ELECTROCHEMISTRY
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References:
1.
2.
3.
Engg.Chemistry by Jain and Jain
Engg.Chemistry by Dr. R.V.Gadag and Dr.
A.Nithyananda Shetty
Principles of Physical Chemistry by Puri and
Sharma
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
Electrochemistry is a branch of chemistry which
deals with the properties and behavior of
electrolytes in solution and inter-conversion of
chemical and electrical energies.
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



An electrochemical cell can be defined as a
single arrangement of two electrodes in one
or two electrolytes which converts chemical
energy into electrical energy or electrical
energy into chemical energy.
It can be classified into two types:
Galvanic Cells.
Electrolytic Cells.
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Galvanic Cells:
A galvanic cell is an electrochemical cell that
produces electricity as a result of the
spontaneous reaction occurring inside it.
Galvanic cell generally consists of two
electrodes dipped in two electrolyte solutions
which are separated by a porous diaphragm or
connected through a salt bridge. To illustrate a
typical galvanic cell, we can take the example
of Daniel cell.
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Daniel Cell.
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At the anode:Zn → Zn 2+ + 2eAt the cathode: Cu 2+ + 2e- → Cu
Net reaction: Zn(s)+Cu 2+ (aq)→ Zn 2+ (aq)+ Cu(s)
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
ELECTROLYTIC CELL
An electrolytic cell is an electro –chemical cell in
which a non- spontaneous reaction is driven by
an external source of current although the
cathode is still the site of reduction, it is now the
negative electrode whereas the anode, the site
of oxidation is positive.
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Representation of galvanic cell.
Anode Representation:
Zn│Zn2+ or Zn ; Zn2+
Zn │ ZnSO4 (1M) or Zn ; ZnSO4 (1M)
 Cathode Representation:
Cu2+/Cu or Cu2+ ;Cu
Cu2+ (1M) ; Cu or CuSO4(1M)/Cu
 Cell Representation:

Zn │ ZnSO4 (1M)║ CuSO4(1M)/Cu
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Liquid Junction Potential.
Difference between the electric potentials
developed in the two solutions across
their interface .
Ej = Ø soln, R - Ø soln,L
Eg: *Contact between two different
electrolytes (ZnSO4/ CuSO4).
*Contact between same electrolyte
of
different concentrations(0.1M
HCl /
1.0 M HCl).

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Salt Bridge.

The liquid junction potential can be
reduced (to about 1 to 2 mV) by joining
the electrolyte compartments through a
salt bridge.
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Function Of Salt Bridge.
 It
provides electrolytic contact between
the two electrolyte solutions of a cell.
 It
avoids or at least reduces junction
potential in galvanic cells containing two
electrolyte solutions in contact.
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Emf of a cell.
The difference of potential, which causes
a current to flow from the electrode of
higher potential to one of lower potential.
Ecell = Ecathode- Eanode
 The E Cell depends on:
 the nature of the electrodes.
 temperature.
 concentration of the electrolyte
solutions.

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
Standard emf of a cell(Eo cell) is defined as the emf
of a cell when the reactants & products of the
cell reaction are at unit concentration or unit
activity, at 298 K and at 1 atmospheric pressure.
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The emf cannot be measured accurately using a
voltmeter :
 As
a part of the cell current is drawn,thereby
causing a change in the emf.
 As
a part of the emf is used to overcome the
internal resistance of the cell.
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
The emf of the cell Ex is proportional to the length AD.
Ex α

AD
The emf of the standard cell Es is proportional to the length AD1.
Es α AD1
Ex
Es
═
AD
AD1
Ex = AD x Es
AD1
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Standard Cell.

It is one which is capable of giving constant and reproducible
emf.

It has a negligible temperature coefficient of the emf.

The cell reaction should be reversible.

It should have no liquid junction potential.
Eg: Weston Cadmium Cell. The emf of the cell is
1.0183 V at 293 K and 1.0181 V at 298 K.
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
Weston Cadmium Cell
Sealed wax
Cork
CdSO4.8/3H2O
crystals
Cd-Hg
12-14%
Cd
Soturated solution of
CdSO4.8/3H2O
Paste of Hg2SO4
Mercury, Hg
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
Cell representation:
Cd-Hg/Cd2+// Hg2SO4/Hg
At the anode:
Cd (s) → Cd2+ + 2eAt the cathode:
Hg2SO4(s) + 2e- → 2 Hg (l)+ SO42-(aq)
Cell reaction:
Cd + Hg22+ → Cd2+ + 2Hg
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Origin of single electrode potential.


Consider Zn(s)/ ZnSO4
Anodic process: Zn(s) → Zn2+(aq)
Cathodic process: Zn2+(aq) → Zn(s)
At equilibrium: Zn(s) ↔ Zn2+(aq)
Metal has net negative charge and solution
has equal positive charge leading to the
formation of an Helmholtz electrical layer.
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Single electrode potential.
Electric layer on the metal has a potential
Ø (M).
Electric layer on the solution has a
potential Ø (aq)
 Electric potential difference between the
electric double layer existing across the
electrode /electrolyte interface of a single
electrode or half cell.
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De-electronation
Electronation
Helmholtz double layer
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MEASUREMENT OF ELECTRODE
POTENTIAL.
 It
is not possible to determine experimentally the
potential of a single electrode.
 It is only the difference of potentials between two
electrodes that we can measure by combining them
to give a complete cell.
 By arbitrarily fixing the potential of reversible
hydrogen electrode as zero it is possible to assign
numerical values to potentials of the various other
electrodes.
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Sign Of Electrode Potential.

The electrode potential of an electrode:
Is positive: If the electrode reaction is reduction
when coupled with the standard hydrogen
electrode
Is negative: If the electrode reaction is oxidation
when coupled with standard hydrogen
electrode. According to latest accepted conventions,
all single electrode potential values represent
reduction tendency of electrodes.
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
when copper electrode is combined with SHE, copper
electrode acts as cathode and undergoes reduction
hydrogen electrode acts as anode.
H2(g) → 2H+ +2e- (oxidation)
Cu2+ +2e- → Cu (reduction)
Hence electrode potential of copper is assigned a
positive sign. Its standard electrode potential is 0.34 V.
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
When zinc is coupled with S.H.E. zinc electrode
acts as anode and hydrogen electrode acts as
cathode.
Zn → Zn2+ +2e2H+ + 2e-→ H2.
Hence, electrode potential of zinc is negative.
The standard electrode potential of zinc
electrode is -0.74 V.
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Nernst Equation.
It is a quantitative relationship between
electrode potential and concentration of the
electrolyte species.
 Consider a general redox reaction:
Mn+(aq) + ne- → M(s) ----(1)
We know that, ΔG =-nFE ----- (2)
ΔGo=-nFEo-----(3)
ΔG =ΔGo +RT ln K

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ΔG =ΔGo +RT ln K
ΔG =ΔGo +RT ln[M]/[Mn+]-----(4)
-nFE= -nFEo + RT ln [M]/[Mn+]----(5)
E= Eo – RT/nF ln 1/[Mn+]------(6)
E=Eo- 2.303 RT/nF log 1/[Mn+]---(7)
At 298K,
E= Eo-0.0592/n log 1/[Mn+]-------(8)
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1. A galvanic cell consists of copper plate
immersed in 10 M solution of CuSO4 and iron plate
immersed in 1M FeSO4 at 298K. If E0cell=0.78 V,
write the cell reaction and calculate E.M.F. of the
cell.
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Solution:
Cell reaction:
Fe + Cu2+
↔
Fe2+ + Cu
ECell= E0Cell-0.0592/2 log [Fe2+ ]/[Cu2+]
ECell= 0.78 + 0.0296 log 10/1
=0.8096V
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Calculate E.M.F. of the zinc – silver cell at 25˚C when
[Zn2+ ] = 1.0 M and [Ag+] = 10 M (E0cell=1.56V at
25˚C). Write the cell representation and cell
reaction
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Solution:
 Cell representation
 Zn/ Zn2+((1M)//Ag+(10M) /Ag
 Cell reaction:
Zn + 2Ag+
↔
Zn2+ + 2Ag
ECell= E0Cell-0.0592/2 log [Zn2+ ]/[Ag+]2

ECell= 1.56 + 0.0592 log 10/1.0
=1.6192 V
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The emf of the cell
Mg│ Mg 2+ (0.01M)║ Cu 2+ /Cu is measured to
be 2.78 V at 298K. The standard eletrode
potential of magnesium electrode is -2.37 V.
Calculate the electrode potential of copper
electrode
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Cell reaction:
Mg + Cu2+
↔
Mg2+ + Cu
E= Eo-0.0592/n log 1/[Mn+]
EMg= EoMg-0.0592/2 log 1/[Mg2+]
=-2.4291V
Ecell=ECu-EMg
2.78 = ECu-[-2.429]
ECu =2.78-2.429
=0.3509 V
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The emf of the cell
Cu│ Cu 2+ (0.02M)║ Ag+ /Ag is measured to be
0.46 V at 298K. The standard eletrode
potential of copper electrode is 0.34 V.
Calculate the electrode potential of silver.
electrode
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Energetics of Cell Reactions.

Net electrical work performed by the cell
reaction of a galvanic cell:
W= QE ------(1)
Charge on 1mol electrons is
F(96,500)Coulombs.
When n electrons are involved in the cell
reaction,
the charge on n mole of electrons = nF
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Q = nF
Substituting for Q in eqn (1)
W = nFE ----------(2)
The cell does net work at the expense of
-
ΔG accompanying. ΔG = -nFE
-
ΔG = nFE
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From Gibbs – Helmholtz equation.
ΔG = ΔH + T [δ(ΔG)/ δT]P ------- (2)
-nFE = ΔH – nFT (δ E/ δT)P
ΔH = nFT (δ E/ δ T)P – nFE
ΔH = nF[T(δ E/ δT)P –E]

We know that,
[δ (∆G)/ δT]P =
- ΔS
ΔS = nF (δE/ δT)P
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Problem: Emf of Weston Cadmium cell is 1.0183
V at 293 K and 1.0l81 V at 298 K.
Calculate ∆G, ΔH and ΔS of the cell reaction at
298 K.
Solution:- ∆G: ∆G = - n FE
n = 2 for the cell reaction; F = 96,500 C
1.0181 V at 298 K
∆G = -2 x 96,500 x 1.0181 J = -196.5 KJ
E=
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∆H: ∆H = nF [ T (δE /δT)P – E]
 (δE/δT)p = 1.0181 – 1.0183 / 298-293 = 0.0002 / 5

= -0.00004VK-1
 T = 298 K
 ∆H = 2 x 96,500 { 298 x (-0.00004) – 1.0181)

= -198. 8 KJ
 ΔS:
ΔS = nF (δE / δT) P

= 2 x 96,500 x (0-00004) = -7.72JK-1

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Classification of Electrodes.
 Gas
electrode ( Hydrogen electrode).
 Metal-metal insoluble salt (Calomel
electrode).
 Ion selective electrode.(Glass electrode).
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Gas electrode.
It consists of gas bubbling over an inert metal
wire or foil immersed in a solution containing
ions of the gas.
 Standard hydrogen electrode is the primary
reference electrode, whose electrode potential
at all temperature is taken as zero arbitrarily.

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Construction.
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
Representation: Pt,H2(g)/ H+

Electrode reaction: H+ + e- 1/2 H2(g)
The electrode reaction is reversible as it can
undergo either oxidation or reduction
depending on the other half cell.

If the concentration of the H+ ions is 1M,
pressure of H2 is 1atm at 298K it is called as
standard hydrogen electrode (SHE).
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Applications.
To determine electrode potential of other
unknown electrodes.
 To determine the pH of a solution.
E=Eo- 2.303 RT/nF log [H2]1/2/[H+]
= 0 -0.0591 log 1/[H+]
= -0.0591pH.
Cell Scheme: Pt,H2,H+(x)// SHE

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
The emf of the cell is determined.

E (cell) = E (c) – E(A)
= 0 – (- 0.0592 pH)
E (cell) = 0.0592 pH
pH = E(cell)/ 0.0592
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Limitations.
Constuction and working is difficult.
 Pt is susceptible for poisoning.
 Cannot be used in the presence of
oxidising agents.

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Metal –metal salt ion electrode.

These electrodes consist of a metal and a
sparingly soluble salt of the same metal
dipping in a solution of a soluble salt
having the same anion.
Eg: Calomel electrode.
Ag/AgCl electrode.
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Construction.
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
Representation: Hg; Hg2Cl2 / KCl
It can act as anode or cathode depending on the
nature of the other electrode.

As anode: 2Hg + 2Cl- → Hg2Cl2 + 2e-

As Cathode: Hg2Cl2 + 2e- → 2Hg + 2 Cl-
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E = Eo – 2.303 RT/2F log [Cl-)2
= Eo -0.0591 log [Cl-] at 298 K
Its electrode potential depends on the concentration of KCl.
Conc. of Cl-
Electrode potential
0.1M
0.3335 V
1.0 M
0.2810 V
Saturated
0.2422 V
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Applications.
Since the electrode potential is a constant
it can be used as a secondary reference
electrode.
 To determine electrode potential of other
unknown electrodes.
 To determine the pH of a solution.
Pt,H2/H+(X) // KCl,Hg2Cl2,Hg
pH = E(cell) – 0.2422/ 0.0592

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Ion Selective Electrode.
It is sensitive to a specific ion present in an
electrolyte.
 The potential of this depends upon the activity
of this ion in the electrolyte.
 Magnitude of potential of this electrode is an
indicator of the activity of the specific ion in the
electrolyte.
*This type of electrode is called indicator electrode.

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
Glass Electrode:`
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Scheme of typical pH glass electrode
1. a sensing part of electrode,
2. a bulb made from a specific glass
sometimes electrode contain small amount
of AgCl precipitate inside the glass
electrode
3 internal solution, usually 0.1M HCl for pH
electrodes
4.internal electrode, usually silver chloride
electrode or calomel electrode
5.body of electrode, made from nonconductive glass or plastics.
6.reference electrode, usually the same type
as 4
7.junction with studied solution, usually
made from ceramics or capillary with
asbestos or quartz fiber.
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
The hydration of a pH sensitive glass membrane
involves an ion-exchange reaction between singly
charged cations in the interstices of the glass
lattice and protons from the solution.
H+ +
Soln.
Na+
glass
Na+ +
soln.
H+
glass
Eg = Eog – 0.0592 pH
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Electrode Potential of glass
electrode.
The overall potential of the glass electrode has
three components:
The boundary potential Eb,

Internal reference electrode potential Eref.

Asymetric potential Easy.- due to the difference in
response of the inner and outer surface of the glass
bulb to changes in [H+].
Eg = Eb + Eref. + Easy.

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Eb = E 1 – E2
= RT/nF ln C1 – RT/nF ln C2
= L + RT/nF ln C1
Eb depends upon [H+]
Eg = Eb + EAg/AgCl + Easy.
= L + RT/nF ln C1 + EAg/AgCl + Easy.
= Eog + RT/nF ln C1
= Eog + 0.0592 log [H+]
Eg = Eog – 0.0592 pH.

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
1.
2.
3.
4.
Advantages:
It can be used without interference in solutions
containing strong oxidants, strong reductants, proteins,
viscos fluids and gases as the glass is chemically robust.
It can be used for solutions having pH values 2 to 10.
With some special glass (by incorporation of Al2O3 or
B2O3) measurements can be extended to pH values up
to 12.
It is immune to poisoning and is simple to operate
The equilibrium is reached quickly & the response is
rapid
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5. It can be used for very small quantities of the
solutions. Small electrodes can be used for
pH measurement in one drop of solution in a
tooth cavity or in the sweat of the skin
(micro determinations using
microelectrodes)
6. If recently calibrated, the glass electrode gives
an accurate response.
7. The glass electrode is much more convenient
to handle than the inconvenient hydrogen gas
electrode.
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Disadvantages:
The bulb of this electrode is very fragile and has to be
used with great care.
The alkaline error arises when a glass electrode is
employed to measure the pH of solutions having pH
values in the 10-12 range or greater. In the presence of
alkali ions, the glass surface becomes responsive to both
hydrogen and alkali ions. Low pH values arise as a
consequence and thus the glass pH electrode gives
erroneous results in highly alkaline solutions.
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The acid error results in highly acidic solutions
(pH less than zero)Measured pH values are
high.
Dehydration of the working surface may cause
erratic electrode performance. It is crucial that
the pH electrode be sufficiently hydrated before
being used. When not in use, the electrode
should be stored in an aqueous solution
because once it is dehydrated, several hours
are required to rehydrate it fully.
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As the glass membrane has a very high
electrical resistance (50 to 500 mΩ), the
ordinary potentiometer cannot be used
for measurement of the potential of the
glass electrode. Thus special electronic
potentiometers are used which require
practically no current for their operation.
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Standardization has to be carried out frequently
because asymmetry potential changes gradually
with time. Because of an asymmetry potential,
not all glass electrodes in a particular assembly
have the same value of EoG . For this reason, it is
best to determine EoG for each electrode
before use.
The commercial verson is moderately expensive
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Limitations.
The bulb is very fragile and has to be used
with great care.
In the presence of alkali ions, the glass surface
becomes responsive to both hydrogen and
alkali ions. Measured pH values are low.
In highly acidic solutions (pH less than zero)
measured pH values are high.
When not in use, the electrode should be
stored in an aqueous solution.
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Applications.

Determination of pH:
Cell: SCE ║Test solution / GE
E cell = Eg – Ecal.
E cell = Eog – 0.0592 pH – 0.2422
pH = Eog -Ecell – Ecal. / 0.0592
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Problems
The cell SCE ΙΙ (0.1M) HCl Ι 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
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Eog= 0.0592pH +Ecell + Ecal.
= 0.0592x2.8 +0.24 + 0.2422
=0.648 V
pH = Eog -Ecell – Ecal. / 0.0592
= 0.648 -0.26-0.2422/0.0592
= 2.46
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CONCENTRATION CELLS.
Two electrodes of the same metal are in
contact with solutions of different
concentrations.
Emf arises due to the difference in
concentrations.
Cell Representation:
M/ Mn+[C1] ║ Mn+/M[C2]

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Construction.
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
At anode: Zn →Zn2+(C1) + 2e-

At cathode: Zn2+(C2) + 2e-→ Zn
 Ecell
= EC-EA
= E0 + (2.303RT/ nF)logC2[E0+(2.303RT/nF)logC1]
 Ecell = (0.0592/n) log C2/C1
Ecell is positive only if C2 > C1
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
Anode - electrode with lower electrolyte
concentration.

Cathode – electrode with higher electrolyte
concentration.

Higher the ratio [C2/C1] higher is the emf.

Emf becomes zero when [C1] = [C2].
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Problems

Zn/ZnSO4(0.001M)||ZnSO4(x)/Zn is
0.09V at 25˚C. Find the concentration of
the unknown solution.
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Ecell = 0.0592/n log C2/C1
0.09 =(0.0592/2) log ( x / 0.001)
x =1.097M
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2. Calculate the valency of mercurous ions
with the help of the following cell.
Hg/ Mercurous
|| Mercurous
/Hg
nitrate (0.001N)
nitrate (0.01N)
when the emf observed at 18˚ C is 0.029
V
Ecell=(2.303 RT/nF) log C2/C1
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Ecell=(2.303 RT/nF) log C2/C1
0.029 = 2.303RT/n) log (0.01/0.001)
0.029 =0.057 x 1/ n
n = 0.057/0.029 =̃ 2
Valency of mercurous ions is 2, Hg2 2+
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Assignment
Answer the following questions:
1.Distinguish between electrolytic and galvanic cells.
2.Explain the origin of electrode potential. What are
the sign conventions for electrode potential?
3.Give reasons for the following.
i) The glass electrode changes its emf over a period
of time.
ii) KCl is preferred instead of NaCl as an
electrolyte
in the preparation of salt bridge
4. What is meant by a standard cell? Give an example
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5. Quote any four limitations of glass electrode
6.Define liquid junction potential. How it can be
eliminated or minimized?
7.Derive Nernst equation for the single electrode
potential.
8.Describe potentiometric determination of emf of a
cell.
9.Writ e construction and working of Calomel
Electrode
10.What are concentration cells? Show that emf of
concentration cell becomes zero at a certain point
of its working.
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