the balanced half–reaction must be determined. The
question refers to the reduction of perchlorate (ClO4–)
to chlorate ion (ClO3–), so the skeletal
reaction is
ClO4– ClO3–
Using the directions for balancing in acid, the reaction becomes
ClO4– + 2 H+ + 2 e–
ClO3– + H2O
With electrons on the reactant side, this is written as a reduction,
so the potential is +1.266 V for this half–reaction.
Since the question asks about concentration, the Nernst equation
is appropriate.
| E |
= |
E° |
– |
|
|
log Q |
For this reaction the Nernst equation is
| E |
= |
1.266 |
– |
|
|
log |
|
Since the reaction is at equilibrium, E = 0. At pH 5.50,
[H+] = 3.2 x 10–6. Using these values
in the equation,
| 0 |
= |
1.266 |
– |
|
|
log |
| [ClO3–] |
|
| [ClO4–][3.2
x 10–6]2 |
|
| 1.266 |
= |
|
|
log |
| [ClO3–] |
|
| [ClO4–][3.2
x 10–6]2 |
|
Since log xy = log x + log y,
| 1.266 |
= |
|
|
log |
|
+ |
|
|
log |
|
| 1.266 |
= |
|
|
log |
|
+ |
|
|
0.326 |
| 0.940 |
= |
|
|
log |
|
>> View
the other Key Equations and Concepts in this chapter
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