766
Lecture #11 of 17
767
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-junction
potentials
● Sections 2.2 & 2.4: Donnan potential; Membrane potentials;
pH meter; Ion-selective electrodes
768
an ISE (for nitrate ions)an SCE
… two general liquid junctions that we care about (the most)…RECALL FROM LAST TIME…
769
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:
770
same salt;
different conc.
Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2
… example “1”: ● 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
771
same salt;
different conc.
Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2
… example “1”: ● 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
… conceptually, 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 large flux) while
at the same time slowing 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.
772
same salt;
different conc.
Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2
FYI, in semiconductor physics this same process
results in a Dember potential … and the transport
process is called ambipolar diffusion
● 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, 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 large flux) while
at the same time slowing 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.
773equivalent ionic
conductivity
774
/////////////////
= F
equivalent ionic
conductivity
775
Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2
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”:
776
… 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.
Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 2.3.2
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”:
777
778Recall that transference number, ti, (or transport 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
779Recall that transference number, ti, (or transport 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 = a(C)1/2
The Kohlrausch law (empirical) and
Debye–Hückel–Onsager equation
(theoretical) predict that the equivalent
molar conductivity is proportional to the
square root of the salt concentration
from Wiki
Physicist
Friedrich Wilhelm Georg Kohlrausch
(1840–1910)
780Recall that transference number, ti, (or transport 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 = a(C)1/2
The Kohlrausch law (empirical) and
Debye–Hückel–Onsager equation
(theoretical) predict that the equivalent
molar conductivity is proportional to the
square root of the salt concentration
from Wiki
P-Chemist & Physicist
Lars Onsager
(1903–1976)
781we use ti values, which are based on kinetic transport to determine
the 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
782
Type 1
Type 2
(α)
(β)
Type 3
we use ti values, which are based on kinetic transport to determine
the liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…
783
Type 1
Type 2
(α)
(β)
Type 3
… just the ratio of the
conductivity attributable
to the dissimilar ions
(with a few assumptions, pg. 72)
we use ti values, which are based on kinetic transport to determine
the liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…
784
Type 1
Type 2… just the ratio of the
conductivity attributable
to the dissimilar ions
(α)
(β)
… sign depends on the charge of the dissimilar ion:
(+) when cations are dissimilar, and (–) when anions are dissimilar
Type 3
(with a few assumptions, pg. 72)
we use ti values, which are based on kinetic transport to determine
the liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…
785
Type 1
Type 2
(α)
(β)
Type 3
the Henderson Eq. (with a few assumptions, pg. 72)
… just the ratio of the
conductivity attributable
to the dissimilar ions
… 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 determine
the liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…
786
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
… just the ratio of the
conductivity attributable
to the dissimilar ions
… 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 determine
the liquid-junction potential (for derivations, see B&F, pp. 70 – 72)…
787example: B&F Problem 2.14d
Calculate Ej for NaNO3 (0.10 M) / NaOH (0.10M)
1. What Type?
788example: B&F Problem 2.14d
Calculate Ej for NaNO3 (0.10 M) / NaOH (0.10M)
1. What Type? Type 2
2. Polarity?
789
790example: B&F Problem 2.14d
Calculate Ej for NaNO3 (0.10 M) | NaOH (0.10M)
1. What Type? Type 2
2. Polarity? Polarity should be clear… compare mobilities;
OH– is higher so NaNO3 side will be (–).
3. Calculate it:
… (–) due to anions moving-
791example: B&F Problem 2.14d
Calculate Ej for NaNO3 (0.10 M) / NaOH (0.10M)
1. What Type? Type 2
2. Polarity? Polarity should be clear… compare mobilities;
OH– is higher so NaNO3 side will be (–).
3. Calculate it:
… (–) due to anions moving-
𝐸𝑗 = −0.05916 logµNO3−
µOH−
𝐸𝑗 = −0.05916 log7.404 x 10−4
25.05 x 10−4= 0.02617 V = +26.17 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 | β
792
… 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!
793
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:
794
Type 3:
Type “2”:
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
Last point: clarifying sign conventions in B&F so that this is crystal clear…
795
Type 3:
Type “2”:
Note: We switched α and β…Type 2:
Last point: clarifying sign conventions in B&F so that this is crystal clear…
796
Type 3:
Type “2”:
… 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)
Last point: clarifying sign conventions in B&F so that this is crystal clear…
797
Type 3:
… 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 really means that the Lewis–
Sargent relation should have a (–) in
front of it when net cations diffuse… so
switch the sign, as ∓, or use ours.
Last point: clarifying sign conventions in B&F so that this is crystal clear…
798
Type 3:
Type “2”:
… since we know that the β side will be (+) in the previous case,
this really means the Lewis–Sargent Eq. should have a (–) sign
in front of it when net cations diffuse…
… if we’re sticking to our convention that the potential is the β
(product/red. phase) versus the α (reactant/ox. phase)…
Last point: clarifying sign conventions in B&F so that this is crystal clear…
799
… anyway…
800
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 to
charged macromolecules
an ionomer film
http://www.futuremorf.com/
Nafion
http://www.williamsclass.com/
801
a film of poly(styrene sulfonate)
Na+
CNaCl
cNa+ cCl-cNa+ cCl-
m m s s
R–
… consider this model which applies to both scenarios…
802
m s
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 ( 𝜇𝑖𝑜) – 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:
803
(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 ( 𝜇𝑖𝑜) – 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:
804Because differences in electrochemical potential ( 𝜇𝑖𝑜) – 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
… 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
805
… 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 ( 𝜇𝑖𝑜) – 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
806
Aside #1: Recall Type 1 case of LJ potential… but now with t– = 0…
(α)
(β)
… with β being the membrane
𝐸Donnan =𝑅𝑇
𝐹ln
𝑎Na+𝑠
𝑎Na+𝑚 = −
𝑅𝑇
𝐹ln
𝑎Cl−𝑠
𝑎Cl−𝑚
𝐸j =𝑅𝑇
𝐹ln
𝑎1 α
𝑎1 β= 𝐸Donnan Na+
807
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 write for this (Donnan) potential…
… Check!
Eqn. (2.4.2)
𝐸Donnan =𝑅𝑇
𝐹ln
𝑎Na+𝑠
𝑎Na+𝑚 = −
𝑅𝑇
𝐹ln
𝑎Cl−𝑠
𝑎Cl−𝑚
808
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...
𝐸Donnan =𝑅𝑇
𝐹ln
𝑎Na+𝑠
𝑎Na+𝑚 = −
𝑅𝑇
𝐹ln
𝑎Cl−𝑠
𝑎Cl−𝑚
809
… 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...
𝐸Donnan =𝑅𝑇
𝐹ln
𝑎Na+𝑠
𝑎Na+𝑚 = −
𝑅𝑇
𝐹ln
𝑎Cl−𝑠
𝑎Cl−𝑚
810
a film of poly(styrene sulfonate)
Na+
CNaCl
cNa+ cCl-cNa+ cCl-
m m s s
R–
… recall the scenario we are analyzing (with R– representing the fixed charges) …
811
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−
s = 𝑐Na+m 𝑐Cl−
m
812
because in the solution, cNa+ = cCl-
for goodness sakes!
… an equation quadratic in cmCl- is obtained as follows…
𝑐Cl−s 2
= 𝑐Cl−m 2
+ 𝑐Cl−m 𝑐R−
m
𝑐Na+s = 𝑐Cl−
s
𝑐Na+s 𝑐Cl−
s = 𝑐Na+m 𝑐Cl−
m
𝑐Na+m = 𝑐Cl−
m + 𝑐R−m
… an equation quadratic in cmCl- is obtained as follows… 813
𝑐Cl−s 2
= 𝑐Cl−m 2
+ 𝑐R−m 𝑐Cl−
m
814
… use the quadratic formula to solve for cmCl- and one gets…
… 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−
s = 𝑐Na+m 𝑐Cl−
m
𝑐Na+m = 𝑐Cl−
m + 𝑐R−m
815
if (which is the typical case of interest), then…
𝑐Cl−𝑚 =
−𝑐R−𝑚 + 𝑐R−
𝑚 2+ 4 𝑐Cl−
𝑠 2
2=𝑐R−𝑚
21 + 4
𝑐Cl−𝑠
𝑐R−𝑚
2
− 1
𝑐Cl−𝑠 ≪ 𝑐R−
𝑚
816
if (which is the typical case of interest), then…
𝑐Cl−𝑚 =
−𝑐R−𝑚 + 𝑐R−
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)
817
… fixed charge sites are responsible for the electrostatic exclusion of mobile
“like” charges (co-ions) from a membrane, cell, etc. This is Donnan Exclusion.
if (which is the typical case of interest), then…
𝑐Cl−𝑚 =
−𝑐R−𝑚 + 𝑐R−
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
𝑐Cl−𝑚 =
𝑐R−𝑚
21 + 2
𝑐Cl−𝑠
𝑐R−𝑚
2
− 1 =
… the larger CR-m, the samller CCl-
m
818
… is a reasonable assumption? What is CR-m?
… 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 lower
than 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
819… so how excluded is excluded?
… an order of magnitude lower
than CCl-s… rather excluded!
… but what if CCl-s is also large (e.g. 1 M)? … No more Donnan exclusion!
… is a reasonable assumption? What is CR-m?
… 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…
=0.1 2
1.0= 0.01 M
𝑐Cl−𝑠 ≪ 𝑐R−
𝑚
820… but Donnan exclusion is “amplified” in Nafion
and some other polymers… how?
Nafion phase separates into a hydrophobic phase, concentrated in
-(CF2)- backbone, and hydrophilic clusters of SO3– solvated by water…
Mauritz & Moore, Chem. Rev., 2004, 104, 4535
821… but Donnan exclusion is “amplified” in Nafion
and some other polymers… how?
Nafion phase separates into a hydrophobic phase, concentrated in
-(CF2)- backbone, and hydrophilic clusters of SO3– solvated by water…
… and, FYI, SO3– clusters are interconnected by channels that percolate
through the membrane, imparting a percolation network for ionic conduction
822
http://www.ocamac.ox.ac.uk/
http://www.msm.cam.ac.uk/doitpoms/tlplib/fuel-cells/figures/dow.png
http://www.chemie.uni-osnabrueck.de/
… but Donnan exclusion is “amplified” in Nafion
and some other polymers… how?
823
So in Nafion, "amplifying" effects operate in parallel…
1) The aqueous volume accessible to ions of EITHER
charge is a small fraction of the polymer’s overall
volume, and…
2) The local concentration of SO3– (CR-
m) is much
larger than calculated based on the polymer's density
and equivalent weight (i.e. the molecular weight per
sulfonic acid moiety).
824the Donnan potential can be used to concentrate metal ions...
cation-exchange membrane
(e.g. PSS)
[H3O+]R = 10 mM
[Cu2+]R = ?? ppm
start
[SO42-]R =
5 mM
[H3O+]L = 1 mM
[Cu2+]L = 30 ppm
[SO42-]L =
0.5 mM
Wallace, I&EC Process Design Development, 1967, 6, 423 from Wiki
P-Chemist
Frederick George Donnan
(1870–1956)
825the Donnan potential can be used to concentrate metal ions...
… there are actually two Donnan
potentials… one at each membrane–
solution interface…
… but that is fine because they are not
equal and opposite… their sum is the
net Donnan potential between the two
solutions wetting the membrane… we
do not need to know each Donnan
potential separately, and in fact, it is
nearly impossible to measure each
individually…
… as long as there is
decent Donnan
Exclusion, the system
will be rather stable
and work as desired
[H3O+]R = 10 mM
[Cu2+]R = ?? ppm
start
[SO42-]R =
5 mM
[H3O+]L = 1 mM
[Cu2+]L = 30 ppm
[SO42-]L =
0.5 mM
Wallace, I&EC Process Design Development, 1967, 6, 423 from Wiki
P-Chemist
Frederick George Donnan
(1870–1956)
cation-exchange membrane
(e.g. PSS)
826the Donnan potential can be used to concentrate metal ions...
[H3O+]R = 10 mM
[Cu2+]R = ?? ppm
start
[SO42-]R =
5 mM
[H3O+]L = 1 mM
[Cu2+]L = 30 ppm
[SO42-]L =
0.5 mM
– ––
Wallace, I&EC Process Design Development, 1967, 6, 423 from Wiki
P-Chemist
Frederick George Donnan
(1870–1956)
827the Donnan potential can be used to concentrate metal ions...
– ––
= 3000 ppm
––
2+
828the Donnan potential can be used to concentrate metal ions...
equilibrium
[H3O+]R = 10 mM
[Cu2+]R = 3000 ppm
[SO42-]R =
5 mM
[H3O+]L = 1 mM
[Cu2+]L = 30 ppm
[SO42-]L =
0.5 mM
… one net Donnan potential for the entire system that satisfies
equilibrium for each ion at each interface… Cool!
cation-exchange membrane
(e.g. Nafion or PSS)
[H3O+]R = 10 mM
[Cu2+]R = ?? ppm
start
[SO42-]R =
5 mM
[H3O+]L = 1 mM
[Cu2+]L = 30 ppm
[SO42-]L =
0.5 mM
829… the less-frequently discussed four-electrode measurement!
Slade, …, Walsh, J. Electrochem. Soc., 2002, 149, A1556
CE/CA1 (S3/P2)WE/CA2 (P1)
WE'/CA2'
(S1)
RE/ref2
(S2)(EC-Lab diagram, from Bio-Logic)