Handouts
Transport numbers
to obtain values of mobility u and of ionic molar
conductivities λ for single, individual ions
t±: transport, transference or migration numbers
t+ for a cation; t- for an anion
t+ is the fraction of current, I+, carried by the cation
total current: I = I+ + I-:
I +_
t+_ =
I+ + I t± means t+ or t-, and I± means the corresponding I+ or I-:
t+ =
I+ ; = I tI+ + I I+ + I -
_I = I + + I - ; thus t+ + t - = 1
and 0  t+_  1
For a general electrolyte
z++
a
A
|z-|b
B
_ a Az++ (aq) + b B|z-|- (aq)
the electro-neutrality condition must hold:
a  z+ = b  | z - |
in dissociation equilibrium for n initial mol of a weak
electrolyte we will have in solution
(1-α)n undissociated molecules
naα cations and nbα anions
transport numbers can be measured for each ion separately
Now, what is their relation to mobilities and ionic molar
conductivities?
F is the charge on 1 mol of charge, so a mol of a cation
with charge z+ have a charge of
Q+ = aFz+ (for 1 mol AaBb) and
Q- = bFz- (for 1 mol AaBb)
So the electricity flowing through a given area A in unit
time is aFz+u+ for 1 mol AaBb in solution and the negative
ion current is bFz-u-, thus
t+ =
aF z+ u+
a z + u+
=
aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u _
bF | z - | u b | z- | u=
t- =
aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u and thus t+ + t- = 1.
Since az+ = b|z-|, we can replace b|z-| in the denominator of
t+ with az+, and az+ in the denominator of t- with b|z-|:
t+ =
t- =
a z + u+
u+
=
a z + u+ + a z + u - u+ + u -
b | z- | uu=
b | z - | u+ + b | z - | u - u + + u -
With the short hand writing
  ( Aa Bb ) ; +  ( Az++ ) ;  -  ( B|z-|- )
we know from before
 + = F z + u+ ; u + =
 - = F | z- | u- ; u- =
+
F z+
F | z- |
 = a  + + b  - = F(a z+ u+ + b | z - | u - )
and therefore
+
t+ =
F z+
u+
=
+ +  u+ + u F z+ F | z - |
t- =
F | z- |
u=
- + u+ + u F z+ F | z - |
Now we can cancel F and since az+ = b|z-| we can replace
b
1 a 1
a
1 b 1
=
|
|
;
=
;
|
|
=
;
=
z+
zzz+
a
b
| z - | a z+
z+ b | z - |
and thus
+
a +

+
z
+
=
=
t+ =
b
+ + b  a + + b  + +  a
z+ a z +
t- =
| z- |
a +
+ b | z- | | z- |
Since (see before)
=
a
+ +  b
=
b a + + b  -
 = a + + b  a +
b thus t+ =
; t- =


For any strong electrolyte, t+ and t- can be measured for
different values of c+ = ac and
c- = bc and then extrapolated to t+o and t-o.
Since t+o and t-o can be measured individually, we can split
Λo into


o
o
o
=
;
=

t+  ta
b
o
o
o
+
For NaCl e.g. a = b = 1 and z+ = |z-| = 1 but
for MgO e.g. a = b = 1 but z+ = |z-| = 2
Methods to measure t+ and tHittorf method
After the anode and cathode solutions are connected, a
small current is sent through the Hittorf cell for a period of
time.
Then the connection is closed, and the anode and cathode
solutions after electrolysis can be collected separately and
undergo chemical analysis, which yields the concentration
changes.
For simplicity let us assume M+A- electrolytes (can be
done for MaAb ones also, but one has to care then for a and
b)
cation current fraction: t+
anion current fraction: tAfter passing of 1 mol electrons through the cell, F
charges have passed through
thus
Ft+ charges are transported to the - electrode (cathode) by
M+ ions and
Ft- charges are transported to the + electrode (anode) by A-
ions
and because 1 mol of charge has passed:
1 mol of M+ was discharged at the cathode and
1 mol of A- was discharged at the anode
Cathode solution
By passage of 1 mol of electrons = F charges
t- mol of A- ions are lost by transport to the anode
1 mol M+ ions are discharged and thus lost
t+ mol M+ ions are transported into the cathode solution:
t+ mol M+ ions are gained = -t+ mol M+ ions are lost
together (1 - t+) mol M+ ions are lost = t- mol M+ ions are
lost
thus t- mol MA are lost in the cathode solution
Anode solution
By passage of 1 mol of electrons = F charges
t+ mol of M+ ions are lost by transport to the cathode
1 mol A- ions are discharged and thus lost
t- mol A- ions are transported into the anode solution:
t- mol A- ions are gained = -t- mol A- ions are lost
together (1 - t-) mol A- ions are lost = t+ mol A- ions are
lost
thus t+ mol MA are lost in the anode solution:
amount MA lost in anode solution t+
=
amount MA lost in cathode solution t -
=
1 - tt+
=
1 - t+
t-
Chemistry 311, Section 02, Lecture01
Instructor: W. Fِörner
Phone: 3553
Office: 4/147-3
email: forner@kfupm.edu.sa
Lectures: Build. 5 - 201; SMW 11:00 - 11:50 am
Office hours: SMW 9:00 - 10:50
in case I am not in office, check right outside Building 4, maybe I am only 5 minutes out for
smoking
You can also make extra appointments for meetings by email or phone
Quizzes: in almost all Review lectures, others will be announced about 1 week before
Unexcused absence in a quiz: 0 out of 10, no makeup
Homework is assigned but will not be collected.
However, if you solve the homework you can get good grades in Quizzes (will be homework
problems only)
Quizzes (out of 100) will be averaged to 75
each unexcused absence (excuses from student affairs must be in 1 week after the absence):
-1 point from classwork; 9 unexcused absences: DN
Major dates, grading of the course, reading assignments: see syllabus
Sections which are not in the syllabus are not covered in class and are not part of exams
Overview
Electrochemistry
Chapter 7 : Solutions of electrolytes (electrolysis)
Chapter 8 : Electrochemical cells (batteries)
1. Major Exam
Chemical Kinetics
Chapter 9 : Elementary (one-step) reactions
Chapter 10 : Kinetics and mechanisms of composite reactions, e.g.:
Step (1) :
A+BC
Step (2) :
C+DE
Overall (C cancels) :
A+B+DE
2. Major Exam
Surface Chemistry
Chapter 18 : Adsorption and adsorption isotherms, heterogeneous catalysis
Transport Properties
Chapter 19 : Viscosity of gases and liquids and diffusion
Final Exam: Chapters 7,8,9,10,18,19 (comprehensive)
ELECTROCHEMISTRY (Solutions of electrolytes)
Objectives general introduction, Faraday's law of electrolysis, Molar conductivity
We are concerned with
1. Properties, like conductivity or dissociation of electrolytes in solutions
Electrolyte: a compound that dissociates into ions when it is dissolved, e.g. in water or other
solvents
2. Processes that happen at electrodes when immersed in electrolyte solutions
Informations can be obtained by the investigation of electrical effects in solutions
Measurement of conductivity versus concentration yields:
extent of ionization (e.g. in water)
association of ions in solutions (movement of ions bound close together)
movement of ions in the solvent (mostly water)
Basic concepts for electricity
Electrical units
Electrostatic force between 2 charges (unit As) Q1 and Q2 at distance r:
Q1 Q 2
F=
4  o r 2
true in vacuum only
εo: permittivity or dielectric constant of the vacuum:
8.854 x 10-12 C2J-1m-1
1 C = 1 As; 1 J = 1 Nm = 1 kgm2/s2 = 1 VAs
Thus 1 C2J-1m-1 = 1 A2s2/(VAsm) = 1 As/(Vm)
force not in vacuum but in a medium with relative dielectric constant ε (no unit):
F=
Q1 Q 2
4  o r 2
 r =   o : relative permittivity
Electric field: force/unit charge
electric field of charge Q at distance r:
E=
Q
4  o r 2
unit: 1 N/C = 1 (J/m)/C = (1 VAs)/(Asm) = 1 V/m
electric potential φ at distance r from charge Q (if the field depends only on distance and not on
angles) is defined such that
d
E=thus Edr = - d
dr
Integration:
 d =  = -  Edr = -
=
Q

dr
4  o r 2
Q
4  o r
The unit is 1 V
Faraday's law of electrolysis
1. law: the mass of the element produced (deposited) at an electrode is proportional to the amount
of electricity (charge) passed through the solution, where
Q = It; unit: 1 As = 1 C (Coulomb)
2. law: the mass of the element produced (deposited) at an electrode is proportional to the
equivalent weight of the element
M+ + e-  M: equivalent weight = molar weight
M2+ + 2e-  M: equivalent weight = 1/2 molar weight
charge of 1 electron: 1.602 x 10-19 C
charge of 1 mol of electrons:
1.602 x 10-19 C x NA (Avogadros's number) =
= 1.602 x 10-19 C x 6.022 x 1023 mol-1 = 96472 C/mol = F
(Faraday's constant)
Ag+ + e-  Ag
H+ + e-  1/2 H2
Thus a charge of 96472 C produces
1 mol Ag, but 1/2 mol H2
Electrolysis:
the negative electrode attracts cations and is called cathode
the positive electrode attracts anions and is called anode
Batteries:
the negative electrode produces electrons and is called anode
the positive electrode takes up electrons and is called cathode
A solution of gold (III) nitrate, Au(NO3)3 is electrolized using a current of 0.0250 A until 1.20 g Au
(molar mass MM = 197.0 g/mol) is deposited at the cathode.
1. What is the amount of electricity (charge) passed through the solution?
Au3+ + 3e-  Au equivalent to 1/3 Au3+ + e-  1/3 Au
1 mol e- yields 1/3 mol Au
96472 C of charge yield 1/3 mol Au
moles of Au deposited:
(1.20 g Au)/(197.0 g/mol) = 6.091 x 10-3 mol Au
6.091  10-3 mol Au 
1 mol e
= 18.27  10-3 mol e
1
mol Au
3
Multiply by Faraday's constant to get the charge:
Q = 96472
C
 18.27  10-3 mol = 1.76  103 C (As)
mol
2. What is the time needed for that?
Q = It and thus the time is t = Q/I:
1.76  103 As
t=
= 7.04  104 s
0.0250 A
= 1.17  103 min = 19.6 h
3. The anode reaction is
2 H2O  O2 + 4H+ + 4e2 oxygen atoms are oxidized from oxidation number -2 in water to 0 in oxygen, and thus 4e- are
released. Divided by 4 the equation for 1 e- is:
1/2 H2O  1/4 O2 + H+ + eHow much (volume) of O2 gas is obtained at STP (0oC,
1 atm) ?
1 mol e-  1/4 mol O2 and thus we have
1 mol O2
-3

18.27

mol
10
e
4 mol e= 4.57  10-3 mol O2
The molar volume of O2 at STP can be obtained from the ideal gas equation PVm = nRT at P = 1
atm, T = 273.15 K and n= 1 mol and is Vm = 22.4 L/mol (1 L = 1 dm3) and thus
V O2 = 22.4
L
 4.57  10-3 mol O 2 = 0.102 L
mol O 2
= 0.102 dm3
Molar Conductivity
The nature of solutions can be obtained from conductivity measurements:
non-electrolytes: no dissociation into ions, no charge carriers, low conductivity
electrolytes: dissociation into ions when dissolved, high conductivity
two types of electrolytes:
strong electrolytes: 100% of the salt is dissociated into ions when dissolved, e.g. NaCl:
NaCl  Na+(aq) + Cl-(aq)
in the solution (no matter if all dissolves or not) there are no "NaCl molecules", only ions
weak electrolytes: only partially dissociated and there are also electrolyte molecules in the solution
not only ions, e.g. acetic acid CH3COOH (HAc).
dissociation equilibrium:
or
CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)
HAc(aq) <=> Ac-(aq) + H+(aq)
in solution all 3 species are present
measurement of conductivity vs concentration of the solute gives informations
Ohm's law: resistance R is proportional to the voltage V and inversly proportional to the current
strength I (proportional to 1/I):
R = V/I
unit of R: 1 Ohm = 1 Ω = 1 V/A (1 Volt/Ampere)
conductance G = 1/R; unit 1 S (Siemens) = 1 Ω-1
When the current flows through a cell of area A (area of the electrodes) and length l (distance
between the electrodes), the conductivity κ is defined as
A
l
G=  ;  = G 
l
A
the conductivity is thus the conductance of a unit cube
For solutions the conductivity is also called the electrolytic conductivity
unit: 1 Sm-1 = 1 Ω-1m-1
more common is the use of cm instead of m:
1 Scm-1 = 1 Ω-1cm-1
S
S
S
1
= 1 - 2 = 100
cm
m
10 m
Kohlrausch: definition of the equivalent conductivity or also called molar conductivity Λ:
=

where c is the molarity of the electrolyte
example: the electrolytic conductivity of 0.1 M
(0.1
mol/L = 0.1 mol/dm3) acetic acid is
5.3 x 10-4 Ω-1cm-1, what is the molar conductivity?
c

5.3  10 -4 -1 cm-1
= =
c
0.1 mol dm-3
_
-1
1 cm-1

= 5.3  10

mol ( 10+1 cm )-3
-3
-1
1 cm-1

= 5.3  10
 -3 -3
mol 10 cm
-3
S cm
cm
= 5.3 
= 5.3
mol
mol
-1
2
2
Λ(c): in all cases Λ slightly decreases when c is increased:
more electrolyte = larger c and smaller molar conductivity
G(c), κ(c): in all cases G and κ increase when c is increased:
more electrolyte = larger c = more ions = larger conductance G and larger conductivity κ
Lecture 02
The electrolytic conductivity at 20oC of 1.00 x 10-2 M
aqueous solution of acetic acid (HAc) was found to be
1.60 x 10-4 S cm-1. Some Λo values are
Λo(HCl) = 426.2 S cm2 mol-1,
Λo(NaAc) = 91.0 S cm2 mol-1 and
Λo(NaCl) = 126.5 S cm2 mol-1.
(a)
Calculate the degree of ionization of acetic acid.
Solution:
c = 1.00  10-2 M = 1.00  10-2 mol(10 cm )-3 = 1.00  10 -5 mol cm-3

1.60  10-4 S cm-1
2
-1
(HAc) = =
=
16.0
S
cm
mol
-5
-3
c
10 mol cm
_
o
o
o
o
 (HAc) =  (HCl) +  (NaAc) -  (NaCl)
= (426.2 + 91.0 - 126.5) S cm2 mol -1
= 390.7 S cm2 mol -1
=


=
o
16.0
= 0.04095 = 0.0410
390.7
(b)
Calculate the equilibrium constant of acetic acid with units in the above solution.
Solution:
2
0.040952

-2
= 1.00  10 M
Ka= c
(1 -  )
1 - 0.04095
= 1.75  10 -5 M
(c)
What is the van't Hoff factor and the osmotic pressure of this solution?
Solution:
v is the theoretical number of particles per dissolved molecule in the solution, if all molecules of
HAc would dissociate (this is not really the case, so theoretical number).
Since HAc, if all molecules would dissociate, would yield H+ and Ac-, and thus v = 2 ions per
molecule.
When the degree of dissociation is α, then in solution we have for each dissolved molecule (1 - α)
undissociated molecules and vα ions. Thus the van't Hoff factor is:
i = 1 -  + v
= 1 - 0.0410 + 2  0.0410 = 1.0410
The concentration is in mol/L and must be changed to mol/m3:
c = 1.00  10-2
mol
dm
-2
=
1.00

10
3
= 10.0
mol
( 10-1 m )3
mol
m
3
and the osmotic pressure Π is
 = icRT
= 1.0410  10.0
mol
m
= 2.54  104
3
 8.315
J
 293.15 K
K mol
kg
4
=
2.54

Pa = 0.254 bar
10
2
ms
= 0.251 atm
units:
1 Pa = 1
N
m
= 10-5 bar =
Lecture 034
=1
2
kg
m s2
1
atm
5
1.01325  10
Alternating Currents (AC)
Concept of ionic atmosphere moving together with the ions:
molar conductivity Λ depends on the frequency of an alternating potential
If the frequency is large enough (very fast change of direction of move)
i.e. so large, that the oscillation time is smaller (faster) than the relaxation time of the ionic
atmosphere:
then the ions are stationary and the atmosphere stays symmetric
The relaxation and electrophoretic effects, which cause Λ to decrease with increasing concentration
c, vanish at large AC frequencies
Thus at large AC frequencies Λ increases with increasing c (more ions = charge carriers at higher c)
Wien effect
large potentials of about 20000 V/cm or larger result in a high speed of the ions
Thus the ions move so fast, that the ionic atmosphere cannot follow them
Thus at very large potentials Λ also increases with increasing c,
because the effects from the ionic atmosphere vanish when the ions are faster than the atmospheres
Dissociation fields
weak electrolytes in large fields (potentials):
Λ becomes large and weak electrolytes behave like strong electrolytes
because the large potentials force the dissolved, undissociated molecules in solutions of weak
electrolytes to dissociate.
lecture 04: linearized plots, independent migration of ions: mobility and ionic molar conductivity of
ions
Question: which kinds of plots must be linear in case of strong and weak electrolytes?
Determination if an electrolyte is strong or weak from linear plots
strong electrolytes
molar conductivity as function of concentration:
 = o - a  c
Thus if the electrolyte is a strong one then a plot of Λ versus c must be linear
(Correlation coefficient r2 close to 1)
the intercept of the straight line gives Λo which is a very good value for strong electrolytes,
the slope is -a (the constant in the formula)
weak electrolytes
Ostwald's dilution law gives
c 2
K=
1-
The degree of dissociation (fraction of dissociated molecules) is

and thus

o
c 2
c 2
c
o
K = o =
= o

1
1( o -  ) o - 
o o
2
2
2
Multiplication by the total denominator Λo(Λo - Λ):
K o ( o -  ) = c  2
K o2 - K o  = c  2
Division by KΛo2Λ:
1 1
c
- =
 o K o2
1 1
1
= +
c
 o K o2
Thus if the elctrolyte is a weak one, then a plot of 1/Λ versus (cΛ) = κ must give a straight line
(Correlation coefficient r2 close to 1) with intercept
1
1
intercept =
; thus o =
intercept
o
But Λo values for weak electrolytes from such plots are not very good.
To obtain correct Λo values in this case the independent migration of ions law must be used
But using intercepts from such plots and the slope gives good K values:
1
intercept
slope =
=
K  o2
K
intercept
K=
slope
2
2
In general: derivations given in labs can be asked in exams!!
Lecture 05
Hittorf Cell
nA: number of moles of MA lost in the anode compartment
nC: number of moles of MA lost in the cathode compartment
nT: total number of moles of MA deposited at the electrodes
nT = charge passed/Faraday's constant, if M+, and Acharge passed = current strength (I) x time (t)
nT = nA + nC
ratio of the transport numbers: t+/t- = nA/nC
since t- = 1 - t+:
t+
n
= A ; t+ nC = (1 - t+ ) n A
1 - t + nC
t+ =
nA
n
= A
n A + nC nT
and therefore
t+ =
nA
n
; t- = C
It/F
It/F
Moving boundary method
for example M+ (can be H+) or AM'+ and A'- must be slower than M+ and Athe indicator solution must change colour when M+ and A- arrive in it
for example an acid/base indicator if M+ is H+ and A- is OH-
A distinct moving boundary is obtained between M'A and MA on one side and between MA and
MA' on the other side
because M'+ is slower than M+ and cannot overtake it and
because A'- is slower than A- and cannot overtake it
moved distances from start (a, b): aa' proportional to u+ and bb' proportional to uto make the boundaries visible indicators must be added that change colour when M+ or A- are there
aa
bb
; t- =
t+ =
aa + bb
aa + bb
Often seen nonsense lab results:
t+ = 360 which is nonsense, because 0  t+  1 must be
and to make t+ + t- = 1, t- = -359 which is also nonsense, because 0  t-  1 must also be
mistake because the students forget that they read their current in mA and not in A
Lab: 1 boundary in acid solution + acid base indicator
at the cathode a
solution is produced,
solution)
neutral, yellow CdCl2
but the H+ (acid in the
ions move faster to
indicator
changes
leave the
- in the beginning -
the cathode, and the
colour when the H+ ions
below
solution
yellow, neutral CdCl2
moved out)
boundary:
(H+ ions
above boundary: red (methylorange) acidic solutions (H+ still there)
acidic solution
Q = It charge transported by the H+ ions, F Faraday's constant, c known concentration of the acid
solution, A diameter of the tube, aa' distance travelled by the boundary in time t:
aa = t+
Q
FcA
Lecture 07
Consider an electrolyte AaBb
Then for electroneutrality we must have az+ = b|z-|
and for ativity coefficients
a+b
a b
y+_ = y+ y-
(a + b)  log 10 y+_ = a  log 10 y+ + b  log 10 yNow we insert the DHLL:
2
=
log 10 yi zi B I
to obtain
(a + b)  log 10 y+_ = - (a z+2 + b z -2 )B I
 a 2 z+2 b2 z -2
= - 
+
b
 a
electroneutrality: (b|z-|)2 = (az+)2:

 B I

 1 1
(a + b) log 10 y+_ = - a 2 z+2  +  B I
 a b 
= - a 2 z+2
a+b
B I
ab
log 10 y+_ = -
a 2
z+ B I
b
electroneutrality: b|z-| = az+: a/b = |z-|/z+ and thus
log 10 y+_ = - z+ | z - | B I
and in water at 25oC:
log 10 y+_ = - 0.51 z+ | z - |
I
mol L-1
ionic strength I for 1 M solutions of different electrolytes:
uni-univalent: NaCl  Na+(aq) + Cl-(aq)
I=
1
1
2
2
2
=
(1
M

(1
+
1
M

(-1
)= 1 M
)
)
c
z
i
i

2 i
2
summation always over ALL ions in a solution
uni-bivalent: Na2SO4  2 Na+(aq) + SO42-(aq)
I=
1
1
2
=
(2  1 M  (1 )2 + 1 M  (-2 )2 ) = 3 M
ci z i

2 i
2
uni-trivalent: Na3PO4  3 Na+(aq) + PO43-(aq)
I=
1
1
2
2
2
=
(3

1
M

(1
+
1
M

(-3
)= 6 M
)
)
c
z
i
i

2 i
2
bi-trivalent: Mg3(PO4)2  3 Mg2+(aq) + 2 PO43-(aq)
I=
1
1
2
2
2
=
(3

1
M

(2
+
2
M

(-3
) = 15 M
)
)
c
z
i
i

2 i
2
Lecture 08
Ionic equilibria
Objectives: activity coefficients from equilibrium constants and solubility products
Equilibrium is obtained very rapidly in ionic solutions
example: HAc <=> H+ + AcHAc: CH3COOH; Ac-: CH3COOactivity: a = cy activity coefficient is written as english lower case y or as greek lower case gamma
(γ)
The dissociation constant Ka of acetic acid is practically
a H+  a Ac- [ H + ]  [ Ac ] y+  y=

Ka=
[HAc]
yu
a HAc
yu: activity coefficient of the undissociated acid about equal to 1 because the molecule has no
charge (not exactly 1 because of the dipole moment)
the activity coefficients of cations (y+) and of anions (y-) are not 1, because of the charges and their
interactions
mean activity coefficient y± for AB compunds:
y± = (y+y-)1/2; (y±)2 = y+yfor AaBb compounds: (y±)a+b = (y+)a(y-)b
Therefore
-
+
[ H ][ Ac ]
2
 y+_
Ka=
[HAc]
taking logarithm basis 10 gives
 c +c log 10 K ua = log 10  H Ac
 c HAc
u

 + 2  log 10 y+_
eq
 c H + c Acu
K c = 
 c HAc
u


eq
superscript u: units taken out
subscript eq: equilibrium
superscript o: standard state: 25oC, 1 atm,
all concentrations equal to 1 M (not equilibrium).
so superscript o at an equilibrium constant means only 25oC, 1 atm
Thus
 c +c log 10  H Ac
 c HAc
u

 = log 10 K ua - 2  log 10 y 
eq
= - pK a - 2  log 10 y+_
Ostwald's dilution law:
 c H + c Ac
 c HAc
2



; =
 = c
1-
o
eq
Thus
 c
log 10 
 1-
2
u

 = log 10 K ua - 2  log 10 y 
eq
On the left and right hand side the same unit must be taken out (mol/L)
DHLL:
 c
log 10 
 1-
2
u

 = log 10 K ua + 2  z+ | z - | B I
eq
B = 0.51 dm3/2mol1/2 in water at 25oC
I=
1
2
ci z i ; c H + = c , c Ac- = c

2 i
_ c HAc = c(1 -  ) (no ion)
1
I =  c 12 + c 12 = c
2
if there are no other ions present, which in case must be added to I.
Thus one has to measure Λ at different concentrations c and obtain Λo.
Then one can calculate ionic strength for each c value and
u
  
 c  
u
2
  o  
 c 
 = log 10 
log 10 


1



 1

o 


2
Now, at small c values only a plot of this quantity versus square root of the ionic strength I1/2 will
give a straight line where c is small enough that DHLL is obeyed.
The straight line at small I (c) has the slope 2B when z+|z-| = 1 as in our case and in water at 25oC:
2B = 1.02 dm3/2 mol1/2
at 25oC and 1 atm, Ku is also written as Ko (without units)
When I and thus c goes to 0, then the equation tells
 c 2
log 10 
 1-
u

 = log 10 K o

Because at c = 0 we have I = 0 and according to DHLL y± = 1 and thus log10y± = 0
At larger I (larger c) where DHLL is not obeyed (no straight line) the difference between the
measured points and log10Ko is equal to -2log10y± and the mean activity coefficient can be obtained
at all c values, since
 c
- 2  log 10 y+_ = log 10 
 1-
2
u

 - log 10 K o

values of y+ or y- cannot be obtained, only the mean value by plotting
 c
f = log 10 
 1-
2
u

 versus

I
When I is very small, the plot can also be done as f versus m (molality)
Solubility products
The solubility product Ks is a constant in saturated (as much electrolyte is dissolved as possible)
solutions of electrolytes.
Example: AgCl <=> Ag+(aq) + Cl-(aq)
The largest amount of AgCl that can be dissolved is the solubility s:
s = [AgCl]max = [Ag+]max = [Cl-]max and
K s = a Ag+  aCl - = [ Ag ]
+
max
 [ Cl - ] max  y+ y -
= s  s y+2 _ = s 2 y+2 _ since y+2 _ = y+ y Example: PbCl2 <=> Pb2+(aq) + 2 Cl-(aq)
The largest amount of PbCl2 that can be dissolved is the solubility s:
s = [PbCl2]max = [Pb2+]max = 1/2 [Cl-]max and thus
[Cl-]max = 2s
2
2+
K s = a Pb2+  aCl - = [ Pb ]
max
 ([ Cl - ] max )2  y+ y-2
= s  (2s )2 y+3 _ = 4 s3 y+3 _ since y+3 _ = y+ y-2
Adding of an inert electrolyte
An electrolyte inert to AgCl is one that has no ions that are also in AgCl
NaBr is inert to AgCl since it has no Ag+ or Cl- ions
NaCl is non-inert to AgCl because it has Cl- ions as also AgCl
then [Cl-] from NaCl is much larger than [Cl-]max from AgCl and
K s = a Ag+  aCl - = [ Ag ]
+
max
 [ Cl - ]  y+ y -
= s  (s + [ Cl - ] NaCl ) y+2 _  s[ Cl - ] NaCl y+2 _
since [ Cl - ] NaCl much larger than s
adding of an inert electrolyte brings about no chemical change, but it changes y± because it changes
I:
no inert NaBr:
1
1
I = ([ Ag+ ] max z+2 + [ Cl - ] max z -2 ) = (s  12 + s  12 ) = s
2
2
inert NaBr present:
1
I = ([ Ag+ ] max 12 + [ Cl - ] max 12 + [ Na+ ]  12 + [ Br - ]  12 )
2
1
= (2s + 2[NaBr])= s + [NaBr]
2
At low I DHLL (at higher I the CI correction must be added) is valid
DHLL:
log 10 y+_ = - z+ | z - | B I
Thus, when I increases because of adding an electrolyte, then the activity coefficients drops
(becomes smaller) because of the - sign, when I is so small that DHLL holds
at high I log10y± increases when I increases, can even become positive, because the CI correction
(positive) must be added
low I y± decreases with increasing I, but Ks must be a constant:
2
2
+_
K s = s y = constant
When Ks must be the same, then when y± decreases, s must increase
thus more AgCl goes into solution than into pure water when NaBr is added to it:
salting in effect
large I y± increases with increasing I, but Ks must be a constant:
2
2
+_
K s = s y = constant
When Ks must be the same, then when y± increases, s must decrease
thus less AgCl goes into solution than into pure water when NaBr is added to it:
salting out effect
Example: Ks(BaSO4) = 9.2 x 10-11 M2 (mol2/L2 = mol2/dm6)
What is y± in a solution (in water at 25oC) that is saturated in BaSO4 and contains also 0.05 M
KNO3 and 0.05 M KCl, salting in or salting out effect?
assume that DHLL applies
In the solution we can neglect the Ba2+ and SO42- concentrations because they are much smaller
than those of KNO3 and KCl:
1
I = ([ K + ] KNO 3  12 + [ NO-3 ]  12 +
2
+ [ K + ] KCl  12 + [ Cl - ]  12 )
1
= (0.05 + 0.05 + 0.05 + 0.05) M = 0.1 M
2
DHLL:
log 10 y+_ = - z+ | z - | B I = - 2  2  0.51
L
mol
0.1
mol
L
= - 0.645
y+_ = 10-0.645 = 0.226
note that for BaSO4 z+ = +2 and z- = -2, thus |z-| = +2
The solubility product is
2
2
+_
Ks= s y
and thus
s=
Ks
= 4.24  10-5 M
y+_
In pure water s is so small that I is almost 0 and thus y± is almost 1:
2
-6
K s = s thus s = K s = 9.6  10 M
In the KNO3/KCl solution more BaSO4 is dissolved than in pure water: salting in effect
If for a salt Ks is larger, i.e. 10-3 or so, the above is only the first step in an iteration:
use the solubility to calculate a new I, a new y and a new solubility. Repeat until there is no more
change
salting out happens only when I is so large that the CI correction becomes important and
log 10 y+_ = -
z+ | z - | B I + CI
1 + aB I
Consider a simple AB electrolyte:
AB(s) <=> A+(aq) + B-(aq), then
2
+
+
K s = a a+  a B- = [ A ][ B ] y+ y - = [ A ][ B ] y+_
u
log 10 K os = log 10 ([ A+ ][ B- ] ) + 2  log 10 y+_
_
u
+
log 10 ([A ] ][ B- ] ) = log 10 K os - 2  log 10 y 
In a saturated solution that contains only AB and inert electrolytes we have
s = [ A+ (aq) ] max = [ B- (aq) ] max = [AB(aq) ] max
and thus
2
log 10 ( su ) = log 10 K os - 2  log 10 y+_
g = log 10 su =
1
 log 10 K os - log 10 y 
2
1
DHLL : g = log 10 su = log 10 K os + B I
2
Plot as before g = log10su versus I1/2 then at low I (DHLL) a straight line with slope B (0.51 L1/2mol-
1/2
in water at 25oC) is obtained and its intercept is 0.5 x log10Kso.
since
- log 10 y+_ = g Lecture 45:
1
1
 log 10 K os = log 10 su -  log 10 K os
2
2
Final Exam from May, 1999
The solubility product of BaSO4 is 9.2 x 10-11 mol2dm-6.
(a)
Calculate the mean activity coefficient of the Ba2+ and SO42- ions in a solution in
water at 25oC, that is 0.05 M in NaNO3 and 0.05 M in KCl, assuming the DHLL to
apply.
Solution
DHLL:
1.
log 10  +_ = - 0.51 z+ | z - | I  M -1 ; z+ | z - |= + 2 | -2 |= 4
ionic strength :
I=

1
0.05 M  12 ( Na+ ) + 0.05M  (-1 )2 ( NO-3 ) +
2
+ 0.05M  12 ( K + ) + 0.05M  (-1 )2 ( Cl - )
=

1
 4  0.05 M = 0.10 M
2
Since the concentrations of Ba2+ and SO42- ions are about the square root of the
solubility product, and therefore about 10-5 to 10-4 M, they can be neglected as
compared to 0.05 M. Then
log 10  +_ = - 0.51  4  0.1 = - 0.645
 +_ = 10-0.645 = 0.226
(b)
Solution
What is the solubility of BaSO4 in that solution, and
The constant solubility product implies that the concentrations of Ba2+ and SO42- in
there are actually each one equal to the solubility of BaSO4 (saturated solution):
2
K sp = a Ba2+ a SO42- = c Ba2+  + cSO42-  - = s  +_
2
Note: the above is true only for AB electrolytes.
K sp
s=
(c)
Solution:
 +_
9.2  10 -11 M 2
=
= 4.24  10-5 M
0.226
in pure water?
Since in pure water the ionic strength contains only the terms for the Ba2+ and for
the SO42- ions, which are very small due to the small Ksp, we can approximate I by
0:
I  0 _ log 10  +_  0 _  +_  1
_ K sp = s 2  +2 _  s 2 _ s = K sp = 9.59  10-6 M
The fact that more BaSO4 dissolves in NaNO3/KCl solution than in pure water is
called "salting in" effect.
2.
The cell
Pt(s), H2 (1 bar) | HCl (m) | AgCl(s) | Ag(s)
gives the following emf values at 25oC:
m/(mol kg-1)
E/V
(E + (2RT/F) ln mu )/V
__________________________________________
3.215 x 10-3
4.488 x 10-3
0.5200
0.5039
5.619 x 10-3
0.4933
-3
7.311 x 10
0.4812
__________________________________________
(a)
Solution
What is the cell reaction?
right side:
reduction
AgCl(s) + e-  Ag(s) + Cl-(aq)
left side :
H2(g)
oxidation
 2H+(aq) + 2eThe cell reaction is the sum of the two with the electrons cancelling out. That can be
done by 2 x reduction + oxidation or by reduction + 1/2 x oxidation:
1
H 2 (g) + AgCl(s)  Ag(s) + HCl(aq) ; z = 1
2
RT
u
Nernst : E = E ln  a H + aCl -  =
zF
o
RT
u RT
=E ln ( c H + cCl - ) ln (  +  - ) =
F
F
o
= Eo -
RT
RT
ln ( c 2HCl )u ln (  +_ )2 =
F
F
= Eo -
_ E+
2RT
2RT
ln mu ln  +_
F
F
2RT
2RT
ln mu = E o ln  +_
F
F
Note: superscript u at the molality m of HCl is not a power, but means "remove the unit
before taking the ln". Further note, that [H+]=[Cl-]=[HCl]=m at all times.
Since E is positive, the reaction goes in the direction as derived above (if it would be reversed, E
would be negative).
It can be shown that the expression for the emf of this cell over a range of concentrations of
HCl is given by
E+
(b)
Solution:
2RT
2RT
ln mu = E o ln  +_
F
F
Complete the table above and determine Eo on the supplied sheet of graph paper
with the axis already marked (plot y versus mu!). Linear regression is also
acceptable.
Trick: we define
2RT
y  E+
ln mu and x  mu
F
A plot of y versus x2 could give us a straight line y=ax2+b with slope a and intercept b,
when such plots are done for small concentrations as here.
But when x is small then ln(1+x) = x NOT lnx = x. Since DHLL applies for small concentrations,
we can replace ln by log10 with the help of 2.303 and insert the DHLL. Also a slope of
2RT/F is not important. Thus
y = Eo -
2RT
ln  +_ ; DHLL : log 10  +_ = - z+ | z - | B I
F
_HCL : z+ | z - |= 1 ; ln  +_ = 2.303 log 10  +_
_ y = Eo +
2RTB
2.303 I ; HCl : I = m
F
_ y = Eo +
2RTB
2.303 m
F
Then a plot of y vs I1/2 is a straight line where DHLL applies.
Since with HCl I=0.5(c(H+)+c(Cl-))=m, thus here a plot of y vs m1/2 must give a straight line.
At the intercept mu is 0. Thus (here: molality not molarity, thus NOT B=0.51M-1/2)
y( mu = 0) = b = E o +
2RTB
2.303 m = E o _ b = E o
F
Completed table (2RT/F = 5.1389 x 10-2 V):
m (10-3 mol/kg)
E(V)
y = E + (2RT/F)ln mu (V)
3.215
4.488
5.619
7.311
0.5200
0.5039
0.4933
0.4812
m (10-3 mol/kg)
0.2250
0.2261
0.2270
0.2284
ln mu
3.215
4.488
5.619
7.311
x = (mu)1/2
-5.7399
-5.4063
-5.1816
-4.9184
0.056701
0.066993
0.074960
0.085504
A plot y vs mu, assuming x1/2=x for small x, is not that good, cause a Taylor series of x1/2
around x=0 is not possible. In linear regression we have
N  xi y i - (  xi )(  y i )
a=
i
i
i
D
; D= N x -(x
2
i
i
i
(  xi2 )(  yi ) - (  xi )(  xi yi )
b=
i
i
i
i
D
where the number of (xi, yi) points N=4. From the table we get:
Σyi = 0.9065 V Σxi = 0.28416 Σxi2 = 0.020633
Σxiyi = 0.064450 V (note that x=mu and thus has no units)
D = 4  0.020633 - (0.28416 )2 = 1.7851  10-3
a=
4  0.064450 V - 0.28416  0.9065 V
= 0.1171 V
1.7851  10-3
b=
0.020633  0.9065 V - 0.28416  0.064450 V
=
1.7851  10-3
= 0.2183 V = E o
So, our final regression function is
2
i
)
y = 0.1171V  mu + 0.2183 V
(b)
Solution:
Calculate the activity coefficient for a 6.000 x 10-3 m solution of HCl.
First we need y(0.006 m) for that concentration:
y(0.006 m)= 0.1171 V  0.006 + 0.2183 V = 0.2274 V
Then using the right hand side of the above given equation we obtain:
y = Eo -
2RT
ln  +_
F
_ y(0.006 m)= 0.2183 V - 5.1389  10- 2 V ln  +_
_ ln  +_ =
(0.2183 - 0.2274) V
= - 0.1771
5.1389  10- 2 V
_  +_ = e-0.1771= 0.8377
3.
A solution of A is mixed with an equal volume of a solution of B containing the same
number of moles (initial concentrations ao and bo are equal) and the reaction
A+BC
occurs. At the end of 1 h, A is 75% reacted.
How much of A (in % of the initial concentration ao) will be left unreacted at the end of 2
h, if
(a) the reaction is first order in A and zero order in B.
a
t
d[A]
d[A]
= - k[A] _ 
= - k  dt
[A]
dt
0
ao
ln
a
= - kt
ao
after 1 h : a = 25% ao = 0.25 ao
75% of ao has reacted, i.e. 25% of ao remains unreacted:
ln 0.25 = - k  1 h _ k = -
a
ao
= e- kt _
a(2 h)
ao
ln 0.25
= 1.386 h-1
1h
-1
= e-1.386 h
2h
= 0.0625
_ a = 0.0625 ao = 6.25% ao unreacted after 2 h
(b) the reaction is both first order in A and first order in B.
d[A]
= - k[A][B]= - k[A ] 2 cause ao = bo thus a = b at all times
dt
a
t
1 1
1 1
d[A]
=
k
dt
_
(
)
=
kt
;
kt
=
 [A ] 2 0
a ao
a ao
ao
_ k 1h=
_
3
1
1
1
3
- _ k=
(4 - 1) =
0.25 ao ao
ha o
ha o
 2 h=
ha o
1 1 6 1 1 1 7
- _ = - _ =
a ao ao a ao a ao
1
_ a = ao = 0.143 a o = 14.3% ao
7
4.
Acetaldehyde decomposes thermally and the main products are methane and carbon
monoxide. A likely mechanism is:
CH3CHO  CH3 + CHO (rate constant k1)
(1)
CH3 + CH3CHO  CH4 + CH3CO (rate constant k2)
(2)
CH3CO  CO + CH3 (rate constant k3)
(3)
CH3 + CH3  C2H6 (rate constant k4)
(4)
Derive the expression for the rate of formation of CO:
d[CO]
k1
3/2
= k2
[ CH 3 CHO ]
dt
2 k4
(To simplify the steady state treatment, further reactions of the radical CHO have been
omitted and their rate equations may be ignored)
Solution:
d[CO]
v=
= k 3 [ CH 3 CO]
dt
CH3CO is an intermediate with unknown, small and constant concentration.
It must be obtained from the steady state approximation for the two intermediates CH3 and CH3CO
in terms of reactant and product concentrations and rate constants.
The reactant is CH3CHO, main products are CH4 and CO, C2H6 is a by-product, CHO is also an
intermediate radical, but its reactions are neglected:
(A) :
d[ CH 3 ]
= 0 = k 1 [ CH 3 CHO] - k 2 [ CH 3 ][ CH 3 CHO] dt
- 2 k 4 [ CH 3 ] 2 + k 3 [ CH 3 CO]
(B) :
d[ CH 3 CO]
= 0 = k 2 [ CH 3 ][ CH 3 CHO] - k 3 [ CH 3 CO]
dt
Note that, looking at reaction (4) alone, we get the following rate term:
1
2
=
v4
vCH 3 ,4 = - k 4 [ CH 3 ]
2
_ vCH 3 ,4 = - 2 k 4 [ CH 3 ] 2
Note, that the terms at k2 and k3 are with opposite signs in both equations. Thus when we
add the equations we get 0 on the left hand side and those terms cancel out:
(A) + (B) : 0 = k 1 [ CH 3 CHO] - 2 k 4 [ CH 3 ] 2
_ [ CH 3 ] =
into (B) : k 2
k1
[ CH 3 CHO]
2 k4
k1
[ CH 3 CHO [ CH 3 CHO] - k 3 [ CH 3 CO] = 0
2 k4
_ [ CH 3 CO] =
k 2 k1
[ CH 3 CHO ] 3/2
k3 2 k4
v = k 3 [ CH 3 CO] = k 2
k1
[ CH 3 CHO ] 3/2
2 k4
5.(A) The density of liquid mercury at 273K is 13.6 g cm-3 and the surface tension is 0.47 N m-1.
If the contact angle is 140o, calculate the capillary depression in a tube of 1 mm diameter.
Solution:
=
rhg
2 cos 
h=
2 cos 
rg
2  0.47 kg s - 2  cos( 140o )
=
0.5  10-3 m  13.6 ( 10-3 kg)( 10- 2 m )-3  9.81 m s - 2
= - 1.08  10- 2 m = - 10.8 mm
(B)
If a molecule dissociates on being adsorbed, the process is referred to as dissociative
adsorption.
(a)
Derive the Langmuir adsorption isotherm for dissociative adsorption:
=
Solution:
K[ A2 ]
1 + K[ A2 ]
Reaction of 1 A2 with 2 surface sites:
AA
| |
| |
A2 + -S-S-  -S-S-
Thus:
adsorption : va = k a [ A2 ](1 -  )2 fraction of free sites squared
_desorption : vd = k d 2 fraction of occupied sites squared
In equilibrium the rates are equal:
2
2
v a = v d _ k a [ A2 ](1 -  ) = k d 

ka
ka
=
[
]
=
K[
]
;
K
=
A2
A2
2
(1 -  ) k d
kd
2

= K[ A2 ] ;  = K[ A2 ] -  K[ A2 ]
1- 
_(1 + K[ A2 ] ) = K[ A2 ] ;  =
(b)
K[ A2 ]
1 + K[ A2 ]
What is the expression for the rate of a reaction, assuming that the mechanism of the
reaction is unimolecular.
Solution:
unimolecular reactions on surfaces after
dissociation:
k K[ A2 ]
v = k =
1 + K[ A2 ]
(c)
Sketch in the following graph the variation of
the rate with [A2]1/2.
Solution:
(d)
What
the reaction,
would be the order of
when
[A2] is
(i)
low
Solution:
1
K[ A2 ] » 1 _ v = k K[ A2 ] : order
2
(ii)
high
Solution:
k K[ A2 ]
K[ A2 ] « 1 _ v =
= k : order 0
K[ A2 ]
6.
(a)
In a normal adult at rest the average speed of flow of blood through the aorta is 0.33
m s-1. The radius of the aorta is 9 mm and the viscosity of blood at body temperature
, 37oC, is about 4.0 x 10-3 kg m-1 s-1.
(i) Calculate the rate of flow of blood through the aorta.
Solution:
area : A =  r 2 = 3.1416  (9  10 -3 )2 m2
= 2.54  10 -4 m2
rate of volume flow = average speed  area A
_
3
m
dV
-4
2
-5 m
= 0.33  2.54  10 m = 8.40  10
=
s
s
dt
(ii) Calculate the pressure drop along 0.5 m length of the aorta.
Solution:
Pouisseuille equation:
dV  r 4 P
8l dV
=
; P = 4
dt
8l
 r dt
_
kg
 0.5 m
3
-5 m
m
s
P =
8.4  10
-3 4
4
s
3.1416  (9  10 ) m
8  4.0  10-3
= 65.2 Pa 
(b)
1 Torr
= 0.49 Torr
133.322 Pa
The diffusion coefficient for glucose in water is 6.81 x 10-10 m2 s-1 at 25oC. The
viscosity of water at 25oC is 8.937 x 10-4 kg m-1 s-1 and the density of glucose is 1.55
g cm-3. Assume the glucose molecule to be spherical and to obey Stokes's law.
(i) Estimate the radius of a glucose molecule.
Solution:
S tokess law : D =
kBT
k T
_ r= B
6r
6D
J
 298.15 K
K
r=
2
kg
-10 m
6  3.1416  6.81  10
 8.937  10-4
s
ms
1.381  10- 23
= 0.359  10-9
Jm
= 0.359 nm
2
kg m
s
2
(ii) Estimate the molar mass of glucose.
Solution:
The radius of a spherical molecule gives the volume of 1 molecule:
4
4
V =  r 3 =  3.1416 (3.59  10-9 )3 m3 = 1.938  10- 28 m3
3
3
The density and the volume give the mass of 1 molecule:
-3
10 kg
m =   V = 1.55 
 1.938  10-28 m3 = 3.00  10-25 kg
3
-2
( 10 m )
The mass of 1 molecule and Avogadro's number give the molar mass MM:
MM = m  N A = 3.00  10-22 g  6.022  1023 mol-1 = 180.7 g mol-1
Transport numbers
to obtain values of mobility u and of ionic molar conductivities λ for single, individual ions
t±: transport, transference or migration numbers
t+ for a cation; t- for an anion
t+ is the fraction of current, I+, carried by the cation
total current: I = I+ + I-:
I +_
t+_ =
I+ + I t± means t+ or t-, and I± means the corresponding I+ or I-:
t+ =
I+ ; = I tI+ + I I+ + I -
_I = I + + I - ; thus t+ + t - = 1
and 0  t+_  1
For a general electrolyte
z++
a
A
|z-|b
B
_ a Az++ (aq) + b B|z-|- (aq)
the electro-neutrality condition must hold:
a  z+ = b  | z - |
in dissociation equilibrium for n initial mol of a weak electrolyte we will have in solution
(1-α)n undissociated molecules
naα cations and nbα anions
transport numbers can be measured for each ion separately
Now, what is their relation to mobilities and ionic molar conductivities?
F is the charge on 1 mol of charge, so a mol of a cation with charge z+ have a charge of
Q+ = aFz+ (for 1 mol AaBb) and
Q- = bFz- (for 1 mol AaBb)
So the electricity flowing through a given area A in unit time is aFz+u+ for 1 mol AaBb in solution
and the negative ion current is bFz-u-, thus
t+ =
t- =
aF z+ u+
a z + u+
=
aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u _
bF | z - | u b | z- | u=
aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u -
and thus t+ + t- = 1.
Since az+ = b|z-|, we can replace b|z-| in the denominator of t+ with az+, and az+ in the denominator
of t- with b|z-|:
a z + u+
u+
=
t+ =
a z + u+ + a z + u - u+ + u -
t- =
b | z- | uu=
b | z - | u+ + b | z - | u - u + + u -
With the short hand writing
  ( Aa Bb ) ; +  ( Az++ ) ;  -  ( B|z-|- )
we know from before
 + = F z + u+ ; u + =
 - = F | z- | u- ; u- =
+
F z+
F | z- |
 = a  + + b  - = F(a z+ u+ + b | z - | u - )
and therefore
+
t+ =
F z+
u+
=
+ +  u+ + u F z+ F | z - |
t- =
F | z- |
u=
- + u+ + u F z+ F | z - |
Now we can cancel F and since az+ = b|z-| we can replace
b
1 a 1
a
1 b 1
=
|
|
;
=
;
|
|
=
;
=
z+
zzz+
a
b
| z - | a z+
z+ b | z - |
and thus
+
t+ =
+ = a +
z+
=
b
+ + b  a + + b  + +  a
z+ a z +
-
t- =
| z- |
a +
+ b | z- | | z- |
=
a
+ +  b
=
b a + + b  -
Since (see before)
 = a + + b  -
thus t+ =
a +
b
; t- = 

For any strong electrolyte, t+ and t- can be measured for different values of c+ = ac and
c- = bc and then extrapolated to t+o and t-o.
Since t+o and t-o can be measured individually, we can split Λo into
 o ; o=  o
=

t+  ta
b
o
o
+
For NaCl e.g. a = b = 1 and z+ = |z-| = 1 but
for MgO e.g. a = b = 1 but z+ = |z-| = 2
Methods to measure t+ and tHittorf method
o
After the anode and
connected, a small current
for a period of time.
cathode
solutions
are
is sent through the Hittorf cell
Then the connection is
closed, and the anode and
cathode solutions after electrolysis can be collected separately and undergo chemical analysis,
which yields the concentration changes.
For simplicity let us assume M+A- electrolytes (can be done for MaAb ones also, but one has to care
then for a and b)
cation current fraction: t+
anion current fraction: tAfter passing of 1 mol electrons through the cell, F charges have passed through
thus
Ft+ charges are transported to the - electrode (cathode) by M+ ions and
Ft- charges are transported to the + electrode (anode) by A- ions
and because 1 mol of charge has passed:
1 mol of M+ was discharged at the cathode and
1 mol of A- was discharged at the anode
Cathode solution
By passage of 1 mol of electrons = F charges
t- mol of A- ions are lost by transport to the anode
1 mol M+ ions are discharged and thus lost
t+ mol M+ ions are transported into the cathode solution:
t+ mol M+ ions are gained = -t+ mol M+ ions are lost
together (1 - t+) mol M+ ions are lost = t- mol M+ ions are lost
thus t- mol MA are lost in the cathode solution
Anode solution
By passage of 1 mol of electrons = F charges
t+ mol of M+ ions are lost by transport to the cathode
1 mol A- ions are discharged and thus lost
t- mol A- ions are transported into the anode solution:
t- mol A- ions are gained = -t- mol A- ions are lost
together (1 - t-) mol A- ions are lost = t+ mol A- ions are lost
thus t+ mol MA are lost in the anode solution:
amount MA lost in anode solution t+
=
amount MA lost in cathode solution t -
=
1 - tt+
=
1 - t+
t-