1. Ridwan Islam Senior Lecturer Dept. of Pharmacy BRAC University, Dhaka-1212

2. Electrolytes A substance which dissociates into ions as a result of passage of electric current through its aqueous solution is known as electrolyte. Example: Sodium chloride, NaCl Copper sulfate, CuSO 4 Types of electrolytes Depending upon the capacity of conducting electricity electrolytes are divided into 5 groups. Strong electrolyte (Strong acids, strong bases, salts of strong acid and base) Weak electrolyte (Weak acid, weak base, HgCl 2 ) Moderately strong electrolytes (Na 2 CO 3, CH 3 COONa, etc) Very weak electrolytes (CH 3 -CO-CH 3 ) Non-electrolytes/insulator (Sugar, alcohol)

3. Electrodes The materials, usually small sheets, plates or wires of metal or may also be of non-metal that are inserted in the electrolytic solution of an electrolytic cell to pass an electric current through the electrolytes are termed as electrodes. Example- Copper electrode Zinc electrode Classification According to the attraction of ions electrodes are classified into two classes- Anode Cathode

4. Anode: The electrode connected to the positive terminal of the battery, attracts the negative anions (anions) and through which the electricity supposed to enter the solution is termed as an anode. In other words, anode is the electrode where oxidation takes place. Some typical anodic reactions are- Cu - 2e - =Cu +2 Fe +2 - e - =Fe +3 Cathode: The electrode connected to the negative terminal of the battery, attracts the positive anions (cations) and through which the electricity supposed to leave the solution is termed as a cathode. In other words, cathode is the electrode where reduction takes place. Some typical cathodic reactions are- Cu +2 + 2e - =Cu 2H + + 2e - =H 2

6. Mechanism of electrolysis Due to the passage of electricity through an electrolyte solution, the electrolyte is dissociated into two ions- cations and anions. The cations migrate to the cathode and form a neutral atom by accepting electrons from it. The anions migrate to the anode and yield a neutral particle by transfer of electrons to it. As a result of the loss of electrons by anions and gain of electrons by cations at their respective electrodes, chemical reaction takes place.

8. Electrochemistry Electrochemistry is the branch of physical pharmacy which elaborately deals with the phenomena of interconversion of electrical and chemical energy, the mechanism of such transformation and other phenomena associated with this. Electrochemistry indicates both the following points: Conversion of electrical energy into chemical energy Conversion of chemical energy into electrical energy

9. Faraday’s law of electrolysis In 1984, Michael Faraday announced his famous laws of electricity which describes the relationship between the amounts of products liberated at the electrodes and quantity of electricity. First law During electrolysis, the amount of an element deposited or dissolved in an electrode is proportional to the product of the amount of current and the time of its flow (i.e. the amount of electricity passed through the electrolyte solution).

10. Mathematical expression: If W is the mass of substance deposited on electrode by passing Q coulombs of electricity, then according to the first law- W α QWhere, or, W α ItI= the strength of current in ampere or,W = ZIt t= time in second Z= constant, known as electrochemical equivalent of electrolyte. Electrochemical equivalent (Z): Electrochemical equivalent is a constant for a particular element. When I= 1 amp and t= 1sec, then Z=W. Therefore Z is equal to the amount of substance deposited due to the passage of 1 amp current through 1 second.

11. Importance of first law By using first law we are able to calculate- The value of electrochemical equivalents of different substances. The masses of different substances produced by passing a known quantity of electricity through their solutions.

12. Second law If the same amount of electricity is passed through different electrolyte solutions during electrolysis, the amount of different elements deposited or dissolved at the electrode are proportional to their respective equivalent weight. Mathematical expression:

13. Let the weight of the substance deposited = W gm the equivalent weight of the substance = E gm According to the second law, W α E when Q is constant If the weights of two substances dissolved by passing the same quantity of electricity is W 1­ and W 2 having the respective chemical equivalent E 1­ and E 2, the second law can be mathematically expressed- W1/W2 = E1/E2……………..(1) From the first law we get, W = ZIt, W 1 = Z 1 It and W 2 = Z 2 It

14. So, W1/W2 = Z1/Z2…………………………..(2) From equation (1) and (2) E1/E2 = Z1/Z2 So, Z α E That is electrochemical equivalent is proportional to the chemical equivalent weight of a substance. Importance of second law The second law helps to determine- The equivalent weights of metals The unit of electric charge The Avogadro’s number

15. Math 96,500 coulombs are required to deposit 1 gm- equivalent of any element. 1 Faraday = 96,500 coulombs Equivalent weight = Atomic weight / valence Electrochemical equivalent, Z = E / 96,500 1. 0.1978 gm of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper?

16. Ans: We know,Here, Q = Itt = 50 minutes = 0.2 × 3000= 50 × 60 seconds = 600 coulombs = 3000 seconds Again, W = ZItI = 0.2 ampere So, Z = W/ It = 0.1978/ 600 = 0.0003296 g So, electrochemical equivalent of copper = 0.0003296 g

17. 2. What current strength in ampere will be required to liberate 10 g of iodine from potassium iodide solution in one hour?

18. Ans: We know, Z = E / 96,500 Here, = 127 / 96,500W = 10 g = 0.0013161t = 1 hour = 3600 seconds Again, W = ZIt Equivalent weight of I = W / Zt iodine = 127 = 10 / (0.0013161×3600) = 2.11 ampere

19. 3. An electric current is passed through three cells in series containing respectively solutions of copper sulfate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited?

21. Conductance may be defined as the ability of the electrolytes in electrolytic solution to conduct electric current. Mathematically conductance of an electrolytic solution is the reciprocal of its resistance. Unit of conductance is Siemens, S. Mathematical expression: The conductance of the electrolytic solution may be measured by Ohm’s law. According to this law- I = E / R Where, I= the current flowing through the solution in ampere R= resistance of the solution E= potential difference in volt Conductance

22. The resistance R of a conductor is directly proportional to its length in cm and inversely proportional to its cross sectional area A in cm 2.

23. Specific conductance The conductance of one centimeter cube (cc) of a solution of an electrolyte. Unit of specific conductance is S cm -1.The specific conductance is denoted by κ (kappa). Thus- κ = 1/ ρ Equivalent conductance The conductance of an electrolyte obtained by dissolving one gram equivalent of it in V cc of water. Unit of equivalent conductance is S cm 2 equivalent -1.The equivalent conductance is denoted by Λ. Thus- Λ = κ × V V= volume of solution in cc containing 1 gm equivalent of material

24. Molar conductance The conductance of all ions produced by one mole (one gram-molecular weight) of an electrolyte when dissolved in a certain volume V cc. Unit of molar conductance is S cm 2 mol -1.Molar conductance is denoted by μ. Thus- μ = κ × V V= volume of solution in cc containing 1 mole of the electrolyte

26. L/A ratio is also called cell- Constant This cell constant (K) is a function of the electrode areas, the distance between the electrodes and the electrical field pattern between the electrodes.

27. Conductance cell The cell which is specially designed for the measurement of conductance of a solution is known as conductance cell.

28. Kohlrausch’s law (Independent migration of ions) The equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductances of the component ions. The law may be expressed as: λ α = λ a + λ c Where, λ a is the equivalent conductance of the anion and λ c that of the cation.

29. 1. The specific conductance of an N/50 solution of KCl T 25˚c IS 0.002765 S. If the resistance of a cell containing this solution is 400ohms, what is the cell constant?

30. = 0.002765 × 400 = 1.106

31. 2. The resistance of decinormal solution of a salt occupying a volume between two platinum electrodes 1.8 cm apart and 5.4 cm 2 in area was found to be 32 ohms. Calculate the specific conductance of the solution. Also calculate the equivalent conductance.

33. Measurement of conductance of a solution Principle: The basic principle of the experimental determination of electrolyte conductance is the Wheatstone’s bridge, (Resistance of the conductance cell, R ÷Known resistance, S) = {L÷(100-L)}

35. Arrangement of apparatus: Wheatstone bridge circuit is a meter bridge consisting of three thick metals (A, B and C). A variable of known resistance is placed in the gap between B and C. The other gap between A and B is filled by a specific conductance cell in which the solution is placed. We have to determine the resistance of the conductance cell solution. The reciprocal of that resistance will be the conductance. A galvanometer is connected with a jockey, J that connects to the bridge wire AB. Procedure: After completing the circuit the jockey is moved along the wire until there is no signal in the galvanometer. This is the null point.

36. If the length of wire AC is 100 cm and null point is at a distance of L cm from point A, then according to the principle of wheatstone bridge gives, R/S = L/(100-L) R= S × L/(100-L) 1/R = (100-L)/(L×S) Therefore, Conductance of the electrolyte solution = (100-L)/(L×S)

37. Transport number During electrolysis the current is carried by the anions and the cations. The fraction of the total current carried by the cation or the anion is termed its transport or Hittorf’s number.

38. Measurement of transport number Transport number can be measured by the following ways- Hittorf’s method Moving boundary method The moving boundary method The moving boundary method is based on the direct observation of migration of ions under the influence of applied potential.

39. Apparatus:

40. The apparatus used consists of a long vertical tube fitted with two electrodes at the two ends. If the transport number of a cation (H + ) is to be determined, the electrolyte HCl solution is taken in the upper part of the apparatus and a layer of another electrolyte CdCl 2 having the common ion (Cl - ) is introduced in the lower part of the apparatus. The electrolyte CdCl 2 is selected so that the velocity of Cd +2 ion is less than that of H + ion.

41. Calculation: Let, c = original concentration of H + ions (gram- equivalents per ml) l = distance through which boundary moves (cm) s = area of cross section of the tube (sq. cm) n = number of faradays of current passed (Q / F) Number of equivalents of H + moving upward = s×l×c The fraction of current carried by H + ions = n×t + Hence, n×t + = s×l×c

42. Math: In a moving boundary experiment with 0.1N KCl using 0.065N LiCl as indicator solution, a constant current of 0.005893 amp was passed for 2130 seconds. The boundary was observed to move through 5.6 cm in a tube of 0.1142 cm 2. Calculate the transport number of K + and Cl - ions

44. Factors affecting transport number Then transport number is affected by the following three factors- 1. Concentration: The transport no. of ions varies with the changes of concentration. Such changes are quite complicated and the extent of change is very small. 2. Temperature: Transport number depends on temperature. t + increases slightly with rise of temperature. t -, therefore generally decreases with increase of temperature. 3. Specificity of each compound: For the same cation or anion the transport number depends on electrolyte. Thus, t + value of k + in KCl, KBr and KNO 3 are different.

45. Conductometric titration The titration in which conductance measurements are employed to determine the end point of acid-alkali reactions, some displacement reactions or precipitation reactions are called conductometric titration. For this purpose, the titrant is added from a burette into a measured volume of the solution to be titrated which is taken in a conductance cell and the conductance readings corresponding to the various additions are plotted against the volume of the titrant. In this way two linear curves are obtained, the point of intersection of which is the end point.

46. 1. Titration of a strong acid against a strong base For example, the reaction in which HCl is titrated against a solution of NaOH. This can be represented by- (H + + Cl - ) + (Na + + OH - ) → Na + +Cl - +H 2 O The highly conducting hydrogen ions present in the solution are replaced by sodium ions having ions having much smaller conductance. The conductance of Cl - remains constant and removed H + combine with OH - to form less ionized water. As a result the conductance of solution decreases with subsequent addition of alkali till the end point is reached.

48. 2. Titration of a weak acid against a strong base For example, the reaction in which CH 3 COOH is titrated against a solution of NaOH. This can be represented by- (H + + CH 3 COO - ) + (Na + + OH - ) → Na + +CH 3 COO - +H 2 O The initial conductance of the solution is low because of the poor dissociation of the weak acid. On adding alkali, highly ionized sodium acetate is formed. The acetate ions at first tend to suppress the ionization of acetic acid further due to common ion effect. But after a while the conductance begins to increase because the conducting power of highly ionizes salt exceeds that of the weak acid.

50. 3. Titration of a strong acid against a weak base For example, the reaction in which HCl is titrated against a solution of NH 4 OH. This can be represented by- (H + + Cl - ) + (NH 4 + + OH - ) → NH 4 + +Cl - +H 2 O The highly conducting hydrogen ions present in the solution are replaced by ammonium ions having ions having much smaller conductance. As a result the conductance of solution decreases with subsequent addition of alkali till the end point is reached.

52. 4. Titration of a weak acid against a weak base The conductometric method is particularly suitable as such titrations do not give a sharp end-point with indicators. For example, the reaction in which CH 3 COOH is titrated against a solution of NH 4 OH. This can be represented by- (H + + CH 3 COO - ) + (NH 4 + + OH - ) → CH 3 COO NH 4 +H 2 O The initial conductance of the solution is low because of the poor dissociation of the weak acid. But it starts increasing as the salt CH 3 COO NH 4 is formed.

54. 5. Precipitation reactions Barium Sulphate + Magnesium Hydroxide

55. In the titration of KCl against AgNO 3 for example, the change in conductance on the addition of AgNO 3 is not much since the mobility of the K + ion is of same order. Thus the curve is nearly horizontal. Ag + + NO 3 - + K + + Cl - → K + + NO 3 - + AgCl

56. Advantages of Conductometric titrations No indicator is required Colored/turbid solution: Where ordinary indicator give poor results can be successfully titrated More accurate results are obtained as the end point is determined graphically The method is useful in case of very dilute solution The method is useful for the titration of weak acids against weak bases It can be used for quantitative estimation of cations and anions in a variety of reaction.

57. Precautions of Conductometric titrations It is necessary to keep the temperature constant throughout the experiment In acid-alkali titrations, the titrant should be about 10 times stronger than the solution to be titrated