• 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

Walther H. Nernst (1864-1941) 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

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 | Zn²⁺ || H⁺ | H₂ | Pt we have a net chemical reaction of Zn(s) + 2 H⁺ = Zn²⁺ + H₂(g) and the standard cell potential DE° = 0.763.

If the concentrations of the ions are not 1.0 M, and the H₂ pressure is not 1.0 atm, then the cell potential DE may be calculated using the Nernst equation:

                 0.0592 V     P(H2) [Zn2+]
     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 H₂ pressure is 1 atm, the term P(H₂) 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) | Zn²⁺ (0.024 M) || Zn²⁺ (2.4 M) | Zn(s)

Solution


Zn2+ (2.4 M)  +  2 e  =  Zn     Reduction
Zn  =  Zn2+ (0.024 M) +  2 e    Oxidation
--------------------------------------------
Zn2+ (2.4 M)  =  Zn2+ (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 Zn²⁺ 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,

Zn²⁺ (0.024 M) = Zn²⁺ (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

Show that the voltage of an electric cell is unaffected by multiplying the reaction equation by a positive number.

Solution
Assume that you have the cell

Mg | Mg²⁺ || Ag⁺ | Ag and the reaction is: Mg + 2 Ag⁺ = Mg²⁺ + 2 Ag Using the Nernst equation

                   0.0592     [Mg2+]
     DE  =  DE°  - ------ log --------
                     2        [Ag+]2

If you multiply the equation of reaction by 2, you will have 2 Mg + 4 Ag⁺ = 2 Mg²⁺ + 4 Ag Note that there are 4 electrons involved in this equation, and n = 4 in the Nernst equation:

                   0.0592     [Mg2+]2
     DE  =  DE°  - ------ log --------
                     4        [Ag+]4

which can be simplified as

                   0.0592     [Mg2+]
     DE  = DE°  - ------ log --------
                     2        [Ag+]2

Thus, the cell potential DE is not affected.

Example 3

The standard cell potential dE° for the reaction Fe + Zn²⁺ = Zn + Fe²⁺ is -0.353 V. If a piece of iron is placed in a 1 M Zn²⁺ solution, what is the equilibrium concentration of Fe²⁺?

Solution
The equilibrium constant K may be calculated using

K = 10^{(n DE°)/0.0592}
    = 10^-11.93
    = 1.2x10^-12
    = [Fe²⁺]/[Zn²⁺]. Since [Zn²⁺] = 1 M, it is evident that
        [Fe²⁺] = 1.2E-12 M.

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 | PbSO₄ | H₂SO₄ | PbSO₄, PbO₂ | Pb would the voltage change if you changed the concentration of H₂SO₄? (yes/no)

  Answer ... Yes!
  Hint...
  The net cell reaction is

  Pb + PbO₂ + 2 HSO₄^- + 2 H⁺ ® 2 PbSO₄ + 2 H₂O and the Nernst equation DE = DE° - (0.0592/2)log{1/{[HSO₄^-]²[H⁺]²}}.

• Choose the correct Nernst equation for the cell Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s).
  1. DE = DE° - 0.0296 log([Zn²⁺] / [Cu²⁺])
  2. DE = DE° - 0.0296 log([Cu²⁺] / [Zn²⁺])
  3. DE = DE° - 0.0296 log(Zn / Cu)
  4. DE = DE° - 0.0296 log(Cu / Zn)

  Answer ... A
  Hint...
  The cell as written has
  Reduction on the Right: Cu²⁺ + 2 e = Cu
  oxidation on the left: Zn = Zn²⁺ + 2 e
  Net reaction of cell is Zn (s) + Cu²⁺ = Cu (s) + Zn²⁺

• The standard cell potential DE° is 1.100 V for the cell, Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s). If [Zn²⁺] = 0.01 M, and [Cu²⁺] = 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{[Zn²⁺]/[Cu²⁺]} (n = 2 in this case).
  Understandably, if the concentration of Zn²⁺ is low, there is more tendency for the reaction,

  Zn = Zn²⁺ + 2 e.

• The logarithm of the equilibrium constant, log K, of the net cell reaction of the cell Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s) . . . DE° = 1.100 V is
  1. 1.100 / 0.0291
  2. -1.10 / 0.0291
  3. 0.0291 / 1.100
  4. -0.0291 / 1.100
  5. 1.100 / 0.0592

  Answer ... A
  Hint...
  Use the Nernst equation in the form

  0 = 1.100 - 0.0296 log ([Zn²⁺] / [Cu²⁺]) The Nernst equation is useful for the determination of equilibrium constants.