Nernst equation

In electrochemistry, the Nernst equation gives the electrode potential (E), relative to the standard electrode potential, (E0), of the electrode couple or, equivalently, of the half cells of a battery. In physiology the Nernst equation is used for finding the electric potential of a cell membrane with respect to one type of ion.



E = E^0 - \frac{RT}{nF} \ln\frac{a_{\mbox{Red}}}{a_{\mbox{Ox}}} $$

The activities of pure solid or liquid phases are taken as unity. For a solution at room temperature (25 °C) the following is true:



E = E^{0'} - \frac{0.0591}{n} \log\frac{[\mbox{Red}]}{[\mbox{Ox}]} $$ (or 0.025679 using ln).

Where 0.0591 is a constant created from the Gas and Faraday constants, the temperature and a factor to convert from ln to log. This means that 0.0591 can only be used at 25°C

For a cell membrane potential with respect to one cation (for an anion the sign before the logarithm is changed to a minus),

E = E^{0'} + \frac{0.0591}{n} \log\frac{[\mbox{ion out of cell}]}{[\mbox{ion inside cell}]} $$

where

E^{0'} = E^0 - \frac{RT}{nF} \log\frac{\gamma_{\mbox{Red}}}{\gamma_{\mbox{Ox}}} $$


 * R is the universal gas constant, equal to 8.314510 J K-1 mol-1
 * T is the temperature in kelvin. (Kelvin = 273.15 + °C.)
 * a the chemical activities on the reduced and oxidized side, respectively
 * F is the Faraday constant (the charge per a mole of electrons), equal to 9.6485309*104 C mol-1
 * n is the number of electrons transferred in the half-reaction.
 * [Red] is the concentration of oxidizing agent (the reduced species).
 * [Ox] is the concentration of reducing agent (the oxidized species).
 * $$E^{0'}$$ is the formal electrode potential
 * $$\gamma$$ is the activity coefficient

Correctly though, the Log Q is the quotient product, and the q can be obtained from placing the concentrations of the products over the reactants, which can alternate between oxidizing and reducing agents.

History
The Nernst equation is named after the German physical chemist Walther Nernst who was the first to formulate it.

Derivation
The Nernst Equation may be derived in several different ways. Chemistry textbooks frequently give the derivation in terms of entropy and the Gibbs free energy, but there is a more intuitive method for anyone familiar with Boltzmann factors.

Nernst Potential
The potential level across the cell membrane that exactly opposes net diffusion of a particular ion through the membrane is called the Nernst potential for that ion. The magnitude of the Nernst potential is determined by the ratio of the concentrations of that specific ion on the two sides of the membrane. The greater this ratio, the greater the tendency for the ion to diffuse in one direction, and therefore the greater the Nernst potential required to prevent the diffusion.

Using Boltzmann factors
For simplicity, we will consider a solution of redox-active molecules that undergo a one electron reaction

\mathrm{Ox} + e^- \rightarrow \mathrm{Red} $$ and which have a standard potential of zero. The chemical potential $$\mu_c$$ of this solution is the difference between the energy barriers for taking electrons from and for giving electrons to the working electrode that is setting the solution's electrochemical potential.

The ratio of oxidized to reduced molecules, [Ox]/[Red], is equivalent to the probability of being oxidized (giving electrons) over the probability of being reduced (taking electrons), which we can write in terms of the Boltzmann factors for these processes:

\frac{[\mathrm{Ox}]}{[\mathrm{Red}]} = \frac{\exp \left(-[\mbox{barrier for losing an electron}]/kT\right)} {\exp \left(-[\mbox{barrier for gaining an electron}]/kT\right)} = \exp \left(\mu_c / kT \right). $$ Taking the natural logarithm of both sides gives

\mu_c = kT \ln \frac{[\mathrm{Ox}]}{[\mathrm{Red}]}. $$ If $$\mu_c \ne 0$$ at [Ox]/[Red] = 1, we need to add in this additional constant:

\mu_c = \mu_c^0 + kT \ln \frac{[\mathrm{Ox}]}{[\mathrm{Red}]}. $$ Dividing the equation by e to convert from chemical potentials to electrode potentials, and remembering that kT/e = RT/F, we obtain the Nernst equation for the one-electron process $$\mathrm{Ox} + e^- \rightarrow \mathrm{Red}$$:

E = E^0 + \frac{kT}{e} \ln \frac{[\mathrm{Ox}]}{[\mathrm{Red}]} = E^0 - \frac{RT}{F} \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]}. $$

Using entropy and Gibbs free energy
Quantities here are given per molecule, not per mole, and so Boltzmann's constant k and the electron charge e are used instead of the gas constant R and Faraday's constant F. To convert to the molar quantities given in most chemistry textbooks, it is simply necessary to multiply by Avogadro's number: $$R = kN_A$$ and $$F = eN_A$$.

The entropy of a molecule is defined as

S \ \stackrel{\mathrm{def}}{=}\ k \ln \Omega, $$ where $$\Omega$$ is the number of states available to the molecule. The number of states must vary linearly with the volume V of the system, which is inversely proportional to the concentration c, so we can also write the entropy as

S = k\ln \ (\mathrm{constant}\times V) = -k\ln \ (\mathrm{constant}\times c). $$ The change in entropy from some state 1 to another state 2 is therefore

\Delta S = S_2 - S_1 = - k \ln \frac{c_2}{c_1}, $$ so that the entropy of state 1 is

S_2 = S_1 - k \ln \frac{c_2}{c_1}. $$ If state 1 is at standard conditions, in which $$c_1$$ is unity (e.g., 1 atm or 1 M), it will merely cancel the units of $$c_2$$. We can therefore write the entropy of an arbitrary molecule A as

S(A) = S^0(A) - k \ln [A], \, $$ where $$S^0$$ is the entropy at standard conditions and [A] denotes the concentration of A. The change in entropy for a reaction

aA + bB \rightarrow yY + zZ $$ is then given by

\Delta S_\mathrm{rxn} = [yS(Y) + zS(Z)] - [aS(A) - bS(B)] = \Delta S^0_\mathrm{rxn} - k \ln \frac{[Y]^y [Z]^z}{[A]^a [B]^b}. $$ We define the ratio in the last term as the reaction quotient:

Q \ \stackrel{\mathrm{def}}{=}\ \frac{[Y]^y [Z]^z}{[A]^a [B]^b}. $$

In an electrochemical cell, the cell potential E is the chemical potential available from redox reactions ($$E = \mu_c/e$$). E is related to the Gibbs free energy change $$\Delta G$$ only by a constant: $$\Delta G = -neE$$, where n is the number of electrons transferred. (There is a negative sign because a spontaneous reaction has a negative $$\Delta G$$ and a positive E.) The Gibbs free energy is related to the entropy by $$G = H - TS$$, where H is the enthalpy and T is the temperature of the system. Using these relations, we can now write the change in Gibbs free energy,

\Delta G = \Delta H - T \Delta S = \Delta G^0 + kT \ln Q, \, $$ and the cell potential,

E = E^0 - \frac{kT}{ne} \ln Q. $$ This is the more general form of the Nernst equation. For the redox reaction $$\mathrm{Ox} + ne^- \rightarrow \mathrm{Red},$$ $$Q = [\mathrm{Red}]/[\mathrm{Ox}]$$, and we have:

E = E^0 - \frac{kT}{ne} \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]} = E^0 - \frac{RT}{nF} \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]}. $$ The cell potential at standard conditions $$E^0$$ is often replaced by the formal potential $$E^{0'}$$, which includes some small corrections to the logarithm and is the potential that is actually measured in an electrochemical cell.

Limitations
When the Nernst equation is expressed in its most convenient form, the activity of the ions is assumed to be equal to their concentrations, however this assumption is only valid for low concentrations. At higher concentrations the true activities of the ions must be used; this complication makes the use of the Nernst equation difficult, as estimation of the activities of ions in their non-ideal state often requires experimental analysis.

The Nernst equation also only applies when there is no net current flow through the electrode. When there is current flow the activity of ions at the electrode surface changes, and there are additional overpotential and resistive loss terms to the measured potential.