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Nernst Equation

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The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).

Introduction

The Nernst Equation is derived from the emf and the Gibbs energy under non-standard conditions.

\[E^o = E^o_{reduction} - E^o_{oxidation}\]

When \(E^o\) is positive, the reaction is spontaneous. When \(E^o\) is negative, the reaction is not spontaneous. Since the change in Gibbs free energy, \(\Delta{G}\), is also related to spontaneity of a reaction, therefore, \(\Delta{G}\) and \(E\) are related. Specifically,

\[\Delta{G} = -nFE\]

with

  • \(n\) is the number of electrons transferred in the reaction (from balanced reaction),
  • \(F\) is the Faraday constant (96500 C/mol), and
  • \(E\) is potential difference.

Under standard conditions, this equation is then

\[\Delta{G}^{o} = -nFE^{o}.\]

Since,

\[\Delta{G} = \Delta{G}^o + RT \ln Q \tag{1}\]

Substituting \(\Delta{G} = -nFE\) and \(\Delta{G}^{o} = -nFE^{o}\) into equation (1), we have:

\[-nFE = -nFE^o + RT \ln Q\]

Divide both sides of the equation above by \(-nF\), we have

\[E = E^o - \dfrac{RT}{nF} \ln Q \tag{2}\]

Equation (2) can be rewritten in the form of \(\log_{10}\):

\[E = E^o - \dfrac{2.303 RT}{nF} \log Q \tag{Generalized Nernst Equation}\]

At standard temperature T = 298 K, the \(\frac{2.303 RT}{F}\) term equals 0.0592 V and this equation turns into:

\[E = E^o - \dfrac{0.0592 V}{n} \log Q \tag{Nernst Equation @ 298 K}\]

The equation above indicates that the electrical potential of a cell depends upon the reaction quotient \(Q\) of the reaction. As the redox reaction proceeds, reactants are consumed, thus concentration of reactants decreases. Conversely, the products concentration increases due to the increased in products formation. As this happens, cell potential gradually decreases until the reaction is at equilibrium, at which \(\Delta{G} = 0\).

Example 1

The \(E^{o}_{cell} = +1.10 \; V\) for the Zn-Cu redox reaction:

\[Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}.\]

What is the equilibrium constant for this reversible reaction?

SOLUTION

Initially, \([Cu^{2+}] = [Zn^{2+}] = 1.0\, M\) at standard T = 298K. As the reaction proceeds, \([Cu^{2+}]\) decreases as \([Zn^{2+}]\) increases. Lets say after one minute, \([Cu^{2+}] = 0.05\, M\) while \([Zn^{2+}] = 1.95\, M\). According to the Nernst equation, the cell potential after 1 minute is:

\[E = E^o - \dfrac{0.0592 V}{n} \log Q\]

\[E = 1.10V - \dfrac{0.0592 V}{2} \log\dfrac{1.95 \; M}{.05 \; M}\]

\[E = 1.05 \; V\]

As you can see, the initial cell potential is \(E  = 1.10\, V\), after 1 minute, the potential drops to 1.05 V. This is after 95% of the reactants have been consumed. As the reaction continues to progress, more \(Cu^{2+}\) will be consumed and more \(Zn^{2+}\) will be generated (at a 1:1 ratio). As a result, the cell potential continues to decrease and when the cell potential drops down to 0, the concentration of reactants and products stops changing. This is when the reaction is at equilibrium.

At equilibrium, the reaction quotient \(Q = K_{eq}\). Also, at equilibrium, \(\Delta{G} = 0\) and \(\Delta{G} = -nFE\), so \(E = 0\).   

Therefore, substituting \(Q = K_{eq}\) and \(E = 0\) into the Nernst equation, we have:

\[0 = E^o - \dfrac{RT}{nF} \ln K_{eq}\]

At standard conditions, the equation above simplifies into:

\[0 = E^o - \dfrac{0.0592}{n} \log K_{eq}\]

This equation can be rearranged into: 

\[\log K_{eq} = \dfrac{nE^o}{0.0592}\]

The equation above indicates that the equilibrium constant Keq is proportional to the standard potential of the reaction. Specifically, when:

  • \(K > 1, E^o > 0\), reaction favors products formation.
  • \(K < 1, E^o < 0\), reaction favors reactants formation.

This result fits Le Châtlier's Principle,which states that when a system at equilibrium experiences a change, the system will minimize that change by shifting the equilibrium in the opposite direction.

References

  1. Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York: W.H. Freeman and Company. p. 214-222.
  2. Sherwood, Lauralee. Human Physiology 6th edition. Thompson Corp. 2007. p. 77
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