new Electrochennnnnnnnnnnnnnnmistry edi-1.ppt

CHAPTER- TWO
 ELECTROCHEMICAL CELLS
 Electrochemistry is the branch of chemistry which deals with
the transformation of electrical energy into Chemical ener
gy and vice-versa.
 It is the study of phenomena at electrode solution interfaces.
 Electrochemistry deals with the relationship between electric
al, chemical phenomena and the laws of interaction of the
se phenomena
 The laws electrochemistry forms the basis of electrolysis and
electro synthesis. Knowledge of electrochemistry is of imme
nse importance to study about the causes of destruction of ma
terials caused due to corrosion.
2

…..cont
 There are two processes where both are interrelated. Elec
trolysis is one process where electrical energy causes che
mical changes.
 The ability to make very precise measurements of curren
ts and potential differences (voltages) means that electr
ochemical method can be used to determine thermodyna
mic properties of reactions that may be inaccessible by ot
her methods.
 An electrochemical cell consists of two electrodes, or m
etallic conductors, in contact with an electrolyte, an ioni
c conductor (which may be solution)
3

…..cont
 E.g. Electrolysis of water yields H2 and O2
 In this process, electricity is passed through water (having f
ew drops of an acid) and H2 and O2 are collected at differe
nt electrodes. (i.e electrolytic cell)
 In the other process, certain chemical reactions take place i
n a vessel and produce electrical energy. The device is calle
d electrochemical cell.
 The specific reaction is called the redox reaction and This
process is also called the electromotive process. Electromot
ive means motion of electron, i.e. the flow of current.
 The two processes above can be summarized as follows:
 1. Electrical energy causing chemical reactions →Electrolys
is (Electrolytic cell
4

…..cont
 In redox reactions, electrons are released (oxidation) and c
onsumed (reduction).
5
Fig. 1: A simple electrochemical cell

…..cont
 The two compartments of this cell are separated by a poro
us barrier that allows ions to pass through while preventin
g gross mixing of the two solutions.
 When the two electrodes are connected, charges flow in th
e directions indicated. Note that the buildup of positive ch
arge on the left side can be offset either by diffusion of Zn2
+ to the right or (less efficiently) by Cl– to the left.
 This arrangement is called a galvanic cell.
 A typical cell might consist of two pieces of metal, one zin
c and the other copper, each immersed each in a solution c
ontaining a dissolved salt of the corresponding metal
6

…..cont
 However, if we connect the zinc and copper by means of a
metallic conductor,
 the excess electrons that remain when Zn2+ions go into sol
ution in the left cell would be able to flow through the exte
rnal circuit and into the right electrode, where they could b
e delivered to the Cu2+ ions which become “discharged”, tha
t is, converted into Cu atoms at the surface of the copper el
ectrode.
7

…..cont
 The net reaction is the same as before— the oxidation of zinc b
y copper (II) ions
 Zn(s) + Cu2+ → Zn2++ Cu(s) but this time, the oxidation and re
duction steps take place in separate locations:
 Left electrode: Zn(s) → Zn2++ 2e– oxidation
 Right electrode: Cu2+ + 2e– → Cu(s) reduction
 Electrochemical cells allow measurement and control of a redo
x reaction.
 The reaction can be started and stopped by connecting or disco
nnecting the two electrodes. If we place a variable resistance in
the circuit, we can even control the rate of the net cell reaction
by simply turning a knob. By connecting a battery or other sour
ce of current to the two electrodes, we can force the reaction to
proceed in its non-spontaneous or reverse direction.
8

…..cont
 By placing an ammeter in the external circuit, we can meas
ure the amount of electric charge that passes through the e
lectrodes,
 The amount of charge carried by one mole of electrons is k
nown as the faraday, which we denote by F.
 Careful experiments have determined that 1 F= 96467 c. Fo
r most purposes, you can simply use 96,500 coulombs as th
e value of the faraday.
 When we measure electric current, we are measuring the r
ate at which electric charge is transported through the circ
uit. A current of one ampere corresponds to the flow of one
coulomb per second.
9

…..cont
 Cell description conventions
 In order to make it easier to describe a given electrochemi
cal cell, a special symbolic notation has been adopted. In
this notation the cell Would be
Zn(s)| Zn2+(aq)|| Cu2+(aq)| Cu(s)
 single vertical bars indicate phase boundaries;
 the double vertical bar in the middle denotes the phase b
oundary between the two solutions.
 As a matter of convention, the chemical species that unde
rgo reduction when the cell reaction proceeds to the right
according to the net equation are shown on the right side,
and those that undergo oxidation are shown on the left.
10

…..cont
 The anode is where oxidation occurs, and the cathode is t
he site of reduction. In an actual cell, either electrode can h
ave either identity, depending on the direction in which th
e net cell reaction is occurring.
 If electrons flow from the left electrode to the right electro
de (as depicted in the above cell notation) when the cell op
erates in its spontaneous direction, the potential of the rig
ht electrode will be higher than that of the left, and the cell
potential will be positive.
11

…..cont
 “Conventional current flow” is from positive to negative, w
hich is opposite to the direction of the electron flow. This
means that if the electrons are flowing from the left electr
ode to the right, a galvanometer placed in the external circ
uit would indicate a current flow from right to left.
 Electrochemical cell reactions can be classified in to two
12

…..cont
Galvanic cell:
 Is an electrochemical cell that produced electricity as a result of t
he spontaneous reaction occurring inside it.
 In redox reaction, electrons are transferred from the oxidized sp
ecies to the reduced species.
 Imagine separating the two 1/2 cells physically, then providing a
conduit through which the electrons travel from one cell to the o
ther.
 Galvanic Cell: Electrochemical cell in which chemical reactions
are used to create spontaneous current (electron) flow.
 convert the energy from spontaneous chemical reactions into ele
ctricity.
 The stored energy changed into electricity
13

…..cont
Electrolytic cell
 Is an electrochemical cell in which a non-spontaneous re
action is driven by an external source of current
 use electricity to drive non-spontaneous chemical reacti
ons.
 Apply external electric energy
 Thus in order to produce the energy first apply electricity
14

…..cont
 Electrode potential (E/E.M.F): is an energy to move one el
ectron in a solution to the other place or external circulate.
 Hydrogen electrode is designated by
Pt/H2 (atm) /H+// Mn+/ M
 E = ER- EL or E = ERed - EOx
 Hydrogen electrode is used to determine electrode potentia
l of cell.
 By convention at 25oC and 1 atm (P) or STP
 EL = 0, E = ER – EL
 E = ER at EL = 0
15

…..cont
Polarity of galvanic electrode
 Ecell = ER – EL
 E > 0, spontaneous process: irreversible (Galvanic cell)
 E< 0, non-spontaneous (electrolytic cell)
 ER> EL, where ER has higher potential (+ve) and EL has lo
wer potential (-ve)
 polarity of anode is negative.
 polarity of cathode is positive.
16

Summery
 E0
cell > 0 DG0 < 0 Spontaneous
 E0
cell < 0 DG0 > 0 Non Spontaneous
 E0
cell = 0 DG0 = 0 Equilibrium
17

Electrolysis
 Redox reactions in which the change in Gibbs energy ΔG is
positive do not occur spontaneously.
 However they can be driven via application of either a kno
wn voltage or a known current.
 Electrolysis is the process of driving a reaction in a non spo
ntaneous direction by using an electric current.
 Electrolysis provides the basis of electro-synthesis and ind
ustrial electrochemistry.
18

…..cont
 A voltaic cell (a Galvanic cell) uses a spontaneous reaction (∆G ne
gative) to generate electrical energy. The reacting system does work
on the surroundings. All batteries are made from voltaic cells.
 An electrolytic cell uses electrical energy to drive a non-spontaneo
us reaction (∆G positive). Here the surroundings do work on the rea
cting system. Chemicals are prepared from electrical energy. This p
rocedure is termed electrolysis or electrochemical synthesis.
 All electrochemical cells have several common features.
They have two electrodes.
Anode: the oxidation half reaction takes place at the anode.
Cathode: the reduction half reaction takes place at the cathode.
The two electrodes are dipped into an electrolyte, a medium that co
ntains a mixture of ions which will conduct electricity.
19

Electrochemical cells:
 electrolytic –electcic energy to chemical energy,
non spontaneus process
 Galvanic cells –
 an electrochemical cell that drives electrons through
an external circuit
 spontaneous redox reaction occurring inside cell.
20

 The Zn/Cu Galvanic Cell
 half-reactions
Cu2+ (aq) + 2 e-  Cu (s) (cathode, RHS)
Zn2+ (aq) + 2 e-  Zn (s) (anode, LHS)

A Schematic Galvanic Cell
21
e-
Reducing Agent
e-
e-
Oxidizing Agent
Anode Cathode
Porous Disk

The Zinc/Copper galvanic cell.
22
a(Zn2+) = 1.00
e-
e-
Anode Cathode
Porous Disk or
Salt Bridge
Zn(s) Cu(s)
e- 1.10 V
e-
a(Cu2+) = 1.00

Cell Reactions
 The difference in the RHS and the LHS reaction
Cu2+ (aq) + Zn (s)  Cu (s) + Zn2+ (aq)
 For each half reaction, we can write the reaction quotient
as follows
Cu2+ (aq) + 2 e-  Cu (s) Q = 1/ a(Cu2+)
Zn2+ (aq) + 2 e-  Zn (s) Q = 1/ a(Zn2+)
Overall  Qcell = a(Zn2+) / a(Cu2+)
23

Cell Diagrams
 A shorthand way of expressing what takes place in
an electrochemical cell.
 For the above electrochemical cell.
24
Pt Cu (s) Cu2+ (aq) Zn2+ (aq) Zn (s) Pt
Note phase boundary , same phase
liquid junction salt bridges

Another Example
 The cell reaction
H2 (g) + Cu2+ (aq)  2 H+ (aq) + Cu (s)
25
Pt H2 (g) H+ (aq) Cu2+ (aq) Cu (s) Pt
 Electrochemical cells
 a cell that has not reached equilibrium can
do electrical work by driving electrons
through an external wire.

2.1.Reversible Electrochemical Cells
 In order for us to make measurements on an
electrochemical cell, it must be operating
reversibly.
 Place an opposing source of potential in the external
circuit
 Cell operates reversibly and at a constant
composition.
 Max work of electro chemical cells
we,max = DG
26

2.2. Reversible electrodes
 A metallic electrode that will dissolve when a current is
passed from it into a solution and that will have plated on
it metal from the solution when the current is passed in
the reverse direction.
27

Reversible electrodes
Electrode type Example Description Electrode reaction (in reduction
direction)
Metal metal-ion
electrode
Cu(s)│Cu2+(aq) Metal bathed in electrolyte
containing its own ions.
Cu2+(aq)+2e→Cu(s)
Ion – ion (redox)
electrode
Pt(s)│Fe3+,Fe2+(aq)
Pt|[Fe(CN)6]3–
,[Fe(CN)6]4–
Noble metal in contact with
solution of a redox couple
Fe3+(aq)+e→Fe2+(aq)
[Fe(CN)6]3– + e→[Fe(CN)6]4–
Metal insoluble salt
electrode
Hg(s) │Hg2Cl2(s)
│KCl(aq)
Ag|AgCl(s) |Cl–
Metal in contact with its
insoluble salt (i.s.) and a solution
containing a soluble anion of the
i.s.
Hg2Cl2(s)+2e→2Hg+2Cl-
AgCl(s) + e → Ag(s) + Cl–
Gas electrode Pt(s)│H2(g) │H+(aq) Noble metal in contact with a
saturated solution for a gas and
contains the reduced or oxidized
form of the gas
H+(aq)+e→1/2H2(g)
Amalgam and
membrane
Na(Hg) solution of a metal in liquid
mercury
28

2.3. Thermodynamics of electrochemical cells
The Measurement of Cell Potentials
 Measure the potential of an electrochemical cell when the
cell is at equilibrium, i.e., the state between the galvanic
and the electrolytic cell.
30
e-
Reducing Agent
e-
e-
Oxidizing Agent
Anode Cathode
Porous Disk
Counter potential
(load)

Derivation of the Nernst Equation
 Consider an electrochemical cell that approaches
the equilibrium state by an infinitesimal amount
d
31



 d
G
d
dG rxn
J
J
J D

 
Reminder
P
T
J
J
J
r
d
dG
G
,









  


D

The Work in Transporting Charge
 The maximum work
32

D d
G
dw rxn
e 
max
,
F = Faraday’s constant = e NA = 96485 C/mole
 For the passage d electrons from the
anode (LHS) to the cathode (RHS)



 d
F
d
eNA 


The Cell Potential
 The work to transport charge
33
cell
e E
d
F
dw 



max
,
G
d
dw
rxn
e
D


max
,
cell
rxn
E
-
F
G


D

Standard Cell Potentials
 From the reaction Gibbs energy
34
cell
cell
o
r
r E
F
Q
RT
G
G 
D
D 


 ln
cell
cell
o
r
r
E
F
Q
RT
F
G
F
G





 

D

D ln
We define
F
G
E
o
r
cell

D




The Nernst Equation
 E represents the standard cell potential, the
potential of the cell when all cell components
are under standard conditions.
 f (all gases) = 1
 a (solutes) = 1
 T = 298.15 K
 P = 1.00 bar pressure
35
cell
cell Q
F
RT
E
E ln


 

Cells at Equilibrium
 When the electrochemical cell has reached
equilibrium
36
cell
cell
cell K
Q
V
0
E 


Kcell = the equilibrium constant for the cell reaction.
RT
FE
K
K
F
RT
E cell
cell

 



 ln
ln
Knowing the E° value for the cell, we can estimate
the equilibrium constant for the cell reaction.

37
 More positive:
Stronger oxidizing agent
More readily accepts e-
 More negative:
Stronger reducing agent
More readily gives e-
 Stronger R.A. + O.A.  Weaker R.A. + O.A.
E0 values

Free Energy and Cell Potential
DG0, E0, and K
38
0
0
max nFE
G 

D

w n: number of moles of e-
F: Faraday’s constant
96485 C
mol of e-
0
0
nFE
RTlnK
G 



D
lnK
nF
RT
E0

so
At equilibrium: DG0 = 0 and K = Q
At 298 K:
logK
n
0.0592
E0


Nernst Equation
 Under nonstandard conditions
40
RTlnQ
nFE
nFE
RTlnQ
G
G
0
0





D

D
lnQ
nF
RT
E
E 0
cell 

lnQ
n
0.0592
E
E 0
298K
cell 


Equilibrium Constant Calculations from Cell Potentials
 Examine the following cell.
42
Pt Sn2+ (aq), Sn4+ (aq) Fe3+ (aq) Fe2+ (aq) Pt
 Half-cell reactions.
Sn4+ (aq) + 2 e-  Sn2+ (aq) E(Sn4+/Sn2+) = 0.15 V
Fe3+ (aq) + e-  Fe2+ (aq) E (Fe3+/Fe2+) = 0.771 V
 Cell Reaction
Sn2+ (aq) + 2 Fe3+ (aq)  Sn4+ (aq) + 2 Fe2+ (aq)
Ecell = (0.771 - 0.15 V) = 0.62 V

2.4. determination of standard electrode potential
Standard Reduction Potentials
 Standard reduction potentials are intensive properties.
 We cannot measure the potential of an individual half-
cell!
 We assign a particular cell as being our reference cell
 Assign values to other electrodes on that basis.
43

a (H+) = 1.00
H2 (g)
e-
Pt gauze
The Standard Hydrogen Electrode
 Eo (H+/H2) half-cell = 0.000 V
44
f{H2(g)} = 1.00

A Galvanic Cell With Zinc and the
Standard Hydrogen Electrode.
45
e-
Zn2+, SO4
2-
a (H+) = 1.00
Anode Cathode
Porous Disk or
Salt Bridge
Source of H+ (e.g.,
HCl (aq), H2SO4 (aq))
a(Zn2+) = 1.00
H2 (g)
0.763 V
e-
Zn(s)
Pt gauze

The Cell Equation for the Zinc-Standard Hydrogen
Electrode.
 The cell reaction
2 H+ (aq) + Zn (s)  H2 (g) + Zn2+ (aq)
46
Pt Zn (s) Zn2+ (aq),a=1 H+ (aq), a=1 H2 (g), f=1 Pt
 When we measure the potential of this cell
Ecell = ERHS - ELHS
but ERHS = E(H+/H2) = 0.000 V
 Ecell = E(Zn2+/Zn) = 0.763 V

The Spontaneous Direction of a Cell Reaction
 Examine the magnitude the of the standard cell
potential!
47
F
G
E
o
rxn
cell

D



 If the standard cell potential is positive, the
DrG is negative!

The Composition Dependence of the Cell Potential
 Nonstandard cell potential (Ecell) will be a
function of the activities of the species in the
cell reaction.
48
cell
cell Q
F
RT
E
E ln


 
 To calculate Ecell, we must know the cell
reaction and the value of Qcell.

Example
 For the following system
49
Pt H2 (g) H+ (aq) Cu2+ (aq) Cu (s) Pt
 Calculate the value of the cell potential when
the f (H2) = 0.50, a(Cu2+) = 0.20, and a(H+) =
0.40.

Obtaining Standard Cell Potentials
 Look at the following cell
51
Pt H2 (g) HCl (aq) AgCl (s) Ag (s) Pt
   












)
(
ln
2
cell
cell
H
f
Cl
a
H
a
F
RT
E
E 
Ecell = E(AgCl/Ag) - E (H+/H2)
= E(AgCl/Ag)

Electrochemical Series
 Look at the following series of reactions
Cu2+ (aq) + 2 e-  Cu (s) E(Cu2+/Cu) = 0.337 V
Zn2+ (aq) + 2 e-  Zn (s) E(Zn2+/Zn) = -0.763 V
 Zn has a thermodynamic tendency to reduce Cu2+ (aq)
Pb2+ (aq) + 2 e-  Pb (s) E(Pb2+/Pb) = -0.13 V
Fe2+ (aq) + 2 e-  Fe (s) E(-Fe2+/Fe) = -0.44 V
 Fe has a thermodynamic tendency to reduce Pb2+ (aq)
54

Thermodynamic Information
 Note
55
G
FE
G
r
P
T
D















,
 And


G
FE r
D
 


Entropy Changes
 To obtain the entropy change for the cell reaction
56
P
P
T
P
T
rxn
T
E
F
G
T
S
S 








































 


D
,
,
 
P
P
T
rxn
rxn
T
E
F
G
T
S 
























D
D ,

Enthalpy Changes
 To obtain the enthalpy change for the cell reaction
57
P
T
P
T
P
T
rxn
S
T
G
H
H
,
,
,




































D
P
T
E
FT
FE 









 

P
rxn
T
E
FT
FE
H 

















D

E.g
The Weston Cadmium Cell has an e.m.f. (E) of 1.01463 V at 298
K and a temperature coefficient of -5.00 × 10–5 V K–
1.calculate the thermodynamic quantities of the overall cell
reaction.
 ΔG = – n F E = – 2 × 96500 C mol–1 × 1.01463 V
= –195824 Jmol-1
ΔS = nF[ ∂ E/∂T]p
= 2 (96500 C mol–1) × (– 5.00 × 10–5) V K–1
= – 9.65 Jmol-1 K–1
 ΔH = ΔG + T ΔS
= – 195824 Jmol-1 + 298 K (– 9.65 Jmol-1 K–1)
= -198700 Jmol-1
58

59
 In galvanic cell e.m.f. produced due to chemical
reaction- called a chemical cell.
 In some cells no overall chemical reaction but the e.m.f.
is generated due to the difference in the concentrations
of either the electrolytes or the electrodes.
 Concentation cell I
 Electrolyte concentration cell
 the electrodes are identical; they simply differ in the
concentration of electrolyte in the half-cells.
2.5.Classes of electrochemical cells

60
 Electrode concentration cells
 the electrodes themselves have different
compositions. This may be due to.
 Different fugacities of gases involved in
electrode reactions (e.g., The H+ (aq)/H2 (g)
electrode).
 Different compositions of metal amalgams in
electrode materials.
Concentration Cells (II)

Electrode concentration cell
 An example of electrode concentration cell is
Na(Hg)(c1)| NaCl| Na(Hg)(c2)
LHS Na(Hg)(c1)—> Na+ + e
RHS Na+ + e —> Na(Hg)(c2)
overall reaction Na(Hg) (c1) —> Na(Hg) (c2)
E cell = – (RT/F ) ln c2/c1
Eg2. Pt/ H2(g)(P1)|HCl Solution | H2(g)(P2)/Pt
overall reaction is ½ H2 (P1) —> ½ H2 (P2)
H2 (P1) → H2 (P2)
Ecell = -(RT/F) ln P2/P1
Note Ecell =0 because the elecetrodes in both sides are same 61

Cells without liquid junction(without transference)
62
 What if we were able to set up a cell so that the
transport at the interface did not contribute to the
overall DG? i.e if we use salt bridge (if no transport of
matter) diffusion potential
 The potential of this cell would be the cell potential
without transference, Ewot.
HCl(a2)  HCl(a1)
 
 2
1
)
(
)
(
ln
HCl
a
HCl
a
RT
G 
D
 
 2
1
)
(
)
(
ln
HCl
a
HCl
a
F
RT
Ewot 


The Liquid Junction Potential
 Examine the following electrochemical cell
 Activity difference of the HCl between
compartment 1 and compartment 2
 There should be a transport of matter from one
cell compartment to the other!
63

Electrolytic Concentration Cell
64
e-
a (HCl ) = 0.0010
Left Right
Porous Disk or
Salt Bridge
a(HCl) = 0.010
0.0592V
e-
Ag(s)
Ag(s)

The Development of Liquid Junction Potentials
 The cell
compartments are
identical except for
the activities of the
electrolyte
solutions.
65
HCl (a1)
HCl (a2)
Ag/AgCl electrode

Cont….
 Note that we now
have the migration
of both cations and
anions through the
liquid junction.
66
Cl-
Ag/AgCl electrode
H+

Cont…
 After a period of
time
67
-----------
- - - - -
Ag/AgCl electrode
+ + + + +

Cont…
 Choose the lower compartment as our LHS
electrode.
68
Ag AgCl HCl(aq)a1 HCl(aq)a2 AgCl (s) Ag (s)
Note: liquid junction
 For the passage of one mole of charge
through the cell
-F Ecell =  DGJ

The Cell Reactions
 For the LHS and RHS electrodes
AgCl (s) + e-  Ag (s) + Cl- (a1) LHS
AgCl (s) + e-  Ag (s) + Cl- (a2) RHS
 Net change
Cl- (a1)  Cl- (a2)
 Note that the charge at the interface is transported
by the anions and cations in the cell reaction!
69

The Transport Numbers
 How is the charge carried at the interface of the cells?
 t+ moles of charge carried by the H+ (cation).
 t- moles of charge carried by the Cl- (anion).
 Passage of one mole of “+” charge through the
interface
 requires the passage of t+ moles of H+ (aq) from the LHS 
RHS, and the passage of t- mole of Cl- charge from the RHS
 LHS.
70

 At the boundary
t+ H+(a1) + t- Cl-(a2)  t+ H+(a2) + t- Cl-(a1)
 For the entire cell
Cl- (a1)+ t+ H+(a1) + t- Cl-(a2)  Cl- (a2)+ t+ H+(a2) +
t- Cl-(a1)
 The cell reaction involves the transport of t+ moles
of HCl from the LHS to the RHs of the cell.
t+(aHCl)1 —> t+(aHCl)2
Note Cl- (a1) =(t+ + t-)Cl- (a1) as t+ + t- =1
71

The Gibbs Energy Changes
 For the above cell reaction, we can write the Gibbs
energy expressions as follows
72
   
 
   
 
1
1
2
2
)
(
ln
)
(
)
(
ln
)
(
)
(
ln
)
(
)
(
ln
)
(


















D
Cl
a
RT
Cl
H
a
RT
H
t
Cl
a
RT
Cl
H
a
RT
H
t
G








 
 1
2
Cl
a
H
a
Cl
a
H
a
RT
t
G
)
(
)
(
)
(
)
(
ln 





D

Cells With Transference
 Note a(H+) a (Cl-) = {a (HCl)}2
73
 
 1
2
HCl
a
HCl
a
RT
t
2
G
)
(
)
(
ln




D
 
 1
2
wt
HCl
a
HCl
a
F
RT
t
2
E
)
(
)
(
ln





Note that the cell potential with
transference, Ewt is determined as follows

Eg. Zn/ZnSO4(0.001M)||ZnSO4(x)/Zn is 0.09V at 25˚C. Find
the concentration of the unknown solution.
Ans. Ewot = 0.0592/n log C2/C1
0.09 =(0.0592/2) log ( x / 0.001)
x =1.097M
Consider : Pt/H2 | HCl(a1) ¦ HCl(a2) | H2/Pt
Cell rxn H+(a2) —> H+(a1)
t– HCl (a±2) —> t– HCl (a±1)
 ΔG = RT ln (a±1/ a±2) 2t– = – nF E
 or, Ewt=-2t-(RT/F)ln(a±1/a±2) (for one electron change)
 In a cell without liquid junction the e.m.f. is given by
Ewot =-2 (RT/F) ln (a±1/a±2) 74

The Liquid Junction Potential
 The liquid junction potential is the difference in
the cell potentials with and without transference!
This is for anion reversible (Cl-)
 L.J potential depend on transport numbers
 What if the following were true?
 t+  t-  0.5  ELJ would be very small and would
only make a small contribution to the overall cell
potential ! 75
wot
wt
LJ E
E
E 

 
 
 1
2
)
(
)
(
ln
2
2
HCl
a
HCl
a
F
RT
t
ELJ






 Comparison of the e.m.f.s of cells with transferance (Ewt) an
d without liquid junction (Ewot) gives a method for the dete
rmination of the transport number.
 For electrode with cation reversible
t– = Ewt/Ewot and t+ = 1 – Ewt/Ewot
For electrode with anion reversible
t+ = Ewt/Ewot and t- = 1 – t+
 Eg. The e.m.f. of the concentration cell: Ag/AgCl(s) /KCl(0.0
5M) /Kx(Hg)/ KCl(0.5M)/ AgCl(s)/Ag is 0.107 V. For the corr
esponding cell with transference the emf is 0.053 V. What is
the transport number of Cl- ion ?
Ans. : The electrodes of the above cell is chloride ion
reversible. t+ = Ewt/Ewot = 0.053/0.107 = 0.495 and t- = 1- t+
= 1- 0.495 = 0.505. 76

2.6 The measurement of pH
 Cell setup: A reference half cell (ion │insoluble salt │metal)
is attached to a hydrogen electrode.
 Pt │ H2 gas │ cH+ ║ satd. KCl sol. │ Hg2Cl2(s) │ Hg(l)
LHS H2 = 2H+ + 2e
RHS Hg2Cl2 + 2e = 2Hg + 2 Cl-
Cell rxn: Hg2Cl2 + H2 = 2H+ + 2Cl- + 2 Hg
77
  2
2
2 0
2
Hg
2
Cl
2
H
/
ln
2 Cl
Hg
H
o
cell
cell
a
p
p
a
a
a
F
RT
E
E







H
calomel
cell E
E
E 


78
0
,
059
.
0
/
0.2802
E
059
.
0
/
0.2802
E
059
.
0
2802
.
0
log
059
.
0
2802
.
0
ln
2802
.
0
0.2802V
a
ln
F
RT
E
ln
ln
ln
0
H
cell
0
H
cell
0
H
cell
H
0
H
cell
H
0
H
cell
caloml
Cl
0
calomel
0
H
0
caloml
0
cell
H
0
cell
cell
H
0
cell
cell




































E
pH
E
pH
pH
E
E
a
E
E
a
F
RT
E
E
E
E
E
E
a
F
RT
a
F
RT
E
E
a
a
F
RT
E
E
Cl
Cl

 The cell SCE // (0.1M) HCl / AgCl(s) /Ag gave emf of 0.24
V and 0.26 V with buffer having pH value 2.8 and
unknown pH value respectively. Calculate the pH value of
unknown buffer solution. Given ESCE= 0.2422 V
 Note the cell representation is reversed and hence sign of
Ecell will be changed
79
46
.
2
059
.
0
/
2422
.
0
26
.
0
648
.
0
059
.
0
/
648
.
0
2422
.
0
24
.
0
8
.
2
059
.
0
059
.
0
0
/
0
/
















calomel
cell
Ag
AgCl
calomel
cell
Ag
AgCl
LHS
RHS
cell
E
E
E
pH
x
E
E
pH
E
E
E
E

Glass electrode
 To avoid using hydrogen and possibility of poisoning
platinum surface glass electrode is used to measure PH
 Made of very thin bulb of glass.The bulb contains a solution
of a constant pH, say 0.1 M HCl, in which a silver-silver
chloride electrode is dipped that has a definite potential.
The bulb is put into the solution, the pH of which is to be
determined together with the reference electrode such as
calomel electrode. A potential difference is developed at the
interface between the glass bulb and the test solution,
which is dependent upon the hydrogen ion concentration,
i.e., the pH of the test solution. It is to be noted that only
H+ ions are transported through the glass, i.e., t+ = 1 and t–
= 0. it is example of ion selective membrane electrode
80

2.7 Membrane potenial
82
 Two electrolyte solutions separated by semi permeable
membrane produces potential deference.
 Consider two KCl solutions separated by a semi permeable
membrane that allows passage of K+ but not Cl-. If solution
1 is more concentrated than solution 2, K+ will diffuse from 1
to 2. this causes solution 2 to have positively charged relative
to 1 and will have higher potential.
K+(1) K+(2)
tential
membranepo
where
a
a
F
RT
K
D






D


1
2
K
)
(
)
(
ln
1
)
(
2
)
(

Unit 3
interfacial electrochemistry
83
 3.1 electric double layer
 when two dissimilar phases come into contact, charge separat
ion occurs in the interfacial region which results in the gener
ation of an interfacial potential difference or electric fiel
d.
• How does this happen?

 Electroneutrality is valid in the bulk but breaks down in sur
face region.
• What about the solution region next to the electrode
surface? In this region forces experienced by ions and solvent
dipoles are no longer isotropic and homogeneous. The
forces are anisotropic because of the presence of the
electrode phase.
 New solvent structure, different from that of the bulk, deve
lops because of the phase boundary.
 Electro-neutrality breaks down on the solution side of the
interface.
 There will be a net orientation of solvent dipoles and a nete
xcess charge in any volume element of the solution adjacent
to the electrode surface. The solution side of the interface b
ecomes electrified. 84

 Interfacial charge separation generates high interfacial E-fiel
d. Once the solution side of the interface becomes electrified
(acquires a net or excess charge), an electric field will operate
across the phase boundary. Since the metallic phase contains
charged particles, the latter will respond to this E field. The f
ree electrons will move away from or move towards the interf
ace depending on the direction of the E field. Thus a net char
ge will be induced on the metal, which will be equal in magni
tude and opposite in sign to that on the solution side of the p
hase boundary. Thus charge separation occurs across the M/S
interface, and this gives rise to an interfacial potential differe
nce. Typically the potential difference is ca. 1.0 V. However th
e spatial dimensions of the interface region are very small, ty
pically 1 nm thick. Thus the electric field strength present at t
he M/S interface will be typically 107 Vcm-1 which is very large
. 85

 –double layer is an electrified interface used to describe the a
rrangement of charges and oriented solvent dipoles at the int
erface between an electrode and an electrolyte solution.
 Three approximate models to describe the properties of the e
lectrified interface:
– Helmholtz model
– Gouy-Chapman diffuse layer model
– Stern model .
Opposite ions can be specifically adsorbed on the electrode
forming rigid and compact layer this layer is called the inner
Helmholtz layer. Also uncharged molecules can be
adsorbed in the inner Helmholtz layer. If only electrostatic
forces are acting, the ions will just approach the electrode to
86

a distance determined by the thickness of the solvation layer at
the electrode surface and of the solvation shell of the ions.
This layer is called “the outer Helmholtz layer.”
 This model has limitatioins: The double layer capacitance is a
constant independent of ion concentration and electrode pot
ential
 Gouy-Chapman :
Helmholtz neglected the disordering effect of the thermal
motion of the ions in the solution especially when there is a
small surface charge and at low electrolyte concentrations.
He introducing a diffuse model of the DoubleLayer. In this
model the charge distribution of ions as a function of
distance from the metal surface allows maxwell-boltzman
statistics to be applied. Thus the electric potential decreases
exponentially away from the surface of the fluid bulk. 87

 In the GC model the solvated ions are modelled as point char
ges. This neglect of ion size is unrealistic. In reality the solvat
ed ion can only approach the electrode surface to a distance e
qual to its solvated radius a.
 Stern model: The Stern model is as follows. Next to the elect
rode we have a region of high electric field and low dielectric
constant with a row of firmly held counter ions. Beyond that
there is an ionic atmosphere (the diffuse layer) where there is
a balance between the ordering electrostatic force and disord
ering thermal motions. The dielectric constant increases rapi
dly with distance in this region.
• The electrical potential varies linearly with distance (hydrated
ion radius) within the inner compact layer and decreases in
an approximate exponential manner with distance within the
diffuse layer, decaying to zero in the bulk solution.
88

How to measure electrode
potential
91

 The Outer Potential: potential outside the charged electrod
e given by the psi (ψ)potential. Defined as the work done in b
ringing a unit test charge from infinity up to a point just outsi
de of the charged electrode or electrolyte solution. (ψm) and
(ψs) are outer potentials for the charged electrode and charg
ed solution in the presence of double layer.
Then the voltage at the interface will be
 The outer potential difference is the contribution to the p
otential difference across an electrified interface arising from
the charges on the two phases.
92
3.2.potential differenced across
interface

 In addition to the charged electrode and solutions, there are
also noncharged solvent molecules will rearrange themselves
to have particular net orientations at the interface, this arran
gement produces dipole potentials. This potential is said to b
e surface potential χ. There is surface potential in the metal(χ
M )and solution side(χS).
 The surface potential at the interface is therefore
 The Inner Potential Difference(φ):The sum of the potenti
al differences due to charges and dipoles.this potential also k
nown as galvani potentntial.
 Outer potential ψ can be measured but surface potential χ no
t measured and so inner potential Δφ can not be measured.
93

3.4 Electrochemical kinetics
94
 For an interfacial electron transfer (ET) process:
– current flow is proportional to reaction flux (rate).
• Reaction rate is proportional to reactant concentration at
interface.
• As in chemical kinetics: – the constant of proportionality
between reaction rate fS (molcm-2s-1) and reactant
concentration c (molcm-3)is termed the rate constant k
(cms-1).
All chemical and electrochemical reactions are activated
processes. An activation energy barrier exists which must
be overcome in order that the chemical reaction may
proceed.

– Energy must be supplied to surmount the activation energy
barrier.
– This energy may be supplied thermally or also (for ET
processes at electrodes) via the application of a potential
to the metallic electrode.
 ET rate effected by:
Applied electrode potential and Temperature
 The electrode potential drives ET processes at interfaces
• Application of a potential to an electrode generates a large
electric field at the electrode/solution interface which
reduces the height of the activation energy barrier and
thereby increases the rate of the ET reaction. • hence the
applied potential acts as a driving force for the ET reaction
and the current should increase with increasing driving
force. 95

Factors effecting the current/potential response at
electrode/solution interfaces.
96
 The current observed at an electrode/solution interface refle
cts two quantities:
– Charging of electrical double layer : non Faradaic charging
current iC.
– Interfacial ET across electrode/solution interface : Faradaic
current iF
• The Faradaic current iF in turn can have components:
– at low potentials : arising from rate determining interfacial
ET,
– at high potentials : arising from material transport (MT) due
to diffusion mechanisms.

 Chemical reactions are activated processes : they requi
re an energy input in order to occur. The relationship b
etween rate constant k and temperature T is given by t
he empirical Arrhenius equation.
98

Unit 4
Kinetic Theory of Gases
The kinetic theory of gases
 A simple (and successful) model to explain the properties o
f an ideal gas is the kinetic molecular theory of gases.
 Kinetic theory of gases, an account of gas properties in term
s of motion and interaction of microscopic particles
 The kinetic theory describes a gas as a large number of su
bmicroscopic particles (atoms or molecules), all of which ar
e in constant, random motion.
 The rapidly moving particles constantly collide with each ot
her and with the walls of the container.
101

….cont
 Kinetic theory explains macroscopic properties of gases, s
uch as pressure, temperature, viscosity, thermal conducti
vity, and volume, by considering their molecular composi
tion and motion
 The theory posits that gas pressure is due to the impacts,
on the walls of a container, of molecules or atoms moving
at different velocities.
102

…..cont
The postulates are:
1. A gas consists of a large number of particles known as
molecules of the same size and mass.
2. The interaction between molecules is negligible unlike
solids.
3. The molecules in a gas are in a state of random motion in
all directions with different velocities ranging from zero to
infinity.
4. The size of the molecules is negligible in comparison
with the volume of a gas.
104

…..cont
5. During the motion, molecules collide with each other
which is perfectly elastic.
the molecules are perfectly spherical in shape,
6. The impact of molecules on the walls of the container
exerts a pressure.
 Pressure of a gas is the average force per unit area of the m
olecules on the walls of the container during impact.
 Larger the number of impacts, greater will be the pressure.
7. Due to random motion of molecules in gas, the molecules
have kinetic energy. The average
kinetic energy of a gas molecule varies as absolute
temperature.
105

4.2. Ideal gas laws
 defined as one for which both the volume of molecules
and forces between the molecules are so small that th
ey have no effect on the behavior of the gas.
 In the limit that P →0, all gases behave ideally.
The ideal gas law is:
PV = nRT,
The ideal gas law has the sub sums of the other gas laws.
106

…cont
 For a fixed amount of gas at a constant temperature, P
V = constant (this is Boyle’s law).
 For a fixed amount of gas at a constant pressure, V incr
eases linearly with T, i.e. V = (nR/P)T (this is Charles’s
law).
 For a fixed volume and temperature, p increases linearl
y with n, i.e. p = (RT/V)n (this is Avogadro’s law).
108

……cont
 A set of standard conditions have been chosen to make it easie
r to understand the gas laws and gas behavior.
 Standard temperature = 0oC = 273.15 K, and
 standard pressure = 1 atmosphere = 760 mmHg = 760 torr.
 1 mole of a gas will occupy a standard molar volume of 22.4 L.
 Many gas law problems involve a change of conditions with no
change in the amount of gas. That is: PV/T = constant. Therefo
re, for a change of conditions,
109

Exercise
E.g. 1: A gas sample in the laboratory has a volume of 45.9 L
at 25.0 oC and a pressure of 743 mm Hg. If the temperature
is increased to 155 oC by compressing the gas to a new
volume of 31.0 L. What is the pressure in atm?
 Answer: p1 = 743 mmHg = 0.978 atm, V1 = 45.9 L, T1 = 25o
C = 298 K; T2 = 155 oC = 428 K,
V2 = 31.0 L, P2 =?
110

Exercise
E.g. 2: Calculate the pressure (in atm) in a container filled
with 5.038 kg of xenon at a temperature of 18.8oC, whose
volume is 87.5 L.
 Strategy:
 (1) Convert all information into the units required, and
 (2) Substitute into the ideal gas equation.
 Answer:
And T = 18.8oC + 273.15 K = 292.0 K
111

Exercise
E.g. 4: Calculate the density of ammonia gas (NH3) in grams
per liter at 752 mm Hg and 55.0 oC.
 Strategy:
(1)Assume one mole and calculate V for given conditions.
(2) Use density = mass/volume to calculate density.
P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm
T = 55.0 oC + 273.15 = 328.2 K
 Answer:
112

113
Figure 1: set up for the determination of Molar mass of
unknown volatile liquid
•A volatile liquid is placed in 590.0
ml flask and allowed to boil until
only vapor fills the flask at a
temperature of 100.0oC and 736
mmHg pressure.
•If the masses of the empty and gas
filled flask were 148.375 g and
149.457 g respectively,
• what is the molar mass of the
liquid?
Use the gas law to calculate the molar mass of the liquid.
Pressure = 736 mm Hg x 1atm/760 mm Hg = 0.968 atm
Mass of gas = 149.457g  148.375g = 1.082 g
Answer:
The molecular mass of the unknown volatile gas is 58.0 g/mol.

4.3. Barometric formula
 The barometric formula sometimes called the exponential
atmosphere or isothermal atmosphere
 formula used to model how the pressure (or density) of the
air changes with altitude.
 Air pressure at any height in the atmosphere is due to the f
orce per unit area exerted by the weight of all of the air lyi
ng above that height.
 Consequently, atmospheric pressure decreases with increas
ing height above the ground.
114

…..cont
 The net upward force acting on a thin horizontal slab of a
ir, due to the decrease in atmospheric pressure with heigh
t, is generally very closely in balance with the downward
force due to gravitational attraction that acts on the slab.
 If the net upward pressure force on the slab is equal to the
downward force of gravity on the slab, the atmosphere is
said to be in hydrostatic balance.
115

116
Cross-sectional area
-dh
pressure = p + dp
dh pressure = p
h
gpdh
Ground
Figure 2: Balances of vertical forces in an atmosphere in hydrostatic balance
The mass of air between h and h + dh in the column of air is h.
The downward gravitational force acting on this slab of air, due to the
weight of the air is g dh
dh.

117
Let the change in pressure in going from height h to height h + dh be dp.
Since we know that pressure decreases with height, dp must be a negative quantity.
The upward pressure on the lower face of the blue labeled block must be slightly greater
than the downward pressure on the upper face of the block.
Therefore, the net vertical force on the block due to the vertical gradient of pressure is
upward and given by the positive quantity −dp as indicated in the figure.
For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical requires that
In a fluid at rest, all frictional stresses vanish and the state of stress of the system is called hydrostatic

4.4. Distribution of molecular velocities
 In a sample of gas the particles do not all have the same ve
locity, rather they exhibit a distribution of molecular spee
ds. The distribution of molecular speeds, f(v), in an idea
l gas at thermal equilibrium is given by the Maxwell-Bol
tzemann distribution:
The distribution depends on the ratio m/T,
 m is the mass of a gas particle
 T is the temperature.
121

…….cont
 The plots below show the Maxwell-Boltzmann speed distri
butions for a number of different gases at two different te
mperatures. As we can see, average molecular speeds for co
mmon gases at room temperature (300K) are generally a fe
w hundred meters per second.
122
Maxwell-Boltzmann speed distributions different gases at
different temperatures

For example, N2 has an average speed of around 500 ms-1,
rising at around 850 ms-1 at 1000 K.
a light molecule such as H2 has a much higher mean speed of
around 1800 ms-1 at room temperature. From the plots, we
can make two observations:
123

….cont
 Increasing the temperature broadens the distribution an
d shifts the peak to higher velocities.
 Decreasing the mass of the gas particles has the same effe
ct as increasing the temperature. (i.e. heavier particles ha
ve a slower, narrower distribution of speeds than lighter p
articles).
124

125
4.5 Molecular collisions
Collisions are one of the most fundamental processes in chemistry, and provide
the mechanism by which both chemical reactions and energy transfer occur in a
gas.
The rate at which collisions occur determines the timescale of these events,
and is therefore an important property for us to be able to calculate.
The rate of collisions is usually expressed as a collision frequency,
defined as the number of collisions a molecule undergoes per unit time. We
will use kinetic theory to calculate collision frequencies for two cases: collisions
with the container walls; and intermolecular collisions.

126
•Collisions with the container walls.
Pressure of a gas arises from collisions of the gas particles with the walls of
the container. By considering these collisions more carefully, we can use
kinetic theory to relate the pressure directly to the average speed of the gas
particles. Firstly, we will determine the momentum transferred to the
container walls in a single collision. The figure below shows a particle of
mass m and velocity v colliding with a wall of area A. Before the collision,
the particle has velocity vx and momentum mvx along the x direction. After
the collision, the particle has momentum -mvx along the x direction (note that
the components of momentum along y and z remain unchanged). Since
momentum must be conserved during the collision, and the momentum of the
particle has changed by 2mvx, the total momentum imparted to the wall must
also be 2mvx.The next step is to determine the total number of collisions with
the wall in a given time interval ∆t. During this time interval, all particles
within a distance, d = vx∆t, of the wall (and travelling towards it) will collide
with the wall. Since the area of the wall is A, this means that all particles
within a volume Avx∆t will undergo a collision. We now need to work out
how many particles will be within this volume and travelling towards the
wall. The number density of the molecules (i.e. the number of molecules per
unit volume) is:

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