Ion activity, mean activity coefficient
Activity (ai) of a molecule or ion in a solution is referred to as effective concentration
and is related to concentration (Ci) by the simple relationship:
ai   i Ci
Where
 i is known as an activity coefficient which may take different forms
depending on the way in which concentrations for a given system are expressed, i.e.
as morality, molality or mole fraction.
For instance, the chemical potential (  i ) of a species i may be expressed
i  i  RT ln  i Ci
For an ideal solutions or infinite diluted solutions (  i  1  ai  Ci ):
i  i  RT ln Ci
Ion activity
Activities of ionic solutions serve us for calculating
accurate chemical potentials
accurate equilibrium constants
Ionic activities are more important than activities for non-electrolytes. Intermolecular
forces are the greatest between ions.
Ions in aqueous solution
When ionic compounds (1:1) dissolve they are completely dissociated into ions.
For example NaCl:
NaCl( aq)  Na(aq)  Cl(aq)
The chemical potentials for the formation of ions,
μNaCl, aq  μNa  ,aq  μCl  ,aq
Since
o
 NaCl ( aq)   NaCl
( aq)  RT ln a NaCl ( aq)
 Na

( aq )
o
  Na
 RT ln a Na  ( aq)

( aq )
 Cl ( aq)   Clo

( aq )
 RT ln aCl  ( aq)
So,
o
o
 NaCl
( aq )  RT ln a NaCl ( aq )   Na

( aq )
  Clo  ( aq)  RT ln a Na  ( aq)  RT ln aCl  ( aq)
Rearrangement
ln a NaCl ( aq)  ln a Na  ( aq)  ln aCl  ( aq)
ln a NaCl ( aq)  ln  Na  C Na   ln  Cl  CCl   ln  Na   Cl  C Na  CCl 
a NaCl ( aq)   Na   Cl  C Na  CCl 
Since,  Na   Cl    2
C NaCl ( aq)  C Na   CCl 
2
aNaCl ( aq)   2CNaCl
The above equation is applicable for 1:1 electrolyte.
Try MgCl2, i.e. 1:2 electrolyte Innovations
MgCl2 → Mg2+ + 2 Cl−
0
0
 salt
 RT ln asalt   Mg
 RT ln C Mg    2 Cl0  2 RT ln CCl  
2
2


0
0
 salt
 RT ln asalt   Mg
 2 Cl0  RT ln C     C 2 2
2

so
asalt  C C2   2 
But for MgCl2 at molality, m, we know that C+ = C and C− = 2C. Further, we define
the geometric mean activity coefficient by,
 2    3
Then
asalt  C  2C    3  4C 3 3
2
Other ionic compounds are done in a similar manner. After some practice you can
probably figure out the expression for asalt just by looking at the compound. Until
then, or when in doubt, go back to the expressions for chemical potentials as we have
done here. (For practice you might want to try Al2(SO4)3.).
General formula
Absolute activities of cations and anions can not be determined experimentally.
The definition of the mean activity coefficient depends on the number of ions into
which a molecule dissociates when it is dissolved.
In the laboratory it is impossible to study solutions which only contain one kind of
ion. Instead, solutions will have at least one positive and one negative type of ion. For
the generic electrolytic compound dissolving in water:
A x By  xA z   yBz 


a salt  aAx aBy  x x y y C  x  y   x  y 
For MgCl2


a MgCl2  a Mg  aCl2 -  11  2 2 C 3 3  4C 3 3
The General Nernst Equation
The general Nernst equation correlates the Gibb's Free Energy ΔG and the emf
of a chemical system known as the galvanic cell. For the reaction
aA  bB  cC  dD
and
ac ad
Q  Ca Db
a A aB
For an ideal solutions a  Concentration , because the activity coefficients equal to
unity
c
d

C  D 
Q
Aa Bb
It has been shown that
ΔG = ΔG° + R T ln Q
and
o
o
Since G  nFEcell and G  nFEcell
Therefore
o
 nFEcell  nFEcell
 RT ln Q
where R, T, Q and F are the gas constant (8.314 J mol-1 K-1), temperature (in K),
reaction quotient, and Faraday constant (96485 C) respectively. Thus, we have
RT C  D 

ln
nF Aa B b
c
Ecell  E
o
cell
d
This is known as the Nernst equation. The equation allows us to calculate the
cell potential of any galvanic cell for any concentrations. Some examples are
given in the next section to illustrate its application.
It is interesting to note the relationship between equilibrium and the Gibb's
free energy at this point. When a system is at equilibrium, Ecell  0 , and
Q  K e . Therefore, we have,
RT
o
Ecell

ln K e
nF
RT aCc aDd
o
Ecell 
ln
nF a Aa aBb
For an ideal solution
RT C  D 

ln
nF  Aa B b
c
E
o
cell
d
o
Thus, the equilibrium constant and Ecell are related.
The Nernst Equation at 298 K
At any specific temperature, the Nernst equation derived above can be reduced
into a simple form. For example, at the standard condition of 298 K (25°), the
Nernst equation becomes
C  D
0.059
log
n
Aa Bb
c
o
Ecell  Ecell

d
For the cell
Zn | Zn2+ || H+ | H2 | Pt
we have the following half cell reactions:
o
Zn( s )  Zn(2aq )  2e (anode)
eZn
 -0.763
2
/ Zn
2 H (aq)  2e  H 2 ( g ) (cathode)
Zn( s )  2 H

( aq )
2
( aq )
 Zn
 H 2(g )
eHo  / H  0
2
E
o
cell
 0.763
If the concentrations of the ions are not 1.0 M, and the H2 pressure is not
1.0 atm, then the cell potential ΔE may be calculated using the Nernst
equation:
Ecell


 
0.059
Zn 2 P( H 2 )
 0.763 
log
2
Zn H  2
with n = 2 in this case, because the reaction involves 2 electrons. The numerical
value is 0.0592 only when T = 298 K. This constant is temperature dependent.
Note that the reactivity of the solid Zn is taken as 1. If the H2 pressure is 1 atm,
the term P(H2) may also be omitted. The expression for the argument of the log
function follows the same rules as those for the expression of equilibrium
constants and reaction quotients.
Indeed, the argument for the log function is the expression for the
equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, Ecell  0 and the expression becomes an
equilibrium constant K, which bears the following relationship:
o
Ecell

RT
ln K e
nF
o
where Ecell is the difference of standard potentials of the half cells involved.
Problem 1
Calculate Ecell for the cell which is shown in the following diagram
Pt H 2 (1.0 atm) HCl (0.5 M) AgCl Ag
Where the temperature is 25 oC and mean activity coefficients of HCl
is 0.758.
Anodic reaction
H 2 ( g )  2H   2e
Cathodic reaction
2 AgCl(s)  2e  2 Ag  2Cl 
2 AgCl(s)  H 2  2 Ag  2Cl   2 H 
Write Nernst equation
Ecell  E
o
cell
2
a Ag
aCl2  aH2 
0.059

log 2
2
a AgCl aH 2
0.059
log aCl2  a H2 
2
2  0.059
o
Ecell  Ecell

log aCl  a H 
2
o
o
o
2 2
Since aHCl  aCl  aH     CHCl and Ecell  ecathod  eanode  0.223
o
Ecell  Ecell

2
Ecell  0.223  0.059 log  2C HCl
Ecell  0.223  2  0.059 log( 0.758  0.5)
Ecell  0.273 V
Problem 2
The standard cell potential dE° for the reaction at 25 oC
Fe + Zn2+ = Zn + Fe2+
is -0.353 V. If a piece of iron is placed in a 1 M Zn2+ solution, what is the
equilibrium concentration of Fe2+?
Solution
The equilibrium constant K may be calculated using
o
Ecell

RT
ln K e
nF
K e  10
K e  10
o 
 nEcell


 0.059 


o 
 nEcell


 0.059 


K e  1.08 10
12
Fe2
2
 1M
, since K e 
2 and Zn
Zn
Fe2  1.08 10 12
Problem
From the standard cell potentials, calculate the solubility product for the
following reaction:
AgCl = Ag+ + ClSolution
There are Ag+ and AgCl involved in the reaction, and from the table of
standard reduction potentials, you will find:
AgCl + e = Ag + Cl-,
E° = 0.2223 V - - - -(1)
Since this equation does not contain the species Ag+, you need,
Ag+ + e = Ag,
E° = 0.799 V - - - - - - (2)
Subtracting (2) from (1) leads to,
AgCl = Ag+ + Cl- . .
. ΔE° = - 0.577
Let Ksp be the solubility product, and employ the Nernst equation,
o
Ecell

RT
ln K sp
nF
log Ksp = (-0.577) / (0.0592) = -9.75
Ksp = 10-9.75 = 1.8x10-10
This is the value that you have been using in past tutorials. Now, you know that
Ksp is not always measured from its solubility.
Variation of e.m.f with temperature
One of the most rewarding aspects of electrochemistry is that ΔS values spring
directly from studies of the temperature variation of e.m.f.
Equation (V-28), to be derived in Chapter V, describes the temperature variation of G:
 G 

  S
 T  P
For G o one can write
 G o 

  S o
 T  P
Since
G o   nFE o
So
dG o
dE o
 nF
, Then
dT
dT
dE o
nF
 S o
dT
dE o S o

dT
nF
Example
Niobium oxides were studied by use of the cell:
The net reaction:
Nb  NbO2  2NbO
Values of e.m.f. at various temperatures are given in the following table:
Temperature (K)
Eo (mV)
1054
213
1086
212
1106
209
1160
201.5
1174
202
1214
200
1244
198
1283
194
5
-1
The temperature coefficient =  9  10 Volt K
(A) Write the two half cell reactions?
S o for the cell reaction.
o
o
(C) Calculate the average value of G and H for the cell reaction?
(B) Calculate
Solution
(A)
Anodic reaction
Nb  O 2  NbO  2e
Cathodic reaction
NbO2  2e  NbO  O 2
The total reaction is
Nb  NbO2  2NbO
(B)
dE o
nF
 S o
dT
S o  2  96485  (9 105 )  17.37 J mol -1 K-1
(C)
o
o
o
o
o
o
o
G o  nFECell
and G  H  TS  H  G  TS
Temperature (K)
Eo (mV)
 G o (kJ mol-1)
 H o (kJ mol-1)
1054
1086
1106
1160
1174
1214
1244
1283
213
212
209
201.5
202
200
198
194
41.10
40.91
40.33
38.88
38.98
38.59
38.21
37.44
39.31
59.41
59.77
59.54
59.54
59.37
59.68
59.82
59.73
59.61
Mean value