1. Electrochemistry Part 2: Conductance of solutions
Presented By:
Dr. Mamta Bhandari
Assistant Professor
Department of Chemistry
University of Delhi

2. Contents Conductance of electrolytic solution
Specific conductance, Equivalent conductance, Molar conductance
Kohlrausch's law
Electrolysis
Faraday’s Laws of electrolysis

3. What are Electrolytes?
Substances whose aqueous solution or molten state conduct electricity.
Conduction takes place by the movement of ions. Examples: salts, acids and bases.
Substances whose aqueous solution does not conduct electricity are called non electrolytes.
Examples: solutions of cane sugar, glucose, urea etc.

4. Types of Electrolytes Strong Electrolytes Weak Electrolytes
Strong electrolytes are highly ionized in the solution at all concentration.
Examples are HCl, H2SO4, NaOH, KOH etc
Weak Electrolytes
Weak electrolytes are only feebly ionized in the solution.
Examples are H2CO3, CH3COOH, NH4OH etc

5. Metallic Vs Electrolytic Conductors
Flow of electricity takes place without the decomposition of substance i.e. no chemical change
Conduction is due to the flow of electrons
Conduction decreases with increase in temperature
Electrolytic
Ions are oxidized or reduced at electrodes i.e. chemical reaction takes place
Flow of electricity is because of movement of ions
Conduction increase with increase in temperature

6. Resistance
Resistance refers to the obstruction to the flow of current.
For a conductor of uniform cross section(a) and length(l); resistance R is given by,
where “ρ” is called resistivity or specific resistance. a l

7. Conductance
The reciprocal of the resistance is called conductance. It is denoted by C.
C=1/R
Conductors allow electric current to pass through them. Examples are metals, aqueous solution of acids, bases and salts etc.
Unit of conductance is ohm-1 or mho or Siemen(S)
Insulators do not allow the electric current to pass through them. Examples are pure water, urea, sugar etc.

8. Specific conductance
Specific conductance or conductivity is defined as the conductance of unit volume of a cell. It is reciprocal of resistivity.
κ = 1/ρ and
Therefore,
or κ = (l/a) x Conductance
The quantity “l/a” is called cell constant .
Specific conductance has units ohm–1cm–1. SI Unit of specific conductance is Sm–1 where S is Siemen.

9. Equivalent Conductance
It is the conductance of all ions furnished when one gram equivalent of an electrolyte is dissolved in V cc of the solution. Equivalent conductance is represented by λeq
Mathematically, λeq= κ x V
Where, κ = Specific conductivity
V = Volume of solution in “cc” containing one gram equivalent of the electrolyte.

10. Molar Conductance
It is the conductance of all ions furnished when one mole of the electrolyte is dissolved in V cc of the solution. Molar conductance is represented by λm
Mathematically, λm= κ x V
Where, κ = Specific conductivity
V = Volume of solution in “cc” containing one gram equivalent of the electrolyte.

11. Factors affecting electrolytic conduction
Nature of electrolyte: Strong electrolytes ionize almost completely in the solution whereas weak electrolytes ionize to a small extent.
Size of ions: larger the ion, smaller will be its conductance.
Nature of solvent and viscosity: Greater the polarity of the solvent, greater is the conductance. Greater the viscosity, lesser will be the conductance.
Concentration of solution: Higher the concentration of the solution, less is the conduction.
Temperature: On increasing the T, the dissociation increases and the conduction increases.

12. Variation of Conductance with dilution
Specific conductivity decreases on dilution.
Equivalent and molar conductance both increase with dilution and reaches a maximum value.
The conductance of all electrolytes increases with temperature.

13. Strong Vs weak electrolytes
Strong electrolytes:
The molar and equivalent conductance at infinite dilution can be obtained by extrapolating the line.
It is denoted by λ∞
Increase in the Molar and equivalent conductance is due to weakening of intermolecular forces upon dilution.
Follow Debye Hückel Theory (λcm = λ∞m – A √C)
Weak electrolytes:
The molar and equivalent conductance at infinite dilution can be obtained by Kohlrausch’s Law
Increase in the Molar and equivalent conductance is due to increase in the number of ions upon dilution.

14. Kohlrausch’s Law
Statement: “At time infinite dilution, the molar conductance of an electrolyte can be expressed as the sum of the contributions from its individual ions” i.e. Λ∞m = v+ λ∞+ + v- λ∞-
where, v+ and v- are the number of cations and anions per formula unit of electrolyte respectively and, λ∞+ and λ∞- are the molar conductivities of the cation and anion at infinite dilution respectively.
For e.g. The molar conductivity of HCl at infinite dilution can be expressed as,
Λ∞HCI = vH+λ∞H+ + vCI-λ∞CI-; For HCI, vH+ = 1 and vCI- = 1.
So, Λ∞HCI = (1 x λ∞H+) + (1 x λ∞H-); Hence,
Λ∞HCI = λ∞H+ + λ∞CI-

15. Applications of Kohlrausch's law
Determination of Λ∞m for weak electrolytes: 
The molar conductivity of a weak electrolyte at infinite dilution (Λ∞m) cannot be determined by extrapolation method. However, Λ∞m values for weak electrolytes can be determined by using the Kohlrausch's equation.
Λ∞CH3COOH = Λ∞CH3COONa + Λ∞HCI - Λ∞NaCI
Determination of the degree of ionization of a weak electrolyte:
  The Kohlrausch's law can be used for determining the degree of ionization of a weak electrolyte at any concentration. If λcm is the molar conductivity of a weak electrolyte at any concentration C and, λcm is the molar conductivity of a electrolyte at infinite dilution. Then, the degree of ionization is given by, α = Λcm = Λ∞m = Λcm/(v+ λ∞+ + v- λ∞-)
Thus, knowing the value of Λcm , and Λ∞m (From the Kohlrausch's equation), the degree of ionization at any concentration can be determined.

16. Determination of the ionization constant of a weak electrolyte : Weak electrolytes in aqueous solutions ionize to a very small extent. The extent of ionization is described in terms of the degree of ionization ( α ). In solution, the ions are in dynamic equilibrium with the unionized molecules. Such an equilibrium can be described by a constant called ionization constant.
For example, for a weak electrolyte AB, the ionization equilibrium is,
AB ⇔ A++ B-;
If C is the initial concentration of the electrolyte AB in solution, then the equilibrium concentrations of various species in the solution are,
[AB] = C (1 - α), [A+] = C α  and [B-] = C α
Then, the ionisation constant of AB is given by
K = [A+] [B-]/[AB] = C α C α /C (1 - α) = C α2/( 1 - α )
We know, that at any concentration C, the degree of ionisation (α ) is given by, α = Λcm / Λ∞m
Then, (K) = C(Λcm/Λ∞m)2/[1 - (Λcm/Λ∞m)] = C(Λcm)2/Λ∞m - Λcm); Thus, knowing Λ∞m and Λcm at any concentration, the ionisation constant (K) of the electrolyte can be determined.

17. Determination of the solubility of a sparingly soluble salt : Salts such as AgCl, BaSO4, PbSO4, etc which dissolve to a very small extent in water are called sparingly soluble salts. As they dissolve very little, even a saturated solution of such a salt can be assumed to be at infinite dilution. Then, the molar conductivity of a sparingly soluble salt at infinite dilution can be obtained from the relationship, Which can be further written as Or λ∞m can be calculated by applying Kohlrausch law (e.g. for AgCl, λ∞m = λ∞Ag+ + λ∞cl-)

18. Problem: A decinormal solution of NaCl has specific conductivity equal to If ionic conductance of Na+ and Cl– ions are 43.0 and 65.0 ohm–1 respectively. Calculate the degree of dissociation of NaCl solution. Solution: Given that Normality of NaCl solution = 0.1 N Equivalent conductance is given by Subsituting all the values, we get λeq = ( x 1000)/0.1 = 92 ohm-1

19. Problem: Equivalent conductance of NaCl, HCl and C2H5COONa at infinite dilution are , and 91 ohm–1 cm2 respectively. Calculate the equivalent conductance of C2H5COOH. Solution: The equivalent conductance of C2H5COOH is given by: λ∞ = λ∞(C2H5COONa) + λ∞(HCl) - λ∞(NaCl) = – = ohm–1 cm2

20. Problem: The conductivity of saturated solution of AgCl at 298 K is found to be x 10-6 S cm-1. Find it solubility. Given that the ionic conductances of Ag+ and Cl-1 at infinite dilution are 61.9 S cm2 mol-1 and 76.3 S cm2 mol-1 respectively.
Solution: λ∞m(AgCl) = λ∞m (Ag+) + λ∞m (Cl-) = = S cm2 mol-1 = (1.382 x 10-6 x 1000)/138.2 = 10-5 mol L-1 = 10-5 x g L-1

21. Electrolysis
Electrolysis may be defined as a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution or molten (fused) state.
The apparatus used to bring about electrolysis is called Electrolytic cell.
Electrolyte is taken in form of aqueous solution or molten state
Cathode: -vely charged i.e. connected to negative pole of battery.
Anode: +vely charged i.e. connected to positive pole of battery.
Current flows from anode to cathode

22. Electrolysis of molten sodium chloride

23. Faraday’s First Law of Electrolysis
Statement: The amount of chemical reaction and hence the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution or melt)
If W grams of the substance is deposited by Q coulombs of electricity, then
W α Q or W = ZQ
Where Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited
If a current of I amperes is passed for t seconds, then Q = I x t
So that W = Z x Q = Z x I x t
Thus, if Q = 1 coulomb or I = 1 ampere and t = 1 second, W = Z.
Hence, Electrochemical equivalent of a substance may be defined as the mass of the substance deposited when a current of one ampere is passed for one second.

25. Problem: On passing 0.1 Faraday of electricity through aluminium chloride what will be the amount of aluminium metal deposited on cathode (Al = 27) Solution:

26. Problem: How many atoms of calcium will be deposited from a solution of CaCl2 by a current of 25 milliamperes flowing for 60 s?

27. Faraday’s Second Law of Electrolysis
Statement: When the same quantity of electricity is passed through different electrolytes the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalence.

28. Example
For AgNO3 solution and CuSO4 solution connected in series, if the same quantity of electricity is passed, Quantitative aspects of Electrolysis: Consider the electrolysis of molten NaCl, i.e. NaCl Na + ½ Cl2 Thus, we have Na+ + e- Na This means that one electron produces one sodium atom. Therefore, passage of one mole of electrons will produce one mole of sodium. Similarly, looking at the reactions Cu2+ + 2e- Cu; Al3+ + 3e- Al Thus, we conclude that two moles of electrons produce one mole of Cu, three moles of electrons will produce one of Al. If “n” electrons are involved in the electrode reaction, the passage of n Faradays of electricity will liberate one mole of the substance/one Faraday of electricity deposits one gram equivalent of the substance.

29. Problem: How much charge is required for the following reaction?
1 mol of Al3+ to Al
1 mol of Cu2+ to Cu
1 mol of MnO4- to Mn2+
Solution: 1) The electrode reaction is
Al3+ + 3e Al
Therefore, quantity of charge required for reduction of 1 mol of Al3+ is given by: 1 mol of Al3+ = 3F = 3 x C = C
2) The electrode reaction is
Cu2+ + 2e Cu
Therefore, quantity of charge required for reduction of 1 mol of Cu2+ is given by: 1 mol of Cu2+ = 2F = 2 x C = C
3) The electrode reaction is MnO Mn2+
or Mn7+ + 5e Mn2+
therefore, charge required is 5F = 5 x = C.

30. Problem: How much Cu will be deposited on the cathode of an electrolytic cell containing CuSO4 solution by the passage of a current of 2 amperes for 30 minutes? (At. Mass of Cu = 63.5)
Solution: Current = 2 amperes Time = 30 min = 1800 s Therefore, quantity of electricity passed = Current x Time = 2 x 1800 =3600 C The reaction at cathode is Cu e- Cu Thus, 2F deposit Cu = 1 mol = 63.5 g Therefore, 3600 C will deposit Cu = (63.5/2x96500) x 3600 = g

32. Problem: What are the products of electrolysis of molten and aqueous sodium chloride?
Solution: (i) Na and Cl2 (ii) H2 and Cl2 Hint: The reduction potential of water is more compared to Na. therefore reduction of water takes place easily and as a result, we get hydrogen instead of Na.
Problem: What is the free energy change for (a) galvanic cell
(b) electrolytic cell?
Solution: (a) ∆G is negative as the reaction is spontaneous
(b) ∆G is positive as the reaction is non-spontaneous.

33. Thank You