Electro Chemistry

 Introduction

Electrochemistry is the study of process involving the interconversion of electrical energy and chemical energy. In brief it deals with the chemical applications of electricity.  Broadly we can classify the cells as electrolytic cells and galvanic cells. In electrolytic cells the conversion of electrical energy  to chemical energy takes place, whereas the chemical energy is converted to electrical energy in a galvanic cell.

Conductance of Electrolytes in Solution 

Electrolytes and Electrolysis :
Electric current is the flow of electrons  generated by a battery when the circuit is completed. Any substance which allows the passage of electric current through it is called an electric conductor. Eg : all metals, graphite, aqueous solution of acids, bases etc.
Electric conductors are of two types :
a)   Metallic conductors or Electronic conductors : Conduct electricity by free, mobile, valence electrons. They involve flow of electrons and there is no chemical decomposition during conductance. Eg: metals, alloys, certain solid salts and oxides.
b)   Electrolytic conductors or electrolytes : These are aqueous solutions or fused (molten) electrolytes which liberate ions and conduct electricity involving the movement of ions, resulting in its chemical decomposition. eg. acids and bases.
Table - 2.1 : Comparison of metallic conductors with electrolytic conductors
  S.No. Metallic conductors  Electrolytic conductors
    1. Conductance is due to the flow of Conductance is due to the movement       free mobile electrons. of ions in a solution of fused electrolyte.
    2. The chemical properties of metallic The conductance involves the
    conductor does not change. chemical reactions of the electrolyte
at the two electrodes.

 S.No.   Metallic conductors Electrolytic conductors
  3. There is no transfer of matter during Transfer of electrolyte in the form  
    conductance. of ions takes place.
  4. The resistance of the conductor The resistance of the conductor
    increases with increasing temperature. decreases with increasing temperature.

The process of decomposition of an electrolyte by passing electric current through its solution is termed electrolysis which is carried in an apparatus called the electrolytic cell. The cell contains aqueous solution of an electrolyte in which two metallic rods, the electrodes,  are dipped which are connected to a battery. The electrode through which current enters the cell is called ANODE denoted as positive electrode. CATHODE is the negative electrode at which current leaves the cell.
“An electrolyte is a substance which in aqueous solution or in molten state liberates ions and allows the  passage of electric current to pass through resulting in chemical decomposition and the phenomenon is called electrolysis.” The products of electrolysis appear only at the electrodes. A substance as such or in aqueous state does not conduct electricity is called non-electrolyte.
Electrolytes obey Ohm’s law to the same extent as metallic conductors. According to Ohm’s law “the current ‘I’ flowing through a conductor is given by the relation E/R where E is electromotive force i.e., the difference of potential at the two ends of the conductor and R is resistance”.

The strength of the current is usually measured in amperes. The ampere is defined as “the current which deposits 0.001118 gms of silver per second from a 15 percent solution of silver nitrate in a voltameter of definite dimensions”.
The quantity of current passing through a conductor is the product of current strength and time and generally expressed as coloumb. A coloumb is defined as “the quantity of current which passes in one second with a current strength of one ampere”.
The unit of electrical resistance is ohm, which is defined as “the resistance at  0 0C of a column of mercury 106.3 cm long and weighing 14.4521 gms”.
The electromotive force is measured in volts. The volt is “the difference in potential required to send a current of one ampere strength through a one ohm resistance”.
The above three units of electrical measurement are therefore related to each other by ohm’s law as

Concepts of ionic COnductance

  The conductance of an electrolyte is due to the movement of ions under electric field and the different types of conductances are given below.
2.3.1  Specific Conductance
The resistance offered by a conductor to the passage of electricity through it is directly proportional to length (l) and inversely proportional to the area of cross section (S). The resistance R is given by the relation
where ‘(rho)’ is a constant known as specific resistance depends on the material of the conductor. When l = 1 cm and S = 1 cm2, R = , hence “the specific resistance of a metallic conductor is the resistance in ohms which a centimeter cube of it offers to the passage of electricity”.
A substance which offers very little resistance to the flow of current allows more current to pass through it, so that the power to conduct electricity is inverse of resistance.
Specific conductivity or specific conductance (K) is the reciprocal of specific resistance.
K =      
K =
Specific conductance is generally expressed in reciprocal ohm or mho or ohm-1. It is measured in units of ohm-1  cm-1 because
K = = ohm-1 cm-1
Internationally recommended unit for ohm-1 is siemens (S), when S is used the specific conductance is expressed in the units S cm-1. Siemens is not plural. The unit is named after a noted electrical engineer Sir Williams Siemens.
Unlike metallic conductors, electrolytic conductors are aqueous solutions, where we cannot measure the length and area of cross section of the conductor to measure the specific conductance.
In such cases to measure the specific conductivity of an aqueous solution of an electrolyte dilution is taken as a basis.  A certain weight of an electrolyte is dissolved in V ml of solvent and the conductance of one ml of the resulting electrolyte solution at a given dilution V is called the specific conductivity  of the electrolytic conductor which is denoted by KV (Fig. 2.1).

2.3.2  Equivalent Conductance
If one gram equivalent weight of an electrolyte is dissolved in V ml of the solvent, the conductivity of all ions produced from one gram equivalent of an electrolyte at the dilution V is known as equivalent conductance, which is denoted by the symbol . Out of the V ml of the electrolyte solution the conductance of 1 ml of the electrolyte solution is called specific conductance.  Hence the equivalent conductance is equal to the product of specific  conductance (Kv) and the volume (V).

Let C be the concentration of a solution containing gm equivalents of an electrolyte per litre and the volume V of the solution will be 1000/C.

Diagramatically we can illustrate the equivalent conductance of an  electrolyte           (Fig. 2.2).  For example a solution having 1 gm equivalent of an electrolyte is dissolved in 6 ml and placed between two plates 1 cm apart, will contain six cubes each of which has a conductivity KV. The total conductivity i.e. V = 6KV
The units of equivalent conductance may be derived as follows.


     =
= ohm-1 cm2 equiv-1 (or) S cm2 equiv-1
Equivalent conductance is expressed in the units of ohm-1 cm2 equiv-1.
2.3.3  Molecular Conductance
The conductance of all the ions produced by dissolving one gram molecular weight (one mole) of an electrolyte when dissolved in a certain volume V ml.  Molecular conductance is denoted by the symbol which  is obtained by multiplying specific conductance Kv by the volume V which contains one mole of an electrolyte.

   
The units of molecular conductance are ohm-1 cm2 mole-1 which may be deduced as follows.
where;


  = ohm-1 cm2  mole-1 (or) S  cm2  mole-1
2.3.4  Effect of Dilution on Conductance
As the dilution increases more and more, the electrolyte ionises more and more i.e. ionisation increases and specific conductance decreases because  specific conductance is the conductance of the ions present in one centimeter cube of the solution. On dilution the number of current carrying particles or ions present per one centimeter cube of the solution becomes less and less.  Hence the specific conductance of an electrolyte DECREASES on dilution.
Equivalent conductance or molecular conductance of an electrolyte INCREASES on dilution because  (1) the above conductances are the product of KV and the volume of the solution  when volume increases, equivalent conductance also, increases. (2) the number of ions of the electrolyte solution increases on dilution contributing to the increase of conductance.
Ionisation increases on dilution, till the whole of the electrolyte substance has ionised.  Further addition of water produces no change in equivalent conductance.  This limiting value is known as equivalent conductance at infinite dilution and it is represented by the symbol .
The conductance ratio  is called the degree of  ionisation () that is

Strong and Weak Electrolytes

Electrolytes may be classified into two types :
a) Strong electrolytes
b) Weak electrolytes
Strong electrolytes :
A strong electrolyte is a substance that gives a solution in which almost all the molecules are ionised, even at low concentrations. Such solutions have increasing value of equivalent conductance  at low dilution. For eg.
Strong acids  : HCl, H2SO4, HNO3, HBr etc.
Strong bases : NaOH, KOH, Ca(OH)2, Mg(OH)2 etc.
The salts       : Practically all salts are strong electrolytes
Weak electrolytes :
The electrolytes which ionise to a small extent on dilution are called weak electrolytes.  They have a low value of equivalent conductance even at higher concentrations and are not completely ionised even at very great dilution. The practical determination of for such electrolytes is not possible. For eg.
Weak acids      : all organic acids like acetic acid, propionic acid and H2SO3
Weak bases     : alkyl amines, NH4OH
Salts                : a few salts such as mercuric chloride and lead acetate

 Measurement of Equivalent Conductance 

The conductance of an electrolyte is determined by measuring the resistance of the electrolyte with the help of a wheatstone bridge  and the reciprocal of resistance gives the conductance of the electrolyte. However the measurement of conductivity gives eratic value by the fact that if a direct current is used, the products of electrolysis collect at the electrode and set up a back e.m.f i.e., back potential which apparently increases the resistance of the electrolyte and the concentration of the solution alters.  The difficulty is overcome by using an alternate current from an induction coil with a frequency of 1,000 times in a second. To detect the flow of current a head telephone earpiece is used instead of a galvanometer.
The electrolyte whose conductance is to be measured is dissolved in conductance water. (The specific conductance of  conductance water is 1  10-6 ohm-1 cm-1 at 18 0C.) The conductance water is obtained by distilling distilled water to which a small quantity of KMnO4 or Nessler’s solution is added in a pyrex glass flask. The vapours are condensed in a pyrex glass or block tin condenser.
The electrolyte solution is placed in a special type of cell known as conductance cell (Fig. 2.3).

The conductance cell is made of highly resistant glass such as pyrex or quartz. The electrodes consist of platinum discs coated with finely divided platinum black. These are called platinized platinum electrodes. Platinum black surface catalyse the union of hydrogen and oxygen which tend to be liberated by the successive pulse of the current and the polarisation e.m.f. is thus eliminated.  The electrodes are welded to platinum wires fused in two glass tubes. The glass tubes contain mercury and are firmly fixed in the ebonite cover of the cell, so that the distance between the electrodes may not alter during the experiment. Contact of the platinum electrodes with circuit is made by  the copper wires of the circuit dipping in the mercury contained in the tube.
The cell is connected to wheatstone bridge (Fig. 2.4) which consists of a wire of platinoid or manganin AB having a uniform thickness so that the ratio of lengths read on the scale gives the ratio of resistance. The wire AB is stretched tightly over a meter scale graduated in millimeters.  A sliding contact H moves along this wire. R is a resistance box. C is the conductance cell containing electrolyte which is placed in the thermostat for  maintaining constant temperature during the measurement of conductance. I is the induction coil to pass alternate current in the circuit.

When the current is flowing any resistance is unplugged in the resistance box R.  The sliding contact H is moved until the sound in the head telephone is minimum. This gives the null-point (or balance point) where resistance of resistance box R and resistance of the electrolyte in the cell C are equal. Thus at null point



Observed conductance =
Cell constant :
Since the electrodes are not exactly 1 cm apart and may not possess a surface area of  1 cm2, the reciprocal of the resistance of solution in cell (C) does not give specific conductivity but a value proportional to it. By measuring the distance between the electrodes and area of cross section the specific conductivity of the cell (C) of an electrolyte can be determined which is very inconvenient.  Hence an indirect method is employed to calculate  specific conductance of an electrolyte from observed conductance of the cell.
If R is the resistance of a column of a liquid enclosed between two electrodes  of  l cm apart, having an area of cross section s cm2.
     R =  

The ratio l/s is called cell constant denoted by the symbol x
      R =  x

                         
The unit of cell constant x is =  cm/cm2 = cm-1
Specific conductance (kv) = x  observed conductance

 Determination of Cell Constant

According to the accurate measurements of Kohlrausch the specific conductance of              KCl solution at 25 0C is 0.002765 ohm-1 cm-1.  An N/50 KCl solution is prepared by dissolving 0.372 gms of pure KCl in 250 ml of conductivity water and its observed conductance is measured as explained previously by placing the conductance cell containing KCl solution in a thermostat at 25 0C. Then
cell constant  () =
Once the cell constant is determined, great care must be taken not to alter the distance between the electrodes during subsequent measurements.
Equivalent conductance of an electrolyte (v) = KV
 kV =    observed conductance
v =   observed conductance
2.3.8  Problems on Conductance
Problem 2.1 :
A conductance cell has two parallel electrodes of 1.25 sq.cm area placed 10.5 cm apart; when filled with a solution of an electrolyte the resistance was found to be 1995 ohms. Calculate the cell constant of the cell and the specific conductance of the cell.
Sol : l = 10.5 cm and s =  1.25 cm2
cell constant (x) =
observed conductance =
Specific conductance (Kv) = x  observed conductance
                 =
Problem 2.2 :
The equivalent conductance of 0.005 N NaOH solution is 240 mho, cm2. What is the specific conductance and electrical resistance if the electrodes are 1 cm apart and each have a surface area of 1 cm2.               (JNTUA Dec 2011)
Sol : The equivalent conductance of 0.005 N NaOH solution () = 240 mho cm2
= KvV Where V =
240 = Kv
Kv =  = 1.210–3 ohm–1 cm–1
Specific conductance of 0.005 N NaOH is = 0.0012 ohm–1 cm–1
  Kv = ,
Kv =
R =
R =
Electrical resistance of the 0.005 N NaOH solution = 833.3 ohm
Problem 2.3 :
A cell whose resistance when filled with 0.1N KCl solution is 192.3 ohm, was found to be 6306 ohm when filled with 0.01N NaCl solution at 25 OC. Calculate (i) the cell constant (ii) the specific conductance of NaCl solution (iii) the equivalent conductance of NaCl solution and (iv) the degree of ionisation of NaCl solution at that dilution. [Given that  specific conductivity of 0.1N KCl is 0.01289 ohm-1, cm-1 at 25 OC and equivalent conductance at infinite dilution of NaCl at 25 OC is 126.450 ohm–1 cm2  equiv-1 ]
Sol : i)   Cell constant (x) =
          =
                  x = 2.478 cm-1
ii) Specific conductance (KV (NaCl) ) = x  observed conductance of NaCl
KV (NaCl) =
      Specific conductance of NaCl = 0.000393 ohm-1 cm-1
iii)  Equivalent conductance (V) =
V(NaCl) = = 39.3 ohm-1 cm2 equiv-1
iv)  Degree of ionisation  =
(NaCl) = 126.45 ohm-1 cm2 equiv–1
        (NaCl) =  = 0.31
Degree of ionisation of NaCl = 0.31
Problem  2.4 :
The resistance of N/2 solution of an electrolyte in a cell was found to be 50 ohm. Calculate the equivalent conductance of the solution if the electrodes in the cell are                        2.2 cm apart with an area of 3.8 cm2. [JNTUA June 2011]
Sol : Length of the electrodes =    2.2 cm
Area of cross section of electrodes =   3.8 cm2
Cell constant (x) =
observed resistance =   50 ohm
observed resistance =  
Specific conductance (KV) =   xobserved inductance
                  KV =   0.57890.02 = 0.01158 mhos
Equivalent conductance (V) = KVV where V =
   C =
V = 0.00158 = 3.16 ohm–1 cm2 eqv–1
Problem  2.5 :
The specific conductance of an N/50 solution of KCl at 25 0C is 0.002765 mho. If the resistance of a cell containing this solution is 400 ohms, what is the cell constant?
Sol : Cell constant (x) =
                 x = 0.002765 x 400 = 1.106 cm-1
Problems for Practice :
1. 0.05N solution of a salt occupying volume between two platinum electrodes of                 1.72 cm apart and area 4.5 cm2  has a resistance of 250 ohm. Calculate the equivalent conductance of the solution.                                               (JNTU 1980)
                                                   [Ans : 30.57 ohm-1 cm-2 equiv–1]
2. The cell constant of a cell is 0.5 cm-1. A solution of an electrolyte taken in the cell showed a resistance of 50 ohms. Calculate the specific conductivity of the solution.
[Ans : 0.01 ohm-1 cm-1]
3. A conductor cell has two parallel plates of 1.25 cm2 area, placed at 10.5 cm apart filled with a solution of an electrolyte. The resistance was found to be 2103 ohms. Calculate the cell constant and specific conductance of the solution.
(Sri Venkateswara IBE 1997)
[Ans : Cell constant : 8.4 cm–1, Specific conductance : 4.210–3 ohm-1 cm-1]
4. A cell whose resistance when filled with 0.1M KCl solution is 192 ohm and was found to be 6300 ohms when filled with 0.008M NaCl solution at 25 0C.  Given that the specific conductance (Kv) of 0.1M KCl is 0.0129 mho at 25 0C. Calculate (a) cell constant (b) specific and equivalent conductivities of NaCl solution (c) the degree of ionisation of NaCl solution at that dilution. The equivalent conductance at infinite dilution (la) of NaCl = 126.4 mhos.       (JNTU Mid Exam 1989)
        [Ans : a) 2.4768 cm-1 b) kV = 3.9310-4 mho cm–1,  v = 49.13 mho cm2 equiv-1 c) 0.389]
5. The specific conductance of N / 50 solution of KCl at 25 0C is 0.002765 mho cm-1. If the resistance of a cell containing this solution is 500 ohm. what is the cell constant ?
     (Anna, May 1995)           [Ans : 0.138 cm-1]
6. The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes 1.8 cm apart and 5.4 cm2 in area was found to be 32 ohm.  Calculate the equivalent conductance of the solution.        (Rajasthan 1997)
  [Ans : 104.1 ohm-1 cm2 equiv-1 ]
7. A conductance cell containing 0.01N potassium chloride was found to have a resistance of 2573 ohms at 25 0C.  The same cell when filled with a solution of 0.2N acetic acid had a resistance of 5085 ohm. Calculate a) the cell constant b) specific conductance of acetic acid c) equivalent conductance of acetic acid d) conductance ratio of 0.2N acetic acid. (Given that KV of 0.01N KCl is 0.00141 ohm-1 cm-1 and la of acetic acid is 390.7 ohm-1 cm2 equiv-1 at 25 0C)
      [Ans : a)  3.627 cm-1 b)  0.00071 mho cm-1 c)  3.567 mho cm2 equiv–1  d) 0.0091]
8. The specific conductivity of a decinormal solution of an electrolyte is 2.5 x 10-4                   ohm-1 cm-1. Calculate the equivalent conductivity of the electrolyte under the same conditions.              (Bharathiar IBE 1984)           [Ans : 2.5 ohm-1 cm2 equiv-1 ]
9. The resistance of N/20 solution of a salt occupying between two platinum electrodes 12 cm apart and 48 cm2 in area was found to be 60 ohms. Calculate the equivalent conductivity of the solution and also the percentage of dissociation if the equivalent conductance at infinite dilution of the salt is 185.18 ohm-1 cm2 equiv-1.
[Ans : V = 83.3 ohm-1 cm2 equiv-1, % of dissociation = 45]
10. Specific conductances of two electrolytic solutions of the 0.1M concentration of the substances A and B are 9.2 x 10-1 and 4.7 x 10-3 ohm-1 cm-1 respectively.  Which of them offers less resistance for the flow of current and which one is a stronger electrolyte ? Explain                    [Ans: A]
11. Resistance of a conductivity cell was found to be 192.3 ohms, and 250 ohms when filled separately with 0.1N KCl and an electrolyte of 0.01N concentration. Calculate the equivalent conductivity of the electrolyte solution, if the specific conductivity 0.1N KCl is 0.01289 ohm-1 cm-1.  All measurements were made at 25 0C.
[Ans : 991.5 ohm-1 cm2 equiv-1 ]
12. The decinormal solution of an electrolyte in a conductivity cell whose electrodes are 2.1 cm apart and 4.2 sq.cm in area offered a resistance of 32 ohms.  Find the equivalent conductance of the solution.                                     (JNTU I. B.Tech ,95)                                     [Ans : 156.25 ohm–1 cm2 equiv–1 ]
13. A 0.05M aqueous solution of a weak monobasic acid HA placed in conductance cell, whose electrodes are 0.04 m apart and 0.08 m2 in area, has a resistance of 1500 ohm at 298K. Ionisation constant of this acid at 298K is 0.018 calculate the equivalent conductance of this acid at infinite dilution at the same temperature.
                      [Ans : 370.3 mho cm2 equiv–1 ]
14. Find the equivalent conductance of 0.1N H2SO4 offering a resistance of 50 ohms when placed in a conductivity cell whose electrodes are 1 cm  apart with a cross-sectional area of 2 cm2 at 25 0C.        (JNTU I/IV B.Tech (CCC) ,97)
            [Ans : 100 mho cm2 equiv–1]
2.4  Ionic Mobilities
The electrolytic conductance is due to the mobility of ions.  In 1884 Arrhenius put forward a comprehensive Theory of electrolytic dissociation, which not only gives the satisfactory mechanism of electrolytic conduction but also a clear explanation for various allied phenomenon like abnormal behaviour of electrolytes in solutions i.e. the higher value of osmotic pressure, depression of freezing point, elevation of boiling point, increase of ionisation with dilution etc. The main points of this theory are summed up as follows:
1) When dissolved in water, salts, bases and acids yield two kinds of particles, one carrying the positive charge and the other carrying the negative charge. These charged particles are called IONS.

When dissolved in water the weak electrostatic forces of attraction of the charged ions in the molecule are considerably weakened by the dielectric constant of water and ionisation takes place.
2) There is a state of dynamic equilibrium between the dissociated and undissociated molecules.

Thus applying law of mass action to the ionic equilibrium the ionisation constant or dissociation constant K is given by
.
The fraction of the total number of molecules present in the solution as ions = K.
3)   When electric current is passed through the solution of the electrolyte, the positive ions move towards the cathode and the negative ions move towards the anode conducting electric current through the solution. Electrical conductivity of solutions depend on the number of ions and the mobility of ions.
4)   The ions behave  like molecules in elevating the boiling point, depressing the freezing point, lowering the vapour pressure and osmotic pressure. Thus if a substance yield two types of ions exerts twice the normal effect of the solute.
5)     The properties of electrolytes are the properties of the ions.  For example in aqueous solutions acid properties of the solution are due to H+ ions and basic properties of the solution are due to OH– ions.
Arrhenius theory does not explain the properties of strong electrolytes.
2.4.1 Factors Influencing Ionisation
The degree of ionisation of an electrolyte in solution is influenced by the following factors.
1.  Nature of solute : The nature of solute is a chief factor which influence the rate of ionisation.  For example strong electrolytes ionise completely whereas weak electrolytes are feebly ionised in solution.
2.  Nature of solvent : The solvent influences the rate of ionisation to a greater extent.  The solvent possess dielectric constant, which is its capacity to weaken the forces of attraction between the electrical charge of the ions present in the electrolyte when the electrolyte is immersed in that solvent.
For example the dielectric constant of water is 80, alcohol is 25 and ether is 4.1. The higher the value of dielectric constant, the greater is the ionisation.
3.  Concentration : The extent of ionisation of an electrolyte is inversely  proportional to the concentration of its solution.  As the dilution increases, the ionisation increases because of the greater number of solvent molecules which will increase the ionisation.
4.   Temperature : The ionisation of an electrolyte increases with increase in temperature.  At higher temperature, the velocities of the molecules increase which will overcome the forces of attraction between the ions in a molecule of the electrolyte.
2.5  Mobility of ions / Migration of ions
The ions exist independently in solution and with the passage of the current, they move through the solution in opposite directions towards the oppositely charged electrodes and usually at a different rate. The phenomenon of ionic migration can be demonstrated by the following experiments.
2.5.1  Lodge’s Moving Boundary Experiment
The apparatus consists of a ‘U’ tube as shown in Fig. 2.5.

The lower horizontal portion of the tube is filled with a jelly of agaragar.  A trace of alkali and phenolphthalein are added during the preparation of the jelly which becomes pink coloured.  The tube is cooled to ena ble the jelly to set.  Dilute sulphuric  acid is added  in the limb containing the anode and sodium sulphate solution to the right limb in which cathode is fixed. On passing the current, the hydrogen ions pass along the solution towards the cathode and their movement can be followed by progressive loss of pink colour in the jelly owing to the neutralization of the alkali by the hydrogen ions and movement of the original boundary.
Relative speeds of ions :
It has been clear earlier that during electrolysis the ions are liberated in equivalent amounts at the two electrodes and they move with different speeds. The ions are always discharged in equivalent amounts no matter what their relative speed is.
The following Fig. 2.6 will make this point clear.
      Case  No.        Anode (+) Anode     Middle                 Cathode (–)
Cathode      
compartment compartment       compartment
  1. + + + + + + + + + + + + +        
-   -  -  - -   -  -  -  - -   -  -  -
  2. + + + + + + + + + + + + +
-   -  -  -  -  - -   -  -  -  - -   -
  3. + + + + + + + + + + + + +
-   -  -  -  -  - -   -  -  -  - -   -
  4. + + + + + + + + + + + + +
-   -  -  -  - -   -  -  -  - -   -  -
                          (a)                              (c)
In the above Fig. 2.6 the cathode and anode compartments are separated by porous diaphragms (a and c) which prevents the convection currents but allow the ions to pass.  The sign + represent a cation while the sign - represents an anion.
Before electrolysis let there be 13 molecules present in the electrolyte, having  4 molecules each in cathodic and anodic compartments and 5 molecules in between the cathodic and anodic compartments i.e. middle compartment (Case No.1). Now consider the following cases.
1.  Let the anions (-) alone be capable of movement (Case No. 2) :
     When only two anions from cathodic compartment move towards anode and cations have not moved, we get the position shown in Case No. 2. The concentration in the anodic compartment has not altered, while in the cathodic compartment has fallen by two molecules.
2. Let the anions and cations move at the same rate (Case No. 3) :
When two ions of each type have crossed the diaphragms towards opposite electrodes, we get the condition as shown in case no. 3. The concentration of both  compartments have also fallen to the same extent i.e. by two molecules.
3.   Let the cations move twice the speed of anions (Case No. 4) :
When two cations have passed the diaphragms to the right and one anion passes to the left, we get the state of affairs as shown in Case No. 4. The concentration  in the cathodic compartment has fallen by one and the concentration in the anodic compartment has fallen by two molecules.
It is evident from the above observations that the loss of concentration around any electrode is proportional to the speed of ion moving away from it.

2.5.2 Transport Numbers
During the electrolysis the current is carried by anions and the cations.  The amount of current carried by each ion is proportional to its speed.
If u represents current carried by the cation and v that of anion,  represents the share taken by the cation in the transport of current and is called transport number of cation, similarly  represents the transport number of onion which is generally represented by n. The transport number of cation is represented by 1-n.
and

= speed ratio which is denoted by  then we have
 or
2.5.3 Determination of Transport Number of Cation (Ag+) and Anion NO–3
The transport number of cations and anions are determined by Hittorf’s method by making use of Hittorf’s apparatus.
Hittorf’s Method : Fig. 2.7 shows the Hittorf apparatus, which consists of two vertical glass tubes joined together through a U tube in the middle and all three are provided with stopcocks at the bottom.  The U tube has stopcocks at the top to control the communication between the solutions in cathode and anode limbs.  The silver anode is sealed in a glass tube as shown and the cathode is a piece of freshly silvered silver foil.  The apparatus is filled with a standard solution of silver nitrate and a steady current about 0.01 ampere is passed for two  to three hours.  It is important to pass the current for a short time so that too large a change in concentration does not take  place. The transport apparatus is connected to a copper or silver voltameter as shown in Fig. 2.7.

After passing the current for about 3 hours, the stopcocks at the top of the U tube  are closed and the whole of the liquid in the anode limb is carefully drained into a weighed flask and the amount of the silver is determined by titration against standard solution of potassium thiocyanate. The weight of silver deposited in the voltameter is also noted.  If a copper voltameter is used the weight of silver equivalent to the copper deposited is calculated. The concentration of the solution in U tube must not change.
Calculations :
Wt. of silver deposited in silver voltameter W gms
Let a gms of anode solution contain b gms of silver nitrate
Wt. of water = (a-b) gms
Let the weight of silver nitrate associated with (a-b) gms of water at the beginning of the experiment be c gms.
Initial weight of AgNO3 = c gms
Final weight of AgNO3 = b gms
Increase in weight of AgNO3 = (b - c) gms
Increase in weight of Ag =  gms = w1 gms
If no Ag had migrated from the anode, the weight of Ag should be = W gms
Fall in concentration due to migration of ions (W – w1) gm
Transport number of Ag+ =
Transport number of NO–3 =
Problems on migration of ions :
Problem 2.6 :
In an experiment the increase in concentration of AgNO3 round the silver anode was 5.8 mgs of silver. 10.8 mgs of silver were deposited by the same current in the silver voltameter placed in series.  Find the speed ratio of the silver and nitrate ions.
 (Delhi 1997)
Sol : Fall in conc. around anode / raise in conc. around cathode = 5.8 gms
If No Ag+ had migrated from anode the increase in conc. around the anode would have been = 10.8 mgs
Fall of conc. of Ag+ around anode due to migration = 10.8 - 5.8 = 5 mgs
Speed ratio () =  = =
0.862
Problem 2.7  :
In an electrolysis of copper sulphate between copper electrodes, the total mass of copper deposited at the cathode was 0.188 gms and the masses of copper per unit volume of the anode liquid before and after electrolysis were 0.79 and 0.96 gms respectively. Calculate the transport number of Cu2+ and  ions.
Sol : Weight of copper in the anode liquid before electrolysis = 0.79 gm
Weight of copper in the anode liquid after electrolysis =  0.96 gm
      Increase in weight of copper = 0.96 - 0.79 = 0.17 gm
      Increase in weight of copper in the voltameter = 0.188 gm
If no copper migrated from anode, the actual amount of increase in copper = 0.188 gm
Fall in concentration of copper due to migration = 0.188 – 0.17 = 0.071 gm
Transport number of Cu2+ ion =
Transport number of  ion = (1 - 0.38) = 0.62
Problem 2.8 :
The speed ratio of Ag+ and  ions in a solution of AgNO3 electrolysed between Ag electrodes is 0.866.  Find the transport number of the two ions.                 (JNTU 1994)
Sol : =

where, n = transport number of anion

Transport number of  ions =
Transport number of Ag+  ions = 1 - 0.536 = 0.464
Problem 2.9 :
A 80% AgNO3 solution was electrolysed using Ag electrodes. After electrolysis                    51 gm of the anode solution was found to contain 0.519 gm of AgNO3 by passing 1 ampere current for 2 minutes. What is the transport number of Ag+ and  ions in AgNO3.
        (Madras 1997)
Sol :   After electrolysis weight of AgNO3 present in 51gm of anode solution (b) = 0.519 gm
Before electrolysis = 100 gms of anode solution contains 80 gms of AgNO3
51 gms of anode solution should contain = = 0.408 gm
Initial wt. of AgNO3 (a) = 0.408 gms
Final wt. of AgNO3 after electrolysis (b) = 0.519 gm
Increase in wt. of AgNO3 (W1) = (b–a) = 0.519–0.408 = 0.11 gms
Amount of current passed = 1A260 = 120 coulombs
Wt. of AgNO3 electrolysed (W) =
Where gm. mol. wt. of AgNO3 = 170
Faraday          = 96500
Amount of current passed = 120 C
Transport No. of Ag+ in AgNO3 =
Transport No. of in AgNO3 = 1 – 0.474 = 0.526
Problem 2.10 :
A dilute solution of CuSO4 was electrolysed using pt. electrodes. The amount of Cu in the anodic solution was found to be 0.635 gm and 0.6236 gm. before and after electrolysis respectively. The wt. of silver deposited in a silver coloumeter placed in series was found to be 0.1351 gms. Calculate the transport number of  and  ions. (At of Ag = 107.88, Cu = 63.6)         (Punjab, 1979)

Sol : Wt. of Cu in anodic solution before electrolysis (a) = 0.635 gm
Wt. of Cu in anodic solution after electrolysis (b) = 0.6236 gm
Loss in  Wt. of Cu in anodic solution (w1) = (a – b) = 0.635 - 0.6236 = 0.0114 gms
Wt. of Silver deposited in voltameter = 0.1351 gms
Equivalents of copper deposited in voltameter (W) =
1 gm. equivalent of Ag (107.88) deposits  0.1351 gms of Ag.
1 gm. equivalent of Cu () deposits           ?           of  Cu
W  =   = 0.0398 gms
Transport No. of  = 0.7136,
Transport No. of Cu2+ = 1 – 0.7136 = 0.2864
Problems for Practice :
1. When an aqueous solution of copper sulphate was electrolysed in a Hittorf’s apparatus, 0.5 gm of copper was deposited at the cathode. The solution in the anode compartment contained 2.0 gms of copper after electrolysis and the same weight of water contained 1.65 gms of copper before electrolysis. Calculate the transport number of sulphate ion.                           (JNTU 1992)
                   [Ans : 0.7]
2. A current which deposits 0.47 gms of copper in a voltameter containing copper sulphate was passed through a solution of silver nitrate using silver electrodes. The liquid from the cathode compartment before experiment contained 2.315 gms of silver and after the experiment 1.231 gms of silver.  Calculate the transport number of nitrate ion.  (JNTU Mid Exam1983)
            [Ans : 0.539]
3. An aqueous solution of copper sulphate containing 2 mgs of copper is subjected to electrolysis between two copper electrodes till 0.22 mgs of copper is deposited.  After electrolysis the solution at the anode contained 2.165 mgs of copper.  Calculate the transport number of copper and sulphate ions. [Ans : 0.25 and 0.75]
4. The speed ratio of silver and nitrate ions in a solution of silver nitrate electrolysed between silver electrodes is 0.916. Find the transport number of silver and nitrate ions.                            (DDIT - 1996)
           [Ans : Ag+ = 0.478, NO–3 = 0.522]
5. A solution of silver nitrate containing 26.5 mg of silver per 25 mg of solution was electrolysed between silver electrodes and the anode solution after electrolysis contained 43 mg. of silver per 25 gm of the solution. In a silver coloumeter in series 31 mg of silver was deposited. Calculate the transport number of silver and nitrate ions.                                (Bangalore 1983)  [Ans : Ag+ = 0.468,  = 0.532]  
2.6  Kohlrausch’s Law of Ionic Mobilities
Kohlrausch had studied the equivalent conductance at infinite dilution () of a number of electrolytes at 25 0C and observed an interesting relationship between conductivity and transport number of electrolytes.  At 25 0C the values of  for some electrolytes are given below.
 Table - 2.2 : Equivalent conductance at infinite dilution of some electrolytes
 S.No. Electrolyte at   Difference S.No. Electrolyte at Difference   25 0C mho 25 0C     mho   mho     mho
    1. NaBr 128.51 2.06 5. NaBr 128.51 23.41
NaCl 126.45 KBr 151.92
   2. KBr 151.92 2.06 6. KCl 149.86 23.41
KCl 149.86 NaCl 126.45

   3. LiBr 117.09 2.06 7. KNO3 144.96 23.41
LiCl 115.03 NaNO3 121.55

   4. NH4Br 151.80 2.06 8. KOH 271.52 23.41
NH4Cl 149.74 NaOH 248.11

It is clear from the above table that when bromide ion is replaced by the chloride ion, a constant difference of 2.06 mhos is obtained in the value of equivalent conductivity at infinite  dilution.  Similarly the replacement of potassium by sodium causes a constant difference of 23.41 mhos in the value of when the anion remains the same. In other words each ion makes certain definite contribution to the conductivity of the solution. This was noticed by Kohlrausch and proposed the Kohlrausch’s law which states that -“The equivalent conductance at infinite dilution of different electrolytes is the sum of the ionic conductances of cations and anions”.

= ionic conductance of anion,  = ionic conductance of cation.
The ionic conductances of the cations and anions remains same, let them be in combination with other ion. Ionic conductance is expressed in ohm–1 cm2 equiv–1
For example :  of Br–       =  78.4 ohm–1 cm2 equiv–1
 Cl-    =  76.34 ohm–1 cm2 equiv–1
NO3   =  71.44 ohm-1 cm2 equiv–1
      of Na+ =  150.11 ohm–1 cm2 equiv–1
 =  73.4 ohm–1 cm2 equiv–1
Ionic conductances are directly proportional to the transport numbers.
  v   (or)   = kv
 u  (or)   = ku
 = k (u + v)



The above expression gives the relationship between ionic conductances and transport numbers and helps to determine the ionic conductances from transport numbers which are experimentally obtained.
2.6.1 Applications of Kohlrausch’s Law
Kohlrausch’s law is applicable to the following calculations.
1.  Calculation of la of weak electrolytes :
Weak electrolytes do not ionise to sufficient extent and not being ionised even at very great dilutions. The calculation of  practically is not possible. With the help of Kohlrausch’s law of weak electrolytes can be calculated in the following way.
– n = n
= n ( + )  [where  =  + ]          
= n
 = (1-n)
The ionic conductance  of an anion can be calculated by multiplying the  of a strong electrolyte in which the anion is present  with the transport number of the anion. In this manner the ionic mobilities of anion and cation are calculated and the sum of  and  gives .
Problems on calculation of of weak electrolytes :
Problem 2.11 :
At 18 OC, the  of HCl and CH3COONa are 383.5 and 78.4 mho cm2 equiv-1 respectively.  If the transport number of hydrogen and acetate ions 0.841 and 0.46 respectively calculate the of acetic acid.
Sol :
          = n
= 0.4678.4 = 36.06  ohm–1 cm2 equiv–1
                = (1-n)
 = 0.841383.5 = 322.52 ohm–1 cm2 equiv–1

= 36.06 + 322.52 = 358.58 ohm–1 cm2 equiv–1
Problem 2.12 :
Calculate the equivalent conductivity at infinite dilution for chloroacetic acid from the following data.  The of HCl, NaCl and ClCH2COONa are respectively 426, 126 and 90 mho cm-1 equiv-1.            [JNTU 1984]
Sol : of ClCH2COOH is to be calculated.
ClCH2COOH ClCH2COO– + H+

 ClCH2COOH     (ClCH2COOH)     (H+)
Given the values of  of the three electrolytes which contain ClCH2COO- and H+ are
H+Cl–, Na+Cl–, ClCH2COO–Na+ = I + III – II  =
             I         II                III   =
  and  are present in electrolytes I and II which is to be added and the excess Na+ and Cl– are to be subtracted.
Add the  values of HCl and ClCH2COONa and subtract the value of NaCl.  The  of ClCH2COOH is obtained.
= 426 + 90 - 126 = 390 mho cm-1 equiv-1
2.  Calculation of absolute ionic mobilities :
The absolute ionic mobility or absolute velocity of an ion represents its velocity in centimeters per second under a potential gradiant of one volt per centimeter.
Potential gradiant =
For example if the velocity of an ion at infinite dilution is u cm per second, when the distance between the two electrodes is 20 cms and voltage is 100. The potential gradiant is 100/20 = 5 volts per cm and the absolute ionic mobility is u/5 cm/sec.
It has been shown that
 ua     = k ua
 uc    = k uc
where, K is the potential gradiant referred above is the charge on 1 gm equivalent of the ion i.e. faraday i.e. 96,500 coulomb
and
ua + uc =  =  Hence the absolute ionic mobility is obtained by dividing ionic conductance by 96500 i.e. faraday.
= ua   96500 = uc96500           = 96500
Problem on absolute ionic mobilities :
Problem 2.13 :
The molecular conductivity at infinite dilution of KCl is 130.1 mho. The transport number of chloride ions in very dilute solution is 0.505.  Calculate the mobilities in cm/sec of potassium and chloride ions. [JNTU Mid Exam]
Sol : Ionic conductance () of Cl– = n = 130.10.505 mho. cm2
Absolute ionic mobility of Cl– =
Ionic conductance (c) of K+    = (1 - n)
        = 130.1(1 - 0.505)
      = 130.10.495 mho
Absolute ionic mobility of uc of K+ =
3.  Calculation of solubility of a sparingly soluble salt :
Substances like AgCl or PbSO4 are called sparingly or partially soluble and possess a definite value of solubility in water, the value of which can be determined by conductometric  titrations. Since a very small amount of solute present must be completely dissociated into ions even in saturated solution so that the equivalent conductivity  is equal to the equivalent conductivity at infinite dilution .
   = KvV  
   =  +
Knowing the values of Kv and  we can find out the volume in ml which certain 1 gram equivalent of the electrolyte.
Problem on solubility of a sparingly soluble salt :
Problem - 2.14 :
The specific conductance of saturated solution of silver chloride at 18 0C is 1.2410-6 mho. The mobilities of Ag+ and Cl- ions at this temperature are 53.8 and 65.3 respectively. Calculate the solubility of silver chloride in grams per litre. (JNTU 1982)
Sol : Kv = 1.210–6 mho
= a + c
= 53.8 + 65.3 = 119.1
= KvV
V =
of water contains 1 gm equivalent of the AgCl  = (143.5 gms of AgCl)
1000 ml of water should contain AgCl =
Solubility of AgCl = 1.494 x 10-3 gms / 1000 cc.
4.   Calculation of apparent degree of ionisation or conductivity ratio
The apparent degree of ionisation or conductivity ratio of an electrolyte () is by
=
v = equivalent conductivity of the electrolyte at a definite dilution
       =  +

Problems on conductivity ratio :
Problem  2.15 :
The conductance of silver ion at 18 0C is 55.7 mho and of the nitrate ion 60.8  mho.  The specific conductivity of decinormal solution of AgNO3 solution at 18 0C is 0.00947 mho cm–1. Calculate the percentage dissociation of the salt at this temperature.
(JNTU MID Exam)
Sol: KV (specific conductivity of AgNO3) = 0.00947 mho.
V = KV x V
V = 1000 / C =  = 10,000 cc
V = 0.00947 x 10,000 = 94.7 mho cm2 equiv–1
     
= 55.7 + 60.8 = 116.5 mho

Problem 2.16 :
The equivalent conductance at 18 0C of a normal solution of KCl is 98.2 mho cm2 equiv–1 and for infinite dilution at the same temperature is 131 mho cm2 equiv–1. Calculate the degree of ionisation of KCl at this dilution.     (JNTU Mid Examination)
Sol :   where, V = 98.2, = 113

The degree of ionisation of a normal solution of KCl = 0.75
5.  Calculation of Ionic Product of Water :
The product of H+ and OH- ions present in water is called ionic product of water which is calculated in the following manner.
  The observed specific conductivity of the purest water at 25 0C is 5.54 10-8 mho cm–1. The conductivity  of one litre of water containing 1 gram equivalent of it would be
= 5.54 10-8  1000 = 5.54 10-5 mho cm2 equiv–1
The ionic conductances of H+ and OH– ions are
= 349.8 mho cm2 equiv–1
= 198.5 mho cm2 equiv–1
= 349.8 + 198.5 = 548.3 mho cm2 equiv–1
The number of H+ = the number of OH–
H2O  H+ + OH–
[H+] =  = 1.01 x 10-7 gm ion/litre
The ionic product of water (Kw) = [H+] = [1.01 x 10-7][1.01 x 10-7]
                            = 1.02 x 10-14
Generally ionic product of water KW is taken as 10-14.
2.6.2 Problems for Practice
1.  The equivalent conductance at infinite dilution of NaCl, KNO3 and KCl are 126.5, 145 and 149 mho cm2  equiv-1 respectively at 25 0C. The transport number of Na+ in NaCl is 0.39. What is the transport number of Na+ in NaNO3 solution ?  [Ans : 0. 406]
                 [JNTU Mid Examination 1990]
2.   At 18 0C the equivalent conductance at infinite dilution for HCl and CH3COONa are 383.5 and 78.4 ohm–1 cm2 equiv–1 respectively. If the transport number of H+ and CH3COO- ions are 0.841 and 0.46 respectively, calculate the ionic mobilities of CH3COO- and H+ ions and equivalent conductivity at infinite dilution of CH3COOH.                                                 [JNTU 1981]
[Ans : 0.00334 and  0.000374 cm/sec, 378 ohm-1 cm2 equiv-1]
3.   The equivalent conductance of a 0.001208 N acetic acid solution is 48.15 ohm-1 cm2 equiv-1 at 25 0C. Given the following data calculate the equivalent conductance of acetic acid at infinite dilution at 25 0C. Also calculate the degree of dissociation at that given concentration.  for HCl, CH3COONa, and NaCl is 426.1, 91.0 and 126.4 ohm-1 cm2 equiv-1 respectively.                            [JNTU 1984]         [Ans : 390.7 ohm-1 cm2 equiv-1 and 0.123]
4.   The molecular conductivity at infinite dilution of KCl is 150 mho. The transport number of chloride ion is 0.502.  Calculate the absolute ionic mobilities of K+ and Cl-
[Ans : For Cl– 0.00078 cm/sec2  and for K+ = 0.000774 cm/sec2]
5.   Calculate the percentage ionisation of AgNO3 at 18 0C from the following data. The conductance of silver ion is 56.7 and nitrate ion is 60.5 mho. The specific conductance of a decinormal solution of AgNO3 at 18 0C is 0.0085 mho, cm–1.       [Ans : 72.52%]
6.  The specific conductance of saturated solution of AgCl at 20 0C is 1.5 x 10-6 mho cm–1. The mobilities of Ag+ and Cl- ions at this temperature are 52.4 and 62.4 respectively.  Calculate the solubility of lead sulphate in grams per litre.                         [JNTU Mid Exam]      [Ans : 1.87 x 10-3 gms / 1000 cc]
2.7  Conductometric titrations
Those titrations in which conductivity measurements are made use of in determining the end point of acid-alkali reactions, some displacement reactions or precipitation reactions are called conductometric titrations. The fact that the conductance of a solution at a constant temperature depends upon the number of ions present in it and their mobility is taken as advantage. The titrant  added from a burette into a measured volume of the solution to be titrated is taken in a conductivity cell and the conductivity readings corresponding to the  various additions are plotted against the volume of the titrant. Two linear curves are obtained in these cases and the point of intersection of two linear curves is the end point. The shapes of curves obtained in certain types of titrations is discussed below.
1) Titration of a strong acid with a strong base :
H+Cl– + Na+OH–  NaCl + H2O
Consider the reaction in which hydrochloric acid is titrated against a solution of sodium  hydroxide. 20 ml of the acid solution is taken in a conductivity cell placed in a thermostat and the conductivity of the solution is measured. One ml of NaOH is added to this solution and the conductance is measured. The process is repeated after every addition of 1 ml of NaOH and the values are plotted as shown below. The point of intersection of the two curves is the end point of the titration.

 Conductometric titration curve for strong acid and strong base
2) Titration of a weak acid against a strong alkali :
CH3COO–H+ + Na+OH–CH3COONa + H2O
When a weak acid like acetic acid is titrated against a strong alkali like sodium hydroxide, a curve shown below is obtained.

Titration curve of a weak acid against strong alkali
The initial conductivity of the solution is low because of the poor dissociation of the weak acid. On adding alkali, highly ionised sodium acetate is formed. The acetate ions at first tend to suppress the ionisation of acetic acid due to common effect. Later the conductivity begins to increase due to the conducting power of the highly ionised salt exceeds that of the  weak acid. After end point, the addition of sodium hydroxide contributes sharp increase in the conductivity of the solution. The point of intersection of the two curves gives the end point of the titration.
3) Titration of a strong acid against a weak base :
The curve obtained for the titration of a strong acid against a weak base is shown in the following figure.

Titration curve of a strong acid against a weak base
In this case the conductivity of the solution will first decrease due to the fixing up of the fast moving H+ ions and their  replacement by slow moving  ions.  After the end
H+ Cl– + NH4OH Cl– + H2O
point has been reached, the addition of excess drop of NH4OH will not cause any appreciable change in conductivity, as NH4OH is a weak electrolyte.
4) Titration of weak acid against weak base :
The titrations of a weak acid with weak base does not give a sharp end point. Consider the titration of acetic acid with ammonium hydroxide.
CH3COOH + NH4OH  CH3COONH4 + H2O
The complete titration curve is as shown below.

Titration curve of acetic acid against ammonium hydroxide
The initial conductance of the solution is low due to the poor dissociation of the weak acid, but starts raising as CH3COONH4 is formed. After the equivalent point, the conductivity remains almost constant, because the free base NH4OH is a weak electrolyte. The end point is quite sharp in this case by conductometric titrations.
5) Precipitation titrations :
During the precipitation titrations the colour change of the indicator is not clear or even impossible to detect when compared with conductometric titrations. Addition of alcohol and using fairly dilute solutions reduces the solubility of the precipitate and prevents absorption. In the titration of KCl against AgNO3, the change in conductivity on addition of AgNO3 is not much, since the mobility of K+  and Ag+  are of the same order and the curve is nearly horizontal. After the end point there is sharp increase in conductance due to an increase in the number of free ions in solution.
K+Cl–+Ag+NO3–K+NO3– + Ag+Cl– (ppt)

 Titration curve of potassium chloride against silver nitrate
Advantages of conductometric titrations :
1. Coloured solutions where no indicator is found can be successfully titrated by this method.
2. This method is more useful for the titration of a weak acid against weak base, where no indicators are available in volumetric analysis.
3. More accurate results are obtained, because the end point is obtained from graph.
4. The end point of the titration is detected graphically, hence it is accurate and no keen observation near the end point is required.
5. Conductometric titrations are very useful for dilute solutions.
Precautions :
1. The temperature of the experiment must be kept constant throughout the experiment.
2. In acid alkali titrations, the titrant must be 10 times stronger than the solution to be titrated.
2.8   Galvanic Cells
An electrochemical cell is a device for converting chemical energy into electrical energy. A redox reaction is utilised to get electrical energy. The electromotive force (emf) of such a cell is directly proportional to the intensity of the chemical reaction taking place in it.  An electrochemical cell is also commonly referred to voltaic or galvanic cell in honour of Volta and Galvani respectively who made the classical discoveries in this field.
The Daniel cell (Fig. 2.8) is a typical example of galvanic cell.  It consists of a copper rod dipping in a solution of copper sulphate, which is separated with the help of a semipermeable membrane or salt bridge from a solution of zinc sulphate in which a zinc rod is dipped. The porous wall prevents the diffusion of the two liquids but allows the passage of ions through it, when the flow of electric current takes place.

Fig. 2.8 Daniel cell
When the two metal electrodes are connected the flow of electric current takes place.  Actually  what happens is that zinc passes into the solution as Zn+2 liberating two electrons because of its higher electrolytic solution pressure, according to the equilibrium reaction. In other words zinc possess higher oxidation  potential than copper.
Zn  Zn+2 + 2
The electrons thus liberated travel along the external circuit (as shown in Fig. 2.8) to the copper electrode where copper ions (Cu++) gain these electrons and converted to metallic copper.
Cu2+ + 2e–  Cu
The movement of electrons from zinc to copper produces  a current in the circuit and the net chemical change described as the cell reaction can be represented as follows.
Zn + Cu++ Zn++ + Cu
The total cell  along with the flow of electrons is described as follows.

The total cell according to the convention is represented as
Zn  |  ZnSO4  || CuSO4  | Cu,   E = 1.09 volt.
The negative electrode is at the extreme left and the positive electrode on the extreme right. The double vertical between the two liquids signifies the salt bridge or semipermeable membrane separating the two half cells.
The process of a metal passing to solution as metal ions by liberating electrons is called de-electronation or oxidation and the electrode where it occurs is called de-electronation electrode or oxidation electrode or Anode.
Zn Zn++ + 2
The reverse process of gaining of electrons by metal ions with the discharge of metal is called electronation or reduction and the electrode where it occurs is called electronation electrode, reduction electrode or cathode.
Cu++ + 2 Cu
It must be noted that the anode of a galvanic cell has negative polarity (–), because the electrons leave the cell from it, whilst the cathode has positive polarity (+) because the electrons enter the cell from it.
The e.m.f.  of the Daniel cell is 1.09 V.
Reversible and Irreversible Cells :
The e.m.f of Daniel cell is 1.09 volt and the cell reaction is
Zn + Cu++  Zn++ + Cu
To such a cell an external e.m.f. which is exactly equal to 1.09 volt is applied the cell reaction stops. This is taken as the basis for the measurement of emf of a cell, but if the external e.m.f is increased infinitismally by small amount, then the cell reaction is reversed  as given below.
Cu + Zn+2Cu++ + Zn
If the external emf applied is less than 1.09 volt, then no change in the cell reaction takes place. A cell of this type is called reversible cell. The electrical energy of such reversible cell is calculated by the free energy decrease of the reaction that takes place in the cell.
       – = nFE
   – = decrease in free energy
         n = valency of ion
          E = emf of the cell
F = faraday
Any cell which does not satisfy the above conditions is irreversible.
For example zinc and silver electrodes, when immersed in a solution of sulphuric acid, When the two electrodes are connected zinc ions go into solution and it is an irreversible cell.
Zn + 2H+  Zn++ + H2(g)
2.9  Electrode Potentials
Each electrochemical cell is made up of two electrodes, at one electrode electrons are evolved and at the other electrons are used up. Each electrode which is dipped in its salt solution is called half cell.
The potential of each electrode is called electrode  potential. The different types of electrode potentials are explained below.
2.9.1 Single Electrode Potential
The potential of half cell i.e, the potential difference  between the metal and its salt solution in which it is dipped is called single electrode potential. It cannot be directly measured.
Zn  |  ZnSO4 || CuSO4  |  Cu
 half cell      half cell
 single electrode       single electrode
(anode) (cathode)
The total cell e.m.f. is equal to the sum of the single electrode potentials.  Each electrode is affixed with a symbol corresponding to the reaction that takes place near the electrode.
Ecell = E(anode)+ E(cathode)
The half cell reactions are as follows :
The half cell reaction which  corresponds oxidation is Zn - 2e  Zn+2.
The electrode is called oxidation electrode and potential of the electrode is oxidation potential or the potential of left hand electrode which is represented as E(ox) or E(L).
The cell reaction of the electrode where reduction  takes place is given below
Cu++ + 2  Cu
The electrode is called reduction electrode or right hand electrode and the potential of this electrode is reduction potential and is represented as E(Red) or  E(R)
The total cell reaction is Zn + Cu++ Zn++ + Cu
E(cell) = E(Ox) + E(Red)
E.M.F of the total cell is equal to the sum of oxidation potential and reduction potential, also expressed  as the reduction potential of the right hand electrode minus reduction potential of the left hand electrode.
E(cell) = E(R) – E(L)
Generally electrode potential refers to the reduction potential.
For example : E(R) of Cu++ / Cu is 0.337 V
E(L) of Zn/Zn++  is - 0.763 V
              E(cell) = E(R) - E(L)
  = 0.337 - (-0.763) = 0.337 + 0.763 = 1.1V
Generally reduction potential can be expressed in terms of oxidation potential with a negative symbol before it. For eg. the oxidation potential of Cu++ / Cu is - 0.337V. But when it is substituted in the place of reduction it must be substituted without negative symbol.
For example :  E(ox) Cu++ / Cu = -0.337V
E(ox) of Zn / Zn++ = + 0.763V
Ecell   = E(ox) + E(Red)
E(cell) = 0.763 - (-0.337) = 1.1V
            = 0.763 + 0.337 = 1.1 V
According to IUPAC convention the standard reduction potentials alone are the standard potentials.
Direction of spontaneous reaction :
If Ecell > 0, the cell reaction has a tendency to go from left to right.  If however the            E(cell) < 0 the cell reaction will have a tendency to go from right to left. The reaction is spontaneous or feasible only when E(cell) is positive for that case the standard free energy of the reaction (G0) would be negative (G0 = - nFE0).
2.9.2  Measurement of Single Electrode Potential
Single electrode potential cannot be directly measured. The single electrode potential of a half cell depends on (a) concentration of ions in the solution (b) tendency to form ions (c) temperature.
The single electrode potential is conveniently measured by combining the half cell with a  reference electrode standard electrode and measuring the total e.m.f. of a cell.
 = E(R) - E(L)
If the standard cell acts as anode the equation becomes
 = ER - E0
where E0 is called the standard electrode potential.
On the other hand if the standard half cell is cathode the equation takes the form
E(cell) = E0 - EL
2.9.3 Types of Electrodes : Reference Electrodes
Reference electrodes are standard electrodes. The following are some important reference electrodes used for the determination of emf of a cell.
2.9.3.1 Standard Hydrogen Electrode (SHE)
Standard hydrogen is otherwise called Normal Hydrogen Electrode (NHE) or the standard hydrogen half cell or the standard hydrogen electrode (Fig. 2.9) is a reference electrode, selected for coupling the half cell.  It consists of a platinum electrode  coated with platinum black immersed in a 1M solution of H+ ions maintained at 25 0C. Hydrogen gas at one atmosphere pressure enters the glass hood and bubbles over the platinum electrode. The H2 gas at the platinum electrode passes into the solution forming H+ ions and electrons.
H2 2H+  + 2e

 Fig. 2.9  Standard Hydrogen Electrode
By convention the standard electrode potential of hydrogen electrode when the hydrogen gas passed at one atmosphere pressure is bubbled through a solution of hydrogen ions of unit concentration (1M) is arbitrarily fixed as zero. Depending on half cell to which it is attached hydrogen electrode can act as a cathode or as an anode. When it acts as anode, oxidation occurs.
                                             H2 (g) (1 atm) H+ (1 M) + e
When it acts as cathode, reduction reaction  takes place.
H+ (1M) + e ½ H2 (g)
2.9.3.2 Nernst Equation
Due to the concentration effect i.e., when H+ conc. is not 1M, the e.m.f. of the electrode (hydrogen) alters.  To calculate the potential of the standard electrode, Nernst equation  is used, as is given below.
E = E0 –  log10 [ H+ ]
       where E0 = standard electrode potential
        R = gas constant,
          T = absolute  temperature
F = faraday (96500 coulomb)
n = valency of the ion and
E = single electrode potential.
Substituting all the values we get the potential of the electrode at 25 0C as
E = E0 -
E = E0 – 0.0591 log10 [H+]
E = E0 – 0.0591 pH
This is the equation for a half cell, where oxidation occurs. In case standard hydrogen electrode is used as reduction electrode. The e.m.f (E) will be measured by reversing the sign of E0 (-E0)
E(red) = - E(ox) = - E0 +  log10 [ H+ ]
2.9.3.3  Standard Calomel Electrode (SCE)
The second type of reference electrode employed for measuring the single electrode potential is calomel electrode. It is a secondary standard electrode and consists of a glass tube, having a side tube on each side as shown in the Fig. 2.10 given below. The mercury of high purity is placed at the bottom of this tube and connected to the circuit by means of a platinum wire, sealed in a glass tube. The surface of mercury is covered with a paste of mercurous chloride (calomel) and mercury in KCl solution. The electrolyte is a solution of potassium chloride. The electrode is connected with the help of a side tube on the left through salt bridge with the other electrode, whose potential has to be measured. The potential of calomel electrode depends upon the concentration of KCl solution.  The electrode is represented as Hg, Hg2Cl2(s) / KCl (sat. sol.). The standard reduction potential of this electrode at 25 OC is
For saturated KCl soln. electrode potential is 0.2415 volts.
For 1N KCl standard reduction potential is 0.28 volts.
For 0.1N KCl solution the reduction potential is 0.3338 volts.
The calomel electrode is only  used as reducing electrode.

Fig. 2.10 Calomel electrode
2.9.3.4  Ion - Selective Electrodes (ISE)
Ion selective electrodes (ISE) use a membrane which is sensitive to a particular chemical species.  ISE respond to certain specific ions present in a mixture while ignoring others and develop potential.  The potential developed is a measure of the concentration of the species of interest.
Measurement of concentration of  cations [Mn+]
The cell constructed for the measurement of concentration of cations [Mn+] is given below.
                      Solution to           Internal
                    Reference be analysed         standard solution Reference
electrode 1                 [Mn+] = C1                   [Mn+] = C2 electrode 2
            membrane      

In ion-selective electrodes, the use of membrane is must. Generally the following four types of membranes are used.
      1)  Glass membrane (Fig. 2.11)
        2)  Solid state membrane (Fig. 2.12)
        3)  Heterogeous membranes
        4)  Liquid membrane (Fig. 2.13)
                   
                                  Figure 2.11                        Figure 2.12                          Figure 2.13
                             1)  Glass membrane        2)  Solid state (pallet)         3)  Liquid membrane                           electrode                            membrane electrode           on a porous polymer electrode
The liquid membrane electrode is usually a large organic molecule capable of interacting with anions and cations and is absorbed on an inert porous support. for eg : tris–1, 10–phe-
nanthraline Fe2+ (ClO4– )2 is used. For solid membrane pressed pallets of Ag2S + AgCl for chloride ions which were doped on lanthanum trifluoride.
The membrane potential is given by the equation
 =
Suppose reference electrode is anode and membrane electrode is cathode then the cell potential E(cell) =
Here is zero if identical electrodes were used as the concentration of Mn+ in the internal reference electrode (C2) is constant. The Ecell is given by
E(cell) = constant –
where C1 is the concentration of Mn+ ions in the solution to be analysed.
Applications of ion-selective electrodes :
1. To determine the concentration of gas levels like CO2 level in blood samples.
2.   To determine the concentration of a number of cations like H+, Li+, Na+, K+, Ag+,  NH+4, Pb2+, Cd2+, Ca2+, Mg2+ etc.
3.  To find the concentration of anions like F–, , CN–, S2– and halide ions.
4.  To determine the pH of the solution.
2.9.3.5  Glass Electrode
Glass electrode is widely employed electrode for pH measurements by combining this electrode with calomel electrode which is used as a second electrode. The glass electrode is made of a special glass, the corning glass containing the composition Na2O(22%), CaO(6%) and SiO2(72%). The corning glass possesses relatively low melting point and  high electrical conducting. The assembly of glass electrode is given below (Fig. 2.14).
            Fig. 2.14 Glass Electrode
The  glass electrode consists of a thin glass bulb filled with 0.1 N HCl and a silver wire coated with silver chloride (Ag/AgCl(s)) is immersed in HCl solution. The Ag/AgCl(s) acts as an internal reference electrode. The glass electrode is represented as Ag/AgCl(s)  glass. When glass electrode is immersed in another solution, whose pH value is to be determined, there develops a potential between the two surfaces of the membrane. The potential difference developed is proportional to the difference in pH value. The glass membrane of the glass electrode undergoes an ion exchange reaction, the Na+ ions on the glass are exchanged for H+ ions.

  (solution)                                     (solution)
The potential difference or the boundary potential difference (Eb) at the interface is the result of the difference in potential (E1-E2) developed across the gel layer of the glass membrane between two liquids. The boundary potential (Eb) is related to the concentration of the solution inside the glass bulb (C1) and concentration of the external solution (C2) into which the glass bulb is dipped which is given by
                                               Eb= E1– E2= ln()
The potential of the glass electrode (EG) is given by relation,
Where  is a constant which is determined by using a solution of known pH.
To determine the pH of an unknown solution, the glass electrode is combined with a saturated calomel electrode and placed in the solution under test. The EMF of the total cell is measured. Because the resistance is very high, special electron tube voltameters are used to measure the emf of the cell.
E(cell) = Eright– Eleft
E(cell) = 0.2422 – [+ 0.0591pH]

Advantages of glass electrode :
a) Glass electrode is easy to operate and not easily poisoned
b) Equilibrium is rapidly reached
c) The results are accurate.
Limitations of glass electrode :
a) Glass electrode can be used in solutions with pH range 0 – 10 because electrodes are composed of special glass that can be used upto pH 12.
b) The resistance is extremely high in the order of 10 to 100 million ohms, which cannot be measured by ordinary potentiometers and special electronic potentiometers have to be used.
2.9.3.6 Fluoride Electrode
The fluoride ion selective electrodes are used to measure quickly, accurately and economically fluoride ions in aqueous solutions. The Fluoride ion electrode consists of a single crystal of Lanthanum fluoride as a membrane bonded to an epoxy body. The electrode is combined with an internal reference electrode  containing KF, KCl/AgCl/Ag/Cu present inside the fluoride selective electrode which minimises the liquid  junction potential. The internal electrolyte is at fixed composition. The fluoride ion electrode contains an internal reference electrode, and internal fluoride standard electrode, and the LaF3 ion exchange crystal as shown in Fig. 2.15. An external reference electrode is used to perform the measurement of the emf of the potentiometrically. Assuming that a saturated calomel electrode is used as the external reference, the potentiometric cell is represented as
Ag/AgCl, KCl/Cl (0.3 m), (F 0.001m) LaF3/test solution/standard calomel electrode (SCE).
Fig. 2.15 Measurement of emf by fluoride electrode
The internal electrolyte of fluoride electrode is at a fixed composition and the EMF of the cell is given by Nernst equation



Where, E  = emf of the cell
        = standard electrode potential of SCE.
R = gas constant
T = absolute temperature
F = Faraday
n = valency
[F] = F ion concentration.
Limitations :
1) There is interference of with LaF3.
2) Fluoride firms complexes with Si4+, Fe3+ and Al3+.
3) The electrode should be calibrated frequently.
4) The electrode potentials are influenced by temperature.
Applications :
1) Majorly used in the determination of fluoride content in water, toothpaste, soil etc.
2) Highly useful in the environmental studies.
3) This electrode is used not only for the determination of fluoride ions but also other ions.
2.9.3.7  Measurement of EMF of a Cell
The e.m.f of a cell cannot be directly measured by a voltmeter, because (1) the internal resistance of the cell may absorb some of the available potential difference. (2) The chemical change involved may result in polarisation.  The e.m.f. can be accurately measured by a potentiometer. To measure the emf of a cell, we measure the external emf which is just equal to the emf of the cell.  At that time no current flows in the cell. We detect the null point and confirm it by comparing this with a standard cell which is weston standard cell (Fig. 2.16). The standard cell consists of a H shaped vessel.  The positive electrode constituted by pure mercury covered with a paste of solid mercurous sulphate and cadnium sulphate in mercury. The negative electrode is 12.5 percent cadmium amalgam. The electrolyte is a saturated solution of cadmium sulphate which is kept saturated by contact with crystals of CdSO48H2O. To prevent the loss of water by evaporation the limbs are scaled with wax.
The cell reaction is given by
Cd2++ Hg2Cd2+ + 2Hg.
The e.m.f of the cell at 20 oC is 1.0183 volt and it varies slightly by 0.0000406 volt per  change of 1 oC.


Fig. 2.16 Weston standard cell
The potentiometer (Fig. 2.17) used for the measurement of e.m.f. consists of a wire AB with uniform thickness and high resistance stretched on a meter scale. The two ends of the wire are connected to a storage battery C having a continuous and large e.m.f. than the cell under measurement. The cell under measurement X is connected to A taking care that the same pole of X and C are linked to A. The other terminal of X is connected to a moving contact O through a galvanometer G. The unknown cell X is replaced by the standard cell S by using a two way key K. The cell C sends a current through the galvanometer in a direction opposite to that in which the current is sent by the cell X. By moving the contact to and fro, the current flows through the galvanometer and the position sliding contact O is so adjusted that no current flows through the galvanometer. The distance AO is measured.  The experiment is repeated with the standard cell S.

Since the wire is of uniform thickness the fall of potential per unit length must be same.  If e represents the fall of potential per unit length the above equation can be given as

e.m.f. of cell X =

Fig. 2.17 Measurement of  EMF of a cell by potentiometer
Mathematically emf of an electrochemical cell is the algebraic sum of the single potentials.     E(R) = (Ecell - E(L)) = (Ecell) (Reduction potential)
E(L) = (Ecell - E(R)) = (E(cell) - E0)
           (Oxidation potential)
Thus the single electrode potentials of the half cells are determined.
Nernst Equation for a Cell Reaction :
An electrochemical cell is obtained by coupling two half cells or electrodes.
For a cell     Zn(s)  | Zn2+(aq)   || Cu2+(aq) | Cu(s) the reaction is
Zn(s) + Cu++(aq)  Zn++(aq) + Cu(s)
E(cell) = E0(cell) -  log10 Q  ; where
At 298 K the Nernst equation can be written as :
Ecell = (E0(Cathode) - E0(anode) +  log10

2.9.3.8 Problems on EMF
Problem 2.17 :
The e.m.f. of the following cell at 25 OC is 0.445 V
Pt, H2(g) (1 atm) | H+ (unknown) || KCl (sat soln) | Hg2Cl2, Hg
Calculate the pH of the unknown solution given that oxidation potential of  saturated colomel electrode is – 0.2415 V.
Sol : Ecell    =  ERight – ELeft
0.445   =  0.2415 – (– 0.0591  pH)
0.445   =  0.2415 + 0.591 pH
pH       =   = 3.64
Problem 2.18 :
What is the single electrode potential of a half cell for zinc electrode dipping in a 0.01M zinc solution ? (EO = 0.763 V)
Sol : According to Nernst equation the reaction is
E = EO –  log10  [ ION ]
            = 0.763 –  log10  [ 0.01]  = 0.763 + 0.0591 = 0.8221 volt.
Problem 2.19 :
Calculate the e.m.f. of the following reaction at 25 O C.
Sol : Cu++ + Zn  Zn++ + Cu
EOZn (ox) = 0.763 V
EOCu (ox) = – 0.337 V
   E(cell)   = E(R) – E(L)
                      = 0.337 – (– 0.763)
                   = 0.337 + 0.763
                    = 1.1 volt.
Problem 2.20 :
The potential of the cell   at 25°C is 1.1 volt. Calculate the potential of cell  at the same temperature.
(JNTUK June 2012)
Sol :

         =
        = 1.1 + 0.0296 log10 0.2
       = 1.1 + 0.02960.69897
        = 1.1 + 0.0297
        = 1.1207V
Problem 2.21 :
What is the emf of the following cell at 25 °C  The standard emf of the cell Eo is 1.54V.               (JNTUK Feb 2011)
Sol :
       
       
       
       
        = 1.54 – 0.139
        = 1.401V
Problem 2.22 :
A Nickel electrode  was combined with an aluminium electrode                                (Eo = 1.66V) to form a cell. What would be the potential of the cell?
  (JNTUK Jan. 2012)
Sol : Since Eo of Aluminium electrode (1.66V) is higher than that Eo of Nickel electrode                             (– 0.13V), Aluminium electrode is cathode and Nickel electrode is anode.

     = 1.66 – (–0.13)
      = 1.66 + 0.13
      = 1.79V
Problem 2.23 :
Calculate the standard electrode potential of Cu2+/Cu, if the electrode potential at              25 °C is 0.296 V when [Cu2+] concentration is 0.015 M.             (JNTUK Feb. 2011)
Sol :
0.296 =
0.296 =
0.296 = Eo – 0.054
    Eo = 0.296+0.054 = 0.3499 V
Problem for practice :
1. Calculate the single electrode potential for copper metal in contact with 0.1M Cu2+ solution given that EOCu+2/Cu is 0.34 volt.                           [Ans : 0.3695 volt]
2. Hydrogen electrode and normal calomel electrode, when immersed in a solution at 25 OC  showed a potential of 0.664 V. Calculate the pH of the solution.        [Ans : 7.07]
3.   What will be the electrode potential of Zn | Zn2+ if C = 0.1N at 25 0C. Std. oxidation potential of zinc is 0.7618 V.                          [Ans : 0.7913 volt]
4. The e.m.f. of the following cell Pt | H2 (g) (1 atm) | H+(?) || Hg2Cl2 | KCl(sat) | Hg  was found to be 0.228 V at 25 0C. Calculate the pH of the solution. [Eocell = 0.2415 V]                                     [Ans : 3.44 volt]
5. Calculate the e.m.f. of the cell Zn | ZnSO4 || AgCl | Ag
given that EO(Zn2+/Zn) = – 0.763 V and EO(Ag+/Ag) = 0.222 V        [Ans : 0.985 V ]
6. Calculate the e.m.f. of a Daniel cell at 25 OC when the concentration of ZnSO4 and CuSO4 are 0.001M and 0.01M respectively. The standard potential of the cell is 1.1 volts.                                    (Amaravathi – 1997) [Ans : 1.1592 volt]
7. Write the half cell and net cell reactions of the following cell at 25 OC
Zn | Zn2+ (1M) || Cu2+ (1M) | Cu
Find the  e.m.f. of the above cell,
Given  EOZn2+/Zn = – 0.76 volt and EOCu2+/Cu = 0.34 volt.
                                                                (Bangalore – 1993)  [Ans : 1.1 volt]
8. The potential of a hydrogen gas electrode in a solution of an acid of unknown strength is 0.29 V at 298 K as measured against normal hydrogen electrode. Calculate the pH of acid solution.                             (Andhra – 1997) [Ans : 4.9 volt]
9. Calculate the e.m.f. of voltaic cell. Fe | Fe2+(aq)  || Cu2+ (aq) | Cu given the electrode potentials of copper and iron are 0.34 volt and – 0.44 volt respectively.                                                                                                                                        (Kuvempu – 1993) [Ans : 0.78 volt]
2.10 Electrochemical series
When the metals are arranged in the order of increasing reduction potentials or decreasing oxidation potentials which are determined with respect to one molar solutions of their ions and measured on the hydrogen scale, a long series or list resulted is called electrochemical series.
Electrochemical Series
   Electrode            Halfcell reaction        Standard oxidation potential at 25o C            Eo volts
Li/Li+ Li  Li+  + e– – 3.045
K/K+ K  K+  + e– – 2.925
Ca/Ca++ Ca  Ca++  + 2e– – 2.87
Na/Na+ Na  Na+  + e– – 2.714
Mg2+/Mg Mg  Mg2+ + 2e– – 2.37
Zn/Zn2+ Zn  Zn2+ + 2e– – 0.763
Fe/Fe2+ Fe  Fe2+  + 2e– – 0.440
Cd/Cd2+ Cd  Cd2+ + 2e– – 0.403
Pb/Pb2+ Pb  Pb2+ + 2e– – 0.126
Pt/H2(g)(1 at)/H+(1M) H2(1 atm) 2H+  + 2e–   0.000
Cu/Cu2+ Cu  Cu2+  + 2e– + 0.337
Ag/Ag2+ Ag  Ag+  + e– + 0.799
Pt/Cl-/Cl2 2Cl-  Cl2  + 2e– + 1.36
Au/Au3+ Au  Au3+  + 3e– + 1.50
The higher a metal is in the series, the greater is its tendency to be oxidised. The metals high up in the series are strong reducing agents and their ions are stable whereas those near the bottom of the series are inactive, stable metals, and their ions are easily reduced to metals. Any metal can reduce any cation below it in the series of one molar solutions are used.
Any metal above hydrogen will displace hydrogen from dilute acid solution. For eg : Na reacts with  water to liberate hydrogen because the EO (Na+ Na) –2.714V is less than                             EO  (H+ / H2) (zero).
The oxidised form of any metal below the hydrogen in the series will be reduced by hydrogen.
The metals above the hydrogen in the series can easily displace the metal below the hydrogen in the series from its salt solution, because EO of Zn+2/Zn (–0.763V) is less than EO of Cu2 / Cu (0.337V).
For eg. Zn + CuSO4  ZnSO4 +  Cu
The metals above the hydrogen in the series can be easily oxidised hence they undergo corrosion easily.
Applications of electrochemical series :
The electrochemical series give more information on
1.  The relative corrosion tendencies of the metals and alloys.
2.  Relative ease of oxidation or reduction of metals.
3.  Replacement tendency of metals.
4.  Calculating the equilibrium constant as given below.
Eo  =
log10 =  =  at 25 oC
5.  Predicting spontaneity of redox reactions.
2.11  Potentiometric Titrations
A titration in which the equivalent or end point of a reaction is determined with the help of the measurement of the potentials of the reaction mixture is known as potentiometric titration.
1)  Acid Base Titrations : The acid solution whose strength has to be determined is taken in a beaker and the hydrogen electrode and calomel electrode were dipped in the solution. The electrodes were connected to the potentiometer and the e.m.f. is measured (Fig. 2.18). A known volume of standard alkali solution is added from a burette, stirred thoroughly and the

Fig.  2.18  Potentiometric titration
emf of the cell is recorded. Like this 10–15 readings are recorded by repeating the procedure of the addition of standard alkali. The volume of alkali added is plotted against emf observed as shown in the Fig. 2.19. The steepest portion of the curve indicates the equivalent point of the titration (Fig. 2.19a). Instead  is plotted against the volume of the base as shown below (Fig. 2.19b).

                                   (a)                                                                   (b)
Fig. 2.19
2) Oxidation – Reduction Titrations : The procedure adopted for oxidation – reduction titrations is the same as in acid-base titrations, the only difference is that the electrode reversible to hydrogen ions is replaced by a bright platinum electrode. The e.m.f. of the electrode is determined by the activity of ratio of the substance being oxidised or reduced. For example Fe2+ titrated against K2Cr2O7. The ferrous iron solution is taken in the beaker, treated with dil. H2SO4 and platinum electrode and calomel electrodes are dipped. The electrodes are connected to the potentiometer and emf of the solution after the addition of K2Cr2O7 is recorded.  A graph is plotted with emf and volume of K2Cr2O7. The steep rise is the end point of the titration.
3) Precipitation reactions : In precipitation reactions also an electrode reversible to one of the ions involved is made use of. For example titration of silver nitrate with sodium chloride where silver chloride precipitates out, silver electrode is used along with calomel electrode. The silver nitrate is placed in the micro burette and added sodium chloride  taken in the beaker, containing electrodes. The e.m.f. of the cell is measured and plotted against volume of silver nitrate added. The steep rise in the curve show the end point of the titration.
4) Determination of pH by e.m.f. method : The e.m.f. of a solution depends on the concentration of H+ ions or pH of the solution. A hydrogen electrode containing solution of unknown pH is paired with a standard calomel electrode. The complete cell may be represented as
Pt, H2 (1 atm) | H+ (unknown) || KCl (sat. soln) | Hg2Cl2. Hg
The e.m.f. of the above cell is measured by potentiometer, and the pH of the unknown solution can be calculated as follows.
The e.m.f. of the cell will be given by the expression
Ecell = ERight  – ELeft
Ecell = 0.2415 – (– 0.0591  pH)
Ecell = 0.2415 + 0.0591  pH
pH =
Advantages of potentiometric titrations :
1. Coloured solutions, where the use of indicator is impossible are estimated                                  by potentiometric titrations.
2. There is no problem with regard to the choice of indicators based on pH value of the solutions.
3. Polybasic acids can be titrated in steps corresponding to different steps of neutralisation.
4. Dependance on colour and external indicators are avoided for redox titration by potentiometric titrations.
5. Solutions containing more than one halide can be analysed in a single titration against silver nitrate.
2.12 Classification of conductors
We can classify the different conductors in the following form

2.13  Concentration Cells
A concentration cell is a Galvanic cell in which electrical energy is produced by the transfer of material from a system of high concentration to one of low concentration.
There are two types of concentration cells.
2.13.1  Electrode Concentration Cells
In these cells, the e.m.f. arises as a result of two like electrodes at different concentrations dipping in the same solution of the electrolyte.
1)   For example two hydrogen electrodes at unequal gas pressure P1 and P2 immersed in the same solution of H+ ions.  The following cell reactions takes place.
Oxidation reaction H2(P1) 2H+ + 2e
Reduction reaction 2H+ + 2eH2(P2)
Total reaction  H2(P1) H2(P2)
It is clear that in this process there is no overall chemical change and there is only transfer of H2 gas from the electrode with pressure P1 to the electrode with pressure P2.  The emf depends only on the two pressures  and independent of the concentration of electrolyte solutions hydrogen ions, in which the electrodes are dipped. The emf of the cell is calculated as follows.
E =
2)  In amalgum cells, two amalgums of same metal at two different concentrations are immersed in the same solution of metal ions for example two unequal concentration of zinc amalgum dipped in a solution ZnSO4.
Zn (Hg)C1 | Zn++ | Zn (Hg) | C2
The overall reaction is
Zn(Hg) C1  Zn (Hg) C2
The emf of the cell is calculated by the following equation :
E =
2.13.2  Electrolyte Concentration Cells
These cells consists of two identical electrodes dipped in a two electrolyte solutions of different concentrations,  due to which a difference in the potential at the electrode  and the electrolyte solution in which it is dipped.  The source of electrical energy in the cell is the tendency of the electrolyte  to diffuse from a solution of higher concentration to that of lower concentration. The emf of the concentration cell falls to zero when the two concentrations of the electrolyte becomes identical.
   
Fig. 2.20  Typical concentration cell where C2 >C1
A concentration cell (Fig. 2.20) with two zinc electrodes dipped in two solutions of ZnSO4 with different concentrations C1 and C2 joined through a salt bridge where the concentration of C2 >C1 can be represented as follows.
(+)Cathode Zn | Zn2+ (C1) || Zn2+ (C2) | Zn(-) Anode
The concentration C2 > C1, the electrode reaction on the left is oxidation.
Zn  Zn++ (C2) + 2e–  (-) anode (oxidation)
Zn++ (C1) + 2e  Zn (+) cathode (reduction)
The e.m.f. of such cell can be calculated by the  following expression based on Nernst equation.

Substituting the value of R = (8.318 107 ergs), T at 25 °C (273 + 25 = 298K) and F = 96,500 coulombs. We can calculate the e.m.f. of the cell as follows.

The EMF of the cell is maximum in the beginning, slowly decreases as the reaction proceeds and becomes zero when the concentration of C1 = C2.
Applications of concentration cells :
1.   To determine the solubility of a sparingly soluble salt.
2.   To calculate the valency of cations.
3.   To determine the transition point.
4.   To calculate the extent of corrosion in metals.
2.13.3  Problems on Concentration Cells
Problem 2.24 :
Calculate the valency of the cation of the following concentration cell, whose observed e.m.f. was 0.029 volt. The two mercury electrodes were dipped in 0.5 N and 0.05 N mercurous nitrate solution at 25 OC.         (Magadh 1996)
Sol:
.
0.029 =  0.029 =      n  =  
valency of cation  = 2
Problem 2.25 :
Calculate the e.m.f. of the following concentration cell which contains two zinc electrodes dipped in two solutions of 0.05 M and 0.5 M concentrations.
Sol: Zn | ZnSO4 || ZnSO4 | Zn      (JNTU Mid Exam 1981)
                      0.05        0.5
       (C1)        (C2)
E(cell) = log 10
E(cell) =  log 10  log 10 = 0.02955 V    E (cell) = 0.02955 volts.
Problems for Practice :
1. Calculate the emf of the following concentration cell at 30 OC.
Zn | ZnSO4 || ZnSO4 | Zn
                        0.1     0.01                                [Ans : 0.03 V] 2. Calculate the valency of the cation of the following concentration cell, whose e.m.f. is found to be 0.58V.
Ag | AgNO3 || AgNO3 | Ag                                            [Ans : 1]
                      0.45      0.045
2.14  Batteries – Primary and Secondary Cells
Battery is an electrochemical cell or often several electrochemical cells connected in series that can be used as a source of direct electric current at a constant voltage. A device which converts chemical energy to electrical energy is called Battery a term usually applied to a group of two or more electric cells, connected together electrically in series. Batteries are commercial electrochemical cells.
Advantages of Batteries :
1.   Batteries act as a portable source of electrical energy.
2.   The portability of electronic equipments in the form of handsets has been made possible by batteries.
3.   A variety of electronic gadgets have been made more useful and popular with the introduction of rechargeable storage batteries having reliability, better shelf life and tolerance to service.
4.   For all commercial applications, batteries are constructed for their service. For example, batteries for automotives and aircrafts, standby batteries etc.
The following requirements should be possessed by the batteries.
1. High capacity which is very small variation of voltage during discharge.
2. High energy efficiency, which is calculated as
% of efficiency =
3. High cycle life is required which is the number of chargings and discharging cycles before failure.
4. Long helf life is required.
5.   Tolerance to different service conditions such as variation in temperature, vibration shock etc.
6.   Reliability is another important criteria.
We can broadly classify the batteries (commercial electrochemical cells) into 3 categories and summarise as follows.
Batteries (Commercial Electrochemical Cells)
       Primary Cells                       Secondary Cells Fuel Cells
  1. Cell reaction is not 1. Cell reaction can be         1. Energy can be withdrawn     reversible.                                   reversed.                     indefinitely as long as     outside supply of fuel is     maintained.
2. Cannot be recharged. 2. Can be recharged.         2. Do not store energy.
3. Can be used as long as   3. Can be used again and   3. Cannot be reused.   the materials are active       again by recharging  
    in their composition   the cell.                      
eg :   eg : eg :
a. Leclanche or dry cell. a. Lead storage cell a. Hydrogen oxygen fuel cell. Zn/NH4Cl(20%), ZnCl2/   emf depends on the   Extensively used in space MnO2 / C emf =1.5V   activity of H2SO4 used   vehicles due to their light
used in radios,     in car batteries.   weight and the bi product     H2O produced is a valuable     source of fresh water for     astronauts.
b. Ruben - Mallory cell. b. ‘Nicol’ or Nickel cadmium     Hg/HgO/KOH/Zn(OH)2/Zn     battery emf =1.4 V has           emf =  1.34 V used     long life when compared        in digital watches,   with others used in   hearing aids, heart     electronic calculators,   makers etc.,     electronic flash & cordless     electronic shavers etc.
2.14.1  The Primary Cells
The cells in which the cell reaction is not reversible. Thus when the reactants have been
converted to products, no electricity is produced and the cell becomes dead and cannot be used after that. These batteries are used as a source of dc power. These cells are convenient to use and cost of discharge is not much. Dry or Leclanch cell, a cell without fluid component is a primary cell. The anode of the cell is zinc can or container, containing an electrolyte of NH4Cl, ZnCl2 and MnO2 to which starch is  added to make a thick paste-like to prevent  leakages. A carbon (graphite) rod serves as a  cathode, which is immersed in the centre of the cell. The net reaction of the cell is as follows.
Zn(S) +  + 2Cl–(aq) + 2MnO2(s) Mn2O3(s) + [Zn (NH3)2]Cl2 (s) + 2H2O.
The dry cell finds its applications in flash lights, transistor radios, calculators etc.
Lithium Cells : Lithium cells belong to primary cells. The cells having lithium anodes are called lithium cells, irrespective of the cathode used. Lithium cells can be classified  into two categories.
a.  Lithium Cells with Solid Cathodes    
b.  Lithium Cells with Liquid Cathodes
a.   Lithium Cells with Solid Cathodes : These systems may have solid or liquid electrolyte. Lithium – manganese dioxide is emerging as most widely used 3 volt solid cathode lithium primary battery. The cathode MnO2 should be heated to >300 OC to remove water before incorporating it in the cathode. This is very important to improve the efficiency of the cell. The anodic and cathodic reactions are as follows.
Anodic reaction     :  Li  Li+ + e–
Cathodic reaction  :  Li+ + xe– + MnO2  LiMnO2
Net reaction           :  Li + MnO2    LiMnO2
The electrolyte is a mixture of propylene carbonate and 1,2 – dimethoxyethane.
Applications :
1.   The cylindrical cells are used in fully automatic cameras.
2.   The coin cells are widely used in electronic devices such as calculators and watches.
b.  Lithium Cells with Liquid Cathode :  Among Lithium cells with liquid cathode is lithium sulphurdioxide cell. The cosolvents used are either acrylonitrile or propylene carbonate or mixture of the two with  50% by  volume of  SO2.  The cell reaction  represents
2Li + 2SO2  Li2S2O4
The second type of this category is Lithium thionyl chloride cells, which consists of a high – surface area carbon cathode, a non-woven glass separator. Thionyl chloride acts as the electrolyte solvent and the active cathode. The reactions of the cell can be represented as follows.
Anodic reaction      : Li  Li+ + e–
Cathodic reaction    : 4Li+ + 4e– + 2SOCl2   4 LiCl + SO2 + S
Net reaction            : 4 Li + 2SOCl2   4 LiCl + SO2 + S
Due to the nature of reaction Li-thionyl chloride cells possess very high energy density. The reaction involves two electrons per thionylchloride. Further the SO2  liberated  as  product is a liquid under the internal pressure of the cell. No cosolvent  is required  for the solution as thionyl chloride is a liquid having moderate vapour pressure. The discharging voltage is                         3.3 – 3.5 volts.
Applications :
1. Owing to the excellent voltage control Li–SOCl2 cells are used  on electronic circuit boards for supplying fixed voltage for memory protection and other standby functions.
2. These cells are also used for military and space applications.
3. These cells are also used in medical devices such as neuro–stimulators, drug-delivery systems.
2.14.2 Secondary Cell or Storage Cells
In secondary cells, the cell reaction is reversed by passing direct current in opposite direction. The secondary batteries can be used through a large number of cycles of discharging and charging. They are used as source of dc power used to supply large, short-term repetitive power requirements such as automotive and air-plane batteries. These batteries have very large capacitance and long periods of low-current-rate discharge. For example Lead acid storage cell, Nickel – Cadmium cell, Lithium cells are widely used storage cells.
a)  Lead Acid Storage Cell :  A storage cell can operate both as a voltaic cell and as an electrolytic. When operating as a voltaic cell, it supplies electrical energy  and as a result eventually becomes ‘rundown’ when being recharged, the cell operates as an electrolytic cell. This storage cell has the great advantage of working both as an electrolytic cell and as a voltaic cell. As shown in the Fig. 2.21, the lead acid storage cell consists of lead anode and lead dioxide (PbO2) cathode, which is made of paste of PbO2, passed into a grid, made of lead. A number of lead plates (–ve plates) are connected in parallel and a number of lead dioxide plates (+ ve plates) are also connected in parallel. The lead plates fit in between the lead dioxide plates. The plates are separated from adjacent one by insulators like strips of wood,  rubber or glass fibre. The entire combination is immersed in 20 – 21% dil. H2SO4, corresponding to a specific gravity of 1.2 to 1.3.
The discharging of the storage cell is operating as a voltaic cell where the oxidation of lead  takes place at anode
Pb  Pb2+  + 2e–  (at lead anode)
The Pb2+ combines with SO42– ion to produce PbSO4
Pb2+ + SO42–  PbSO4
The 2e-- released at anode flows to the PbO2 electrode and causes reduction of PbO2  to produce Pb2+ which finally combine with  to produce PbSO4 (at cathode)
PbO2 + 2H+ + Pb2+ + 2H2O
Pb2+ +    PbSO4

Fig. 2.21 A lead accumulator
A lead accumulator for car consists of six lead acid storage cells in series and it is capable of delivering 12 V. The net reaction is
Pb + PbO2 + 4H+ +   2PbSO4 + 2H2O + Energy
Charging : During charging / recharging of the battery an external emf greater than 2 volts is passed, so that the cell reactions are reversed as shown below.
PbSO4 + 2e--  Pb +  (Reaction at the cathode)
PbSO4 + 2H2O + 2e–  PbO2 (Reaction at the anode)
The net reaction is
2PbSO4 + 2H2O + Energy  Pb + PbO2 + 4H+ +
The +ve pole of the generator is attached to +ve pole of the battery and –ve pole of the generator is attached to the –ve pole of the battery. During discharging operation the concentration of H2SO4 decreases, while the concentration of acid is restored during charging.
The battery containers are made of hard rubber and plastic. For cars six lead-acid storage cells in series are used.
Applications of Lead Storage Cells
The lead storage cells are used to supply current for electrical vehicles, gas engine ignition, in telephone exchanges, electric trains, mines, laboratories, hospitals, broadcasting stations, automobiles and power stations.
Advantages of  Lead Storage Batteries :
1)   The lead storage battery is both a voltaic cell and an electrolytic cell. When electricity is being drawn from the cell, to start the car it acts as an voltaic cell, when the car is running, the cell is being recharged as an electrolytic cell.
2)     It has relatively constant potential i.e., 12 volts.
3)   The electrolyte density signals its state of change.
4)     It is portable and inexpensive.
b)  Nickel – Cadmium Cell / Ni–Cd Cell : This battery consists of a cadmiun anode and a cathode composed of a paste of NiO(OH)(S). The cell reactions are given below:
Anode :  Cd(s) + 2OH(aq) Cd (OH)2(s) + 2e–
Cathode : 2NiO(OH) + 2H2O(l) + 2e–   2Ni(OH)2(S) +2OH--(aq)
Net reaction : 2NiO(OH)(s) + Cd(s)  + 2H2O(l)  Cd (OH)2(s) + 2Ni(OH)2
Recharging Ni-Cd Battery : The reaction can be easily reversed because the reaction products Ni(OH)2 and Cd(OH)2 adhere to the electrode surface. Ni – Cd cell gives a voltage of 1.4 volts and they can be connected in series to give Ni – Cd storage battery.
There are two types of Ni – Cd batteries.
i)   Pocket Plate Ni – Cd Battery : The materials used for cathode and anode are filled into pocket plates (two perforated Nickel coated steel strips) and connected in series as shown below (Fig. 2.22). These cells have long shelf life >20 years with capacities between 10 to 1000 ampere and maintains a voltage of 1.4 – 1.45 volts.

Fig. 2.22 Pocket Plate Ni-Cd battery
ii)  Sintered – Plate Ni - Cd battery : The anode materials - fine Ni sintered in a mold around a nickel screen impregnated with Nickel nitrate and processed to produce Nickel hydrate in the pores. For cathode the molds are impregnated with cadmium salt processed to get  hydrated oxides inside the pores (Fig. 2.23). The electrodes are assembled with suitably placed separators such as porous polymer membrane. The electrolyte solution is a solution of KOH with 1.24 to 1.3 specific gravity. The oxygen liberated during electrolysis, liberated  through the porous membrane to Cd cathode to produce Cd (OH)2, which helps in charging.
       
Fig.2.23 Sintered-plate Ni–Cd battery
The sealed cells are made in three different designs - plates, horizontal  discs (button cells) and spirally wound cylindrical cells.
Advantages of Ni–Cd batteries :
1. The potential of Cd anode is below the hydrogen potential hence the Cd electrode  is completely inert to electrolyte. It requires no float current to keep charged. The water consumption and float charge currents  are extremely low.
2. These batteries are suitable to very high rate discharge and low temperature operations.
3. They have long shelf life without any maintenance.
Applications of Ni–Cd Batteries :
The Ni – Cd batteries are used for aircraft and diesel engine starting, lighting of trains, emergency power supply and for many military applications.
2.15  Fuel CELLS
A fuel cell is an electrochemical cell which converts chemical energy contained in an easily available fuel oxidant system into electrical energy. The basic principles of fuel cells are identical to those of the well known electrochemical cells, the only difference is  that in fuel cell the chemical energy is provided by a fuel and an oxidant stored outside  the cell.
The fuel and the oxidising agents are continuously and separately supplied to the electrodes of the cell, at which they undergo reactions. Fuel cells are capable of supplying current as long as the reactants are supplied.
Fuel cells are characterised by :
1.  high efficiency
2.  low noise levels
3.  free from vibration, heat transfer and thermal pollution and
4.  have built in wide range of power requirements.
1.  Hydrogen – Oxygen Fuel Cell : The best example of fuel cells is hydrogen – oxygen fuel cell (Fig. 2.24).


Fig. 2.24  Hydrogen oxygen fuel cell
The cell consists of two inert porous electrodes (made of graphite impregnated with finely divided Pt or a 75/25 alloy of Pb with Ag or Ni) and an electrolyte solution 2.5% KOH. Through the anode hydrogen gas is bubbled and through cathode oxygen gas is bubbled. The following cell reactions take place.
At anode       : 2H2(g)+ 4OH–   4H2O(l) + 4e–
At cathode    : O2(g) + 2H2O(l) +  4e–  4OH– (aq)
Net reaction  : 2H2(g) + O2(g) 2H2O(l)
The product discharged is water and the standard emf of the cell is EO = 1.23 volts.
A number of such fuel cells are stacked together in series to make a battery.
Applications of Oxygen Fuel Cells :
1. They are used as auxillary energy source in space vehicle, submarines or other military vehicles.
2. Because of the light weight these fuel cells are preferred for space crafts, and product H2O is a valuable fresh water source for astronauts.
Advantages of Fuel Cells :
The following are the advantages of fuel cells.
1. The energy conversion is very high (75 – 82%)
2. The product H2O is a drinking water source for astronauts.
3. Noise and thermal pollution are low.
4. Fuel cells offer an excellent method for use of fossil fuels.
5. The maintenance cost is low for these fuels.
6. The regenerative hydrogen – oxygen fuel cell is an energy storage system for space applications.
Limitations :
1. The life time of fuel cells are not accurately known.
2. Their initial cost is high.
3. The distribution of hydrogen is not proper.

exercise
I. SHORT Answer QUESTIONS :
1. What is a metallic conductor?
2. What are the units of specific conductivity?
3. What is specific conductance?
4. What is equivalent conductance?
5. Define molar conductivity.
6. What is the unit of molar conductivity?
7. What is the unit of equivalent conductivity?
8. How do you measure equivalent conductivity from specific conductivity?
9. What is cell constant, give its units?
10. How is molar conductivity related to molarity?
11. Define kohlrausch law.
12. What is the effect of temperature on molar conductivity?
13. What is electrode potential?
14. What is an electrochemical cell?
15. Define reduction potential.
16. What is a fuel cell?
17. How can we increase the value of reduction potential of an electrode?
18. Why do concentration cells stop working after sometime?
19. In conductometric estimation only A.C. current is used. why?
20. Why specific conductance increase on dilution ?
21. Why does a dry cell become dead after sometime even though it is not used?
22. How do you calculate the membrane potential of a liquid membrane electrode?
23. What is the standard electrode potential of hydrogen electrode at 25 °c, when the concentration of the hydrogen ions is 1M and H2 gas is passed at 1 atm. pressure?
24. Why salt bridge is used in Hittorf’s apparatus for measuring transport number?
25. Give any two examples of secondary cells.
26. Write Nernst equation for the calculation of electrode reaction.
27. Calculate the emf of a cell if the oxidation and reduction potentials of the cell are 0.763V and –0.337V.                                                              [Ans. 1.1V]
28. Why do ion-selective electrodes (ISE) make use of a membrane?
29. How do you represent a glass electrode schematically?
30. What type of membrane is used in flouride ion electrode?
31. How are the electrodes made of in weston standard cell?
32. What is the difference between the potentiometric acid base titration and red-ox titrations?
33. What are conductometric titrations?
34. What is ionic product of water?
35. Calculate the degree of ionisation of an electrolyte if the equivalent conductance of the electrolyte is 89 mhos and the equivalent conductance at infinite dilution is 112 mhos.  [Ans. 0.795]
36. What is meant by emf series?
37. Explain electrolyte concentration cells.
38. Write the cell reaction of Weston standard cell.
II.  ESSAY Answer Questions :
1. What are conductors? How are they classified? Differentiate a metallic conductor
    from electrolytic conductor.
  2. Define the following terms and explain their relationship.
    a) Specific conductance
b) Equivalent conductance  
c) Molecular conductance
  3. What is specific conductance of an electrolyte? How does it vary with dilution and
temperature?
  4. What is cell constant? How do you measure the conductance of an electrolyte by Wheatstone bridge?
  5. Explain the following conductivities, along with their units, and effect of temperature and dilution.
  a) Specific conductance
b) Equivalent conductance
c) Molar conductance
  6. What are transport numbers ? How do you determine the transport number of cation and anion by Hittorf’s method?
  7. State and explain Kohlrausch’s law of ionic mobilities. What are the applications of the law ?                          [JNTU-2010]
  8. How are the following calculated by making use of Kohlrausch’s law ?
  a) Equivalent conductance of an electrolyte at infinite dilution.
  b) Absolute ionic mobilities
  c) Solubility of a sparingly soluble salt
  d) Ionic product of waters
  9. What are conductometric titration? Give their significance.
10. How do you estimate a strong acid by using a strong base conductometrically?
11. What are electrochemical cells? Differentiate an electrochemical cell from an electrolytic cell.
12. What are single electrode potentials? How do you determine the electrode potential Zn / ZnSO4 ?
13. Explain the standard electrode potential by taking calomel electrode as an example.
14. How do you calculate the pH of a solution by making use of hydrogen electrode and  calomel electrode ?
15. What are reference electrodes? Explain the application of Quinhydrone electrode?
16. What is pH of a solution? How do you measure the pH of an acid solution by making use of glass electrode?
17. What are batteries? How are they classified?
18. Explain the composition, applications and advantages of the following cells.
      a) Ni-Cd cell b) Lithium cell and c) Pb acid cell
19. What are fuel cells? Explain the hydrogen-oxygen fuel cell and its advantages.
20. Differentiate primary, secondary and fuel cells with suitable examples.
21. Give an account of e.m.f series and its applications.
22. How do you differentiate e.m.f. series from galvanic series?
23. Give a brief account on the following :
      a) Glass electrode       b) Lead-acid cell c) Transport numbers
24. Differentiate the following with suitable examples:
      a) Metallic from electrolytic conductors
b) Primary cells from secondary cells
      c) Equivalent conductance from molecular conductance
      d) Strong electrolytes from weak electrolytes
25. Give a comparative account of the following:
      a) Primary, secondary and fuel cells.
      b) Electrochemical cells and electrolytic cells.
    c) Galvanic series and e.m.f series.
26. a) Explain the Arrhenius theory of electrolytic dissociation.
      b) What are the factors influencing ionization?
27. How is the phenomenon of ionic migration demonstrated by moving boundary  experiment?
28. Explain the relative speeds of ions during electrolysis.
29. What is speed ratio ? How do you determine the speed ratio of silver and nitrate ions?
30. i) State Kohlrausch law. Derive an expression for the relationship between ionic conductance and transport number.
ii) Give the following application of kohlrausch law.
      a) Apparent degree of ionization or conductivity ratio
      b) Calculation of  of a weak electrolyte
c) Solubility of sparingly soluble salt
31. a) Give reasons for the following statements :   [June-2010]
i)  When a zinc rod is dipped in a solution of aq. copper sulphate, copper is precipitated   out.      
ii) Nernst equation is applicable for the determination of emf of a concentration cell.
32. Write a brief account on the construction, working and advantages of the following electrodes.
a) Ion selective electrode b) Glass electrode or quinhydrone electrode.
[JNTUK-2014]
33. a) How do you determine the pH of a solution by making use of fluoride ion selective electrode?
b) Explain the different types of conductors.
34. a) By making use of quinhydrone electrode how do you determine the pH of a solution?
b) The potential of a hydrogen gas electrode in a solution of an acid of unknown strength is 0.29V at 298K, as measured against normal hydrogen electrode. Calculate the pH of the acid solution.           (Andhra University 1997) [Ans : 4.9]
35. a) Give an account of the electrolyte concentration cells.
b) Consider the cell Fe|Fe2+(0.01M)||Cu2+(0.5m)|Cu. The standard electrode potential of iron and copper are –0.44V and 0.34V respectively. Write the cell reaction and calculate the emf of the cell.   (Karnataka 1997) [Ans : 0.8007V]
36. Calculate the emf of a Daniel cell at 25 °C when the concentration of ZnSO4 + CuSO4 are 0.01 and 0.1 M respectively. The standard potential of the cell is 1.1. Volts.                        (JNTUK - 2014)  [Ans : 1.103 V]
37. Explain the construction and working principle of glass electrode.        (Gitam  2014)
III. Multiple Choice Questions :
1. Electrolytes can conduct electricity because
a) their molecules contain  unpaired electrons which are mobile.
b) their molecules contain loosely held electrons which become free under the      influence of voltage.
c) their molecules are broken up into ions, when the electrolyte is fused or dissolved    in a solvent.
d) their molecules break up into ions when voltage is applied.
2. HCl is called an electrolyte because
      a) its molecules are made of electrically charged particles.
      b) it breaks up into ions when current is passed through it.
      c) it ionises when electric current is passed through it.
      d) it ionises when dissolved in a proper solvent.
3. Which of the following is a weak electrolyte?
    a) NH4OH b) NaOH c) Hcl d) Nacl
4. Ionisation of an electrolyte in aqueous solution is due to
a) hydrolysis of the electrolyte
b) increase in electrostatic forces of attraction between the ions
c) stability of the compound in aqueous medium
d) decrease in the electrostatic forces of attraction between the oppositely charged ions  
5. A good ionising solvent should possess
a) low value of dielectric constant b) a dielectric constant equal to one
c) a high melting point d) a high value of dielectric constant
6. Which of the following does not conduct electricity?
a) molten Nacl b) solution of Nacl in H2O
c) Nacl crystal d) none of the above
7. The specific conductance of a solution increases with
a) increase in concentration b) decrease in concentration
c) decrease in temperature d) None of the above
8. The ionisation of a strong electrolyte increases when the solution is diluted and the relation is given by
a) Nernst equation b) Ostwald’s law
c) Arrhenius equation d) Law of mass action
9. Acetic acid is a weak electrolyte because
a) its molecular weight is high b) it is weakly ionized
c) it is a covalent compound d) it is highly unstable
10. The degree of dissociation of acetic acid in an aqueous solution of the acid is practically unaffected
a) by adding a pinch of Nacl           b) by adding a drop of concentrated HCl
c) by diluting with water d) by raising the temperature
11. Solid Nacl is a bad conductor of electricity because
a) it contains only molecules            b) the ions present in it are not free to move
c) it does not possess ions                d) it does not contain free molecules
12. The equivalent conductance of 0.1N Nacl at 25 0c is 25.2 ohm-1 cm2eq-1.
The equivalent conductance at infinite dilution of Nacl is 126 ohm-1 cm2eq-1. The degree of ionization of the 0.1N Nacl is
      a) 0.5 b) 0.2                        c) 0.4                           d) 0.1
13. Pure water does not conduct electricity because, it is
      a) acidic b) high boiling         c) almost unionized d) decomposed easily
14. The molar conductivity of a solution of any electrolyte is the product obtained by multiplying
a) specific conductivity with molecular weight
        b) specific conductivity with the volume of the solution containing 1 gm mole of the      electrolyte
  c) reciprocal of conductivity with volume
    d) Specific conductivity with equivalent weight
15. A galvanic cell converts
a) electrical energy into chemical energy        b) chemical energy into electrical energy  
c) electrical energy into heat energy  d) chemical energy into heat energy
16. The potential of standard hydrogen electrode dipped in a solution of 1M  concentration and hydrogen gas is passed at 1 atm pressure
a) 1 volt b) 10 volt c) 0 volt d) 100 volt
17. The potentials of two metal electrodes used  in a cell are 0.35 V and 0.8 V. The e.m.f  of the cell formed by combining  them is
a) 0.45 V b) 1.15 V c) –1.15 V d) –0.45 V
18. In electrochemical series the elements are arranged in the
a) decreasing order of standard reduction potentials
b) increasing order of standard reduction potentials
c) increasing order of equivalent weights
d) increasing order of oxidation potentials
19. Calomel electrode is reversible with respect to
a) mercury ion   b) chloride ion
c) both mercury and chloride ions d) mercurous ions
20. The electrode potential is the tendency of a metal
a) to gain electrons b) to lose electrons
c) either to lose or gain electrons d) none of the above
21. Calomel electrode is constructed using a solution of
a) saturated KCl b) saturated CaCl2
c) saturated MgCl2 d) saturated NaCl
22. The standard reduction potential at 298K for Zn+2 Cr+3 , H+ and Fe+3 are -0.76V, -0.74V, -0.0V and 0.77V respectively. The strongest reducing agent among them is
a) H+        b) Cr+3 c) Zn+2 d) Fe+3
23. Calomel is
a) mercuric sulphate b) mercurous sulphate
c) mercurous chloride d) mercuric chloride
24. A storage cell is a device that can operate
a) both as voltaic cell and electrical cell b)  as voltaic cell
c) as electrical cell d)  concentration cell
25. An electrochemical cell or several electrochemical cells connected in series, that can be used as a source or direct electric current at a constant voltage is called
a) battery b) voltaic cell c) electrolytic cell d) metal conductor
26. The cathode of Ni-Cd battery is composed of
a) cadmium b) nickel
c) paste of Nio(OH) d) paste of Cd(OH)2
27. A fuel cell converts
a) chemical energy of fuels directly to electricity
b) chemical energy of fuels directly to heat
c) electrical energy of fuels directly to heat
d) electrical energy of fuels directly to chemical energy
28. When hydrogen is used as  fuel in hydrogen-oxygen fuel cell, the electrodes are made of
a) an alloy of palladium and silver b)  aluminium and palladium
c) iron and cadmium d)  cadmium and silver
29. When storage cell is operating as voltaic cell it is said to be
a) charging b) discharging
c) neutral d) None of the above
30. In lead-acid storage cell during discharging operation the concentration of H2SO4
a) increases b)  decreases
c) remains same   d)  gradually increasing
31. When standard hydrogen electrode is used as reduction electrode, the emf of the cell is calculated by the formula
a) b)
c) d)
32. The ion-selective electrodes make use of
a) Membrane b) Mercury c) KCl solution d) Platinum electrode
33. The glass electrodes are made of special corning glass whose composition is
a) K2O (22%), Na2O (6%) and SiO2 (72%)    b) Na2O (6%) MgO (22%) SiO2(72%)
c) Na2O (22%) CaO (6%) and SiO2 (72%)    d) Na2O (22%) CaO(16%) SiO2(62%)
34. Fluoride ion electrode consists of
a) A single crystal Germanium fluoride      
b) A single crystal Germanium trifluoride
c) A single crystal of Lanthanum trifluoride
d) A single crystal of Lanthanum fluoride
35. The emf of weston standard cell is
a) 1.0183V at 25 °C   b) 1.0183V at 20 °C  
c) 1.0813V at 25 °C   d) 1.0813V at 20 °C
36. The conductance of an electrolyte solution at constant temperature does not           depend upon its
a) number of ions b) delocalised electrons
c) dilution d) ionisation
37. The speed ratio is calculated by making use of one of the following expressions :
a) b) c) d)
38. The metals below the hydrogen in electrochemical series can be
a) reduced b) oxidised
c) neutralised d) neither oxidised nor reduced

IV. Fill in the blanks :
  1. The resistance of a metallic conductor  as the temperature is increased.
  2. A substance which in aqueous solution or in molten state liberates ions and allows electronic current to pass through is called
  3. The substance which conducts electricity without decomposition is called
4. Graphite is a  conductor.
5. A substance which allows the electric current to pass through it is called
6. On dilution the specific conductivity of an electrolyte
7. Specific conductivity of an electrolyte is calculated by the  formula.  
8.   Specific conductance is expressed in  units.
9.   The units of resistivity are
10. The total conductance of 1 gm equivalent of an electrolyte at a given dilution is called
11. The unit of equivalent conductance is
12. The total conductance of all ions is present in one mole of an electrolyte in the solution is called
13. The unit of molar conductivity is
14. The equivalent conductivity is related to normality by  equation.
15. The equivalent and molar conductivities are related by equation.
16. A device which converts electrical energy to chemical energy is called
17. Nernst equation for electrode reaction is
18. The equivalent conductance  on dilution.
19. The relation between the electrode potential E and concentration of an ion is given as  equation
20. The standard electrode potential of saturated calomel electrode at 25 0C is
21. The equivalent conductance at infinite dilution of a weak electrolyte is calculated
by  law.
22. The transport number of an anion is calculated by .
23. Speed ratio of the cation and anion is given by .
24. A cell whose reaction is not reversible is called .
25.   are the cells which do not store energy.
26. Practically all salts are  electrolytes.
27. When two identical electrodes are dipped in two electrolyte solutions of different concentrations, the resulting cell obtained is called  .
28. The specific conductance of a 0.1 N NaOH is 0.0025 ohm–, cm–1. Its equivalent conductance is  ohm–1, cm2, equiv–1.
29. The potential of glass electrode is given by relation .
30. Any metal above hydrogen in electrochemical series can be easily oxidised. Hence they undergo  easily.
31. ________ converts chemical energy contained in an easily available fuel oxidant system into electrical energy.
32. A storage cell can operate both as _______ cell and as ________ cell.
33. The cathode of Nickel - Cadmium cell is composed of a paste of _______.
V. indicate True or False FOR the FOLLOWING :
1. Molecular conductance increases on dilution.         [T/F]
2. The e.m.f  of a concentration cell gradually decreases.    [T/F]
3. Oxidation potential is the tendency of an electrode to gain electrons.    [T/F]
4. Fuel cell is a device for converting the energy of a fuel directly into electrical energy.      [T/F]
5. On increasing the temperature the equivalent conductivity decreases.        [T/F]
6. The specific conductance is measured by the units siemens / cm.        [T/F]
7. Measurement of conductance is carried out by using Wheatstone bridge.
           [T/F]
8. The molar conductivity at infinite dilute of any electrolyte is not equal to the sum of the molar conductances of its cation and anion.     [T/F]
9. The fraction of the total current carried by a cation or anion is called transport  number.     [T/F]
10. Conductance bridge is used for the measurement of transport number.         [T/F]
11. An acid is strong when it produces large concentration of H+ ions in its aqueous solution.    [T/F]
12. Sodium hydroxide is a weak base.      [T/F]
13. An acid has a tendency to give up proton.         [T/F]
14. Standard hydrogen electrode is a reference electrode.         [T/F]
15. The electrical energy is converted into chemical energy in a galvanic cell.  [T/F]
16. The electrode where oxidation occurs is called anode in electrochemical cells.           [T/F]
17. A standard cell is capable of giving constant and reproducible e.m.f.           [T/F]
18. The unit of e.m.f. is ohms.      [T/F]
19. Potentiometer is used to measure transport number.           [T/F]
20. Calomel is mercuric chloride.       [T/F]
21. The potential of calomel electrode on the hydrogen scale varies with concentration of KCl solution.          [T/F]
22. The pH of the solution is determined by using glass electrode.    [T/F]
23. A device that stores chemical energy for later release of electricity is called a              battery.     [T/F]
24. Fuel cells store chemical energy.      [T/F]
25. A storage cell is one that can operate both a voltaic and as electrical cell.    [T/F]
26. The cell reaction is not reversible in primary cells.   [T/F]
27. The conversion of chemical energy to electrical energy takes place in electrolytic cell.    [T/F]
28. There is no change in chemical properties of a metallic conductor during conductance.   [T/F]
29. An electrolytic cell does not involve chemical reactions.   [T/F]
30. The process of spontaneous splitting up of an electrolyte into ion when  dissolved in a solution is called ionisation.                  [T/F]
 Answers
III. Multiple Choice Questions :
 1) c  2) d  3) a  4) d  5) d
 6) c  7) a  8) c  9) b 10) c
11) b 12)   b 13) c 14) b 15) b
16) c 17) b 18) b 19) b 20) c
21) a 22) c 23) c 24) a 25) a
26) c 27) a 28) a 29) b 30) b
31) b 32) a 33) c 34) d 35) b
36) b 37) a 38) a
iV. fill in the blanks questions :
1) decreases 2) electrolyte 3) metallic conductor
4) metallic 5) conductor 6) decreases
7) .     8) ohm-1 cm-1 or S.cm-1 9) ohm
10)  equivalent conductance  11) ohm-1 cm2  gm eq-1 (or) S cm2 gm eq-1
12)  molar conductivities
13)  ohm-1 cm2  gm mol-1  (or) S cm2 gm mol-1
  14)     15) 16)  electrolytic cell 17)   log10 (ION) or  
 or 18) increases
19) Nernst           20) 0.2415V
21) Kohlrausch’s law 22)   23)    
24) Primary cell 25)  Fuel cell 26) Strong
27) Concentration cell   28)  25 ohm–1 cm2 equiv–1   29)  or
30) Corrosion 31) Fuel cell 32) Voltaic, electrolytic
33) NiO(OH)
 V. True or false questions :
 1) T  2) T  3) F  4) T  5) F  6) T  7) T    8) T    9) T   10) F
11) T 12) F 13) T 14) T 15) F   16) T     17) T   18) F   19) F   20) F
21) T 22) T 23) T 24) F 25) T 26) T 27) F 28) T   29) F   30) T