Electrochemistry for b.tech first year sem 1/2

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1. ELECTROCHEMISTRY

2. Conductance Transport Number Nernst Equation Contents Instrumental titrations Kohlrausch law ECS Galvanic cells

3. Learning points Electrolytic conductance Equivalent conductance Effect of dilution Numericals Conductance Specific conductance Molar conductance

4. Conductance Ease with which electricity flows through a solution. Dependence: Number of ions Mobility of ions

5. Electrolytic conductance Conductance of an electrolytic solution. It is due to movement of ions. C = 1/R Units: ohm -1 or Siemen (S) Effect of dilution: It increases with dilution.

6. Specific conductance Conductance / unit volume κ = 1/ ρ = C.l /A Units: Ω -1 cm -1 Effect of dilution: It decreases. Specific conductanceof N/50 KCl = 0.002765ohm-1cm-1

7. Molar conductance Conductance of all the ions present in one mole of electrolyte in the solution. λ m = κ .V = 1000 κ / C Units: Ω -1 cm 2 mole -1 Effect on dilution: It increases.

8. Equivalent conductance Conductance of all the ions present in one gram equivalent of electrolyte in the solution. λ eq = κ .V = 1000 κ / C Units: Ω -1 cm 2 equivalent -1 Effect on dilution: It increases.

9. Cell constant x = l /A x = Observed resistance / Specific resistance x = Specific conductance / Observed conductance Units: cm -1

10. Degree of dissociation α = λ c λ ∞ Equivalent conductance at any dilution Equivalent conductance at infinite dilution α dilution Strong electrolyte Weak electrolyte

11. Kohlrausch law λ ∞ = λ + + λ _ Applications: Calculation of absolute ionic mobility. λ = F U Determination of solubility of sparingly soluble salt. K V = λ͚ = λ a + λ c Calculation of degree of dissociation α = λ c λ ∞

12. Problems Specific conductance of a decimolar solution of KCl at 18 o C is 1.12 Sm -1 . The resistance of a conductivity cell containing the solution at 18 o C was found to be 55 ohm. What is the cell constant? The resistance of 0.01M solution of an electrolyte was found to be 210 ohm at 25 o C. Calculate the molar conductance of the solution at 25 o C. Cell constant is 0.88 cm -1 . Answers 1. 61.6 m -1 2. 0.0419 Sm 2 mole -1

13. Electrochemical cells

14. Learning points Single electrode Nernst’s equation Galvanic cells ECS & its applications Electrochemical cells Numericals

15. Redox Reactions-the source of energy OXIDATION : Loss of electron(s) R O + ne - ‘OR’ increase in oxidation number RE DUCTION : Gain of electron(s) O + ne - R ‘OR’ decrease in oxidation number OXIDIZING AGENT : electron acceptor REDUCING AGENT : electron donor  

16. Oxidation: Loss of electrons Source of electrons-anode

17. Reduction: Addition of electrons Consumer of electrons - cathode

18. Redox reaction Flow of electrons from anode to cathode

19. - + M M +n + ne -1 Oxidation removal of electrons at anode In to Solution In To Circuit Anode

20. - + M +n + ne -1 M At Electrode From Solution From Circuit Addition of electrons at cathode Reduction Cathode

21. Electrochemical cells Device for converting chemical energy into electric current. Important electrochemical cells: Galvanic/voltaic cell Daniel cell

22. Galvanic Cells spontaneous redox reaction anode oxidation cat hode red uction - +

23. Representation of cell reaction Zn ( s ) | Zn 2+ (1 M ) || Cu 2+ (1 M ) | Cu ( s ) anode cathode

24. Salt Bridge Device used to connect the oxidation and reduction half-cells of a galvanic cell. Allows the flow of ions to maintain a balance in charge between the oxidation and reduction vessels while keeping the contents of each separate. Types: Glass tube bridges Filter paper bridges Filled with inert electrolyte.

25. Representation of single electrodes Non metal electrodes SHE or NHE : Pt|H 2 (g, 1 bar)|H + (a = 1) || Chlorine electrode : Pt|Cl 2 (g, 1 bar)| Cl - (M) || Metal electrodes Copper electrode : Cu(s)|Cu 2+ (1 M)|| Zinc electrode : Zn(s)|Zn 2+ (1 M)||

26. Standard Electrode Potential Potential of eletcrode during reaction under standard condition (when all solutes are 1 M and all gases are at 1 atm ). Standard reduction potential Standard oxidation potential

27. Determination of standard electrode potential W.r.t . Standard Hydrogen Electrode. 2e - + 2H + (1 M ) H 2 (1 atm ) E = 0 V H 2 (1 atm ) → 2e - + 2H + (1 M ) E = 0 V

28. Examples anode anode cathode cathode

29. Determination of EMF E = E cathode - E anode E cathode : Reduction potential of electrode undergoing reduction E anode : Reduction potential of electrode undergoing oxidation

30. Nernst Equation

31. Consider a reduction reaction, Cu 2+ + e → Cu E = E + 0.05691] log[Cu 2+ ] n

32. TABLE OF STANDARD REDUCTION POTENTIALS 2 E o (V) Cu 2+ + 2e- Cu +0.34 2 H + + 2e- H 0.00 Zn 2+ + 2e- Zn -0.76 oxidizing ability of ion reducing ability of element To determine an E ox from a reduction table, just take the opposite sign of the reduction! ( Red. Pot )

33. Consider a reduction reaction, Red1 + Ox2 Red2 + Ox1 Eg ., Zn + Cu2+ Cu + Zn2+

34. •Electrons travel through external wire. Salt bridge allows anions and cations to move between electrode compartments. Zn --> Zn 2+ + 2e- Cu 2+ + 2e- --> Cu <--Anions Cations--> Oxidation Anode Negative Reduction Cathode Positive

35. Galvanic Cells 19.2 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell equation Zn ( s ) + Cu 2+ ( aq ) Cu ( s ) + Zn 2+ ( aq ) Zn ( s ) | Zn 2+ (1 M ) || Cu 2+ (1 M ) | Cu ( s ) anode cathode

36. Criteria for reversibility Thermodynamic criteria Driving force = opposite force Rate of discharging = Rate of charging E(cell) = o volts While discharging: Cu 2+ + Zn -----  Cu + Zn 2+ While charging: Cu + Zn 2+ ---- Cu 2+ +Zn Net reaction: Cu 2+ +Zn <-----  Cu + Zn 2+

37. Applications of EMF measurements Determination of activity coefficients of electrolytes Determination of Transport number of ions Determination of Valency of ions in doubtful cases Determination of solubility product Determination of pH of an unknown solution by using a) HE || SCE system c) GE || SCE b) SCE || QH 2 system

38. Determination by HE|| SCE system Construct a cell by using HE and SCE as shown below: Pt, H 2 ( 1 atm ), H + ( M = ?)|| KCl (sat),HgCl 2 (s),Hg E cell = E RHS - E LHS = 0.2422 - ( -0.05915pH)

39. Determination by HE||GE system Construct a cell by using HE and GE as shown below: Pt, 0.1M HCl|glass | H+(M=?)|| KCl (sat),HgCl 2 (s),Hg E cell = E RHS - E LHS = 0.2422 - ( Eg ⁰ -0.05915pH)

40. Determination of pH by SCE||QH 2 system Construct a cell by using SCE and QH2 as shown below: Hg,Hg 2 Cl 2 (s), KCl (sat)|| H + (M =?) Q,QH 2 ,Pt E cell = E RHS - E LHS = ( 0.6996-0.0591pH) - 0.2422

41. Instrumental analysis

42. Learning points Conductometric titrations SA Vs. SB SA Vs. WB WA Vs. SB Instrumental analysis

43. HCl + NaOH Na + Cl - + H 2 HCl Base N1 = ? N2 = Given V1 = 20ml(say) V2 = from graph N1V1 = N2V2 N1 ca be evaluated Consider an acid base reaction

44. Strong acid vs. strong base λ Vol. of base HCl + NaOH Na + Cl - + H 2 End point?

45. Strong acid vs. weak base λ Vol. of base HCl + NH 4 OH NH 4 + Cl - + H 2 End point?

46. Weak acid vs. strong base λ Vol. of base CH 3 COOH + NaOH CH 3 COO - Na + + H 2

47. Mixture of two acids vs. strong base NaOH CH 3 COOH λ Vol. of base HCl + CH 3 COOH + NaOH CH 3 COO - Na + + NaCl + H 2 HCl

48. Mixture of two acids vs. strong base NaOH CH 3 COOH λ Vol. of base HCl + CH 3 COOH + NaOH CH 3 COO - Na + + NaCl + H 2 HCl