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Nernst Equation
Skills to develop
 Explain and distinguish the cell potential and standard cell potential.
 Calculate cell potentials from known conditions (Nernst Equation).
 Calculate the equilibrium constant from cell potentials.
Nernst Equation
Electrochemistry deals with cell potential as well as energy of
chemical reactions. The energy of a chemical system drives the charges
to move, and the driving force give rise to the cell potential of a
system called galvanic cell. The energy
aspect is also related to the chemical equilibrium.
All these relationships are tied together in the concept of
Nearnst equation.
Walther H. Nernst (18641941) received the
Nobel prize in 1920 "in recognition of his work in
thermochemistry". His contribution to chemical thermodynamics
led to the well known equation correlating chemical energy and
the electric potential of a galvanic cell or battery.
Electric Work and Gibb's Free Energy
Energy takes many forms: mechanical work (potential and kinetic energy),
heat, radiation (photons), chemical energy, nuclear energy (mass), and
electric energy. A summary is given regarding the evaluation of electric
energy, as this is related to electrochemistry.
Electric Work
Energy drives all changes including chemical reactions. In a redox reaction,
the energy released in a reaction due to movement of charged particles give
rise to a potential difference. The maximum potential differenc is
called the electromotive force, (EMF), E
and the maximum electric work W is the product of charge q in
Coulomb (C), and the potential DE in
Volt (= J / C) or EMF.
W J = q DE
C J/C (units)
Note that the EMF DE is determined by the
nature of the reactants and electrolytes, not by the size of the cell or
amounts of material in it. The amount of reactants is proportional to
the charge and available energy of the galvanic cell.
Gibb's Free Energy
The Gibb's free energy DG is the
negative value of maximum electric work,
DG =  W
=  q DE
A redox reaction equation represents definite amounts of reactants
in the formation of also definite amounts of products.
The number (n) of electrons in such a reaction equation,
is related to the amount of charge trnasferred when the reaction
is completed. Since each mole of electron has a charge of 96485 C
(known as the Faraday's constant, F),
q = n F
and,
DG =  n F DE
At standard conditions,
DG° =  n F DE°
The General Nernst Equation
The general Nernst equation correlates the Gibb's Free Energy DG
and the EMF of a chemical system known as the galvanic cell.
For the reaction
a A + b B = c C + d D
and
[C]^{c} [D]^{d}
Q = 
[A]^{a} [B]^{b}
It has been shown that
DG = DG° + R T ln Q
and
DG =  n FDE.
Therefore
 n F DE =  n F DE° + R T ln Q
where R, T, Q and F are the gas constant
(8.314 J mol^{1} K^{1}), temperature (in K),
reaction quotient, and Faraday constant (96485 C) respectively.
Thus, we have
R T [C]^{c} [D]^{d}
DE = DE°   ln 
n F [A]^{a} [B]^{b}
This is known as the Nernst equation.
The equation allows us to calculate the cell potential of any galvanic
cell for any concentrations. Some examples are given in the next section
to illustrate its application.
It is interesting to note the relationship between equilibrium and the
Gibb's free energy at this point. When a system is at equilibrium,
DE = 0, and Q_{eq} = K. Therefore, we have,
R T [C]^{c} [D]^{d}
DE° =  ln , (for equilibrium concentrations)
n F [A]^{a} [B]^{b}
Thus, the equilibrium constant and DE° are related.
The Nernst Equation at 298 K
At any specific temperature, the Nernst equation derived above
can be reduced into a simple form. For example, at the standard
condition of 298 K (25°), the Nernst equation becomes
0.0592 V [C]^{c} [D]^{d}
DE = DE°   log 
n [A]^{a} [B]^{b}
Please note that log is the logrithm function based 10, and ln, the natural
logrithm function.
For the cell
Zn  Zn^{2+}  H^{+}  H_{2}  Pt
we have a net chemical reaction of
Zn(s) + 2 H^{+} = Zn^{2+} + H_{2}(g)
and the standard cell potential DE° = 0.763.
If the concentrations of the ions are not 1.0 M, and the H_{2}
pressure is not 1.0 atm, then the cell potential DE may be
calculated using the Nernst equation:
0.0592 V P(H_{2}) [Zn^{2+}]
DE = DE°   log 
n [H^{+}]^{2}
with n = 2 in this case, because the reaction involves 2 electrons.
The numerical value is 0.0592 only when T = 298 K. This constant is
temperature dependent. Note that the reactivity of the solid Zn is
taken as 1. If the H_{2} pressure is 1 atm, the term P(H_{2}) may also be omitted.
The expression for the argument of the log function follows the same rules
as those for the expression of equilibrium constants and reaction quotients.
Indeed, the argument for the log function is the expression for
the equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, DE = 0.00 and
the expression becomes an equilibrium constant K, which bears the
following relationship:
n DE°
log K = 
0.0592
where DE° is the difference of standard potentials of
the half cells involved. A battery containing any voltage is not
at equilibrium.
The Nernst equation also indicates that you can build a battery simply
by using the same material for both cells, but by using different
concentrations.
Cells of this type are called concentration cells.
Example 1
Calculate the EMF of the cell
Zn(s)  Zn^{2+} (0.024 M)  Zn^{2+} (2.4 M)  Zn(s)
Solution
Zn^{2+} (2.4 M) + 2 e = Zn Reduction
Zn = Zn^{2+} (0.024 M) + 2 e Oxidation

Zn^{2+} (2.4 M) = Zn^{2+} (0.024 M), DE° = 0.00   Net reaction
Using the Nernst equation:
0.0592 (0.024)
DE = 0.00   log 
2 (2.4)
= (0.296)(2.0)
= 0.0592 V
Discussion
Understandably, the Zn^{2+} ions try to move from the concentrated
half cell to a dilute solution. That driving force gives rise to
0.0592 V. From here, you can also calculate the energy of dilution.
If you write the equation in the reverse direction,
Zn^{2+} (0.024 M) = Zn^{2+} (2.4 M),
its voltage will be 0.0592 V. At equilibrium concentrations in the
two half cells will have to be equal, in which case the voltage
will be zero.
Example 2
0.0592 [Mg^{2+}]
DE = DE°   log 
2 [Ag^{+}]^{2}
If you multiply the equation of reaction by 2, you will have
2 Mg + 4 Ag^{+} = 2 Mg^{2+} + 4 Ag
Note that there are 4 electrons involved in this equation, and
n = 4 in the Nernst equation:
0.0592 [Mg^{2+}]^{2}
DE = DE°   log 
4 [Ag^{+}]^{4}
which can be simplified as
0.0592 [Mg^{2+}]
DE = DE°   log 
2 [Ag^{+}]^{2}
Thus, the cell potential DE is not affected.
Example 3
Example 4
From the standard cell potentials, calculate the solubility
product for the following reaction:
AgCl = Ag^{+} + Cl^{}
Solution
There are Ag^{+} and AgCl involved in the reaction, and from the table of
standard reduction potentials, you will find:
AgCl + e = Ag + Cl^{}, E° = 0.2223 V    (1)
Since this equation does not contain the species Ag^{+}, you need,
Ag^{+} + e = Ag, E° = 0.799 V       (2)
Subtracting (2) from (1) leads to,
AgCl = Ag^{+} + Cl^{} . . . DE° =  0.577
Let K_{sp} be the solubility product, and employ the Nernst equation,
log K_{sp} = (0.577) / (0.0592) = 9.75
K_{sp} = 10^{9.75} = 1.8x10^{10}
This is the value that you have been using in past tutorials.
Now, you know that K_{sp} is not always measured from its solubility.
Confidence Building Questions
 In the lead storage battery,
Pb  PbSO_{4}  H_{2}SO_{4}  PbSO_{4}, PbO_{2}  Pb
would the voltage change if you changed the concentration of H_{2}SO_{4}?
(yes/no)
Answer ... Yes!
Hint...
The net cell reaction is
Pb + PbO_{2} + 2 HSO_{4}^{} + 2 H^{+}
® 2 PbSO_{4} + 2 H_{2}O
and the Nernst equation
DE = DE°  (0.0592/2)log{1/{[HSO_{4}^{}]^{2}[H^{+}]^{2}}}.

Choose the correct Nernst equation for the cell
Zn(s)  Zn^{2+}  Cu^{2+}  Cu(s).
 DE = DE°  0.0296 log([Zn^{2+}] / [Cu^{2+}])
 DE = DE°  0.0296 log([Cu^{2+}] / [Zn^{2+}])
 DE = DE°  0.0296 log(Zn / Cu)
 DE = DE°  0.0296 log(Cu / Zn)
Answer ... A
Hint...
The cell as written has
Reduction on the Right: Cu^{2+} + 2 e = Cu
oxidation on the left: Zn = Zn^{2+} + 2 e
Net reaction of cell is Zn (s) + Cu^{2+} = Cu (s) + Zn^{2+}

The standard cell potential DE°
is 1.100 V for the cell,
Zn(s)  Zn^{2+}  Cu^{2+}  Cu(s).
If [Zn^{2+}] = 0.01 M, and [Cu^{2+}] = 1.0 M, what is
DE or EMF?
Answer ... 1.159 V
Hint...
A likely wrong result is 1.041 V.
The term that modifies DE is
(0.059/n)log{[Zn^{2+}]/[Cu^{2+}]} (n = 2 in this case).
Understandably, if the concentration of Zn^{2+}
is low, there is more tendency for the reaction,
Zn = Zn^{2+} + 2 e.

The logarithm of the equilibrium constant, log K, of the net cell
reaction of the cell
Zn(s)  Zn^{2+}  Cu^{2+}  Cu(s) . . .
DE° = 1.100 V
is
 1.100 / 0.0291
 1.10 / 0.0291
 0.0291 / 1.100
 0.0291 / 1.100
 1.100 / 0.0592
Answer ... A
Hint...
Use the Nernst equation in the form
0 = 1.100  0.0296 log ([Zn^{2+}] / [Cu^{2+}])
The Nernst equation is useful for the determination of equilibrium constants.
Understanding is the key. Take time to understand it, there is no point
in rushing.
© cchieh@uwaterloo.ca