Lecture #16 of 26 - chem.uci.edu

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Lecture #16 of 26

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Liquid-Junction Potentials

Chapter 2

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Q: What’s in this set of lectures?A: B&F Chapter 2 main concepts:

● “Section 2.1”: Salt; Activity; Underpotential deposition

● Section 2.3: Transference numbers; Liquid-junctionpotentials

● Sections 2.2 & 2.4: Donnan potentials; Membrane potentials;pH meter; Ion-selective electrodes

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an ISE (for nitrate ions)an SCE

Now on to two general liquid junctions that we care about (the most)…

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when two ionic solutions are separated across an interface that prevents bulk mixing of the ions, but has ionic permeability, a potential (drop) develops called the liquid junction potential.

Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2

same salt;different conc.

one ion in common;same conc.

everything else

… liquid junctions:

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same salt;different conc.


● starting at the side with larger ion concentration● the ion the with larger mobility will impart its

charge to the opposite side of the junction

… example “1”:

… conceptually, let’s think about a condition in the limit where tH+ → 1 (say tH+ ≈ 0.9)…

… as H+ diffuses down its concentration gradient, an electrostatic force is exerted on Cl– to pull it along (at a larger flux) while at the same time slowing down transport of H+

… this happens until ti–effective = 0.5 for both H+

and Cl–, and at which time the system has attained steady-state mass transport and has generated a maximum liquid-junction potential.

… FYI, in semiconductor physics this same process results in a Dember potential… and the transport process is called ambipolar diffusion

556Recall that transport number, ti, (or transference number) is based on …

… and that the ionic conductivity, κ or σ, is defined as…

… so ti is the fraction of the solution conductivity attributable to ion "i"

units: S/cm or 1/(Ω cm)

units: cm2/(s V)(Stokes' law)

Siemens

Λ α (C)1/2

The Kohlrausch law (empirical) and Debye–Hückel–Onsager equation (theoretical) predict that the molar conductivity is proportional to the square root of the salt concentration

from Wiki

Physicist

Friedrich Wilhelm Georg Kohlrausch

(1840–1910)

557Recall that transport number, ti, (or transference number) is based on …

… and that the ionic conductivity, κ or σ, is defined as…

… so ti is the fraction of the solution conductivity attributable to ion "i"

units: S/cm or 1/(Ω cm)

units: cm2/(s V)(Stokes' law)

Siemens

from Wiki

P-Chemist & Physicist

Lars Onsager

(1903–1976)

The Kohlrausch law (empirical) and Debye–Hückel–Onsager equation (theoretical) predict that the molar conductivity is proportional to the square root of the salt concentration

Λ α (C)1/2

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/////////////////

= F

… and recall the Einstein–Smoluchowskiequation to calculate Di,

𝝁𝒊 =𝒛𝒊 𝑭𝑫𝒊

𝑹𝑻

"equivalent" molar (ionic) conductivity

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… the sign of the liquid-junction potential is obvious for Types 1 and 2 (but not Type 3) based on the mobilities of the individual ions…

… and so when in doubt, think logically about the sign of the potential to verify answers…

… and yes, Cl– will migrate/drift based on the electric potential formed by cation transport


one ion in common;same conc.

● compare dissimilar ions (cations or anions)● the ion with the larger mobility will impart its

charge to the opposite side of the junction

… example “2”:

Cl–

561we use ti values, which are based on kinetic transport to determinethe liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…

Type 1 Type 2 Type 3

same salt;different concentrations

same cation or anion;different counter ion;same concentration

no common ions,and/or one common ion; different concs

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Type 1

Type 2

(α)

(β)

Type 3

the Henderson Eq. (with a few assumptions, pg. 72)

… as written, these equations calculate Ej at β vs α

… use the conductivity due to all ions, even the common one

… sign depends on the charge of the dissimilar ion:(+) when cations are dissimilar, and (–) when anions are dissimilar

(with a few assumptions, pg. 72)

we use ti values, which are based on kinetic transport to determinethe liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…

… use the activity of the entire salt

563example: B&F Problem 2.14d

Calculate Ej for NaNO3 (0.10 M) | NaOH (0.10M)

1. What Type?

2. Polarity?

Type 2α | β

565example: B&F Problem 2.14d

Calculate Ej for NaNO3 (0.10 M) | NaOH (0.10M)

1. What Type?

2. Polarity? Polarity should be clear… compare mobilities… µOH– is larger…… so NaNO3 side will be (–)… and so β vs α will be (+)

3. Calculate it:

… (–) due to anions moving-

𝐸𝑗 = −0.05916 logµNO3− + µNa+

µOH− + µNa+

𝐸𝑗 = −0.05916 log7.404 + 5.913

20.5 + 5.913= 0.0176 V = +17.6 mV

as predicted, a (+) LJ potential correlates with the compartment in the denominator, β, vs α

… a rather large Ej (at β vs α)!

α | βType 2

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… so, why do trained electrochemists prefer to use saturated KCl (or KNO3) as the salt to fill reference electrodes?

… similar mobilities and thus, similar ti’s and thus,…

… vanishingly small LJ potentials!

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Type 3:

Type “2”:

Last point: clarifying sign conventions in B&F so that this is crystal clear…

where Λ (the equivalent conductivity) is defined as follows:

… and where Ceq is the concentration of positive or negative charges associated with a particular salt in solution, and so rearranging 2.3.40…

… and by comparing this to the form of our Type 2 equation, one sees…

, where

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Type 3:

Type “2”:


Note: α and β are switched…Type 2:

… they are off by a factor of (-1)… let’s look at that pre-factor…… for a specific example:

0.1M HCl (α) | 0.1M KCl (β)

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Type 3:


… they are off by a factor of (-1)… let’s look at that pre-factor…… for a specific example:

0.1M HCl (α) | 0.1M KCl (β)

… so this means that the Lewis–Sargent relation should have a (–) in front of it when net cations diffuse based on our convention… so switch the sign, as ∓… or use our equation

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… since we know that the β side will be (+) in the previous case, this really means the Lewis–Sargent relation should have a (–) sign in front of it when net cations diffuse…

… if we’re sticking with our convention that the potential is the β(product/reduced phase) versus the α (reactant/oxidized phase)…

Type 3:

Type “2”:


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… anyway…

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a cell

Donnan potential: A special liquid-junction potential due to fixed charges… here are two systems in which Donnan potentials play a prominent role:

semipermeable membrane

membrane impermeable tocharged macromolecules

an ionomer film

http://www.futuremorf.com/http://www.nafion.mysite.com/

Nafion

http://www.williamsclass.com/

http://www.futuremorf.com/

http://www.nafion.mysite.com/

http://www.williamsclass.com

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Na+

CNaCl

cNa+ cCl-cNa+ cCl-

m m s s

R–

a film of poly(styrene sulfonate)

… consider this model which applies to both scenarios…

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(for ion "i"… its electrochemical potential in the membrane… is the same as in solution… this is the definition of something that has equilibrated!)

𝜇𝑖𝑜,𝑚 + 𝑅𝑇 ln γ𝑖

𝑚 + 𝑅𝑇 ln 𝑐𝑖𝑚 + 𝑧𝑖𝐹𝜙

𝑚 = 𝜇𝑖𝑜,𝑠 + 𝑅𝑇 ln γ𝑖

𝑠 + 𝑅𝑇 ln 𝑐𝑖𝑠 + 𝑧𝑖𝐹𝜙

𝑠

Because differences in electrochemical potential ( ҧ𝜇𝑖𝑜) – think free energy

– drive net mass transport (of unstirred solutions), mobile Na+ and Cl–

partition between the membrane and the solution in compliance with their electrochemical potentials:

m s

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… so we can express EDonnan, an equilibrium electric potential difference, in terms of any ion that has access to both the membrane and the solution:

𝐸Donnan =𝑅𝑇

1 𝐹ln

𝑎Na+𝑠

𝑎Na+𝑚 =

𝑅𝑇

−1 𝐹ln

𝑎Cl−𝑠

𝑎Cl−𝑚

… Assuming that standard state chemical potentials (𝜇𝑖𝑜) are the same inside and

outside of the membrane, we can easily solve for the ("Galvani" / inner) electric potential difference, ϕm – ϕs

… which is exactly what was required to calculate liquid-junction potentials!

𝜙𝑚 − 𝜙𝑠 =𝑅𝑇

𝑧𝑖𝐹ln

γ𝑖𝑠𝑐𝑖

𝑠

γ𝑖𝑚𝑐𝑖

𝑚 = 𝐸Donnan

Because differences in electrochemical potential ( ҧ𝜇𝑖𝑜) – think free energy

– drive net mass transport (of unstirred solutions), mobile Na+ and Cl–

partition between the membrane and the solution in compliance with their electrochemical potentials:

𝜇𝑖𝑜,𝑚 + 𝑅𝑇 ln γ𝑖

𝑚 + 𝑅𝑇 ln 𝑐𝑖𝑚 + 𝑧𝑖𝐹𝜙

𝑚 = 𝜇𝑖𝑜,𝑠 + 𝑅𝑇 ln γ𝑖

𝑠 + 𝑅𝑇 ln 𝑐𝑖𝑠 + 𝑧𝑖𝐹𝜙

𝑠

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Aside #1: Recall Type 1 case of LJ potential… but now with t– = 0…

(α)

(β)

𝐸j =𝑅𝑇

𝐹ln

𝑎1 α

𝑎1 β= 𝐸Donnan Na+

… with β being the membrane

Aside #2: This is what B&F writes for this (Donnan) potential…… Check!

Eqn. 2.4.2


𝐹ln

𝑎Na+𝑠

𝑎Na+𝑚 = −

𝑅𝑇

𝐹ln

𝑎Cl−𝑠

𝑎Cl−𝑚

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… or…

𝑎Na+s 𝑎Cl−

s = 𝑎Na+m 𝑎Cl−

m

Anyway… now divide both sides by RT/F and invert the argument of the “ln()” on the right to eliminate the negative sign, and we have...


𝐹ln

𝑎Na+𝑠

𝑎Na+𝑚 = −

𝑅𝑇

𝐹ln

𝑎Cl−𝑠

𝑎Cl−𝑚

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Na+

CNaCl

cNa+ cCl-cNa+ cCl-

m m s s

R–

a film of poly(styrene sulfonate)

… recall the scenario we are analyzing…… with R– representing the fixed charges…

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now, there is an additional constraint: the bulk of the solution and the bulk of the membrane must be electrically neutral:

… an equation quadratic in cmCl– is obtained as follows…

… if these are dilute electrolytes, we can neglect activity coefficients…

𝑐Na+s 𝑐Cl−

s = 𝑐Na+m 𝑐Cl−

m

𝑐Na+m = 𝑐Cl−

m + 𝑐R−m𝑐Na+

s = 𝑐Cl−s

𝑎Na+s 𝑎Cl−

s = 𝑎Na+m 𝑎Cl−

m

𝑐Na+s = 𝑐Cl−

s

𝑐Na+s 𝑐Cl−


m

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because in solution, cNa+ = cCl– for goodness sakes!

𝑐Cl−s 2

= 𝑐Cl−m 2

+ 𝑐Cl−m 𝑐R−

m

… an equation quadratic in cmCl– is obtained as follows…

581… an equation quadratic in cmCl– is obtained as follows…

𝑐Cl−s 2

= 𝑐Cl−m 2

+ 𝑐R−m 𝑐Cl−

m

0 = 𝑐Cl−m 2

+ 𝑐R−m 𝑐Cl−

m − 𝑐Cl−s 2

𝑐Cl−𝑚 =

−𝑐R−𝑚 + 𝑐R−

𝑚 2+ 4 𝑐Cl−

𝑠 2

2=𝑐R−𝑚

21 + 4

𝑐Cl−𝑠

𝑐Cl−𝑅

2

− 1

𝑐Na+s = 𝑐Cl−

s

𝑐Na+s 𝑐Cl−


m

𝑐Na+m = 𝑐Cl−

m + 𝑐R−m

… use the quadratic formula to solve for cmCl– and one gets…

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if (which is the typical case of interest), then…

𝑐Cl−𝑚 =


𝑚 2+ 4 𝑐Cl−

𝑠 2

2=𝑐R−𝑚

21 + 4

𝑐Cl−𝑠

𝑐R−𝑚

2

− 1

𝑐Cl−𝑠 ≪ 𝑐R−

𝑚

1 + 4𝑐Cl−𝑠

𝑐R−𝑚

2

≈ 1 + 2𝑐Cl−𝑠

𝑐R−𝑚

2

(Taylor/Maclaurin series expansion to the first 3 (or 4) terms)

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… fixed charge sites are responsible for the electrostatic exclusion of mobile “like” charges (co-ions) from a membrane, cell, etc. This is Donnan Exclusion.

𝑐Cl−𝑚 =


𝑚 2+ 4 𝑐Cl−

𝑠 2

2=𝑐R−𝑚

21 + 4

𝑐Cl−𝑠

𝑐R−𝑚

2

− 1

1 + 4𝑐Cl−𝑠

𝑐R−𝑚

2

≈ 1 + 2𝑐Cl−𝑠

𝑐R−𝑚

2

𝑐Cl−𝑚 =

𝑐R−𝑚

21 + 2

𝑐Cl−𝑠

𝑐R−𝑚

2

− 1 =

… the larger is CR–m, the smaller is CCl–

m

if (which is the typical case of interest), then…𝑐Cl−𝑠 ≪ 𝑐R−

𝑚

… is a reasonable assumption? What is CR–m?

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… well, for Nafion 117, the sulfonate concentration is 1.13 M…… for CR61 AZL from Ionics, the sulfonate concentration is 1.6 M…

so, as an example, if CCl–s = 0.1 M…

… so how excluded is excluded?

… an order of magnitude lowerthan CCl–

s… rather excluded!

Source: Torben Smith Sørensen, Surface Chemistry and Electrochemistry

of Membranes, CRC Press, 1999 ISBN 0824719220, 9780824719227

=0.1 2

1.0= 0.01 M

𝑐Cl−𝑠 ≪ 𝑐R−

𝑚

… but what if CCl–s is also large (e.g. 1 M)?

… No more Donnan exclusion!

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