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Nernst Equation

Introduction

The standard reduction potentials provide potentials for solutes at 1 M or 1 atm partial pressure. At other conditions the next section gives us some idea as to what we might predict for non-standard conditions.


Effect of Concentration on E

You'll see the Nernst equation:

    nernst1.gif

also written as:

    nernst2.gif

where RT/F at 298.15 K = (8.3145 J/mol·K)(298.15 K)/(96485.34 C/mol) = 0.02569 V and 2.303log replaces ln. (Note that J·C is equivalent to a Volt.) We can use either of these forms to:

Let's consider the reaction of tin and bromine again when the tin(II) concentration is 0.050 M and the tin(IV) and bromide concentrations = 0.00010 M.

Sn2+(aq) + Br2 (l) ---> Sn4+(aq) + 2 Br-(aq)

Q = [Sn4+][Br-]2 / [Sn2+]

(b) The Nernst equation for this reaction is:

             0.0591V      [Sn4+] [Br-]2
E = 0.92 V - ------- log( ------------ )
                2            [Sn2+]

             0.0591V      (0.00010 M)(0.00010 M)2
E = 0.92 V - ------- log( ----------------------- )
                2                  0.050 M

             0.0591V
E = 0.92 V - ------- log(2.0x10-11)
                2

E = 0.92 V - (-0.316 V)

E = 1.24 V


Example of a Concentration Cell

Consider two beakers containing silver nitrate solutions of different concentrations, say 0.010 M and 0.50 M. If we connect the two beakers do we have a system at equilibrium?

No, there is a concentration gradient and there is a driving force for diffusion to attain equal concentrations in both beakers. We can set up an electrochemical cell to capture the chemical potential of this non-equilibrium system.

|         |     |         |
|---------|_____|---------|
|          _____          |
| 0.01 M  |     |  0.5 M  |
|   Ag+   |     |   Ag+   |
|_________|     |_________|

The two half-reactions are:

Ag(s) ---> Ag+(aq) + e-          Ag+(aq) + e- ---> Ag(s)

Q = [Ag+] / [Ag+]    (To determine which concentration goes where, on which side of the half-reactions are the silver ion.)

(b) The Nernst equation for this reaction is:

             0.0591V     0.010 M
E = 0.0 V - ------- log( ------- )
                1         0.50 M

E = 0.0 V - (- 0.100 V)

E = 0.10 V


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