Nernst equation

In electrochemistry, the Nernst equation is an equation that relates the reduction potential of a half-cell (or the total voltage, i.e. the electromotive force, of the full cell) at any point in time to the standard electrode potential, temperature, activity, and reaction quotient of the underlying reactions and species used. When the reaction quotient is equal to the equilibrium constant of the reaction for a given temperature, i.e. when the concentration of species are at their equilibrium values, the Nernst equation gives the equilibrium voltage of the half-cell (or the full cell), which is zero; at equilibrium, $Q = K$ , $Δ G = 0$ , and therefore, $E = 0$ . It is named after the German physical chemist who first formulated it, Walther Nernst.^[1]^[2]

The Nernst equation gives a formula that relates the numerical values of the concentration gradient to the electric gradient that balances it. For example, if a concentration gradient is established by dissolving KCl in half of a divided vessel that is full of H₂O, and then a membrane permeable to K⁺ ions is introduced between the two halves, then after a relaxation period, an equilibrium situation arises where the chemical concentration gradient, which at first causes ions to move from the region of high concentration to the region of low concentration, is exactly balanced by an electrical gradient that opposes the movement of charge.

Expression

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

E_\text{red} = E^{\ominus}_\text{red} + \frac{RT}{zF} \ln\frac{a_\text{Ox}}{a_\text{Red}}

(half-cell reduction potential for Ox +

z

e⁻ → Red)

E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q_r

(total cell potential)

where

$E red$ is the half-cell reduction potential at the temperature of interest
$E o red$ is the standard half-cell reduction potential
$E cell$ is the cell potential (electromotive force) at the temperature of interest
$E o cell$ is the standard cell potential
$R$ is the universal gas constant: $R = 7000831447200000000♠ 8.314 472 (15) J K -1 mol -1$
$T$ is the temperature in kelvin
$a$ is the chemical activity for the relevant species, where $a Red$ is the reducing agent and $a Ox$ is the oxidizing agent. $a X = γ X c X$ , 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: $F = 7004964853399000000♠ 9.648 53399 (24) \times 104 C mol -1$
$z$ is the number of moles of electrons transferred in the cell reaction or half-reaction
$Q r$ is the reaction quotient.

At room temperature (25 °C), $RT / F$ may be treated like a constant and replaced by 25.693 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{0.05916}{z} \log_{10}\frac{a_\text{Ox}}{a_\text{Red}}.

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

Nernst potential

Main article: Reversal potential

The Nernst equation has a physiological application when used to calculate the potential of an ion of charge $z$ across a membrane. This potential is determined using the concentration of the ion both inside and outside the cell:

E = \frac{R T}{z F} \ln\frac{[\text{ion outside cell}]}{[\text{ion inside cell}]} = 2.3026\frac{R T}{z F} \log_{10}\frac{[\text{ion outside cell}]}{[\text{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 the Goldman equation:

E_\mathrm{m} = \frac{RT}{F} \ln{ \left( \frac{ \sum_{i}^{N} P_{\mathrm{M}^{+}_{i}}[\mathrm{M}^{+}_{i}]_\mathrm{out} + \sum_{j}^{M} P_{\mathrm{A}^{-}_{j}}[\mathrm{A}^{-}_{j}]_\mathrm{in}}{ \sum_{i}^{N} P_{\mathrm{M}^{+}_{i}}[\mathrm{M}^{+}_{i}]_\mathrm{in} + \sum_{j}^{M} P_{\mathrm{A}^{-}_{j}}[\mathrm{A}^{-}_{j}]_\mathrm{out}} \right) }

$E m$ is the membrane potential (in volts, equivalent to joules per coulomb)
$P ion$ is the permeability for that ion (in meters per second)
$[ion] out$ is the extracellular concentration of that ion (in moles per cubic meter, to match the other SI units, though the units strictly don't matter, as the ion concentration terms become a dimensionless ratio)
$[ion] in$ is the intracellular concentration of that ion (in moles per cubic meter)
$R$ is the ideal gas constant (joules per kelvin per mole)
$T$ is the temperature in kelvins
$F$ is Faraday's constant (coulombs per mole)

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.

A similar expression exists that includes $r$ (the absolute value of the transport ratio). This takes transporters with unequal exchanges into account. See: sodium-potassium pump where the transport ratio would be 2/3. The other variables are the same as above. The following example includes two ions: potassium (K⁺) and sodium (Na⁺). Chloride is assumed to be in equilibrium.

V_{m} = \frac{RT}{F} \ln{ \left( \frac{ rP_{\mathrm{K}^+}[\mathrm{K}^+]_\mathrm{out} + P_{\mathrm{Na}^+}[\mathrm{Na}^+]_\mathrm{out}}{ rP_{\mathrm{K}^+}[\mathrm{K}^+]_\mathrm{in} + P_{\mathrm{Na}^+}[\mathrm{Na}^+]_\mathrm{in}} \right) }

When chloride (Cl⁻) is taken into account, its part is flipped to account for the negative charge.

V_{m} = \frac{RT}{F} \ln{ \left( \frac{ P_{\mathrm{K}^+}[\mathrm{K}^+]_\mathrm{out} + P_{\mathrm{Na}^+}[\mathrm{Na}^+]_\mathrm{out} + P_{\mathrm{Cl}^-}[\mathrm{Cl}^-]_\mathrm{in}}{ P_{\mathrm{K}^+}[\mathrm{K}^+]_\mathrm{in} + P_{\mathrm{Na}^+}[\mathrm{Na}^+]_\mathrm{in} + P_{\mathrm{Cl}^-}[\mathrm{Cl}^-]_\mathrm{out}} \right) }

Derivation

Using Boltzmann factors

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

Ox + e⁻ ⇌ Red

and that 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 factor for these processes:

\frac{[\mathrm{Red}]}{[\mathrm{Ox}]} = \frac{\exp \left(-[\mbox{barrier for losing an electron}]/kT\right)} {\exp \left(-[\mbox{barrier for gaining an electron}]/kT\right)} = \exp \left(\frac{\mu_\mathrm{c}}{kT} \right).

Taking the natural logarithm of both sides gives

\mu_\mathrm{c} = kT \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]}.

If $μ c \neq 0$ at [Ox]/[Red] = 1, we need to add in this additional constant:

\mu_\mathrm{c} = \mu_\mathrm{c}^0 + kT \ln \frac{[\mathrm{Red}]}{[\mathrm{Ox}]}.

Dividing the equation by $e$ to convert from chemical potentials to electrode potentials, and remembering that $k / e = R / F$ ,^[3] we obtain the Nernst equation for the one-electron process Ox + e⁻ → Red:

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

Using thermodynamics (chemical potential)

Quantities here are given per molecule, not per mole, and so Boltzmann 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 $Ω$ is the number of states available to the molecule. The number of states must vary linearly with the volume $V$ of the system (here an idealized system is considered for better understanding, so that activities are posited very close to the true concentrations. Fundamental statistical proof of the mentioned linearity goes beyond the scope of this section, but to see this is true it is simpler to consider usual isothermal process for an ideal gas where the change of entropy $Δ S = nR ln(V 2 / V 1)$ takes place. It follows from the definition of entropy and from the condition of constant temperature and quantity of gas $n$ that the change in the number of states must be proportional to the relative change in volume $V 2 / V 1$ . In this sense there is no difference in statistical properties of ideal gas atoms compared with the dissolved species of a solution with activity coefficients equaling zero: particles freely "hang around" filling the provided volume), 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 2 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(\mathrm{A}) = S^0(\mathrm{A}) - k \ln [\mathrm{A}], \,

where $S 0$ is the entropy at standard conditions and [A] denotes the concentration of A.

The change in entropy for a reaction

a

A +

b

B →

y

Y +

z

is then given by

\Delta S_\mathrm{rxn} = \big(yS(\mathrm{Y}) + zS(\mathrm{Z})\big) - \big(aS(\mathrm{A}) + bS(\mathrm{B})\big) = \Delta S^0_\mathrm{rxn} - k \ln \frac{[\mathrm{Y}]^y [\mathrm{Z}]^z}{[\mathrm{A}]^a [\mathrm{B}]^b}.

We define the ratio in the last term as the reaction quotient:

Q = \frac{\prod_j a_j^{\nu_j}}{\prod_i a_i^{\nu_i}} \approx \frac{[\mathrm{Z}]^z [\mathrm{Y}]^y}{[\mathrm{A}]^a [\mathrm{B}]^b}.

where the numerator is a product of reaction product activities, $a j$ , each raised to the power of a stoichiometric coefficient, $ν j$ , and the denominator is a similar product of reactant activities. All activities refer to a time $t$ . Under certain circumstances (see chemical equilibrium) each activity term such as $a ν j j$ may be replaced by a concentration term, [A].

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 energy change $Δ G$ only by a constant: $Δ G = - nFE$ , where $n$ is the number of electrons transferred and $F$ is the Faraday constant. There is a negative sign because a spontaneous reaction has a negative free energy $Δ G$ and a positive potential $E$ . The Gibbs 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 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 Ox + $n$ e⁻ → Red,

Q = \frac{[\mathrm{Red}]}{[\mathrm{Ox}]},

and we have:

\begin{align} 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. \end{align}

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

Or at standard temperature,

\log_{10} K = \frac{nE^0}{0.05916\text{ V}} \quad\text{at }T = 298.15\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 the 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 approaches toward $\pm\infty$ . This is physically meaningless because, under such conditions, the exchange current density becomes very low, and there is no thermodynamic equilibrium necessary for Nernst equation to hold. The electrode is called to be unpoised in such case. Other effects tend to take control of the electrochemical behavior of the system.

Time dependence of the potential

The expression of time dependence has been established by Karaoglanoff.^[4]^[5]^[6]^[7]

Significance to related scientific domains

The equation has been involved in the scientific controversy involving cold fusion. The discoverers of cold fusion, Fleischmann and Pons, calculated that a palladium cathode immersed in a heavy water electrolysis cell could achieve up to 10²⁷ atmospheres of pressure on the surface of the cathode, enough pressure to cause spontaneous nuclear fusion. In reality, only 10,000–20,000 atmospheres were achieved. John R. Huizenga claimed their original calculation was affected by a misinterpretation of Nernst equation.^[8] He cited a paper about Pd–Zr alloys.^[9]

References

↑ Orna, Mary Virginia; Stock, John (1989). Electrochemistry, past and present. Columbus, Ohio: American Chemical Society. ISBN 0-8412-1572-3. OCLC 19124885.
↑ Wahl (2005). "A Short History of Electrochemistry". Galvanotechtnik 96 (8): 1820–1828.
↑ $R = N A k$ ; see gas constant
$F = eN A$ ; see Faraday constant
↑ Karaoglanoff, Z. (January 1906), "Über Oxydations- und Reduktionsvorgänge bei der Elektrolyse von Eisensaltzlösungen" [On Oxidation and Reduction Processes in the Electrolysis of Iron Salt Solutions], Zeitschrift für Elektrochemie (in German) 12 (1): 5–16, doi:10.1002/bbpc.19060120105
↑ Bard, Allen J.; Inzelt, György; Scholz, Fritz (eds.), "Karaoglanoff equation", Electrochemical Dictionary, Springer, pp. 527–528
↑ Zutshi, Kamala (2008), Introduction to Polarography and Allied Techniques, pp. 127–128
↑ The Journal of Physical Chemistry, Volume 10, p 316. http://books.google.com/books?id=zCMSAAAAIAAJ&pg=PA316&lpg=PA316&hl=en&f=false
↑ Huizenga, John R. (1993). "Cold Fusion: The Scientific Fiasco of the Century" (2 ed.). Oxford and New York: Oxford University Press: 33, 47. ISBN 0-19-855817-1.
↑ Huot, J. Y.; et al. (1989). J. Electrochem. Soc. 136: 631 http://jes.ecsdl.org/content/136/3/630.abstract. Missing or empty |title= (help)

External links

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