Nernst Equation

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1 Physiological application: the Nernst potential
2 Derivation
- 2.1 Using Boltzmann factors
- 2.2 Using entropy and Gibbs free energy
3 Relation to equilibrium
4 Limitations
5 See also
6 External links

In electrochemistry, the Nernst equation is an equation which can be used (in conjunction with other information) to determine the equilibrium reduction potential of a half-cell in an electrochemical cell. It can also be used to determine the total voltage (electromotive force) for a full electrochemical cell. It is named after the German physical chemist who first formulated it, Walther Nernst.

The two (ultimately equivalent) equations for these two cases (half-cell, full cell) are as follows:

$E_{red} = E^0_{red} - \frac{RT}{zF} \ln\frac{a_{\mbox{Red}}}{a_{\mbox{Ox}}}$ (half-cell reduction potential)

$E_{cell} = E^0_{cell} - \frac{RT}{zF} \ln Q$ (total cell potential)

where

$E r e d$ is the half-cell reduction potential
$E^0_{red}$ is the standard half-cell reduction potential
$E c e l l$ is the cell potential (electromotive force)
is the standard cell potential
- All of the above are measured in volts.
R is the universal gas constant: R = 8.314472 J K^-1 mol^-1
T is the absolute temperature in kelvins: T = T_°C + 273.15. (So, for example, at 25 °C, T = 298.15 K.)
a is the chemical activity for the relevant species. $a X = γ X X$ , where $γ X$ is the activity coefficient of species X. (Since activity coefficients tend to unity at low concentrations, activities in the Nernst equation are frequently replaced by simple concentrations.)
F is the Faraday constant, the number of coulombs per mole of electrons: F = 9.6485309×10⁴ C mol^-1
z is the number of electrons transferred in the cell reaction or half-reaction
Q is the reaction quotient.

At room temperature (25 °C), RT/F may be treated like a constant and replaced by .025679 V or 25.679 mV for cells.

The Nernst equation is frequently expressed in terms of base 10 logarithms (i.e., common logarithms) rather than natural logarithms, in which case it is written, for a cell at 25 °C:

$E = E^0 - \frac{.0591\mbox{ V}}{z} \log_{10}\frac{a_{\mbox{Red}}}{a_{\mbox{Ox}}}$ .

It can also be written in terms of millivolts using 59.1 mV instead of .0591 V.

The Nernst equation is used in physiology for finding the electric potential of a cell membrane with respect to one type of ion.

Physiological application: the Nernst potential

The Nernst potential, of a ion of charge $z$ , across a membrane, is determined by the concentration ratio in and outside the cell:

$E = \frac{R T}{z F} \ln\frac{[\mbox{ion outside cell}]}{[\mbox{ion inside cell}]}$

When the membrane is in thermodynamic equilibrium, i.e. no net flux of ions, the membrane potential must be equal to the Nernst potential. However, in physiology, due to active ion pumps, the inside and outside of a cell are not in equilibrium. In this case the resting potential can be determined from e.g. the Goldman equation.

The potential across the cell membrane that exactly opposes net diffusion of a particular ion through the membrane is called the Nernst potential for that ion. As seen above, 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.

Derivation

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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.

Using Boltzmann factors

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

$\mathrm{Ox} + e^- \rightleftharpoons \mathrm{Red}$

and which have a standard potential of zero. The chemical potential $μ c$ 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: $R = k N A$ and $F = e N A$ .

The entropy of a molecule is defined as

$S \ \stackrel{\mathrm{def}}{=}\ k \ln \Omega,$

where $Ω$ 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 ( $E = μ c / e$ ). E is related to the Gibbs free energy change $Δ G$ only by a constant: $Δ G = - n e E$ , where n is the number of electrons transferred. (There is a negative sign because a spontaneous reaction has a negative $Δ G$ and a positive E.) The Gibbs free energy is related to the entropy by $G = H - T S$ , 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 $\mathrm{Ox} + ne^- \rightarrow \mathrm{Red},$ $Q = [Red] / [Ox]$ , and we have:

$E = E^0 - \frac{kT}{ne} \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]} = E^0 - \frac{RT}{nF} \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]} = E^0 - \frac{RT}{nF} \ln Q.$

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.

Relation to equilibrium

At equilibrium, E = 0 and Q = K. Therefore

$\begin{align} 0 &= E^o - \frac{RT}{nF} \ln K\\ \ln K &= \frac{nFE^o}{RT} \end{align}$

Or at standard temperature,

$\log_{10} K = \frac{nE^o}{59.2\text{ mV}} \quad\text{at }T = 298 \text{ K}.$

We have thus related the standard electrode potential and the equilibrium constant of a redox reaction.

Limitations

In dilute solutions, the Nernst equation can be expressed directly in terms of concentrations (since activity coefficients are close to unity). But at higher concentrations, the true activities of the ions must be used. This complicates the use of the Nernst equation, since estimation of non-ideal activities of ions generally requires experimental measurements.

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

At very low concentrations of the potential determining ions, the potential predicted by Nernst equation tends to ±infinity. This is physically meaningless because, under such conditions, the exchange current density becomes very low, and then other effects tend to take control of the electrochemical behavior of the system.

External links

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