Fig. 1: Ion transport across an area DA Fig. 2: Ionic migrations within a Daniel cell
The Story of Ionic
Conduction of Electricity
(May download
in printer-friendly Microsoft Word Form)
by Rituraj
Kalita
at Dept. of Chemistry, Cotton College, Guwahati (India)
1. Electrical conduction in response to an electric field
Whenever there arises an electric field,
as a result there always occurs a flow of electricity, which flow may be in the
form of either electronic conduction (associated with flow of electrons, as
happens in metals and semiconductors) or ionic conduction (associated with flow
of ions, i.e. of cations and anions, as happens in aqueous electrolyte solutions
and also in fused salts). The electric field is measured in terms of its electric
field strength E, which equals –∂f/∂x
(the more general mathematical expression is –Ñ.f
or –Gradf),
where f
is the electric potential and x is displacement in the direction
along which f
changes in the fastest rate (the negative sign indicates that direction of E is
opposite to the one along which f
increases). The familiar potential difference V (= f2–f1,
expressed in volt unit) between two parallel, identically shaped electrodes
placed within the liquid electrolyte medium (you should note that the electric
potential would change in the fastest rate in the direction perpendicular to the
electrodes) is a integral of E with respect to x (i.e., |V| = ∫E
dx), which, assuming the uniformity of field strength E within the two
electrodes, means that E equals* V/ l
(thus E is in Volt cm–1 or in Volt m–1 unit).
The current density j is the electrical current per unit area
along the (aforesaid) x-direction: it equals the current I divided by the area A
(measured perpendicular to the direction of current) through which the current
is passing, and is expressible in A cm–2 (or in A m–2)
unit. The current itself (in Ampere, A = C s–1 unit) is measured in
terms of electric charge (in Coulombs) passing per unit time through the
cross-sectional area of the conducting material (perpendicular to the direction
of current).
* The direction of E should be carefully chosen as the one along
which the potential difference V falls: we may ignore this distinction
here.
The most fundamental form of the Ohm's law
states a proportionality between the electrical field strength E and the
resulting current density j, i.e., j = kE or, in
other words, rj = E where k
is the specific conductivity and r is the specific
resistivity of the material (obviously k =
1/r), both properties dependent on the temperature.
As V = E l and
j = I/A, we get I/A = kV/ l,
implying that I = (kA / l)V
and V = (rl /A)
I
{where (kA / l)
is the conductance C (measured in S, i.e. Siemens = mho) and (rl
/A) is the resistance R}, which are more familiar forms of the
Ohm's law. For ionic conduction through an electrolyte medium, we consider the
passage of electricity through the region within the two parallel electrodes, so
that in such cases the area A is the area of either electrode, and l
is the distance between the two electrodes. For a given conductometric cell, the
electrode area A and their fixed distance l
are constants, so that (l/A) is a
constant known as the cell constant k,
further implying that (k /k)
= C (i.e., k
= Ck
).
2. Ionic conduction of electricity
An electrolyte medium, whether a fused
(molten) salt, or a solution of an electrolyte in water (or in another polar
liquid, say, in ammonia), invariably contains ions (both cations and anions --
the electrolyte system is always electrically neutral) that can carry
electricity. Upon application of the electric field (of strength E = V/ l)
between the two electrodes, the ions get forced by the electric field by a force
qiE along the electric field, where qi = zie is the electrical charge*
of each (i-th) ion (zi, +2 for Ca2+ or –1 for Cl–,
is the ionic charge in atomic unit), and starts getting accelerated in its
direction. Without an applied electric field, the ions keep moving randomly in zigzag
motions along all possible directions, similar to motions of gas molecules in an
ideal gas or to Brownian motions of colloidal particles. However, upon
application of the electric field the each type of ion soon show a net overall
velocity (it is, obviously, in opposite directions for the cations and the
anions) superimposed over such random motions (it would be nice if we had a
movie file depicting this -- try to imagine one). This net velocity, also called
the (terminal) drift speed vi,
is observed to be a constant for each (i-th) type of ions (say, for Ca2+)
within the given solution under a given electric field as well as temperature,
as the net motion for each ion get opposed by the viscous force resisting that
motion, so that a terminal net velocity is soon established for each ion (as in
the case of the familiar Stokes' law experiment of a small metal ball falling
through a long vertical column of viscous liquid), and that terminal net
velocity is observable as its drift speed
vi.
* Note: e, the atomic unit of charge, is actually the charge of a
proton (not of an electron), i.e., 1.602 x 10–19
C.
It may be easily shown that the ionic
drift speed should be proportional to the electric
field strength E. The electrical force on
the ion is qiE = zieE, which gets balanced by a viscous force that may be
approximated as the Stokes' law value (6phrivi).
Thus we find that vi = {(zie)/(6phri)}.E,
implying that vi is proportional to E (as the charge zie,
the viscosity coefficient h of the medium, and the
ionic radius ri are all constants for a given ion within a given
solution at a given temperature). The proportionality constant
ui
= vi /E {= (zie)/(6phri)}is
called the ionic mobility for the i-th (type of) ion: we thus have vi
= ui E.
Let us now relate the electrical current
with the drift speeds of the ions. For simplicity, let us assume a
electrolyte medium with only one type of cation and anion each (e.g., an aqueous
NaCl solution containing only Na+ and Cl– ions). Let v+
& v– be their respective drift speeds*, z+ &
z– be their ionic charges (in atomic unit), and N+ &
N– be their number densities** (i.e.,
numbers of ions per unit volume). Let an electric field be imposed with the help
of an anode placed at the left and cathode at the right (Fig. 1) corner of the
electrolyte. Perpendicular to the direction of current, let us consider a small
cross-sectional area DA. During an
infinitesimal time interval dt, both cations and anions would cross the said
cross-sectional area in opposite directions carrying their respective electrical
charges, and thus both would contribute to the electrical
current. The number of cations that would cross
this area in time dt is N+(DA.v+dt),
carrying electrical charges N+(DA.v+dt).z+e
along, as all the cations contained within the infinitesimal volume element dV =
(DA.v+dt)
lying left of the said area up to a distance dx = v+dt
would cross that area because of their drift speed
(their number is, obviously, N+dV). Similarly, the anions
would carry the charge N–(DA.v–dt).z–e
across the said area in opposite direction, meaning that the total electrical
charge carried is N+(DA.v+dt).z+e – N–(DA.v–dt).z–e
which equals N+(DA.v+dt).|z+|e
+ N–(DA.v–dt).|z–|e,
as z– is intrinsically negative i.e., z–
= – |z–|.
* As they are speeds, not velocities, both v+
& v– are positive even though the drift velocities are
oppositely directed.
** For NaCl, |z+| = |z–|
and N+ = N–. But for electrolytes of different
stoichiometry, these relations may, obviously, differ.
Note: This picture of the ions themselves traveling
through the electrolyte medium carrying current is not exactly valid for the
speedy conduction by hydrogen and hydroxyl ions in aqueous solutions (though
valid for other ions), for which ions there is an alternative pathway of
carrying electric charge (known as Grotthus conductance) that involves solvent water
molecules. Do learn about that!
Fig. 1: Ion transport across an area DA
Fig. 2: Ionic migrations within a Daniel cell
The current through the area is obtainable
by dividing the charge carried by the time dt, and then the current density is
obtainable by dividing this current by the area DA.
This means that the current density j is:
j = N+v+|z+|e
+ N–v–|z–|e
= e (N+v+|z+|
+ N–v–|z–|)
The number densities are related to the moles per unit volume (c+ & c–)
by the obvious relations
(here NA is the Avogadro number): N+
= NA c+ , N–
= NA c–. Care should be taken
to distinguish these moles per unit volume quantities (involving cm3
or m3 unit of volume) from the molarities, which uses liter (litre)
as the unit of volume, and so wouldn't obey the above simple relations with the
number densities. These relations give (note that e.NA =
F, the Faraday i.e., 96485 C mol–1):
j = NA e (c+v+|z+|
+ c–v–|z–|)
= F (c+v+|z+|
+ c–v–|z–|)
Using the relations v+ = u+E and
v– = u–E for the drift
speeds, we get:
j = F.(c+u+|z+|
+ c–u–|z–|).E
Comparing Ohm's law j = kE
stated above, we arrive at a simple expression for the specific conductivity k
for the electrolyte
medium: k = (c+
u+ |z+|
+ c– u– |z–|).F
Introducing the ionic stoichiometric coefficients u+
& u–
for the electrolyte, and noting that for a strong electrolyte such as Al2(SO4)3,
c+ = n+ c and c–
= n– c {for Al2(SO4)3,
n+ = 2 and n–
= 3 whereas z+ = 3 and z– = –2}, we get:
k = (n+ |z+|
u+ + n–
|z–|
u–).c F
We know that k /c = Lm,
the molar conductance (here concentration c involves cm3
or m3 for unit of volume
depending on length units of k and Lm,
and so Lm equals k
/c, without a factor of 1000 arising for c expressed in mol L–1).
Thus we get:
Lm
= (n+ |z+| u+ + n–
|z–|
u–).F , this
relation relating the molar conductance with the ionic mobilities.
The last relation is better expressible
as Lm =
n+ l+
+ n– l–
, where the ionic molar conductances l+
and l–
are defined* as: l+ =
|z+|
u+ F & l–
= |z–|
u– F. Thus the molar conductance of a single
electrolyte is always expressible as a sum of its constituent ionic conductance
contributions. However, the familiar Kohlrausch's law of independent migration
of ions, mathematically expressible as the similar-looking relation Lmo =
n+ l+o
+ n– l–o,
goes a step further by implying that at (yes, only at) infinite dilution the
contribution to Lm
(at a given temperature) from each ion (say, contribution n+ l+o
from the Ca2+ ion) is a definite quantity independent of the identity
of the co-ion (i.e., irrespective of the co-ion being Cl– or NO3–).
* The ionic molar conductances li
are related to the ionic equivalent conductances le,i
(sometimes found tabulated in books) as: li =
|zi|
le,i -- thus the ionic equivalent
conductances of Na+(aq) and Cu2+(aq) are almost the same
in value, whereas the ionic molar conductance of Cu2+(aq) is almost
double that of Na+(aq). Obviously, le,i
= ui F. Sum of the ionic equivalent conductances le+
and le- for the cation and anion of an
electrolyte is called the equivalent conductance Le
of the electrolyte (i.e., le+ + le-
= Le).
Problem: Calculate the drift
speed of Na+ ions in an aqueous solution where the pair of
electrodes placed 2 cm apart bears a potential difference of 2.4 V, at a
temperature for which its ionic molar conductance is 50.2 S cm2 mol–1.
Ans: Here |zi| = 1, so that ui
= li/(1.F) = (50.2
S cm2 mol–1)/(96485 C
mol–1) = 5.203 x 10–4
cm2 s–1 V–1.
(noting that 1 S = 1 A V–1 whereas 1 A = 1 C s–1)
As field strength E (= V/ l)
= 1.2 V cm–1, so drift speed vi = ui E =
6.243 x 10–4 cm
s–1 (= 6.243 mm
s–1, looks so tiny!).
Problem: At 298.15 K, the molar conductance at infinite dilution Lmo
for NaCl, NaNO3 and KCl are 126.5, 121.6
and 149.9 S cm2 mol–1
respectively. Using Kohlrausch's law of independent migration
of ions, calculate Lmo of KNO3.
Ans: From Kohlrausch's law, Lmo (NaCl) = l+o
(Na+) + l–o (Cl–),
Lmo (NaNO3) = l+o
(Na+) + l–o (NO3–),
Lmo
(KCl) = l+o (K+) + l–o
(Cl–) and Lmo
(KNO3) = l+o (K+)
+ l–o (NO3–)
So we have, Lmo
(KNO3) – Lmo (KCl)
= Lmo (NaNO3) – Lmo
(NaCl) = l–o (NO3–)
– l–o (Cl–)
This gives, Lmo
(KNO3) – Lmo (KCl)
= –4.9 S cm2 mol–1
So we have, Lmo
(KNO3) = Lmo (KCl) –
4.9 S cm2 mol–1 =
145.0 S
cm2 mol–1
3. The compulsorily arising interface of ionic conduction with electronic one
Unlike the situation for purely electronic conduction (as in the case of a light bulb connected in series with a direct-current toy dynamo) where electrons move around the conducting wires in continuous looping motion, starting nowhere and ending nowhere, ionic conduction of electricity must begin at one electrode and end at the other one (i.e., conduction by cations begins at the anode and ends at the cathode). This is because the electrode surfaces are the interface at which ionic conduction must hand over the responsibility of conduction of electric charge to electronic conduction, so that the Kirchoff's-law obeying looping motion of net electrical current becomes complete. It's something like the water cycle in atmosphere, with water going up from the ground in its gaseous form, and coming down back from the clouds in its liquid form as rain. Thus, considering the electrolytic refining process of copper using CuSO4 solution, we find that the along the outer electronic circuit, electronic current moves towards the (impure copper) anode, or in other words electrons flow away from the anode (thus in an electrolytic cell, anode is the positive electrode). Similarly electronic current moves away from the (pure copper) cathode, i.e., electrons flows towards the cathode. Completing the loop of electric current, inside the electrolyte solution cations (Cu2+) move from anode to cathode whereas anions (SO42–) move from cathode to anode, implying that the electrical current within the electrolyte moves from anode to cathode.
At the two electrodes, chemical reactions
(called electrode reactions) happen that generates (at anode, an oxidation
reaction happens) or consumes (at cathode, a reduction reaction) electrons. Thus in case
of the above example, at anode the reaction
Cu(s) → Cu2+(aq) + 2e- occurs
(these electrons generated at the anode are the ones that would flow away from
the anode via the outer electronic circuit), whereas at cathode the reaction Cu2+(aq) + 2e- → Cu(s)
happens, consuming the electrons that are flowing to the cathode*. Thus it is at the
pair of electrodes where the electrons take over the responsibility of electrical conduction
for the part of the circuit beyond the electrolyte medium (Fig. 1). Another interesting
set of examples would be those for the galvanic cells. Thus in the case of Daniel cell,
at the anode the reaction Zn(s) → Zn2+(aq) + 2e- happens,
whereas at the cathode Cu2+(aq) + 2e- → Cu(s)
happens. Here within the electrolyte medium, cations move from anode to cathode
and anions from cathode to anode, implying (here also) that ionic current
flow is from anode to cathode (Fig. 2). Completing the
Kirchoff's-law current loop, electronic current flows (here also) from cathode to
anode (the electrons themselves, however, moving from anode to cathode) via the
outer, electronic part of the circuit. This means that in the galvanic cell,
cathode (with lack of electrons, due to reduction occurring) is the positive
electrode and anode (with excess of electrons) is the negative one, or in other
words the polarity of the electrodes are opposite in galvanic cells compared to
electrolytic cells. However, in both kinds of cells, anode is the electrode at
which oxidation occurs and cathode is the electrode at which reduction occurs.
* The amount of ions discharged at an electrode is given by the Faraday's
laws, implying that for passage of Q amount (in Coulombs) of electricity
(i.e., electric charge) Q/F equivalent i.e., Q/(|zi|.F) mole of ions
will get discharged. This quantitative relation arises because transfer of
charge Q implies consumption or generation of Q/e number, i.e., Q/(NAe)
= Q/F mole of electrons, thus finally implying that Q/(|zi|.F) mole
of ions will get converted to neutral atoms or molecules (i.e., discharged).
Are you wondering how, within the galvanic cell, the cations paradoxically move from the negative electrode (anode) to the positive electrode (cathode)? [Something similarly paradoxical happens about for the anions also!] In Section 7, we'll come across the answer to this enigma.
One interesting, and important, point about electrode reactions is that the kind of ion that's rushing towards an electrode, carrying the current, may not actually take part (i.e., get liberated) in the electrode reaction. Thus in both of the examples above, it is the SO42– ions that flow to the anode, but in case of the Cu-refining process it is Cu atoms that reacts to become Cu2+ ions at the anode (in Daniel cell, it is Zn atoms that become Zn2+ ions at the anode). The reader probably knows that electrolysis of an aqueous H2SO4 solution using Pt electrodes, it is (mainly) the SO42– ions that rush towards the anode, but at the anode water molecules react to give oxygen gas {the anode reaction is 2H2O(l) → 4H+(aq) + O2(g) + 4e- -- may better rewrite this reaction using the hydronium (H3O+) ion formalism}. On the other hand, during electrolysis of an aqueous KNO3 solution using Pt electrodes, even at the cathode the K+ cations rushing towards it do not react, it is H2O that reacts to liberate H2 {the cathode reaction is thus 2H2O(l) + 2e- → H2(g) + 2OH–(aq) -- do guess what is the anode reaction!}. A useful rule of thumb is that an electrode-reaction product that is unstable in the electrolytic medium (e.g., elemental K during electrolysis of aqueous KNO3) do not gets liberated [it is (most generally) wrong to think that such an unstable product first gets liberated and then reacts with the medium -- rather the case is that the corresponding ion (e.g., K+) doesn't react at all and that some other species (e.g., water or H+ ion) reacts instead].
Because of occurrence of
transport of ions limited within the region between the two electrodes, ionic
conduction is invariably associated with transport of matter, unlike the case of
electronic current. This matter transport, in association with the particular
electrode reactions that are occurring, causes concentration changes (in some
cases even chemical changes) at the regions near the two electrodes. Thus, in
the
electrolytic refining example, the concentration
of
CuSO4 increases in the region near the anode (due to generation of Cu2+ ions therein and inflow of SO42– ions)
and decreases at the vicinity of the cathode (why?). On the other hand, during
electrolysis of aqueous KNO3 using Pt
electrodes, the vicinity of the cathode becomes increasingly alkaline (turns
into a mixture of aqueous KNO3 and KOH) whereas
the vicinity of the anode becomes increasingly acidic* (turns into a mixture of aqueous KNO3 and HNO3)
-- do find out why! Such concentration changes and/or chemical changes are
measured in the Hittorf method for determination of the ionic transport
numbers, discussed in the next section.
* Such changes are distinctly observed in some
electrolytic cells shaped so as to prevent diffusion of ions, such as in the
Hittorf's apparatus.
4. Electricity carried in part both by the
anions and the cations
"What brings in
September? The cloud roars to say it's him. Weeping, the rain says it's her. The
kadam plant, smiling, says it's both of them." -- A popular Assamese
song
While deriving the expression j = (c+v+|z+|
+ c–v–|z–|)F in Section 2 about current
density j of an electrolyte solution, we find that part of the current density (c+
|z+ | v+ F) is due to the transport of
electric current by the cations, whereas the rest part
(c– |z– | v– F) is due to that
by the anions. Thus both the cations and anions generated from an electrolyte
contribute to the conduction of electricity, and to different extents in
general. Thus, for an aqueous 0.1 M CuSO4
solution
at room temperature, around 0.37 fraction (i.e., 37%) of the current gets
carried by the migration of Cu2+ cation, while the rest 0.63 fraction
(63%) gets carried by the migration of SO42–
anions (in this case, only these two kinds of ions are there in appreciable
concentration).
The transport number ti of the i-th kind of ion is defined
as the fraction of total current carried by the ions of that kind. Thus, for the
cations and anions of a single electrolyte in solution (as in the example just
above), the transport numbers t+
or t– are as follows (obviously, here t+
+ t– = 1 -- verify this yourself from the following expression):
t+
= c+ |z+ | v+ F /
(c+ |z+ | v+ F + c– |z–
| v– F)
t– = c– |z–
| v– F / (c+ |z+ | v+ F + c–
|z– | v– F)
For this case, assuming the electrolyte to be a strong (i.e., completely
ionizing) one, we find
c+ = n+ c and c–
= n– c as in Section 2, so that
t+
= n+
|z+ | v+ / (n+
|z+ | v+ + n–
|z– | v– ) and
t– = n– |z–
| v– / (n+
|z+ | v+ + n–
|z– | v– ).
For electroneutrality of the single electrolyte compound, the relation n+
|z+ | = n–
|z– | {e.g., 2x3 = 3x2 in case of Al2(SO4)3}
must however be satisfied. This finally
gives:
t+
= v+ / ( v+ + v–
) and t– =
v– /
( v+ + v–
). As the drift speeds are proportional to the ionic mobilities, we
have
v+ = u+E and
v– = u–E. This gives (check how!) another equivalent relation for
the two transport numbers:
t+
= u+ / ( u+ + u–
) and t– = u– /
( u+ + u–
). For a solution containing several dissolved electrolytes
(e.g., KNO3 and
CuSO4 simultaneously kept dissolved in water),
there are more than two kinds of ions with a transport number for each ion, and
so neither of the above two relations remain valid {the relation ti
= ci |zi | vi F / (Si
ci |zi | vi F), however, remains valid}.
As the ionic molar conductances are related to the
mobilities as l+ =
|z+|
u+ F and l–
= |z–|
u– F, the following relation is also valid only for the
special case of
|z+|
= |z–| (i.e., for electrolytes such as HCl or
CuSO4, but not for CaCl2): t+ = l+
/ ( l+
+ l– )
and t– = l– / (
l+
+ l– ). [However, for the ionic
equivalent conductances le+ and le-
(obeying le–
= u– F and le+
= u+ F) the corresponding relations are t+ = le+
/(le+ + le–)
and t– = le–
/ (le+ + le–),
obeyed for any electrolyte (e.g., even for CaCl2).] For electrolytes obeying n+
= n–
= 1 (e.g., HCl,
CuSO4 etc.), this simply means l+
= t+ Lm
and l– = t– Lm
(and for all electrolytes, le+ = t+ Le
and le– = t– Le).
Fig. 3: Hittorf's apparatus for transport number determination
In the Hittorf's apparatus (Fig. 3) used for determination of transport numbers, there is a peculiarly shaped electrolytic cell made of long glass tubes with a vertical anode compartment, a U-shaped central compartment and a vertical cathode compartment with these three compartments joined via glass tubes at top part (the joining tubes are fitted with stopcocks). The shape of the apparatus minimizes intermixing of solutions in the three compartments via diffusion. This electrolytic cell is connected to a direct current source (a battery) via a Coulometer and an ammeter in series (as in figure). The electrolytic cell is fitted with electrodes of particular materials depending on the electrolyte, then filled with the desired electrolyte solution, and then a controlled current (monitored with ammeter) is allowed to pass through it for a suitable duration of time (say for several minutes) to perform electrolysis. The net charge passing through the cell is measured using the Coulometer. The concentration of the electrolyte solution within the cathode and the anode compartments is measured before and after electrolysis by methods such as chemical titration.
The exact calculations that would give the
transport numbers from the measured concentration changes within
the cathode and the anode compartments depend on the electrolyte solution and
the electrode materials used. As an example, the transport numbers t+
and t– of Cu2+ and SO42–
ions in a
CuSO4 solution may be determined using cathode and anode made of
copper (resembles the above electrolytic refining example, isn't
it?). In this case, for charge Q (say, 645.6 C)
passing through the cell (and measured by the Coulometer), Q/(|z+|F)
mole of Cu2+ ions have been discharged
at the cathode, and so are lost from the cathode compartment. However, t+
fraction of the total charge carried Q were carried by the cations, which means
that t+Q/(|z+|F)
mole of Cu2+ ions moved into the
cathode compartment (from the central compartment), partially compensating the
aforesaid loss. Thus the decrease in amount of Cu2+ ions from the cathode
compartment is {Q/(|z+|F) – t+Q/(|z+|F)
= (1–t+)Q/(|z+|F)
= t–Q/(|z+|F)
mol}. Now, t–
fraction of the total charge were carried by the SO42–
anions, implying that t–Q/(|z–|F)
mol of SO42– ions also moved away from the cathode
compartment (towards the central compartment). It may be noted that the charge of the
cations and anions decreasing or increasing from any compartment are equal and
opposite {z+F.t–Q/(|z+|F) = t–Q
and z–F.t–Q/(|z–|F) =
– t–Q respectively}, so that electroneutrality of the solution in
any compartment remains preserved -- this is a basic requirement in
electrochemistry. As |z+|
= |z–| = 2, this together means that t–Q/(2F)
moles of CuSO4 decrease from the cathode
compartment. By a similar argument, it can be shown that t–Q/(2F)
moles of CuSO4 increase {yes, t–Q/(2F)
moles, not t+Q/(2F) moles -- please find out how!}
in the anode compartment. Now by measuring the
concentration changes during electrolysis, and multiplying that with the
solution volume of the respective compartment, the change (decrease or increase)
in the amount of CuSO4 at the cathode or
at the anode compartment may be calculated. As this
change equals t–Q/(2F)
moles in this case involving CuSO4,
so by knowing Q from the Coulometer reading, the transport numbers t–
and t+ = 1 – t–
may be easily calculated (see problem below).
Problem: During electrolysis of aqueous
CuSO4
solution using copper electrodes within a Hittorf's apparatus, the concentration
of CuSO4 at
the anode compartment changed from 0.102 M to 0.112 M for passage of 645.6 C of
electricity. If the volume of solution at the anode compartment was 105 mL,
calculate the transport numbers of the Cu2+ and SO42–
ions in the CuSO4 solution.
Ans.: Initial amount of
CuSO4 = 0.102 mol/L x (105/1000) L = 0.01071
mol. Final amount = 0.01176 mol. Increase in amount = 0.00105 mol.
As here t–Q/(2F)
= 0.00105 mol, so t–Q
= 0.00105 mol x (2 x 96485 C mol–1)
= 202.6185 C
As total charge passed Q = 645.6 C, so t–
= 202.6185 C / 645.6 C = 0.314. So, t+
= 1 – t– = 0.686 (Huh, imaginary values for CuSO4!)
5. Interrelation of ionic mobilities and diffusion
coefficients
The aforesaid electrical mobility ui
of the ion i, as well as its mechanical mobility wi
(the ratio of its drift velocity vi to the applied force fi
on it, i.e., vi/fi), are related to its diffusion
coefficient or diffusivity Di. The diffusivity Di of an
i-th species (whether an ion or a neutral molecule) expresses the constant of
proportionality of the diffusion flux Ji (the diffusion flux is
expressed in moles per unit area per unit time) for the species in response to
its concentration gradient ∂ci/∂x
(where
ci
is its molar concentration, in moles per
unit volume), the direction of diffusion being opposite
to this concentration gradient, as:
Ji
= –Di ∂ci/∂x
-- It is just the Fick's law of diffusion.
The Einstein relation, also known as the Einstein-Smoluchowski relation,
states that the diffusivity Di is
related to the mechanical mobility wi
as:
Di = wikT , where k is the Boltzmann constant and T the absolute (i.e.,
Kelvin-scale) temperature.
As the applied force fi
on an ion equals qiE
= zieE, so
we get wi
= vi/fi
= vi/(qiE)
= (vi/E)/qi
= ui/qi,
implying
ui = qi.wi
which is the relation relating the mechanical and the electrical mobilities of
an ion. Thus we get
Di = uikT/qi , which is the Einstein relation for electrical mobility of ions.
Thus we find that just as the electrical ionic
mobility ui is the ratio of its drift velocity vi to
the applied electric field strength E (i.e., ui
= vi/E),
the mechanical mobility wi
(of an ion or a
molecule) equals vi/fi
(i.e., wi
= vi/fi)
-- wi
is reverse of the drag coefficient gi
(i.e., wi
= 1/gi
implying gi = fi/vi).
However, for streamline flow involving nearly-spherical particles, the Stoke's
law is applicable, which gives fi = 6phrivi
implying that gi = 6phri
and wi = 1/(6phri).
This gives the following relation (the Stokes-Einstein relation):
Di = kT/(6phri)
.
This relation is useful for estimating the
radius ri of (the
nearly-spherical shaped) globular proteins. If there is a good estimate for the
radius ri of a spherically shaped ion, then its diffusion coefficient
Di may also be estimated using this relation.
Problem: Within a
non-uniform colloidal emulsion of spherical protein molecules each having
diameter 6.2 nm, there remains a concentration gradient of 0.00012 M cm–1
along one direction regarding the molar concentration of the protein. At
298 K temperature, calculate the rate of diffusion of these protein molecules
along that direction (at 298 K, h for water or for
dilute aqueous solutions is 8.91 x 10–4
Pa s).
Ans.: Radius ri of the spherically shaped ion = 6.2/2 nm = 3.1 nm =
3.1 x 10–9 m.
At 298 K, its
diffusion coefficient Di = (1.381 x 10–23
J K–1
x 298 K)/(6 x 3.1416 x 8.91 x 10–4
x 3.1 x 10–9 Pa
s m) = 7.90 x 10–11 m2 s–1 (= 7.90 x 10–7
cm2 s–1)
Conc. Gradient ∂ci/∂x
= 0.00012 mol dm–3
cm–1
= 0.00012 mol (10–1
m)–3
(10–2
m)–1
= 0.00012 x 105 mol m–4
= 12 mol
m–4.
So, magnitude of rate of diffusion, |J| = Di.|∂ci/∂x|
= 7.90 x 10–11 m2 s–1
x 12
mol m–4
= 9.48 x 10–10 mol
m–2
s–1
6. The dependence of ionic
mobilities on the ion concentrations
The electrical ionic mobility ui
(and also the ionic molar conductance li =
|zi| ui F) of a particular type of ion (say, of Na+) within an electrolyte
medium is, however, dependent on the concentrations of various ions within the
medium. This happens because at higher ionic concentrations, interionic interactions become
more, thus increasingly obstructing free passage of ions to carry electricity
through the medium. Thus the ionic mobilities decrease with increasing ionic
concentrations. For dilute solutions of a strong electrolyte, the change in ionic mobilities
ui with electrolyte concentration c approximately obeys a simple expression
(namely ui
= uio –
b.c1/2, where uio
and b are two constants) offered by the Debye-Huckel Theory of interionic
interactions, the basic aspects of which theory are as mentioned below.
Note: You should have noted that uio
is the mobility ui
for ionic concentrations limiting to zero (this condition called as at
infinite dilution), because from the above relation we
get ui
= uio when c = 0,
isn't it? So, uio
is called the limiting ionic mobility at infinite dilution.
According to this theory, any ion j
within an electrolyte solution can be considered to be surrounded, on the
average, by an ionic atmosphere that consists of the same type of ions
and also the ions of opposite charge, with the net charge of the ionic atmosphere
being equal and opposite (naturally, isn't it?) to the charge of the ion j. In
the absence of an external electric field, the ionic atmosphere is spherically
symmetric around the given ion j, and so does not offer any net force on the
ion. However, introduction of an external field leads to retardation of the
moving ion in two distinct ways, both of which are related to the ionic
atmosphere concept.
Upon application of an external electric field, the ion atmosphere no longer
remains symmetric: as the given ion moves in one direction in response to the
field, the ionic atmosphere takes some time (called the time of relaxation) to
rearrange itself spherically around that new position (i.e., to undergo relaxation).
Thus at every moment, the ionic atmosphere around the given ion is actually asymmetric, and so offers a net force on the moving ion, this force being always
in opposite direction to its motion. This retarding effect because of the
actually asymmetric ionic atmosphere is known as the relaxation effect or
the asymmetry effect.
On the other hand, the ions in a solution always remain solvated, i.e.,
surrounded by solvent molecules sticking to it. So, the given ion moving forward
in response to the electric field encounters the solvent molecules attached to
the ions of the ionic atmosphere which is, as a whole, oppositely charged and so
are moving, on the average, in the opposite direction. Thus the given ion need
to move across a number of nearby solvent molecules moving, on the average, in
the opposite direction. Because of this the ion's movement gets further
retarded: this retarding effect is known as the electrophoretic effect.
Because of both the aforesaid effects,
the mobility of an ion gets lowered with increasing ionic concentration. Higher
is the ionic (i.e., electrolyte) concentration, denser is the ionic atmosphere,
and stronger are the retarding effects. Thus we see that the lowering of ionic
mobility (i.e., uio – ui
= b.c1/2, upon
increase of ionic concentration from zero is proportional to the square root of
the molar electrolyte concentration c (b being a
constant).
Converting the ionic mobilities u+ and u–
of both ions to the ionic equivalent conductances le+
and le–
(using le,i = ui F) we get the
familiar expression in terms of equivalent conductance Le
of the electrolyte, known as the Debye-Huckel-Onsager relation: Le
= Leo – b.c1/2.
For solutions containing multiple dissolved electrolytes (e.g., NaCl, KNO3
and ZnSO4 together), concentration c must be replaced with the ionic
strength I = 0.5 Si ci zi2
(zi being the ionic charge as before), giving
Le
= Leo – b.I0.5.
It has been shown by Onsager that b = q Leo
+ s, where the values of q
and s depends on the temperature, dielectric
constant, ionic charges, the individual ionic conductances (only in case of q)
and the coefficient of viscocity h of the medium
(only in case of s). The part q Leo
of the constant b arises from the relaxation effect, whereas the other part s
arises from the electrophoretic effect. For the simplest case of a 1-1
electrolyte (e.g., NaCl or KNO3) dissolved in water at a low
concentration, we have, at 298.15 K, q = 0.2273 (for
1-1 electrolytes, q is independent of the individual
ionic conductances) and s
= 59.78 (both values are obtained considering S
cm2 equiv–1 unit everywhere).
Problem: At 298.15 K, what'll
be the equivalent conductance of KCl in an aqueous solution containing 0.01 M
KCl and 0.01 M CaCl2 dissolved in it (given that the equivalent
conductance at infinite dilution, Leo
for KCl is 149.9 S cm2 equiv–1 at 298.15 K).
Ans.: Here constant b = 0.2273 x 149.9 + 59.78 = 93.85.
As [K+] = 0.01, [Cl–] = 0.01 + 0.01 x 2 = 0.03,
[Ca+] = 0.01, the ionic strength I = 0.5 x [ 0.01 x 12 +
0.03 x 12 + 0.01 x 22 ] = 0.04
So the equivalent conductance of KCl, Le
= Leo – b.I0.5 =
131.1 S cm2 equiv–1
7. How do the ions apparently migrate upstream within a
galvanic cell?
As we found in Section 3, for a galvanic cell the anode is the negative electrode and the cathode is the positive one, yet the positively charged cations migrate towards the cathode (i.e., in the Daniel cell, the Zn2+ cations generated at the anode migrate away from it whereas the Cu2+ cations migrate towards the cathode to get discharged there) and the negatively charged anions migrate towards the anode (as does the SO42– anions within the Daniel cell). It seems paradoxical why the ions would migrate in a direction opposite to the electrical polarity of the two electrodes. True, the two electrode reactions that goes on in the two electrodes sum up to produce a spontaneous chemical reaction which is the driving force behind the galvanic cell processes, and also these electrode reactions tend to produce imbalance of charges in the vicinity of the electrodes which are best resolved by such migrations of the ions, but these two explanations don't fully explain the causation of such migrations in a galvanic cell. It is something like: the common interests of a market-based society are best served when the peasants, the factory workers, the knowledge workers and the entrepreneurs all would work hard at their productive activities, but that hardly explains why all the people keep working hard at their jobs -- to fully understand that we must also look at the incentives (wages or profit) that lies before each of them.
Thus, the ions within the electrolyte medium in a galvanic cell actually do not move against an electric field, they rather move along an electric field which they experience (i.e., they have just a usual incentive to undergo such migrations). The field that the ions experience is actually oppositely aligned compared to the aforesaid apparent polarity of the two electrodes. This can happen, because the electrode polarities that we observe for a galvanic cell are observed from the electronic conductor (i.e., metallic) side of the electrode: because of the presence of the electrical double layer at each electrode, the electrical potential f within the electrolyte solution a little distance away from the metallic electrode surface is greatly different (say, of the order of a volt different) from that inside the metallic electrode! Thus, in case of the Daniel cell, we say that the Zn-anode has a lower f than the Cu-cathode, but that is the case when we compare the metallic part of the two electrodes. A little distance (equal to width of its electrical double layer) away from the Zn-anode within the aq-ZnSO4 electrolyte, f would have sharply increased (Fig. 4), whereas at a little distance away from the Cu-cathode within the aq-CuSO4 solution f would have sharply decreased, so that within the electrolyte medium, the electrical potential f close to the anode actually becomes more than the f value close to the cathode. This results in the cations moving away from the anode and towards the cathode, and vice-versa for the anions! Fig. 4 shows the schematic variation of the electrical potential f within the electrolytic region (assuming an included salt bridge) and within the electronic conductor regions (assuming a resistive load, such as a heating element, connected in series, as was in the case illustrated in Fig. 2).
Fig. 4: Schematic variation of potential f along
electrolyte media and electronic-conductor load for a Daniel Cell
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