POTENTIALS 1
Liquid junction potentials are almost unavoidable in electrochemical measurement systems. However, understanding the conceptual model for their origin helps one make choices in experimental design that minimize the errors caused by junction potentials. A mathematical model helps one calculate the magnitude of the junction potential in order to decide whether it is a concern. If the error is small enough to be acceptable or it is constant for all sample and standard solutions so that it is accounted for by the calibration procedure, then no change in procedure is warranted.
The formation of a potential difference at the interface between two solutions stems from the difference in mobilities of the ions in the electrolyte. The boundary between a salt bridge and the sample solution is usually a porous frit made of glass, ceramic, or polymer material. Movement of an ion across the boundary is driven by a concentration gradient and/or an electric field and can be described by the flux of the ion, in units of mol/cm2/s.
It is sometimes helpful to remember that a flux has the same dimensions as the product of a concentration and a velocity. (Convection or bulk flow of liquid is another mechanism of material transport, but is assumed to have a negligible influence on the liquid junction potential and is not included in the model here.) The flux for an ion is described by the Nernst–Planck equation.
Ji= −Di𝜕Ci
𝜕x −DiCiziF RT
𝜕𝜙
𝜕x (C.1)
In Eq. (C.1),Diis the diffusion coefficient,Ciis the concentration, andziis the charge of the ion, i. The electrical potential energy is𝜙,Fis Faraday’s constant (96,485 C/mol),R is the molar gas constant (8.314 J/(K mol)), and the temperature,T, is given in Kelvin.
In this discussion, it is assumed that the movement (along thex-coordinate) is perpen- dicular to a flat plane that forms the boundary between the two solutions.
Although the Nernst–Planck equation is unfamiliar to most people, most chemists recognize the phenomena represented by the two terms on the right-hand side. The first term indicates that a concentration gradient will drive movement of a species by diffu- sion. The diffusion coefficient is merely the proportionality constant relating the flux of an uncharged species to the concentration gradient. (The equation without the second term
1This discussion is based on the derivation of Bakker [1].
Electroanalytical Chemistry: Principles, Best Practices, and Case Studies, First Edition. Gary A. Mabbott.
© 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.
k k is often called Fick’s first law of diffusion.) The second term on the right accounts for an
additional driving force for migration of charged species. The driving force is the potential gradient (or electric field),𝜕𝜙𝜕x. The movement of charge under the influence of an electrical potential energy difference is also an intuitively reasonable mechanism. It is the basis for separation of charged molecules by electrophoresis, for example.
The goal of this analysis is to find the difference in the potential between the two phases. This potential energy difference is the liquid junction potential. There are two keys to solving this problem. The first of these is the fact that the Nernst–Planck equation applies to each ion. The second key is an assumption that the boundary conditions reach a steady state, namely, a situation where the sum of all the fluxes equals zero. The last idea is a conse- quence of the restriction that the measurement must be made without drawing significant current. The zero-current condition was already a requirement imposed on all potentio- metric measurements in order to avoid distorting the conditions at the sensor/sample solution interface. (Of course, a finite current does exist during real experiments, but it is usually on the picoampere level or lower and is usually negligible.)
An analysis of a very simple (but also very common) set of conditions reveals some important guidelines for experimenters to follow. Consider a salt bridge separating a ref- erence electrode with a solution of 1 M KCl from a sample solution that was spiked with an ionic strength adjusting buffer so that the sample solution electrolyte is essentially 0.1 M KCl. (Assume that the other ionic components in the original sample solution are less than 1 mM so that their influence on the liquid junction potential can be neglected.) A further reasonable assumption is that the concentration gradients for K+ and Cl−are linear and that at any point the total cation concentration and total anion concentration are equal to each other. That isCK+ =CCl− =C=Cspl+mx, whereCspl is the ion concentration in the sample solution and the concentration at the maximum value ofxisCref, the concentration of each ion in the reference solution.
In the first moment that the boundary is formed, there is no charge separation at the interface, so the electric field is zero. The concentration gradient drives the ion movement.
The first term in Eq. (C.1) indicates that the ion with the bigger diffusion coefficient will have a bigger flux; its velocity across the boundary is greater than the other ion. In this case, chloride ions are faster than potassium ions. An excess of negative charge develops on the sample solution side and that creates an electric field which attracts potassium ions and slows down the flux of chloride. In a very short time, the forces associated with the electric field and the concentration gradient balance each other, and the anion flux equals the cation flux:
−DK+
𝜕CK+
𝜕x −DK+CK+
zK+F RT
𝜕𝜙
𝜕x = −DCl−𝜕CCl−
𝜕x −DCl−CCl−zCl−F RT
𝜕𝜙
𝜕x or
−DK+
𝜕CK+
𝜕x −DK+CK+ F RT
𝜕𝜙
𝜕x = −DCl−𝜕CCl−
𝜕x +DCl−CCl− F RT
𝜕𝜙
𝜕x (C.2)
SinceCK+ =CCl− =C=Cspl+mx,𝜕C𝜕xK+ = 𝜕C𝜕xCl− =m, Eq. (C.2) simplifies to
(DCl−−DK+)m+ (−DK+−DCl−) {
C F RT
𝜕𝜙
𝜕x }
=0 (C.3)
k k Rearranging gives
m (DCl−−DK+) (−DK+−DCl−)
{RT CF
}
= 𝜕𝜙
𝜕x (C.4)
Integrating Eq. (C.4) gives the potential difference between the two liquids.
∫
xmax 0
m (DCl−−DK+) (−DK+−DCl−)
{RT CF
}𝜕x=
∫
𝜙ref 𝜙spl
𝜕𝜙 (C.5)
=∫
xmax 0
m (DCl−−DK+) (−DK+−DCl−)
{
RT (Cspl+mx)F
}
𝜕x=∫
𝜙ref 𝜙spl
𝜕𝜙 (C.6)
The right-hand side of Eq. (C.36) is the junction potential,𝜙ref−𝜙spl=Ej: Ej=𝜙ref−𝜙spl=m (DCl−−DK+)
(−DK+−DCl−) {RT
mF }
ln(Cspl+mx)⌉x0max (C.7) Recalling thatCspl+m(xmax)=Cref,
Ej=𝜙ref−𝜙spl= −(DCl−−DK+) (DK++DCl−)
{RT F
}
{ln(Cref) −ln(Cspl)} (C.8) Ej=𝜙ref−𝜙spl= −(DCl−−DK+)
(DK++DCl−) {RT
F } {
ln (Cref
Cspl )}
(C.9)
In this case, Cspl = 0.1 M, Cref = 1 M, DCl− = 2.03×10−5 cm2/s, and DK+ = 1.96× 10−5cm2/s. Evaluating the junction potential at 298 K gives
Ej=𝜙ref−𝜙spl= −(2.03−1.96)(10−6) (2.03+1.96)(10−6)
{(8.314)(298) 96 485
} { ln
(0.1 1
)}
=0.001 04 V (C.10) The conclusion is that a salt bridge between a sample solution with a supporting elec- trolyte of 0.1 M KCl and a reference solution of 1.0 M KCl is only 1 mV. If a voltage error of that size were operating in an experiment with an ion selective electrode, it would lead to an error in the concentration of a singly charged analyte of about 0.3%. That seems quite reasonable. Of course, if the sample and standard solutions have the same electrolyte con- ditions, then the liquid junction potential is constant and becomes a part of they-intercept in the calibration curve and the error is compensated for.
Other supporting electrolytes are less well matched in terms of the cation and anion diffusion coefficients. Note the difference in the diffusion coefficients for Na+and Cl−, for example, in Table C.1. If one were to use NaCl in a similar arrangement, the junction poten- tial would be−0.012 V or−12 m which corresponds to an error in the analyte concentration of 3% on an ISE experiment.
A simple 1 : 1 salt at different concentrations on the two sides of the boundary is the only combination that leads to a simple analytical solution of the Nernst–Planck equation for calculating the liquid junction potential. Even the situation where the junction between
k k two phases share the same anion but have different cations is more difficult to solve.
(The integral does not have an analytical solution.) Fortunately, Henderson introduced an equation (C.11) that yields an approximate solution for the junction potential for other combinations involving linear concentration gradients.
Ej=
∑
j
zjuj(Cj,s−Cj,r)
∑
j
z2juj(Cj,s−Cj,r) (RT
F )
ln
⎧⎪
⎨⎪
⎩
∑
j
z2jujCj,s
∑
j
z2jujCj,r
⎫⎪
⎬⎪
⎭
(C.11)
In Eq. (C.11), terms are summed for all ions, j, using their respective concentrations,Cj,r andCj,s, in the separate phases, r and s. The termujis the mobility of an ion. It is also called the electrophoretic mobility. It is the proportionality constant relating the velocity,𝜐, that an ion reaches under the strength of an electric field,. Whenever an ion is accelerated the drag of the solution acts as a force in opposition to the electric force limiting the velocity of the ion. The ion mobility can be calculated by setting the forces equal to each other and solving for𝜐/ε.
uj= 𝜐
ε = ⌊zj⌋e
6𝜋𝜂r (C.12)
In Eq. (C.12), e represents the charge on an electron,𝜂is the viscosity of the solution, andris the effective radius of the hydrated ion. Because data may be available for only the diffusion coefficient rather than the mobility of the ions of interest, it is helpful to be able to estimate one from the other. Here is Einstein’s expression relating the electric mobility and the diffusion coefficient.
Dj= (RT
F ) ( uj
⌊zj⌋ )
oruj= ( F
RT
)⌊zj⌋Dj (C.13)
Thus, the Henderson equation can also be written using diffusion coefficients:
Ej=
∑
j
zj⌊zj⌋Dj(Cj,s−Cj,r)
∑
j
z2j⌊zj⌋Dj(Cj,s−Cj,r) (RT
F )
ln
⎧⎪
⎨⎪
⎩
∑
j
z2j⌊zj⌋DjCj,s
∑
j
z2j⌊zj⌋DjCj,r
⎫⎪
⎬⎪
⎭
(C.14)
Here is an example calculation. Consider a reference bridge containing 1 M KCl and a sample solution containing 0.1 M HCl. From Table C.1,DH+ =96.6×10−6cm2/s,DK+ = 19.57×10−6cm2/s, andDCl− =20.32×10−6cm2/s.
Ej= (96.6)(0.1−0) +19.57)(0−1) + (−1)(20.32)(0.1−1) (96.6)(0.1−0) +19.57)(0−1) + (1)(20.32)(0.1−1)
(8.314×298 96 485
)
×ln
{(96.6)(0.1) + (20.32)(0.1) 19.57)(1) + (20.32)(1)
}
=0.008 57 V or 8.57 mV
k k TABLE C.1 Diffusion coefficients for
selected ions
Ion Diffusion
coefficient (cm2/s)
OH− 52.73×10−6
Na+ 13.34×10−6
K+ 19.57×10−6
SO42− 10.65×10−6
Ca2+ 7.92×10−6
Cl− 20.32×10−6
Mg2+ 7.06×10−6
H+a 96.6×10−6
NO3−a 19.46×10−6
aCalculated from the electric mobility given by Bakker [1].
Source: From Bakker [1].
TABLE C.2 Junction potential as a function of [HCl] in the sample solution for a KCl salt bridge
Reference Junction potentials (mV)
M KCl Sample=0 M HCl 0.05 M HCl 0.1 M HCl 0.2 M HCl
1 −2.7 −5.7 −8.57 −12.5
2 −2.8 −3.8 −5.7 −8.57
3 −2.8 −3.1 −4.5 −6.8
Table C.2 shows the calculated junction potential for three different reference salt bridge solutions and different concentrations of HCl in the sample solution (Solutions varying in acid concentration represent situations where junction potentials are likely to vary). It is apparent that the junction potential for the more concentrated KCl solution is less influenced by the changes in the sample solution.
REFERENCE
1. Bakker, E. (2014). Fundamentals of Electroanalysis: 1. Potentials and Transport. Apple iBook Store (ebook) https://itunes.apple.com/us/book/fundamentals-electroanalysis-1-potentials- transport/id933624613?mt=11.
k k
APPENDIX
D