ELECTROCHEMISTRY
CHAPTER-1ELECTROCHEMISTRY
1.1 Important Terminologies
Current: Flow of electrons through any conducting material is
known as the current.
Oxidation:The tendency to lose electrons
Reduction: The tendency to gain electrons
Electrode:When a metal rod is dipped in its salt solution, it
develops a positive or negative potential. This assembly is called
as an electrode.
Anode:The electrode at which oxidation occurs is called as
anode.
Cathode:The electrode at which reduction occurs is known as
cathode.
Electrolytes:Electrolytes are the conductors which are in the
form of solution or fused state in which conductance takes place
due to the movement of ions present in it.Electrochemical cell: The
electrochemical cell consists of two conductors called electrodes
that are immersed in an electrolyte and are connected externally by
means of a metal conductor.
Electrolysis:The process that occurs in the electrochemical cell
is called as electrolysis. It is the breakdown of electrolyte into
ions by electricity.
Half cell:A part of a cell containing electrode dipped in an
electrolytic solution is called as a half cell.
Oxidation half cell : The electrode where oxidation takes place
i.e. where electrons are lost
Reduction half cell : The electrode where reduction takes place
i.e. where electrons are gained
Conductors:Material that permits electric current to pass
through.
Example: Copper, aluminium, silver, fused salts, bases, aqueous
solutions of acids etc.,
Insulators or non-conductors: Materials which do not allow the
electric current to pass through it.
Example: Wood, plastics etc.,
Standard electrode potential:It is the potential of the
electrode which is in contact with its solution of 1M concentration
which is measured at 298K and 1atm pressure.
Single electrode potential:The measure of tendency of a metallic
electrode to lose or gain electrons, when it is in contact with a
solution of its own salt.
Representation Notation and sign conventions:
1.Molecules, elements, gases and electrode material are
represented by the usual chemical symbols. Concentration of ions
and molecules and partial pressures of gases are given in
brackets.
2.Interface between an electrode and electrolyte or between two
electrolytic solutions is represented by a semicolon (;) or by a
single vertical line (l). Example Zn ; Zn2+ or Zn l Zn2+
3.Example: Ag, AgCl l Cl-In the example the (,) indicates that
the Ag and AgCl should be taken together always to constitute an
electrode.
4.The electrode on the left is written in the order the
electrode and then the ion (Zn l Zn2+)
5.The electrode on the right is written in the order, ion and
then electrode (Zn2+l Zn)
6.A double vertical line (ll) represents a salt bridge which has
zero potential difference.Example:Zn l Zn2+ ll Cu2+ l Cu7.The
combination of two single electrodes or two half cell constitute an
cell
The left hand electrode is anode The right hand electrode is
cathodeThe left hand half cell is cathode half cellThe right hand
half cell is anode half cell
Example: Zn l Zn2+ ll Cu2+ l Cu(anode)(Cathode)8.At anode
oxidation takes place Zn Zn2+ + 2eAt cathode reduction takes place
Cu2+ + 2e Cu9.If E0cell of the reaction is positive, the cell
reaction takes place and is said to be spontaneous.If E0cell of the
reaction is negative, the cell reaction is not feasible and is said
to be non-spontaneous.
1.2 Role of Salt bridge
The salt bridges are U-shaped and contain agar-agar gel
containing KCl or MNO3. Some times NH4NO3 is also used. The reason
for using these salts, is that the cations and anions have almost
same speed and same transport number.
The salt bridge provides a passage for the flow of charges in
the internal circuit and thus prevents the accumulation of charges.
In the presence of salt bridge, the charges present flow towards
the oppositely charged electrodes. The negative ions move from the
copper electrode passes through the CuSO4 solution, salt bridge and
zinc sulphate solution to zinc rod. The positive ions move in the
reverse direction i.e., zinc rod to copper rod. Hence flow of
electricity is maintained when the salt bridge is used.But when the
salt bridge is not used the electrons released by the anode flow to
the cathode through the external circuit. During this there occurs
an accumulation of charges near the two electrodes, which prevents
further flow of current. Hence the electrochemical change stops and
current drops to zero.
Introduction
. Electrochemistry is branch of chemistry. This is based on
properties of a solution when it is made part of an electrochemical
cell. The principle of electrochemical cell deals with the
conversion of energy from electrical to chemical and vice versa.
This principle is used in the construction of cells and batteries
which is a source of electrical energy. Refining of copper
involving electrolysis is done to obtain high quality electric
cables. The electroplated materials are now used in large scale in
automobile industries, in jewelry, etc.. These provide both
protection of metals from corrosion and aesthetic appeal. The
extraction of the metals from its ores like sodium from common salt
also involves electrolysis.
Not only the electrochemical concept is used for protecting the
material but also used in explaining the cause of destruction of
metal. For example, corrosion of steel beams used in the
construction works like bridges, houses etc., occur through this
principle. But methods like cathodic protection has been developed
involving the same principle, so as to counteract and protect
metals from corrosion.
Galvanic cellThese cells convert chemical energy into electrical
energy.
These cells generate an electric current due to chemical
reaction occurring within them. The redox reaction occurring in a
galvanic cell is spontaneous and involves a decrease in the free
energy system.
Example:Dry cell, Lechlanche cell, Daniel cell.
Electrolytic cellThese cells convert electrical energy into
chemical energy.
In these cells chemical change is brought about when from outer
source the current is passed into the electrolyte. This reaction is
non-spontaneous but is feasible.
Example: Used in electroplating, electrotyping, electro-refining
of metals.
Differences between Electrolytic cells and electrochemical
cells.S.NoElectrolytic cellElectrochemical cell
1.In electrolytic cell electrical energy is converted into
chemical energyIn electrochemical cell Chemical energy is converted
into electrical energy
2.The negative charge is carried by cathodeHere the negative
charge is carried by anode.
3.The positive charge is carried by anodeHere the positive
charge is carried by cathode
4.The chemical reaction takes place during electrolysis is
governed by the Faradays law of electrolysisThe e.m.f produced in
the cell depends on the type of the electrode used and on the
concentration of the electrolytes
5.Calumniator is used for the measurement of electricity passed
during the electrolysis process Potentiometer is used for the
measurement of e.m.f produced in the cell
6.Here the external energy is used for the supply of electrons
to the cellHere electrons are drawn from the cell.
1.3 Reversible cell
A cell is said to be reversible when
1.there is infinitesimally small difference in the driving and
opposing force.
2.it is possible to reverse any change taking place by applying
a force infinitesimally greater than one acting on it.Therefore, in
a reversible cell the reaction occurs only when the current is
drawn from connector.
A reversible cell satisfies the following conditions:
1.If an external EMF equal to that of the cell is applied in
opposite direction, no reaction should take place on either of the
two electrodes.
2.If the external EMF is slightly greater than the EMF of the
cell, the overall cell reaction gets reversed. In this case the
current flows in the opposite direction.
ExampleThe Daniel cell is a reversible cell. It maybe
represented by
Zn(s) Zn2+(1M) Cu2+(1M) Cu(s)
At standard conditions, in the Daniel cell with emf 1.1. volts,
the cell reaction is Zn(s) + Cu2+(1M) Zn2+(1M) + Cu(s)
When an external emf is slightly greater than 1.1 volts the cell
reactions gets reversed as follows
Cu(s) + Zn2+(1M) Zn(s) + Cu2+(1M)
The current will now flow in opposite direction.
1.4 Irreversible cell
Cells which do not obey the conditions of thermodynamic
reversibility are called as irreversible cells.
An irreversible cell satisfies the following conditions.
1.Even if external EMF applied is equal to emf of the cell the
reaction does not stop.2.Even when the external EMF applied is
slightly greater than that of the cell, the reaction occurs in the
same direction and does not get reversed.
Example:
Zn(S) H2SO4(soln) Cu(s)
The following cell reaction occurs
Zn + H2SO2 ZnSO4 + H2
Liberation of hydrogen occurs which results in
irreversibility.
1.5 Electrode Potential
When a metal is in contact with a solution of its own ions at
25C, it may either undergo oxidation reaction or reduction
reaction.
Fig 1.1
If it is a oxidation reaction
M(s) Mn+ (aq) + ne
The Mn+ leaves the metal electrode and enters into solution.
This leaves the free electrons on metal electrode resulting in a
negative charge on the electrode. This negative charge attracts the
positively charged ions in the solution leading to the formation of
a layer of positive ions around the metal electrode.
Similarly, if it is a reduction reaction Mn+(aq) + ne M(s)
The Mn+ ion from solution enter into the metal electrode
resulting in positive charge on the electrode. This attracts the
negative charged ions in the solution leading to the formation of a
layer of negatively charged ions around the electrode.The formation
of a layer of negative or positive ions around the metal electrode
is called Helmholtz electrical double layer. Because of this layer
a potential difference is set up between the metal and the
solution. At equilibrium, this potential difference will become
constant and this is known as standard electrode potential of
metal.
Single Electrode Potential
Definition:Single electrode potential is the measure of tendency
of a metallic electrode to lose or gain electrons, when it is in
contact with a solution of its own salt.
The tendency of a metallic electrode to lose electrons is known
as oxidation potential and the tendency of an electrode to gain
electrons is known as reduction potential.
It is impossible to measure absolute value of single electrode
potential because neither oxidation nor reduction takes place
independently. It is possible to measure the potential difference
between two electrodes with the help of the potentiometer. In other
words, we can only measure electrode potential with reference to a
reference electrode.
Standard Electrode Potential
DefinitionStandard electrode potential is the tendency of a
metallic electrode to lose or gain electrons when in contact with
1M concentration of its own salt solution at 250C.
Electromotive Force and its Measurements
DefinitionElectromotive force or EMF of a galvanic cell is
defined as the difference of electrode potential which causes the
flow of current from one electrode to another when virtually no
current is drawn from the cell.(NOTE: the difference between cell
potential and EMF is, cell potential is positive while EMF can have
positive as well as negative value.) MeasurementIt is necessary
that no current should be drawn from the cell while measuring its
EMF. This is possible only when the EMF of the cell under study is
balanced by EMF of some other cell. This is done in Poggendorfs
compensation method
Poggendorfs compensation methodIn this method the EMF of the
given cell is compensated by the EMF of a known standard cell so
that no current could flow in the circuit.
ConstructionIt consists of a potentiometer wire AB stretched
over a meter scale. The cell C is a source of direct current. Its
positive electrode is connected to end A while the negative
electrode to the end B of the wire. X is the cell whose EMF is to
be measured while the cell S is standard cell of known EMF. The
positive electrodes of both the cells X and S are connected to the
end A of the wire while their negative electrodes are connected to
the two terminals of a double pole double throw switch P. The third
terminal of the plug is connected to a sliding contact D through a
galvanometer G.
GCell for which emf is to be measuredABDDGalvanometerSliding
contactSXStandard cell-+C (Storage battery)RAdjustable resistance
Measurement of emf of a cellP
Fig 1.2
WorkingStandard cell S is first brought into the circuit with
the help of the double pole double throw switch and the null point
D is found when no current passes through the circuit and
galvanometer does not record any deflection. At this point, the EMF
of the cell C is balanced by that of the cell S. If Ec is the EMF
of the cell C and Es is the EMF of the cell S, then
(Ec / Es) = AB/ADNow, the cell S is removed from the circuit and
the cell X is connected in the circuit. The bridge is again
balanced and the null point D is found for no current flowing in
the circuit. At this stage, the EMF of the cell C is balanced by
that of the cell X. If EX is the EMF of the cell X, then
(Ec / Ex) = AB/AD
From the above equations
i.e.,
Since the EMF of the standard cell S is known and the length AD
and AD can be directly read with the help of meter scale, the EMF
of cell X under study can be obtained.
Standard cellPoggendroffs compensation method requires a
standard cell whose EMF is standard and is precisely known. For
this a saturated Weston cadmium cell is used. The standard cell has
a very low temperature coefficient and its EMF does not change with
time. These unique properties make it a standard cell.
RepresentationCd(12.5% in Hg) 3CdSO4.8H2O(Satd soln) Hg2SO4(s),
Hg(l)
Electrode reactions
At anode :Cd(s) + SO42-(aq) CdSO4(s) + 2e
At cathode :Hg2SO4(s) + 2e 2Hg(l) + SO42- (aq)
Overall cell reaction :
Cd(s) + Hg2SO4(s) CdSO4(s) + 2Hg(l)
Construction
This type of cell consists of a H-shaped glass vessel. The
bottom of the glass limbs are sealed with platinum wire. The
cathode contains mercury and mercurous sulphate is placed over it.
The anode consists of an amalgam of cadmium (Hg-Cd) over which some
crystals of CdSO4. (8/3)H2O are placed. The rest of the part of the
vessel is filled with saturated solution of CdSO4. The upper end of
the arms is closed with corks. The purpose of solid crystals is to
keep the electrolyte saturated at all temperatures. The emf of this
cell is 1.01807 V at 20C.
+ SaturatedCdSO4 CorkHgSO4 HgCd-HgFig 1.3 The Weston Standard
cellCdSO4 8/3 H2O SaturatedCdSO4
Applications of EMF measurements
1.Determination of solubility of sparingly soluble salts and its
solubility products2. Determination of the valency of an
ion3.Determination of pH4. Determination of standard free energy
change and equilibrium constant.5.In potentiometric titrations.
1.Determination of solubility of sparingly soluble salts and its
solubility productsThere are salts that are sparingly soluble in
water, whose solubility is difficult to be determined by other
methods. Example :Silver chloride (AgCl)
The solubility of these sparingly soluble salts can be
determined by the EMF measurement by constructing a concentration
cell.
Construction of the cell
Ag(s) AgCl(s) KCl (0.01N) AgNO3(0.01 N) Ag(s) (unknown)
In this, one of the Ag electrodes is placed in contact with
0.01N AgNO3 solution and the other electrode is in contact with
0.01N KCl solution. The two solutions are connected through a salt
bridge containing saturated solution of ammonium nitrate. A drop of
AgNO3 solution is added to the KCl solution. This forms a small
amount of AgCl which is sufficient to give a saturated
solution.
WorkingOne of the electrodes of this cell thus is in contact
with a solution of silver ions of known concentration (0.01N). The
other electrode is in contact with a solution of unknown
concentration of silver ions furnished by the ionisation of
sparingly soluble silver chloride formed.
The EMF of the above cell is given as
i.e.,
where, the only unknown factor is the C, the concentration of
Ag+ ions furnished by AgCl in KCl solution. The emf of the cell can
be measured and thus the concentration of AgCl C is calculated.
Multiplying the concentration of AgCl with the equivalent weight
of AgCl (143.5), the solubility of AgCl is calculated in
gm/litre.(OR)
The solubility product of silver chloride is given by
Ksp(AgCl) = C 0.01
The solubility (S) of the silver chloride is then given by
S =
2 Determination of the valency of an ionThe valency of the ions
could be determined from the EMF of the concentration cells.
ExampleThe valency of the mercurous ion could be obtained as
follows
Construction of the cell
Mercury / Mercurous nitrate (C1) // Mercurous nitrate (C2) /
Mercury solution solution
The salt bridge contains the saturated solution of ammonium
nitrate.
The EMF of the cell is given by
C2 and C1 are known concentration of mercurous nitrate
electrolyte. The emf is obtained experimentally. Substituting these
values in the above equation the n the valency of the mercurous ion
is calculated.
3.Determination of pHThe pH could be obtained using any one of
the indicator electrodes.
Example :Using a standard hydrogen electrode:
ConstructionA cell with a reference electrode and hydrogen
electrode is constructed using a salt bridge.
Pt, H2(1atm), H+(c= unknown) // KCl(satd), Hg2Cl2(s), Hg
CalculationThe potential of the reference electrode i.e. calomel
electrode is +0.2422 volts is known. The emf of the cell is
obtained experimentally. Substituting the values in the following
equation Ecell = Eright Eleft(1)
Thus the potential of hydrogen electrode used is calculated as
follows.Eleft = Eright Ecell (2)
To obtain the pH, the potential of the hydrogen electrode
calculated from eqn (2) is substituted in the following Nernst
equation
By convention E0 of hydrogen is zero, therefore
Thus, substituting the value of the hydrogen electrode
calculated from equation(2), the pH could be determined from the
above equation.
By using glass electrode
A cell with a reference electrode and glass electrode is
constructed using a salt bridge as follows.
Construction Pt, HCl(0.1N) / glass / test soln // KCl(satd),
Hg2Cl2(s), Hg(l)
The salt bridge consists of KCl solution.
CalculationThe potential of the reference electrode i.e. calomel
electrode is 0.2422volts is known. The emf of the cell is obtained
experimentally. Thus substituting the values in the following
equation
Ecell = Eright Eleft(1)
The potential of a glass electrode used is calculated as
follows.
Eleft = Eright Ecell (2)
To obtain the pH, the potential of the glass electrode obtained
from eqn (2) is substituted in the equation (3) or (4).
EG = E0G + 0.0591 log[H+]
EG = E0G - 0.0591 pH
. (3)
(OR) ..(4)
[ Where EG = Ecalomel Ecell and glass electrode is anode].
Thus the pH is calculated using the glass electrode.
4.Determination of standard free energy change and equilibrium
constantThe standard free energy change (G0) of a reaction can be
calculated from the using the standard emf from the following
expression ( )G0 = nFE0Where, F = 96,500 coulombsE0 = Standard emf
of the cell n = number of electrons involved
The equilibrium constant k of a reaction can be calculated from
the following equation:
Where, E0 is the standard EMF of the cell k is the equilibrium
constantn is the number of electrons involved in the reaction.
1.6 Electrochemical series
The electrochemical series is the arrangement of various
electrode systems in the increasing order of their standard
reduction potentials.
Importance of Electrochemical series1.The most active metals are
at the top of the series.
2. The electrode system having negative values of standard
reduction potentials act as anode when connected to a standard
hydrogen electrode, while those having positive values act as
cathode.3.On moving from the top to the bottom in the series,
tendency to gain electrons i.e. the tendency for reduction to occur
increases.Li will act as the strongest reducing agent.
Standard reduction potentials
Electrode reaction
Oxidised formReduced formStandard reductionPotential, E0, V
PotassiumK+ + e- K-2.925
Calcium Ca2+ + 2e- Ca- 2.87
Sodium Na+ +e- Na-2.714
Magnesium Mg2+ +2e- Mg-2.37
Aluminum Al3+ 3e- Al-1.66
Zinc Zn2+ + 2e- Zn-0.763
Chromium Cr3+ + 3e- Cr-0.74
Iron Fe2+ + 2e- Fe-0.440
Lead Pb2+ +2e- Pb-0.126
Hydrogen 2H+ +2e- H20.00
Copper Cu2+ +2e- Cu+0.337
Iodine I2 + 2e- 2I-+0.5355
Mercury Hg22+ + 2e- 2Hg+ 0.789
Silver Ag+ +e- Ag+0.7991
Bromine Br2(l) + 2e- 2Br-+1.0652
PlatinumPt2+ + 2e- Pt+1.2
GoldAu+ + e- Au+1.68
Increasing ease of reduction of reactantFluorine F + 2e- 2F-+
2.87
Increasing ease of oxidation of reactant
1.92 Engineering Chemistry1.91Electrochemistry
4.On moving from the bottom to the top in the series, the
tendency to lose electrons i.e tendency for oxidation increases.F2
will act as the strongest oxidizing agent.5.The systems placed
above hydrogen are stronger reducing agents, while the one below
the hydrogen are weaker reducing agents.
APPLICATIONS OF ELECTROCHEMICAL SERIES1.To predict the relative
oxidizing and reducing powerIn general, Oxidizing agents have
positive Eo Values Higher the positive value, stronger will be the
oxidizing agent that is placed below the hydrogen.
Reducing gents have negative Eo values Higher the negative value
stronger will be the reducing agent that are placed above the
hydrogen
2.To predict the spontaneity of any redox reactionFor any
spontaneous reaction G = -veSince G = -nFEcell Ecell should be
positive for spontaneous reaction.
The Ecell is calculated from the standard redox potentials by
using the relationEcell = Ecathode Eanode
3.Displacement of hydrogen by metals from dil. acidsThe
displacement of hydrogen from an acid is a reduction process
i.e.
2H+ + 2e H2(g) Eo = 0.00 volts
This is brought about by a metal whose reducing power is greater
than that of hydrogen. Hence any metal placed above hydrogen in
electrochemical series i.e. with negative std. reduction potential
can displace hydrogen from a dil. acid.Example :Zn, Mg, Ni.
4.Displacement of metals from salt solutionA metal with lower
value of reduction potential can displace another metal with higher
value of reduction potential.
Example: Zinc displaces copper from its solution spontaneously,
but reverse reaction is not feasible.
Zn + Cu2+ Cu + Zn2+
5.Electropositive character of metals (tendency to form positive
ions)The metals at the top of the electrochemical series have the
greater tendency to lose electrons. So, these metals at the top,
possesses the highest electropositive character. The value of the
std. reduction potentials increases in going down the series, which
implies that the electropositive character of metals
decreases.Example: Alkali metal > Hg, Ag, Au
6.Thermal stability of the metal oxideHigher the electro
positive character, greater is the stability of its oxide. Hence
the stability of oxides of metals decreases in going from top to
bottom of the series. Example: Alkali metal oxides like Na2O is
more stable than Ag2O, HgO etc.
7.Calculation of standard cell potentials (EoCell)The EoCell is
calculated using
EoCell = EoCathode - EoanodeWhere EoCathode is the standard
reduction potentials of cathodeEoanode is the standard reduction
potentials of anode.
8.To calculate equlilbrium constants Go = - RTlnKeqm Go = -
nFEoEo = (RT/nF) ln keqm ln keqm = nFEo/RTFrom E0 equilibrium
constant can be calculated.
9. Prediction of correct metallurgical methodEo values of Cu,
H2O and Al are (+)0.34, (-)0.83and (-) 1.66V. Cu gets easily
reduced when compared to H2O and H2O get reduced more easily than
Al. Thus, copper is produced by electrolysis of aqueous CuSO4 and
aluminium cannot be produced form aqueous solution of aluminium
This is because when Al3+(aq) is electrolyzed, only H2O will be
electrolysed and not Al3+
1.7 Thermodynamics of a reversible cell
NERNST EQUATION FOR ELECTRODE POTENTIAL
Nernst equation is used to calculate the electrode potential of
the electrode assembly under any given condition. It tells us the
effect of electrolyte concentration on electrode potential.
Nernst equation is given by
This is applicable to a general redox reaction. Consider a
equation
aA + bB cC +dD
For a general reversible chemical system, according to
Vant-Hoffs isotherm and Gibbs free energy, the relation between the
free energy change G and its equilibrium constant is expressed
as
. (1) But , G0 = ( ) RTln K (2)
Substitute equation (2) in (1)
. (3) where, G0 = standard free energy change .
For a reversible reaction, the decrease in free energy (G ) is
given by G = () nFE ...(4)
G0 = () nFE0
(or) ()G0 = nFE0 (5)
Where , G0 is the standard free energy change . n is the number
of electrons F is Faraday = 96,500 Coulombs of electricity E0 is
the standard potential
Substituting the equation (4) and (5) in equation (3), we
get
Rearranging the above equation
(6)
Consider a reduction reaction Mn+ + ne M Applying eqn. (6) to an
electrode reduction equation
Since [M] = 1 for the solid metal
The above equation may be written as,
Where, R = 8.314 J/K/mole; T = 298C; F = 96500 coulombs.
Substituting the value of R, T and F
The above equation is known as Nernst equation.
Nernst equation for a reduction reaction
Consider an oxidation reaction M Mn+ + ne
Similarly for an oxidation reaction Nernst equation is
Nernst equation for an oxidation reaction
Applications of Nernst equation1. To study the effect of
electrolyte concentration on electrode potential2. For calculation
of the potential of a cell under non-standard conditions3.
Determination of unknown concentration of one of the ionic species
in a cell when a concentration of other ionic species is known.4.
The pH of a solution can be calculated.5. To find the valency of an
ion or the number of electrons involved in the electrode
reaction.
1.8 Types of Electrodes
Electrodes are of two types namely:Reference electrode Indicator
electrode
We know that absolute value of a single electrode potential
cannot be measured, because the oxidation or reduction does not
take place independently. The potential difference between two
electrodes can be measured only by combining two electrodes to form
a complete cell and by using the potentiometer.Thus electrode
potential can be measured with reference to a reference
electrode.
Reference electrodes:Reference electrode is an electrode which
has a constant emf or constant potential, with which we can compare
the potentials of other electrodes.
There are two types of reference electrode 1)Primary reference
electrode Example : Standard hydrogen electrode (SHE)
2)Secondary reference electrodeExample : Calomel electrode,
Ag-AgCl electrode, Glass electrode, Quinhydrone electrode etc.
PRIMARY REFERENCE ELECTRODE Standard Hydrogen Electrode
(SHE)
Fig 1.4
Standard hydrogen electrode is an example of primary reference
electrode. It consists of platinum foil or wire coated with
platinum black dipped into a molar (1M) solution of H+ ions. The
platinum wire is surrounded by an outer tube into which the
hydrogen enters through a side inlet and escapes at the bottom
through the test solution. Hydrogen gas at 1 atmosphere pressure is
passed through it continuously at 2980K. Although an open vessel as
shown in diagram, in practice the electrode will be used in a
stoppered flask with a suitable exit for hydrogen. Its potential is
arbitrarily taken as zero at all temperatures. SHE is a reversible
electrode i.e, depending on the nature of another electrode to
which it is connected, i.e., this electrode may act as anode or
cathode.Representation: Pt, H2(g), (1atm) /H+(1M)
Reaction As Anode : H2(g) 2H+(aq) + 2e
As cathode : 2H+(aq) + 2e H2(g) Electrode Potential of SHE :
arbitrarily considered as zero
SECONDARY REFERENCE ELECTRODEDue to the difficulties encountered
with SHE, secondary reference electrode is developed
Definition:The electrodes whose electrode potentials are found
using the primary reference electrode and are always constant at
given conditions are used instead of primary reference
electrode.
As at first the potential is obtained from the primary reference
electrode which is then being used as a reference to another (to a
third) electrode in order to obtain its (third electrodes)
electrode potential this is called as secondary reference
electrode.
Example:Calomel electrode, Ag-AgCl electrode, quinhydrone
electrode etc.
Calomel electrode It is a secondary reference electrode
ConstructionIt consists of a glass tube provided with a bent
side tube A and another side tube B containing a rubber tubing that
could be closed with a screw clip. The glass tube consists of layer
of pure mercury in the bottom. The surface of the mercury is
covered with a paste of mercurous chloride and mercury in potassium
chloride.
Fig 1.5 Calomel ElectrodeABA saturated solution of KCl is filled
in the tube. A platinum wire sealed through a glass tube C, is
dipped in the mercury layer for electrical contact. The glass tube
C contains a little mercury into which an amalgamated copper wire
dips. The side tube is used to make electrical contact with other
electrode through a salt bridge.
Representation: Hg(l), Hg2Cl2(S) / KCl(satd)Reaction Anode2Hg(l)
Hg22+(aq) + eHg22+(aq)+ Cl-
Hg2Cl2(S)------------------------------------2Hg(l) + Cl-(aq)
Hg2Cl2(S) + e
Cathode
Hg2Cl2(S) + e
2Hg(l)Electrode potential : E = E0Cl-/Hg2Cl2/Hg - (RT/nF) ln
[Cl-]
Working
Electrode As anodeHg liberates electrons and sends Hg22+ ions
into solution. The Hg22+ would combine with Cl- ions furnished by
the KCl forming sparingly soluble Hg2Cl2. The result is a fall in
concentration of the chloride ions in the solution with electrons
transferred to metal.
Electrode As cathodeIf the electrode involves reduction
reaction, the Hg22+ ions furnished by the sparingly soluble Hg2Cl2
would be discharged at the cathode. Hence more and more of calomel
would pass into solution. This results in an increase in
concentration of the chloride ions. Thus the potential of the
calomel electrode depends on the activity of the chloride ions and
increases as the activity of the chloride ions decreases. The
electrode is reversible with respect to chloride ions.
Potential of saturated calomel electrode (SCE) is (+)0.2422
volts, Normal calomel electrode (NCE) i.e. 1N KCl is (+)0.28 volt
and Decinormal calomel electrode (DNCE ) i.e. 0.1N KCl is (+)0.3338
Volts
Advantages of calomel electrode1.It is simple to construct2.
Results of cell potential measurements are reproducible and stable
over a long period and do not vary with temperature.
Silver Silver chloride electrode
ConstructionIt consists of a silver wire or a silver plated
platinum wire coated electronically with a thin layer of silver
chloride, dipping into a potassium chloride solution of known
concentration. The potentials of the 0.1M and saturated Ag-AgCl
electrode at 25C with respect to the normal hydrogen electrode are
0.29. and 0.199 volt respectively.
Representation : Ag(S), AgCl(s) / KCl(satd)Reactions : At
anode
Ag(s) Ag+ + e-
Ag+ + Cl-(aq) AgCl(s)
Ag(s) + Cl-(aq) AgCl(s) + e-
Electrode potential = E = E0 +
At cathode
AgCl(s) Ag+ + Cl-(aq)
Ag+ + e- Ag(s)
AgCl(s) + e- Ag(s) + Cl-(aq)
This result in the removal of chloride ions from the solution
with electrons transferred to the metal. The overall reaction
involves only the concentration of Cl- ions as the variable. Thus
the electrode is reversible with respect Cl- ion.
Electrode potential: E = E0Cl + (RT/nF) ln [Cl-]
Working
Electrode as anode
Since the sparingly soluble salt AgCl is in contact wit Cl- ion
solution, the solution is saturated. Some Ag passes from the
electrode into the solution
Ag Ag+ + e
The Ag+ ions produced combine with Cl- of KCl to form AgCl(s).
This result in the removal of chloride ions from solutions with
electrons transferred to metal.
Electrode as CathodeIf the electrode involves reduction
reaction, the Hg22+ ions furnished by the sparingly soluble Hg2Cl2
would be discharged at the cathode. Hence more and more of calomel
would pass into solution. This results in an increase in
concentration of the chloride ions.
Thus the potential of the calomel electrode depends on the
activity of the chloride ions and increases as the activity of the
chloride ions decreases. The electrode is reversible with respect
to chloride ions.
UsesThis electrode is used to measure the concentration of the
chloride ions.
Quinhydrone electrode :Quinhydrone is a compound of quinine and
hydroquinone and in solution is decomposed into equimolar
quantities of these substances.C6H4O2.C6H4(OH)2 = C6H4O2 +
C6H4(OH)2 Quinhydrone quinine hydroquinoneThis consists of platinum
wire dipped in an equimolar solution of hydroquinone and quinone.
The electrode is represented as Pt(s) QH2(aq), Q(aq)_, H+(aq)Where
QH2 and Q represent hydroquinone C6H4(OH)2 and Quinone C6H4O2
respectively .The electrode reaction involved is
C6H4O2(aq) + 2H+(aq)+ 2e C6H4(OH)2(aq)
Q(aq) + 2H+(aq) + 2e- QH2(aq)
The potential developed on platinum electrode is given by Nernst
equation as
In practice the [Q] / [QH2] is maintained constant at unity by
saturating the solution with the substance quinhydrone, which is
equimolar mixture of quinone and hydroquinone. The middle term
reduces to zero.
(since ln[H+]2 = 2ln[H+] )
E = E0 + 0.0591 log [H+] at 250 C
(or) E = E0 0.0591 pH
Where E0 is the standard electrode potential of quinhydrone
electrode = 0.6998V.
(or) E = 0.6998 0.0591 pH
Determination of pH of a solution using quinhydrone electrodeTo
find the pH of a solution, quinhydrone electrode is connected to a
saturated calomel electrode.
This can be represented as SCE // H+ (unknown) / Q(aq) , QH2(aq)
/ Pt(s) Ecell = Eright - Eleft
Ecell = 0.6998 - 0.0591pH 0.2422
Merits of quinhydrone :1. It attains equilibrium rapidly2. It is
not so readily incapacitated as the hydrogen electrode3. It has low
internal resistance4. It is free of salt errors and non-reducing
errors.5. It can be used for pH measurements in non-aqueous
media.
Demerits of quinhydrone1. It cannot be used in solution of pH
greater than 8.2. It is not stable for long time at high
temperatures.
Indicator electrode or Ion selective electrodeThe electrode in
which the potential changes with change in concentration is known
as indicator electrode. i.e. the indicator electrode of a cell is
one whose potential is dependent upon the activity of a particular
ionic species whose concentration is to be determined.Example :
glass electrodeGlass electrodeThe glass electrode is the most
widely used hydrogen ion responsive electrode
Principle : When a glass membrane is immersed in a solution, a
potential is developed between the two surfaces of the membrane
which is a linear function of the hydrogen ion concentration of the
solution i.e. pH value
ConstructionThe glass tube of the glass electrode is made of
Lithium based glasses of composition SiO2 63%, Li2O 28%, Cs2O 2%,
BaO 4%, La2O3 3%. Its melting point should be low and electrical
conductivity is high.
Fig 1.6 Glass electrode
The electrode consists of a glass bulb. This is filled with a
solution of hydrochloric acid (usually 0.1M). A silver- silver
chloride electrode is inserted into it. The upper end of the
electrode must be sealed. This is done in order to ensure a
constant concentration of the inner hydrochloric acid solution. The
internal hydrochloric acid is maintained at constant concentration,
so that 1. the potential of the silver silver chloride electrode is
constant 2. the potential between the hydrochloric acid and the
inner surface of the glass bulb is also constant. Hence the only
potential which can vary is that existing between the outer surface
of the glass bulb and the test solution into which it is
immersed.
So the overall potential of the electrode is governed by the
hydrogen ion concentration of the test solution.
Representation : Ag(s) , AgCl(s) /HCl (0.1M) /Glass
WorkingTo measure the hydrogen ion concentration of a solution
the glass electrode is immersed in the test solution and is
combined with a reference electrode a calomel electrode. Thus the
cell representation is
Internal std soln/ ref. electrode 2Ref. electrode1 / test
soln
External reference electrodeIon selective electrode
Ag(s),AgCl(s) /HCl(0.1M) /Glass/ Test soln // KCl(satd)
/Hg2Cl2(S), Hg(l)
Measurement of pH using glass electrode
Fig 1.7
For pH measurements, E0G is first measured by dipping glass
electrode in buffer solutions of known pH values. Once E0G for a
particular arrangement is determined, the electrode is then placed
in the solution of unknown pH, the emf of the cell is measured and
using the equation, its pH can be calculated. Ecell = Eright Eleft
= 0.2422 (E0G - 0.0591 pH) = 0.2422 E0G +0.0591 pH
(or)
The emf of the cell is expressed as E = K + (RT/F) ln[H+]E = K +
0.0591pH
In these equations K is constant and partly dependent on the
nature of the glass used, individual character of each electrode
and its value may vary with time. This variation of K with time is
related to the existence of an asymmetry potential. The asymmetric
potential is due to the differing responses i.e. the strain due to
changes of hydrogen ion activity that is developed at the inner and
outer surfaces of the glass bulb. So glass electrode has to be
standardized frequently placing in a buffer solution of known
hydrogen in activity.
Advantages of Glass Electrode 1. It can be used in any
solutionsExample: in alkaline, oxidizing, turbid, coloured
solutions etc2. It is most convenient and simple to use3. It is not
easily poisoned4. Equilibrium is easily achieved5. A small quantity
of solution is sufficient for the determination of pH.
Disadvantages1. It gives erroneous results when used in poorly
buffered solutions which are nearly neutral.2. It is found to be
too sensitive to Na+ ions at very high pH values greater than pH =9
resulting in alkaline error
Types of electrodeType (I) electrodeMetal - Metal ion
electrodeMetal amalgam electrodeGas ion electrodeType (II)
electrodeMetal Metal insoluble salt electrode reversible to anion
Type (III) electrodeMetal Metal insoluble salt electrode reversible
to cation Type (IV) electrodeRedox electrode
Type (I) electrode
1.Metal - Metal ion electrode
Pure metal (M) is in contact with a solution of its cation
(Mn+)
Representation : Mn+(aq) / M (s)
Reaction : Mn+(aq) + ne M (s)
Electrode potential :
Example Zinc electrode: Zn(s) /ZnSO4(aq) Copper electrode:Cu(s)
/CuSO4(aq)
2.Metal amalgam electrodeMetal amalgam is in contact with a
solution of metal ion
Representation : Mn+(aq) / M(Hg)(s)
Reaction : Mn+(aq) + ne M(Hg)(s)
Electrode potential :
The active metals which cannot form electrodes when dipped in
their salt solution are converted into metal amalgam and used. This
is because the activity of the metal is lowered when mixed with
mercury.
Example : Sodium electrode : Na(Hg)(s) /Na+(aq)
3.Gas Ion electrodeInert metal is dipped in a solution
containing ions to which the gas is reversible. The gas is
continuously bubbled through the solution.
Representation : Pt(s) , X2 (p = y atm) / X-(aq)Reaction : X2
(p) + ne X-(aq)
Electrode potential :
Example Std. hydrogen electrode: Pt(s) , H2(g) /H+(aq) Chloride
electrode : Pt(s) , Cl2(g) /Cl-(aq)
Metal Metal insoluble salt electrode reversible to anion (type
II)Metal (M) is covered by layer of sparingly soluble salt (MX)
immersed in a solution containing a common anion.
Representation: X-(satd) // MX(S)/ M(s)Reaction MX(S) + e M(s) +
X-(aq)
Electrode potential :
Example : Calomel electrode : Hg(l) /Hg2Cl2(S), KCl(satd)Silver
electrode : Ag(s) /AgCl(S), KCl(satd)
Metal Metal insoluble salt electrode reversible to cation (type
III)Metal is in contact with one of its sparingly soluble salt
together with another insoluble salt having the same anion, in the
solution of a salt having a common cation as that of the latter
salt.
Representation: M2X2 (soln), M2X1(S), M1X1(S)/ M1(s)
Reaction M1X1(S)+ M2n+(aq) + ne M1(s)+ M2X1(S)
Electrode potential :
E = E0 M2X2 (soln), M2X1(S), M1X1(S)/ M1(s) + RT ln [M2n+(aq)]
nF
Example Pb(s) PbSO4(s), SrSO4(s), SrCl2(soln)
Redox electrode (type IV)An inert material is (eg. Pt) is dipped
in a solution containing ions in two oxidation states of the
substances.
Representation:Pt/Mn1+(C1), Mn2+(c2)Reaction : Mn1+(C1)
Mn2+(c2)+ e
E Pt/Mn1+(C1), Mn2+(c2) = E0 Pt/Mn1+(C1), Mn2+(c2) - RT ln
[Mn2+] nF [Mn1+]Electrode potential :
Example : Iron electrode : Pt/Fe2+(C1), Fe3+(C2)
GALVANIC CELL (or) VOLTAIC CELLThese cells convert chemical
energy into electrical energy on account of some chemical or
physical change taking place within the cell. The driving force to
cause the physical or chemical change is the decrease in free
energy.
Example : Daniel cell can be obtained by joining an oxidation
electrode Zn/Zn2+ to a reduction electrode Cu2+/Cu as shown in Fig
.
1.1VZn strip (Anode)Cu strip (Cathode) + Zn SO4 solutionCu SO4
solutionPorous plateDaniel cellFig 1.8 ZnSO4 and CuSO4 solution in
direct contact through porous plate
1.1VZn strip (Anode)Cu strip (Cathode) + Zn SO4 solutionCu SO4
solutionGalvanic cellFig 1.9 ZnSO4 and CuSO4 solution are in
Electrical communication through a salt bridgeSalt bridgeKCl
solution
The solutions in which two electrodes are immersed are made in
communication with each other by a direct contact through a porous
diaphragm or through a salt bridge .The two electrodes are
connected at the outer circuit to a device which utilize the
electrical energy produced . The cell is represented as
Zn(s) ZnSO4(aq) CuSO4(aq) Cu(s)
Anodic reaction : Zn Zn2+ + 2e- (Oxidation )
Cathodic reaction: Cu2+ + 2e- Cu (Reduction )
Cell reaction: Zn + Cu2+ Zn2++Cu (redox reaction )
The zinc metal acts as anode because it has greater potential
(0.76 V ) than copper (+0.34 V ) . The emf of the cell is 1.1 volts
.
Galvanic cells are of the following two types
1. Chemical cell2. Concentration cell
CHEMICAL CELL OR FORMATION CELLIn this type, the electrical
energy is produced on account of a chemical change within the
cell.
Ex.Daniel cell, Lechlanche cell, dry cell
Classification1.Chemical cells with transference2.Chemical cell
without transference.
CHEMICAL CELL WITH TRANSFERENCEThe two electrodes are dipped in
the electrolyte containing their ions. These electrolytes are in
direct contact but are separated by a porous diaphragm. This
creates a liquid Junction. Since the ions on the two sides of the
liquid junction are of equal velocities, their moving across the
junction sets up potential called liquid junction potential. It is
represented by Ej. This type of cells involve three types of
potential oxidation potential at the anode, reduction potential at
the cathode and liquid junction potential. However the Ej could be
minimized or even eliminated through the use of a KCl salt
bridge.
Representation : M1/M1n+(aq) M2n+(aq) /M2
Reaction Anode : M1(s) M1n+ (aq) + 2e
Cathode :M2n+ (aq) + 2e M2(s)
Cell reaction : M1(s) + M2n+ (aq) M1n+ (aq) + M2(s)
Cell potential Ecell = Eanode + Ecathode+ Eliquid junction
Example : Zn(s) /Zn2+(aq) Cd2+(aq) / Cd(s)
CHEMICAL CELLS WITHOUT TRANSFERENCEThis type of cell consists of
two electrodes which are dipped in same solution. One of the
electrode is reversible to cation of the electrolyte and the other
electrode is reversible to the anion of the electrolyte. Since both
electrodes are dipped in same solution the liquid potential is
absent. There is no transport of ions so it is called as chemical
cell without transference.Representation:Pt(s) , X2(g) / MX(aq) /
M(s)
Example : Pt(s) , H2(g) / HCl(aq), AgCl(s) /Ag(s)H2 is
reversible to the H+ ions and Ag/AgCl electrode is reversible to
the Cl- ions.
ReactionsAnode :(1/2)H2(g) H+(a1) + eCathode :AgCl(s) + e Ag(s)
+ Cl-(aq)
Cell reaction: (1/2)H2(g) + AgCl(s) H+(aq) + Ag(s) + Cl-(aq)
Cell Potential
ECell = E0Ag / Agcl(s) / Cl- - RT ln aH+ . a Cl- nF (pH2)1/2
1.9 Concentration cell
The electrical energy is produced due to the differences in the
concentration of either the electrolyte or the electrode.The
physical transfer of matter involves a decrease in the free energy
of the system. Thus the process is spontaneous. The electrical
energy arises due to the transfer of matter from the solution /
electrode of the higher activity to that of lower activity
Types of concentration cell1.Electrode Concentration
cell2.Electrolyte concentration cell
ELECTRODE CONCENTRATION CELLImmerse two electrodes of the same
material but of different activities (concentrations) in the same
electrolyte. The electrical energy is produced due to the transfer
of matter from one electrode to other. (i.e. higher activity to
that of lower activity electrode)
In these cells there is no liquid junction potential as these
cells use only one electrolyte.
Types of electrode concentration cell 1.Amalgam electrode
concentration cell2.Gas electrode concentration cell
i.e in electrode concentration cells the electrodes used are
either amalgam electrodes or gas electrodes.
Amalgam electrode concentration cellTwo amalgam electrodes of
different activities of the same metal is immersed in a single
common electrolyte thus forming this cell.
RepresentationM(Hg)(a1) Mn+(aq) M(Hg)(a2)
Cell reactionAnode :M(Hg) (a1) Mn+(aq) + e
Cathode :Mn+(aq) + e M(Hg) (a2)
Cell reaction : M(Hg) (a1) M(Hg) (a2)
Thus a transfer of metal occurs from the anode to cathode giving
rise to emf . Cell potential Ecell = Eright Eleft
Thus the emf of the amalgam electrode concentration cell depends
on the ratio of the activities of the metal amalgam taken in two
electrodes. It does not depend on the concentration of H+ ions in
the solution. Thus, the electrode with higher gas pressure should
be made anode.
Example : Ag(Hg) (a1) / AgNO3(aq) / Ag(Hg) (a2)
GAS ELECTRODE CONCENTRATION CELL
Two gas electrodes of different activities of the same gas are
immersed in a single common electrolyte thus forming this cell.
RepresentationPt(s) , H2 (g) (p1) / HCl(aq)/ H2(g) (p2),
Pt(s)
Cell reaction Anode : (1/2)H2(g) (p1) H+(aq) + e
Cathode : H+(aq) + e (1/2)H2(g) (p2)
Cell reaction : (1/2)H2(g) (p1) (1/2)H2(g) (p2)
Cell potential Ecell = Eright Eleft
Thus the emf of the gas electrode concentration cell depends on
the ratio of the partial pressure of the gas taken in two
electrodes. It does not depend on the concentration of H+ ions in
the solution. Thus, the electrode with higher gas pressure should
be made anode.
Electrolyte concentration cells It consists of two identical
electrodes immersed in into two solutions of same electrolyte at
different concentrations. The electrodes of such cells are
reversible to one of these ions of the electrolyte. In this there
is transfer of matter from electrolyte of higher activity to
electrolyte of lower activity. Due to this transfer emf is
generated.
Example :
Electrolysis of HCl is carried out by passing electricity in an
aqueous solution of HCl through two platinum electrodes. During
this electrolysis process, chlorine is evolved at the anode and
hydrogen is evolved at the cathode. ( Fig ).
1 N HClPlatinum electrodes Fig 1.10 Electrolysis
When electric current is passed in HCl solution the following
reactions take place at the electrodes.
At cathode : The electrode at which reaction occurs is called
cathode.
H+ions will move towards the cathode, where H+ ions gets reduced
to hydrogen.
2H+ + 2e H2
At anode :The electrode at which oxidation takes place is called
the anode.
2Cl Cl2 + 2e
Types of Electrolyte concentration cells
1. Electrolyte concentration cells without transference2.
Electrolyte concentration cells with transferenceElectrolyte
concentration cells without transferenceTwo chemical cells with no
liquid junction potential are combined together in such a way that
they oppose each other. The two electrolytes are thus not in direct
contact with one another.
Example:Pt, H2(g)(1atm) HCl(a1), AgCl(s) Ag(s)
Pt, H2(g)(1atm) HCl(a2), AgCl(s) Ag(s)
These two cells are combined together so that they oppose one
another.
Representation
Pt,H2(g)(1atm) HCl(a1),AgCl(s) Ag(s) - Ag(s) AgCl(s),HCl(a2)
H2(g)(1atm),Pt
Half cell reactions
Left half cell reaction (1/2) H2 + AgCl(s) Ag(s) + HCl(a1)
Right half cell reaction Ag(s) + HCl(a2) (1/2) H2 + AgCl(s)
________________________________Overall reaction HCl(a2)
HCl(a1)
This indicates there is transfer of HCl from one concentration
to other. Since the two half cells oppose each other, the net EMF
of the cell is given by
Ecell = Eright Eleft
Thus the emf of the cells depends on the ratios of the
concentrations/ activities of the two electrolyte solutions used in
the cell.
Electrolyte concentration cells with transferenceIt consists of
two identical electrodes immersed in into two solutions of same
electrolyte at different concentrations. The electrodes of such
cells are reversible to one of these ions of the electrolyte. The
two electrolytes are kept in direct contact with one another. Thus
a liquid junction potential develops at the interface of two
electrolytes.
Electrolyte concentration cells with transference and having
cation as the reversible ion.
RepresentationM(s) MX(a1) MX(a2) M(s)Reactions Left half cell
reactionM(s) Mn+(a1) + ne
Right half cell reaction Mn+ (a2) + ne M(s)
___________________Overall cell reaction Mn+ (a2) Mn+ (a1)
The overall reaction indicates that only the Mn+ from the
electrolyte of concentration C2 is transferred from the electrolyte
of concentration C1.
But in actual as the two electrolytes are in direct contact,
both the ions present in the electrolyte also migrate. Since the
transport numbers of cation and anions are not equal, the relative
migration of the both the ions across the liquid junction
contribute significantly to emf.
EMF of the electrolyte concentration cells with transference and
having cation as the reversible ion
Pt, H2(g)(1atm) HCl(a1) HCl(a2) H2(g)(1atm), Pt
EMF of the electrolyte concentration cells with transference and
having anion as the reversible ion Ag (s), AgCl(s), HCl(a1) HCl(a2)
AgCl(s), Ag
Hence in both the cases, when the electrodes are reversible to
cation or anion, the emf of the cell depends on the transport
number of the ion other than to which the electrode is reversible.
Therefore, such cells are used in the determination of transport
number of the ions.
1.10 Conductometric Titrations
Conductance measurements are frequently employed to find the
endpoints of acid-alkali and other titrations.
(NOTE: The titrating solution in the burette is usually five
times stronger than the solution taken in the conductivity vessel.
This is done inorder to minimize the change in the volume in the
conductivity cell due to addition of burette solution)Experimental
methodsThe measurement of electrical conductance of a solution
amounts to determination of electrical resistance of the solution
because conductance is just the reciprocal of resistance. The
wheatstone bridge is generally employed for this purpose. This is
shown in figure. If the ratio of the arms is equal i.e., R3 = R4 ,
then the resistance of the cell is equal to that taken from the
resistance box R2, when the bridge is balanced. It is then possible
to read directly the resistance of the cell.
The reciprocal of this reading is proportional to the specific
conductance. This conductance is then plotted against the volume
added and by titration curves, the end point may be calculated.
ApparatusThe solution whose conductivity is to be determined, is
placed in the conductivity cell. The cell is connected to the
resistance box R on one side and to a long wire AB on the other
side. I is the induction coil from which the current is led. On
moving the sliding contact along the wire AB, a point is reached at
which the galvanometer reading is zero.
Resistance of the solution = Length BDResistance of RLength
AD
Conductance = (1/Resistance of the solution)
Fig 1.11
In order to use the titrations for high precision work the wheat
stone bridge is replaced in the form of visual detector or
(telephone earpiece). In this apparatus the alternating current is
used. The apparatus shown in fig. is used.
Here the A.C is reduced to about 3 to 5 volts by means of
transformer T. The galvanometer G and a rectifier (D) is used. The
800 ohm resistances A and B are used. The rectifier D may be
rectifying crystal, a copper oxide rectifier or suitable vacuum
tube circuit giving rectification and amplification. The solution
to be titrated is placed in the cell. The resistance A and B
adjusted and then the current is recorded in G. The titration is
now carried out and the galvanometer readings are plotted against
the volume of the titrant added.
Fig 1.12
The end point is determined by the point of intersection of two
parts of the curves.
ACID - BASE CONDUCTIVITY TITRATIONS
Principle of acid base conductivity titrationElectrical
conductance depends upon the number and mobility of ions.
ProcedureThe acid is taken in the conductivity vessel and alkali
in the burette. Alkali is added gradually and the conductance is
noted. The conductance is plotted against the volume of alkali
added. This plot will have two straight lines AB and CD. The point
of intersection X of these two lines gives the volume of alkali
required for neutralisation .Theory of Strong acid vs Strong base
reaction
Increase in conductance due to addition of OH-Decrease in
conductance due to removal of H+End pointInitial conductance due to
H+Volume of alkali added Titration of a strong acid vs strong
baseBCDAConductance
Fig 1.13
The initial conductance is of hydrochloric acid which is due to
the presence of hydrogen and chloride ions. As alkali is added
gradually, the hydrogen ions are replaced by slow moving sodium
ions.
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
So on continued addition of sodium hydroxide, the conductance
will sharply decrease until the acid has been completely
neutralised.
Subsequent addition of alkali will result in introducing the
fast moving OH- ions. The conductance therefore after reaching a
certain minimum value, will begin to sharply increase.
NaOH Na+ + OH-
Theory of Weak acid vs Strong base reaction
Volume of alkali added Titration of a weak acid vs strong
baseConductanceBCDA Increase in conductance due to addition of
OH-Increase in conductance due to formation of highly ionized
saltEnd pointInitial conductance is low due to poor dissociation of
weak acids
When a weak acid (acetic acid) is used the initial conductance
is low because the weak acids have poor dissociation. As alkali is
added gradually, the acid combines with alkali to yield highly
ionized salt. (Acetic acid forms sodium acetate). This contributes
to the increase in conductance.
CH3COOH + Na+ + OH- Na+ + CH3COO- + H2O
On continued addition of sodium hydroxide, the conductance will
slightly increase until the acid has been completely
neutralised.
Subsequent addition of alkali will result in introducing the
fast moving OH- ions. So there is sharp increase in conductance.
NaOH Na+ + OH-
Theory of Strong acid vs Weak base reaction
Titration of a strong acid vs weak baseConductance No change in
conductance due to poor dissociation of weak base Decrease in
conductance due to removal of H+End pointInitial conductance due to
H+Volume of alkali added BCDA
Fig 1.14
The initial conductance is of hydrochloric acid which is due to
the presence of hydrogen and chloride ions. As weak alkali is added
gradually, the hydrogen ions are replaced by slow moving ammonium
ions.
H+ + Cl- + NH4OH NH4+ + Cl- + H2O
On continued addition of ammonium hydroxide, the conductance
will sharply decrease until the acid has been completely
neutralised.
Subsequent addition of weak alkali will not have appreciable
change in the conductance as the weak base has poor
dissociation.
Theory of Weak acid vs Weak base reaction
No change in conductance due to poor dissociation of weak
baseIncrease in conductance due to formation of highly ionized
saltEnd pointInitial conductance is low due to poor dissociation of
weak acidsBCDA
Conductance
Volume of alkali added
Fig 1.15Titration of a strong acid vs strong base
When a weak acid (acetic acid) is used the initial conductance
is low because the weak acids have poor dissociation. As alkali is
added gradually, the acid combines with alkali to yield highly
ionized salt. (acetic acid forms ammonium acetate). This
contributes to the increase in conductance.
CH3COOH + NH4OH CH3COO- + NH4+ + H2O
On continued addition of ammonium hydroxide, the conductance
will increase until the acid has been completely neutralized.
Subsequent addition of weak alkali will not have appreciable
change in the conductance as the weak base has poor
dissociation.
NaOH Na+ + OH- Theory of Mixture of weak and strong acids vs
strong base reactions
BCDA Increase in conductance due to addition of OH-Decrease in
conductance due to removal of H+End pointInitial conductance due to
H+ Increase in conductance due to formn. of highly ionized saltEnd
pointVolume of alkali added Fig 1.16 Titration of a strong acid vs
strong baseConductanceEF
The plot resembles the combination of the strong acid against
strong base and weak acid against strong base.
The strong acid first gets titrated first. Only after the strong
acid is neutralised, the weak acid will be neutralised.
Hence the initial conductance is due to the presence of hydrogen
and chloride ions of hydrochloric acid. As alkali is added
gradually, the hydrogen ions are replaced by slow moving sodium
ions.
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
On continued addition of sodium hydroxide, first there will be
decrease in the conductance due to the removal of fast moving H+
ions. This is followed by slight increase in conductance due to the
highly ionised salt formed from weak acid and strong base reaction.
This increase is noticed until the weak acid has been completely
neutralized. CH3COOH + Na+ + OH- Na+ + CH3COO- + H2O
Subsequent addition of alkali will result in introducing the
fast moving OH- ions which results in a sharp increase in
conductance.
NaOH Na+ + OH-
The plot of conductance against the volume of alkali added will
have three straight lines AB, EF & CD. The point of
intersection of AB and EF gives the volume of alkali required for
neutralization of strong acid. The point of intersection of lines
EF and CD gives the volume of alkali required for neutralisation of
weak acid.
Precipitation reactions
Example : Consider the precipitation of silver chloride from
potassium chloride (KCl) and silver nitrate (AgNO3)
Principle Electrical conductance depends upon the number and
mobility of ions.ProcedureKCl is taken in the conductivity vessel
and AgNO3 in the burette. AgNO3 is added gradually and the
conductance is noted. The conductance is plotted against the volume
of AgNO3 added. The point of intersection of two lines is noted
which indicates the complete precipitation of AgCl.
Theory behind the reaction
Volume of AgNO3 added Fig 1.17 Titration of a weak acid vs
strong baseConductanceBCDA Increase in conductance due to
Ag+NO3-Conductance is constant because Mobility of Cl- and NO3- are
same. End point
The Ag+ ions are replaced by the K+ ions. The mobility of K+
ions and ag+ is more or less same. So the conductance will remain
more or less constant. After complete precipitation the value
increases because of the mobility of K+ ions.
The point of intersection X of these two lines gives the volume
of AgNO3 required for precipitation of AgCl.
Replacement titration
ProcedureA strong acid is added to a salt of weak acid. The
conductance observed is plotted against the volume of strong acid
added.
Theory of Replacement titration
Volume of HCl addedconductanceFig 1.18 Replacement titrationsEnd
pointABDC
When a strong acid is added to a salt of weak acid, the weak
acid is replaced first.
The chloride ions have higher conducting power than acetate
ions. So there will be a slight increase in the conductance upto
the end point. But on continued addition of the strong acid,
conductance increases more rapidly due to H+ ions
The plot of conductance against the volume of strong acid added
will have two straight lines AB and CD. The point of intersection X
of these two lines is the end point required for complete
replacement of ions.Redox Titration : ( Fe2+ Vs Cr2O72- + H+)
ProcedureIron solution is added to an acidified solution of
dichromate. The conductance observed is plotted against the volume
of Iron added.
Theory of Redox titration
End pointVolume of Fe2+ addedconductanceFig 1.19 Redox
titrationsABCD
With the solution of acidified dichromate, if we add iron
solution, due to decrease in H+ ions concentration, the conductance
will decrease up to end point.
6Fe2+ +Cr2O72- + 14H+ 6Fe3+ +2Cr3+ + 7H2O
After the equivalent point there will not be any change in the
conductance.
The plot of conductance against the volume of iron added will
have two straight lines AB and CD. The point of intersection X of
these two lines is the end point required for complete oxidation of
ions.
1.11 Over voltage
The excess potential that is required in order to operate a cell
than the theoretical potential value of the same cell is called
overvoltage.
Example:The theoretical value required for Zn2+ to deposit on
the cathode as Zn is 0.76 . The experimental value is just the
same. So no overvoltage is observed.But in case of evolution of
gases say hydrogen or oxygen at anode, the experimental potential
required to achieve the cell reaction is much greater than the
theoretical potential. This difference in the experimental
potential and the theoretical potential is the overvoltage.
ElectrodeHydrogen overvoltage (in volts)
Platinised platinum0.01
Smooth platinum0.26
Nickel0.33
Copper0.54
Mercury1.04
This overvoltage is sometimes referred to as the bubble
overvoltage, as it is observed just at the point at which the gas
bubbles begin to appear. The value varies largely with nature of
the metal.
Hydrogen overvoltageThe hydrogen overvoltage is the difference
of potential of the electrode at which hydrogen gas is actually
evolved and the potential of the reversible hydrogen electrode. At
zero current density and with platinised platinum as electrode, the
over voltage of hydrogen is zero.
Determination of hydrogen overvoltageConstructionThe
electrolytic solution is taken in the H shaped vessel. There are
two electrodes one of which is platinum dipped in electrolyte. The
cell is then connected to a reference electrode generally calomel
electrode through a salt bridge containing the saturated solution
of a salt. The whole apparatus is then connected to a variable
resistance D through a galvanometer. The potential of the
electrodes is determined by connecting it with a potentiometer.
Fig 1.20
WorkingThe electrolyte used is a dilute Sulphuric acid. The
current density is maintained at a desired level. The potential of
the cathode at a given current density is measured
potentiometrically by combining it with a calomel electrode. The
difference between this potential and the theoretical potential
gives the hydrogen overvoltage at the particular metal at the given
current density.
Factors affecting the phenomenon of over voltage:1.Current
density : The over voltage increases with increasing current
density (I) according to the following equation = a + b log Iwhere
a and b are constants.
The value of b may be represented byb = 2 x 2.303 RT/F
Its value is equal to 0.118 at ordinary temperatures.
2.Hydrogen ion concentration In the absence of strongly absorbed
ions the over voltage is independent of hydrogen ion concentration.
In the case of strongly acidic or alkaline solutions some
deviations occur.3.Temperature Over voltage decreases with increase
in temperature.4.Impurities Impurities present in cathode material
greatly influence the over voltage.5.Pressure At lower pressure,
the over voltage increases rapidly on the cathodes like copper,
mercury or nickel.At high pressures, the over voltage is affected
very slightly.
Applications of overvoltage1.For solutions of equivalent ionic
concentration, the higher the electrode potential involved, the
more easily the cations will be discharged.2.Used in the
electrolysis of a solution which is molar w.r.t zinc sulphate as
well as sulphuric acid. Where electroplating of zinc from acidic
solution is achieved due to the overvoltage only.3.It is used in
electrolysis of neutral solution of cadmium salts. Since the
overvoltage of hydrogen over cadmium is 0.99V more, cadmium
deposition could be achieved before the evolution of hydrogen. 4.
Metal plating is achieved because of the high value of hydrogen
overvoltage.5.In lead accumulators due to over voltage the lead is
deposited on the negative plate accumulators while changing the
cell.6.In the electrolytic reduction of organic compounds. In the
reduction of nitrobenzene, the electrodes with high over voltage
i.e. lead are used instead of platinum.7.It is utilized in the
industrial production of chlorine and NaOH by the electrolysis of
NaCl solution.
1.12 Polarisation
The phenomenon of back EMF brought about by the product of
electrolysis is termed as polarization.
Explanation of polarizationIn a voltaic cell, zinc and copper
electrodes are connected externally by a wire. Hydrogen ions
(carrying a positive charge) will move towards the copper
electrode. They give their charges at the electrode and escape
through the solution as bubbles. In this process, a film of neutral
hydrogen is gradually deposited at the copper electrode. This with
time increases in thickness. So it provides a greater resistance to
the flow of the current through the cell. At this stage the current
is diminished.
As the process proceeds, the film of neutral hydrogen increases
in thickness. As hydrogen is positive towards zinc, an electric
field from the layer of hydrogen to that of zinc is set up. This is
known as back emf. When the field is sufficiently high, it stops
further traveling of positively charged hydrogen ions towards the
copper electrode, and hence the current stops altogether. This is
the stage at which the cell is said to be completely polarized.
Causes of polarization1. The liberated gases offer resistance to
the normal flow of current through the cell which requires high
voltage to keep the normal flow.2. The product of electrolysis may
convert the inert electrode into active electrodes which can
exercise a back emf.3. The deposited metals may also form a cell
functioning in the opposite direction similar to the gas
electrodes.MODERN VIEW1. Polarisation occurs due to the slowness of
one or more processes taking place at the electrodes.2. Due to
slowness of the diffusion of ions in the solution. Concentration
polarization.
Factors which affect electrode polarization :
1.Concentration of electrolyte High concentration of electrolyte
increases the polarization effects.2.Nature of ions deposited on
electrodes : Extent of polarization increases when there is
formation of the adherent and non-porous film on the
electrodes.3.Size of the electrode Small surface area of the
electrode increases the polarization effects.4.Nature of electrode
surface Rough surface of the electrode decreases polarization
effects than smoother of surface.For example, platinum black
electrode exhibits less polarization effect than the smooth
platinum electrode.5.Stirring of the electrolytic solution
Polarization effects can be minimized by stirring.6Temperature The
rate of diffusion of ions increases with increase in temperature,
which reduces the polarization effects.
Effects of electrode polarization1.Polarized galvanic cells
develop smaller potential than theoretically predicted.2.In an
polarized electrolytic cell, more than theoretically required
potential has to be used to maintain a given current strength.
Elimination of polarization 1.Mechanical method By brushing off
the hydrogen from time to time from the electrode.By roughening the
surface of the electrode, so the bubbles may not stick to the
surface.
2.Electro-chemical methodIn these electro-chemical methods, the
two solutions are taken such that the hydrogen meets with a second
solution, from which ions of the same metal as that of the positive
plate or gas is liberated.
3.Chemical methodUsage of depolarizers. The chemicals which are
used to eliminate polarization are known as depolarizers.
The depolarizers are usually strong oxidizing agents (like
nitric acid or chromic acid). They convert evolved hydrogen into
water thus not allowing the hydrogen to get absorbed on the
electrode.
1.13 Decomposition potential
The minimum voltage necessary to produce continuous electrolysis
of an electrolyte is called the decomposition potential of the
electrode.
ExampleWhen two smooth platinum electrodes are placed in dilute
Sulphuric acid and a low voltage is applied, practically no current
flows through the circuit.
ABEDCApplied Voltage (V)Current(C)Fig 1.21 Decomposition
Potential
But as the applied voltage is gradually increased by means of
the external battery, to electrolyse the solution, the current
increases in the manner as shown in the current density vs
potential plot.
It could be seen that an appreciable voltage is required for
free flow of electricity through the cell. The emf applied at the
point D at which the electrolysis restarts and proceeds
continuously, is called the decomposition voltage
Measurement of Decomposition potentialPrincipleThe decomposition
potential can be measured by plotting current density against the
applied voltage. The point where there is sudden increase in the
current gives the measure of decomposition potential.
ConstructionThe apparatus consists of two platinum electrodes A
and A dipped in an electrolyte as shown in fig. S is a stirrer for
stirring the solution constantly. The two electrodes are connected
to a voltmeter V through ammeter C to measure the strength of
current. B is the external source of E.M.F and R is the variable
resistance. The voltage may be measured by voltmeter V.
First of all a resistance is applied from R and the ammeter and
voltmeter readings are noted. Then at another reduced resistance
from R, again the values of V and C are taken. The process is
continued until the electrolysis in the cell is not visible.A
series of current density is taken versus the applied voltage above
and below the decomposition potential. i.e., The current (C ) is
plotted against the applied voltage (V).
Importance and significance of decomposition potential1.It is of
greater importance in the controlled deposition of metals or other
electrolytic products from an solution containing different
electrolytes. The one having the lowest decomposition potential is
electrolysed first.2.The knowledge of decomposition potential are
of use in electro refining, electroplating and electrometallurgy of
metals.3.Used in separation of copper and zinc electrically.
1.14 Kohlrauschs law
Kohlrauschs law states that at infinite dilution when
dissociation is complete and all interionic effects vanish, each
ion moves independently of its co-ion and contributes to the total
molar conductance of an electrolyte a definite share.
Thus the equivalent conductivity of an electrolyte at infinite
dilution (0) is the sum of two parts one due to the cation + and
other due to the anion
0 = 0+ + 0In terms of molar conductivities the Kohlrauschs law
may be mathematically stated as
0m = 0+ + 0where 0+ = contribution of cation towards the molar
conduction at dilution. 0 = contribution of anion towards the molar
conduction at dilution.
Applications of Kohlrauschs law
i) Determination of molar conductivity of weak electrolytesThe
equivalent conductivity for a weak electrolyte does not become
constant even at very high dilutions, so it is not possible to
determine the value of 0 for these directly. Kohlrauchs law can be
used to determine 0 for weak electrolytes.Example : The equivalent
conductance of a weak electrolyte like acetic acid can be
determined by determining the equivalent conductances of strong
electrolytes like HCl , CH3COONa and NaCl at infinite dilutions as
below :
i.e.,
Similarly
The molar conductance at infinite dilution of a sparingly
soluble substances like silver chloride, can also be obtained from
similar considerations.
ii) To find net Ionic product of water
Water dissociates feebly. Dissociation of water may be written
as:H2O H+ + OH-
Its equilibrium is given by the relation
Since the dissociation of water is very low, concentration of
undissociated molecules [H2O] may be considered to remain constant
. Hence
[H+] [OH] is also a constant.This constant is called Ionic
product of water and is represented as Kw . Thus ,
Kw = [H+] [OH]
In pure water the product of the concentration of hydrogen and
hydroxide ion is constant. The conductivity of purest distilled
water is first determined experimentally. Let it be k.The
equivalent conductivity of water is given by
Applying the formula = k V (Volume in cm3 containing 1 eq of
electrolyte)where k = conductivity of water = 18 k ( V = 18 in case
of water)
The degree of dissociation of water ,
From the dissociation of water H2O H+ + OH [H+] = c = [OH]where
c is moles of H2O per litre of water = 1000/18 . Knowing the values
of c and , [H+] and [OH] is computed. And from this
Kw = [H+] [OH] = c2 2 is worked out.
iii) Determination of solubility of sparingly soluble salt :
There are number of salts , such as silver chloride, barium
sulphate, lead sulphate etc., which are so sparingly soluble in
water.
The conductance of the solution containing sparingly soluble
salt in water is determined using conductivity meter
The conductance of water used in the preparation of the solution
is also determined
Specific conductance cell constant
Let the specific conductance be z Sm-1.
Suppose the solubility of silver chloride = x mole /m3Then
concentration of AgCl in aquous solution = x mol/m3
Molar conductance is given by
= At infinite solution, from Kohlrausch law,
or
Solubility of AgCl in water is =
iv) Degree of Dissociation of Weak electrolyte : The degree of
dissociation of weak electrolyte such as NH4OH , acetic acid can be
determined by measuring the equivalent conductivity , of the
solution of the electrolyte at any given dilution. Since 0 is the
equivalent conductance when electrolyte is fully dissociated , the
ratio /0 would give the degree of dissociation of the weak
electrolyte /0 = 0 could be obtained from Kohlrauschs law.
Similarly, if 0 and the degree of dissociation are known, can be
calculated.
v) Limiting (or absolute )mobilities of ions . The mobility of
an ion u is its speed under potential gradient of 1 volt/cm .At
infinite dilution,
Example 1 : The molar conductance of sodium acetate,
hydrochloric acid and sodium chloride at infinite dilution are
91104, 426.16104 and 126.45 104 S m2 mol1 respectively at 250C
Calculate the molar conductance at infinite dilution for acetic
acid.Solution :
= 91104 S m2 mol1
= 426.16104 S m2 mol1
= 126.45 104 S m2 mol1
(or)
= (91 + 426.16 126.45 ) 104= 390.71 104 S m2 mol1
Example 2: The molar conductance of a 0.01 M solution of acetic
acid was found to be 16.30 104 S m2 mol1 at 250C. The molar ionic
conductances of hydrogen and acetate ions at infinite dilution are
349.8104 and 40.9104 S m2 mol1 respectively, at the same
temperature. What percentage of acetic acid is dissociated at this
concentrationSolution :
= ( 40.9 +349.8) 104 = 390.7 104 S m2 mol1
0.01 M acetic acid is 4.172 percent dissociated.
Example 3 : The conductivity of a saturated solution of barium
sulphate at infinite dilution is 3.06 10-6 ohm-1 cm-1 The
equivalent conductance of BaSO4 at infinite dilute is 143 ohm-1 cm2
eq-1. What is the solubility of BaSO4 at 250C in gm equivalent per
litre. ( AU Jan 2005)Solution :Conductivity of saturated solution
of Barium sulphateK = 3.06 106 ohm1 cm1Equivalent conductivity of
BaSO4 =
.
Concentration of BaSO4 = C =
= = 2.14105 gm/equ/litwe know equivalent weight of BaSO4 =
116.5Solubility of BaSO4 = C equivalent weight of BaSO4 = 2.14105
116.5 = 2.49 103 gm lit1
Example 4 : A conductivity cell , when filled with an aqueous
solution of 0.02 M KCl at 250C, had a resistance of 250 ohm. Its
resistance, when filled with 6105 M NH4OH solution was 105 ohm .
The specific conductance of 0.02 M KCl was 0.277 S m-1 . The molar
conductances at infinite dilution of NH4+ and OH ions are 73.4104
and 198104 S m2 mol-1, respectively. Calculate the degree of
dissociation of 6105 M NH4OH solution.Solution :Since specific
conductance = k = cell constant/R , henceCell constant = kR = 0.277
Sm1 ( 250 ) = 69.2 m1For NH4OH solution,c = 6 105 M = 6 105 mol dm3
= 6 102 mol m3
= 115 104 S m2 mol1
According to Kohlrauschs law,0m = 0+ + 0 = (73.4 +198) 104 S m2
mol1 = 271.4 104 S m2 mol1
Example 5: At 250C the specific conductance of a saturated
solution of AgCl is 2.68 104 S m-1 and that of water with which the
solution was made is 0.86 104 Sm-1. If molar conductances at
infinite dilution of AgNO3 , HNO3 and HCl are respectively, 133
104, 421 104 and 426 104 S m2 mol1, calculate the solubility of
AgCl in grams per dm3 in water at the given temperature.Solution :
ksolution = kAgCl + kwater kAgCl = ksolution kwater = (2.68 0.86)
104 S m1 = 1.82 104 S m1Since AgCl is formed according to the
reaction
AgNO3 + HCl AgCl + HNO3
Hence using kohlrauschs law,
= (133.0 + 426.0 421.0 ) 104 S m2 mol1 = 138.0 104 S m2 mol1
and for the saturated solution of the salt,
= 1.32 105 S dm3Mm(AgCl) = 143.5 g mol1Solubility of AgCl =
(1.32 105 mol dm3)(143.5g mol1) = 1.89 103 g dm3
Example 6 : At 250C the specific conductance of a 0.01M aqueous
solution of acetic acid is 1.63 102 S m1 and the molar conductance
at infinite dilution is 390.7 102 S m2 mol1. Calculate the degree
of dissociation and hence the dissociation constant of the
acid.Solution : k = 1.63 102 S m1, c = 0.01 mol dm1 = 0.01 103 mol
m3
The degree of dissociation is given by
The dissociation constant of the acid =
Review Questions1)i) What is single electrode potential ii)
Derive Nernst equation . Explain the various terms involved ( Anna
University May 2003 )2)How is emf of electrochemical cell
determined by potentiometry . (Anna Univ Campus Nov2002 )3)Define
the term single electrode potential .Derive the Nernst equation and
give its applications . (Anna Univ. Model Q.P)4)How is calomel
electrode constructed ? . Discuss how this electrode may be used
for the determination of pH of a solution ( Anna University Model
Q.P )
5) What is emf series . Explain its significance .6)What are
galvanic cells and concentration cell .7)The emf of the cell
Zn/ZnCl2 // AgCl /Ag is 1.005 V at 250C . Calculate the heat of
reaction H at the given temperature.
8)Explain different kinds of single electrodes with examples
.9)What are reversible and irreversible calls . Give examples
10)What is electrochemical series , what are its applications .
11)Define e.m.f , Derive an expression for emf of a cell
12)Explain sacrificial anode cell
13)Determine the emf or Zn-Fe cell at 2980K . The oxidation
potentials of zinc and iron are +0.76V and +0.44V respectively
.14)Determine the potential developed when a zinc rod is kept
immersed in zn(NO3)2 solution of strength 0.5M at 250C E0Zn2+/Zn =
-0.76 V
15) Determine the emf of the cell Zn/Zn2+// Ag+/Ag , when Zn2+ =
0.1 M and Ag+ = 10M , Ecell at 250C = 1.56 V.
16. What do you understand by a half cell ? What are the half
cells present in a Daniel cell ?
17. What for you understand by a standard hydrogen electrode and
what its significance ?
18.What is a salt bridge and what is its role in a Galvanic cell
?19. What is electrochemical series and what are its important
features ? 20. Write short notes on the Following :i) Potential
decompositionii) Over voltage
Example Problems
Example 1: ( Anna SE , Jan2001)Determine the emf of a Daniel
cell at 250C , when the concentration of ZnSO4 and CuSO4 are 0.001M
and 0.1 M respectively . The standard emf of the cell is 1.1 volt
Solution:Given : E0 = 1.1 Volts C2 = 0.1 , C1 = 0.001The emf of the
cell is given by
= 1.1591 VoltsExample 2Determine the emf of a concentration cell
at 250 C consists of two Zn electrodes immersed in a solution of
zinc ions of 0.1M and 0.01M concentration.
Solution :Given : n = 2 C1 = 0.01 C2 = 0.1
The reaction isZn2+ + 2e- ZnThe emf of the cell is given by
= 0.0296 Volts
Example 3 ( Anna University Dec2001 )Determine the reduction
potential of Cu/Cu2+ is 0.5M at 250 C E0Cu/Cu2+ = 0.337 V .Solution
:E0 = 0.337 Volts , [Cu2+] = 0.5The reduction potential is given
by
log [Cu2+] E = 0.0591 n
log 0.5 = 0.0591 2 = 0.328 Volts
Example 4Determine the emf of the concentration cell given
below: Cu Cu2+ Cu2+Cu (0.2M) (2M)
Solution :Given : n = 2 C1 = 0.2 M C2 = 2 M
The reaction isCu2+ + 2e- CuThe emf of the cell is given by
= 0.029 Volts
Example 5Determine the emf of the following cellZn Zn2+ Ag+ Ag ,
[Zn2+ ] = 0.1 M , [Ag+] = 0.1M , E0Ag = +0.8Volts , E0Zn = 0.76
Volts
Solution :Given :[Zn2+ ] = 0.1 M , [Ag+] = 0.1M , E0Ag =
+0.8Volts , E0Zn = 0.76 Volts
The emf of the cell is given by
Ecell = Ecathode Eanode ( EAg+/Ag) (EZn2+/Zn )
= (0.8 (0.76) ) 0.02955 = 1.53045 VoltsExample 6 Determine the
emf of the cell at 250C concentration of ZnSO4 and CuSO4 are 0.01 M
and 0.1 M respectively . The standard e.m.f of the cell is 1.1
volts .Solution:
Given :[ZnSO4] = 0.01 M , [CuSO4] = 0.1 M , Ecell = Ecathode
Eanode = ECu2+/Cu EZn2+/Zn
( We have E0Cu2+/Cu - E0Zn2+/Zn = 1.1 volts )
= 1.129 Volts
Example 7:Determine the emf of a concentration cell consisting
of silver electrodes immersed in 0.01M and 0.1 M solutions of its
ions at 250 C .Solution : For the concentration cell with silver
electrode ,The electrical reaction involved is Ag Ag+ + en = 1 , C2
= 0.1 M and C1 = 0.01MThe emf of the cell is given by
= 0.0591 Volts
Example 8 :The standard electrode potentials of lead and silver
are 0.18V and + 0.8V respectively . Determine the emf .Solution :
E0Pb2+/pb = 0.18 V E0Ag+/Ag = + 0.8 V Ecell = E0cathode E0anode =
+0.8 ( 0.18 ) = 0.98 Volts
Example 9 ( Anna University Nov2001)Determine the concentration
of H+ in the following cell . Pt, H2 ( P= 1 atm) H+ (C = 10-6
M)