Electrochemistry full chapter notes .pptx

Notes of electrochemistry

1. Electrochemical Cells The device or cell that converts the chemical energy liberated during the redox reaction to electrical energy is called a galvanic or a voltaic cell. Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

2. Daniell Cell

3. Daniell Cell (CONSTRUCTION) In this the spontaneous redox reaction between Zn and Cu 2+ produces an electric current. It consists of two half-cells. The half-cells on the left contains a Zn metal dipped in ZnSO 4 solution. The half-cell on the right consists of Cu metal in CuSO 4 solution. The half-cells are joined by a salt bridge that prevents the mechanical mixing of the solution. The overall redox reaction is: Zn(s) + Cu 2+  → Cu(s) + Zn 2+ ( aq ) At the zinc rod, oxidation occurs so it is the anode of the cell and is negatively charged while at copper electrode, reduction takes place, it is the cathode of the cell and is positively charged . Electrons flow from anode to cathode in the external circuit. The zinc rod loses its mass while the copper rod gains in mass. 8. The concentration of ZnSO 4 solution increases while the concentration of CuSO 4 solution decreases.

4. Daniell Cell

5. Uses of a salt bridge It completes the circuit. It helps in prevention of accumulation of charges at the electrodes.

6. L eft ( L oss of e - s) O xidation A node N egatively charged

7. Cell Representation For the cell reaction, Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

8. Cell Representation For the cell reaction, Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Half-cell reactions : Left electrode (ANODE) : Zn(s) → Zn 2+ ( aq , 1 M) + 2 e – Right electrode ( CATHODE) : Cu 2+ ( aq , 1 M) + 2 e – → Cu(s) The cell can be represented as: Zn(s)|Zn 2+ ( aq )||Cu +2 ( aq )|Cu(s)

9. Question 1. For the Cell representation: Cu(s)|Cu 2+ ( aq )||Ag + ( aq )|Ag(s) Find the Anode Find the positive electrode. Electrode at which oxidation is taking place Metal ion getting reduced Write the reaction taking place at the anode and cathode Write the net reaction.

10. Question 1. For the Cell representation: Cu(s)|Cu 2+ ( aq )||Ag + ( aq )|Ag(s) Find the Anode Find the positive electrode. Electrode at which oxidation is taking place Metal ion getting reduced Write the reaction taking place at the anode and cathode Write the net reaction. Answers: Cu Ag Cu Ag+ Anode (oxidation): Cu(s) → Cu 2+( aq ) + 2e – Cathode (reduction): 2Ag + ( aq ) + 2e – → 2Ag(s) vi) Cu(s) + 2Ag + ( aq ) → Cu 2+( aq ) + 2 Ag(s)

11. At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. A potential difference develops between the electrode and the electrolyte which is called electrode potential. ELECTRODE POTENTIAL

12. Cell Potential The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts . The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell.

13. Measurement of Standard Electrode Potential The potential of individual half-cell cannot be measured. We can measure only the difference between the two half-cell potentials that gives the emf of the cell. According to convention, a half-cell called standard hydrogen electrode (SHE) represented by Pt(s)⎥ H 2 (g)⎥ H + ( aq ) is assigned a zero potential at all temperatures corresponding to the reaction H + ( aq ) + e – →1/2H 2 (g)

14. Construction of SHE The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity. The pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the solution is one molar. The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. bar

15. At 298 K the emf of the cell, standard hydrogen electrode ⎜⎜ second half-cell constructed by taking standard hydrogen electrode as anode (reference half-cell) and the other half-cell as cathode, gives the reduction potential of the other half-cell. As E o L for standard hydrogen electrode is zero. E cell o = E o R – 0 = E o R

16. STANDARD CELL POTENTIAL OR EMF OF A CELL: 1.The difference between the reduction potentials of two half cells is called cell potential. 2. It is known as electromotive force (EMF) of the cell if no current is drawn from the cell. E o cell = E o cathode - E o anode E o cell = E o REDUCTION - E o OXIDATION E cell = E o R - E o L

17. Calculation of Standard electrode potential of a cell EXAMPLE 1 . Find the EMF of the cell for the reaction, Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) a) Write reactions at the electrode. b) Cell representation c)EMF of the cell ( E Cu2+/Cu = 0.34 V , E Zn2+/Zn = - 0.76 V )

18. Calculation of Standard electrode potential of a cell EXAMPLE 1 . Find the EMF of the cell for the reaction, Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) SOLUTION: Left electrode : Zn(s) → Zn 2+ ( aq , 1 M) + 2 e – Right electro de: Cu 2+ ( aq , 1 M) + 2 e – → Cu(s) Cell Representaion will be Zn(s)|Zn 2+ ( aq )||Cu +2 ( aq )|Cu(s) Emf of the cell = E cell = E o R - E o L = 0.34V – (– 0.76)V = 1.10 V

19. Calculation of Standard electrode potential of a cell EXAMPLE 2 . Find the EMF for the cell: Cu(s) + 2Ag + ( aq ) → Cu 2+( aq ) + 2 Ag(s) Half-cell reactions : Cathode (reduction): 2Ag + ( aq ) + 2e – → 2Ag(s) Anode (oxidation): Cu(s) → Cu 2+( aq ) + 2e – The cell can be represented as: Cu(s)|Cu 2+ ( aq )||Ag + ( aq )|Ag(s) and we haveE cell = E o R - E o L = E Ag +⎥Ag – E Cu 2+ ⎥Cu

20. Points to Remember 1. Positive value of EMF ( E cell ) means that the redox reaction is feasible or spontaneous. 2. The value of E cell is the minimum voltage required to bring about the electrolysis. 3. More the Reduction potential, stronger it is an OXIDISING AGENT 4. Better OXIDISING AGENT means lower REDUCING POWER Because Reduction potential α ____ 1_______ Reducing power 5. As we go from top to bottom a) Reduction potential decreases b) Reducing Power increases c) Oxidation potential increases d) Oxidising Power decreases

21. If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species. Points to Remember

22. The standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F 2 ) has the maximum tendency to get reduced to fluoride ions (F – ) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.

23. Sign Conventions for the Reduction Potentials The electrode at which reduction takes place w.r.t SHE has a POSITIVE sign for the reduction potential. The electrode at which oxidation takes place w.r.t SHE has a NEGATIVE sign for the reduction potential.

24. EMF of a cell Points to remember: a) If the E cell is positive, it indicates that the choice of anode and cathode is right.

25. Intext Questions 3.1 How would you determine the standard electrode potential of the system Mg 2+ |Mg? 3.2 Can you store copper sulphate solutions in a zinc pot? 3.1 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn. Hint : A metal with lower electrode potential would displace another with a higher electrode potential from a solution of its salt. So the order is Mg, Al, Zn, Fe, Cu and Ag. EXERCISES

26. 3.2 Given the standard electrode potentials, K + /K = –2.93V, Ag + /Ag = 0.80V, Hg 2+ /Hg = 0.79V Mg 2+ /Mg = –2.37 V, Cr 3+ /Cr = – 0.74V Arrange these metals in their increasing order of reducing power. Hint: as we go from top to bottom in Table the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction. 3.3 Depict the galvanic cell in which the reaction : Zn(s)+2Ag + ( aq ) →Zn 2+ ( aq )+2Ag(s) takes place. Further show: ( i ) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

27. 3.2 Given the standard electrode potentials, K + /K = –2.93V, Ag + /Ag = 0.80V, Hg 2+ /Hg = 0.79V Mg 2+ /Mg = –2.37 V, Cr 3+ /Cr = – 0.74V Arrange these metals in their increasing order of reducing power.

28. Effect of Conc. on electrode and cell potentials: Nernst Equation If the concentration of all the species involved in the electrode reaction is not unity, Nernst showed that for the electrode reaction: M n + ( aq ) + ne – → M(s), the electrode potential at any concentration measured with respect to SHE can be represented by: E ( Mn +/M) = E o ( Mn +/M) – RT ln [M] nF [ Mn + ] but concentration of solid M is taken as unity E ( Mn +/M) = E o ( Mn +/M) – RT ln [1] nF [ Mn + ]

29. Effect of Conc. on electrode and cell potentials: Nernst Equation If the concentration of all the species involved in the electrode reaction is not unity, Nernst showed that for the electrode reaction: For Eg. Zn 2+ ( aq ) + 2e – → Zn(s) E (Mn+/M) = E o (Mn+/M) – RT ln [M] nF [Mn + ] E (Mn+/M) = E o (Mn+/M) – 2.303RT log 1__ nF [Mn + ] 2.303 RT = 0.0591 F (as R = 8.314 JK –1 mol –1 , F is Faraday constant = 96487 C/mol), T is 298K, concentration of solid M is taken as unity) E cell = E o cell – 0.0591 log [M] n [Mn + ]

30. Effect of Conc. on electrode and cell potentials: Nernst Equation If the concentration of all the species involved in the electrode reaction is not unity, Nernst showed that for the electrode reaction: For Eg. Cu 2+ ( aq ) + 2e – → Cu(s) E (Mn+/M) = E o (Mn+/M) – RT ln [M] nF [Mn + ] E (Mn+/M) = E o (Mn+/M) – 2.303RT log 1__ nF [Mn + ] E (Cu 2+ /Cu) = E o (Cu 2+ /Cu) – 2.303RT ln [Cu] 2F [Cu +2 ] 2.303 RT = 0.0591 (as R = 8.314 JK –1 mol –1 , F is Faraday constant = 96487 C/mol), T is 298K F and concentration of solid M is taken as unity) E (Cu 2+ /Cu) = E o (Cu 2+ /Cu) – 0 .0591 log [Cu] 2 [Cu +2 ]

31. For the reaction Zn(s) + Cu 2+ C 1 (aq) → Zn 2+ C 2 (aq) + Cu(s) For Cathode: E (Cu2+/Cu) = E o (Cu2+/Cu) – 0.0591 log 1    2 [Cu 2+ ] For Anode: E (Zn2+/Zn) = E o (Zn2+/Zn) – 0.0591 log __ 1__ 2 [Zn 2+ ] E (cell) = E (Cu2+/Cu) – E (Zn2+ /Zn) = E o (Cu2+/Cu) – 0.0591 log 1 – E o (Zn2+/Zn) + 0.0591 log __ 1__ 2 [Cu 2+ ] 2 [Zn 2+ ] = ( E o (Cu2+/Cu) – E o (Zn2+/Zn) ) – 0.0591 log 1    - log 1__    2 [Cu 2+ ] [Zn 2+ ] E (cell) = E o (cell) – 0.0591 log [Zn 2+ ] 2 [Cu 2+ ] E (cell) = E o (cell) – 0.059 log [Oxidized] n [Reduced]

32. Points to remember: 1. If a cell is given: For eg. Zn/Zn 2+ // Cu 2+ / Cu E (cell) = E o (cell) – 0.059 log [left] 2 [right] E (cell) = E o (cell) – 0.059 log [Zn 2+ ( aq )] 2 [Cu 2+ ( aq )] 2. But if a reaction is given Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) E (cell) = E o (cell) – 0.059 log [right] 2 [left] E (cell) = E o (cell) – 0.059 log [Oxidized] 2 [Reduced] 3. 2.303 RT = 0.0591 ( ONLY AT 298 K) F ( IF T is given, then it is taken as it is)

33. and for a general electrochemical reaction of the type: a A + bB ⎯⎯⎯→ cC + dD Nernst equation can be written as: E (cell) = E o (cell) – RT ln Qc nF E (cell) = E o (cell) – 0.0591 log [C] c [D] d n [A] a [B] b

34. For Practice Ni(s)⎥ Ni 2+ (aq) ⎥⎥ Ag + (aq)⎥ Ag Write the cell reaction Write the Nernst Equation Answers: a) The cell reaction is Ni(s) + 2Ag + ( aq ) →Ni 2+ ( aq ) + 2Ag(s) b) The Nernst equation can be written as E (cell) = E o (cell) – 0.059 log [Ni 2+ ] 2 [Ag + ] 2

35. Example 3.1 Represent the cell in which the following reaction takes place Mg(s) + 2Ag + (0.0001M) → Mg 2+ (0.130M) + 2Ag(s) Calculate its E (cell) if E o (cell) = 3.17 V. Solution: The cell can be written as Mg⎥ Mg 2+ (0.130M)⎥⎥ Ag + (0.0001M)⎥ Ag E (cell) = E o (cell) – 0.059 log [Mg 2+ ] 2 [Ag + ] 2 = 3.17 – 0.059 log 0.130 2 [ 0.0001 ] 2 = 3.17 V – 0.21V = 2.96 V

36. NCERT Exercises 3.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd 2+ (aq) → 2Cr 3+ (aq) + 3Cd (ii) Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag(s) Calculate the ΔG r and equilibrium constant of the reactions. 3.5 Write the Nernst equation and emf of the following cells at 298 K: Mg(s)|Mg 2+ (0.001M)||Cu 2+ (0.0001 M)|Cu(s) Fe(s)|Fe 2+ (0.001M)||H + (1M)|H 2 (g)(1bar)| Pt(s) Sn (s)|Sn 2+ (0.050 M)||H + (0.020 M)|H 2 (g) (1 bar)|Pt(s) Pt(s)|Br 2 (l)|Br – (0.010 M)||H + (0.030 M)H 2 (g) (1bar)|Pt(s)

37. Extra Numericals Question: A cell is prepared by dipping a Cu rod in 1 M CuSO 4 soln. and a Ni rod in 1M NiSO 4 soln. The standard reduction potentials of Cu and Ni electrode are 0.34V and -0.25 V resp. What will be the cell reaction? What will be the Std. EMF of the cell Which electrode will be positive How will the cell be represented ?

38. SPONTANEITY OF A REACTION If the concentrations are 1M, E o cell should be positive and ΔG o should be negative. If the concentrations are not 1M, E cell should be positive and ΔG should be negative.

39. TYPES OF ELECTRODES: Metal-Metal Ion electrodes: Anode: M ==== M n+ + ne– Cathode: M n+ + ne– ==== M Gas Electrodes: Electrode gases like H 2 , Cl 2 etc. are used with their respective ions. For example, H 2 gas is used with a dilute solution of HCl (H+ ions). The metal should be inert so that it does not react with the acid.

40. TYPES OF ELECTRODES: 3. Metal-Insoluble salt electrode: a) We use salts of some metals which are sparingly soluble with the metal itself as electrodes. b) For example, if we use AgCl with Ag 4. Calomel Electrode : Mercury is used with two other phases, one is a calomel paste (Hg 2 Cl 2 ) and electrolyte containing Cl – ions. 5. Redox Electrode: In these electrodes two different oxidation states of the same metal are used in the same half cell. For example, Fe2+ and Fe3+ are dissolved in the same container and an inert electrode of platinum is used for the electron transfer.

41. Equilibrium Constant from Nernst Equation If the circuit in Daniel cell is closed then the reaction Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) takes place and as time passes, the concentration of Zn 2+ keeps on increasing while the concentration of Cu 2+ keeps on decreasing. After some time there will not be any change in the concentration of Cu 2+ and Zn 2+ ions and at the same time, voltmeter gives zero reading. This indicates that eqm . has been attained. In this situation the Nernst equation may be written as: E (cell) = 0 = E o (cell) – 0.059 log [Zn 2+ ( aq )] 2 [Cu 2+ ( aq )] Or E o (cell) = 0.059 log [Zn 2+ ( aq )] 2 [Cu 2+ ( aq )]

42. Equilibrium Constant from Nernst Equation But at equilibrium, [Zn 2+ ] = K c [Cu 2+ ] E o (cell) = 0.059 log K c 2 In general, E o (cell) = 2.303RT log K c nF

43. Example 3.2 : Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag+( aq ) → Cu2+( aq ) + 2Ag(s) E o (cell) = 0.46 V Solution : E o (cell) = 0.059 log K c = 0.46 V 2 log K c = 0.46 X 2 = 15.6 0.059 K C = 3.92 × 10 15

44. Electrochemical Cell and Gibbs Energy of the Reaction When a cell reaction takes place, electrical energy is produced and Electrical work done = Decrease in free energy Electrical work done = electrical energy produced = quantity of electricity flowing x EMF = n F E cell Hence Δ r G = - n F E cell Δ r G o = - n F E o cell Also, E o cell = 2.303RT log K c nF Δ r G o = - 2.303RT log K c

45. SPONTANEITY OF A REACTION: For a spontaneous cell reaction ΔG should be negative and cell potential should be positive. Δ r G o = - n F E o cell CONCENTRATION CELLS If two electrodes of the same metal are dipped separately into two solutions of the same electrolyte having different concentrations and the solutions are connected through salt bridge, such cells are known as concentration cells. H 2 ⎥ H + (c 1 )⎥⎥ H + (c 2 )⎥ H 2 E (cell) = 0.059 log [c 2 ] n [c 1 ] Cu (s)| Cu 2+ ( c 1 )||Cu 2+ ( c 2 )|Cu(s) H 2 (p 1 )⎥ H + (c)⎥⎥ H + (c)⎥ H 2 (p 2 ) E (cell) = 0.059 log [p 2 ] n [p1]

46. Example 3.3 : The standard electrode potential for Daniel cell is 1.1V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu 2+ (aq) ⎯→ Zn 2+ (aq) + Cu(s) (–21.227 kJ mol –1 ) Intext Questions 3.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. 3.5 Calculate the emf of the cell in which the following reaction takes place Ni(s) + 2Ag + (0.002 M) → Ni 2+ (0.160 M) + 2Ag(s) Given that E o (cell) = 1.05 V 3.6 The cell in which the following reaction occurs: 2Fe 3+ ( aq ) + 2I − ( aq ) → 2Fe 2+ ( aq ) + I 2 (s) has E o (cell) = 0.236 V at 298 K. Calculate Δ r G o and K c of the cell reaction.

47. Resistance of Electrolytic Solutions The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross section, A. R ∝ l/A or R = ρ l/A The constant of proportionality, ρ (rho), is called resistivity (specific resistance) SI units are ohm metre ( Ω m ) and ohm centimetre ( Ω cm). R = ρ l A The resistivity for a substance is its resistance when it is one metre long and its area of cross section is one m 2 . 1 Ω m = 100 Ω cm or 1 Ω cm = 0.01 Ω m

48. The inverse of resistance, R, is called conductance, G G =1/R = A/ ρ l = κ A/l The SI unit is siemens ,‘S ’ and is equal to ohm –1 or Ω –1 . Conductivity of Electrolytic Solutions The inverse of resistivity, called conductivity (specific conductance) represented by κ (Greek, kappa) SI units of conductivity are S m –1 or S cm –1 . G = κ A/l Conductivity of a material in S m –1 is its conductance when it is 1 m long and its area of cross section is 1 m 2 . 1 S cm –1 = 100 S m –1 . Conductance of Electrolytic Solutions

49. Conductance Depending on the magnitude of their conductivity, materials are classified into : Conductors : Metals and their alloys have very large conductivity and are known as conductors. Certain non-metals like carbon-black, graphite and some organic polymers are also electronically conducting. Insulators : Substances like glass, ceramics, etc., having very low conductivity are known as insulators. Semiconductors : Substances like silicon, doped silicon and gallium arsenide having conductivity between conductors and insulators are called semiconductors and are important electronic materials. Superconductors : Substances having zero resistivity or infinite conductivity. Earlier, only metals and their alloys at very low temperatures (0 to 15 K) but nowadays a number of ceramic materials and mixed oxides are also known to show superconductivity at temperatures as high as 150 K.

50. Types of Conductance Conductance is of two types: Electronic conductance E lectrolytic or ionic conductance 1 conductance through metals is called metallic or electronic conductance and is due to the movement of electrons The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance 2 the nature and structure of the metal the nature and conc. of the electrolyte added 3 the number of valence electrons per atom size of the ions produced and their solvation 4 temperature (it decreases with increase of temp.) temperature (it increases with the increase of temperature). 5 the nature of the solvent and its viscosity

51. Measurement of the Conductivity of Ionic Solutions R = ρ l/A but 1/ ρ = κ R = l/ κ .A The quantity l/A is called cell constant denoted by the symbol, G* . It depends on the distance between the electrodes and their area of cross-section and has the dimension of length –1 G* =l/ A G* = R κ Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is denoted by  ᴧ m (lambda).

52. Measurement of the Conductivity of Ionic Solutions Molar conductance is related to specific conductance as, ᴧ m = κ /C If κ is expressed in S m –1 then the units of Λ m are in S m 2 mol –1 . Λ m (S m 2 mol –1 ) = κ (S m –1 ) 1000 × M If κ is expressed in S cm –1 then the units of Λ m are in S cm 2 mol –1 . Λ m (S cm 2 mol –1 ) = κ (S cm –1 ) X 1000 M

53. Example 3.4 Resistance of a conductivity cell filled with 0.1 mol L –1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L –1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L –1 KCl solution. The conductivity of 0.1 mol L –1 KCl solution is 1.29 S/m.

54. Example 3.4 Resistance of a conductivity cell filled with 0.1 mol L –1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L –1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L –1 KCl solution. The conductivity of 0.1 mol L –1 KCl solution is 1.29 S/m. Solution : G* = R κ = conductivity × resistance = 1.29 S/m × 100 Ω = 129 m –1 = 1.29 cm –1 κ = G* = 1.29 cm –1 R 520 Ω κ = 0.248 × 10 –2 S cm –1 Λ m = κ × 1000 = 0.248×10 –2 ×1000 = 124 Scm 2 mol –1 molarity 0.02

55. Conductance : conductance increases on dilution. Reason: Conductance of a solution is due to the presence of ions in the solution. Greater the no. of ions , greater is the conductance. As on dilution, more ions are produced, so the conductance increases. 2. Conductivity ( κ ) : The conductivity of a solution at any given concentration is the conductance of one unit volume of solution kept between two platinum electrodes with unit area of cross section and at a distance of unit length. It ↓ ses with dilution both, for weak and strong electrolytes. Reason: This is because the number of ions per unit volume that carry the current in a solution decreases on dilution. 3. Molar Conductivity : ↑ ses on dilution Λ m = κ V Reason: Decrease in κ on dilution of a solution is more than compensated by increase in its volume. Variation of Conductance, Conductivity and Molar Conductivity with Concentration

56. Λ m increases slowly with dilution and can be represented by the equation: Λ m = Λ ° m – A c ½ Where Λ ° m is the limiting molar conductivity or molar conductivity at infinite dilution (when conc. approaches zero, the molar conductivity is called as limiting molar conductivity) Variation in Λ m with conc. for strong electrolytes: Debye Huckel – Onsager equation

57. Λ m = Λ ° m – A c ½ The value of ‘A’ for a given solvent and temp. depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in the solution. Thus, NaCl, CaCl 2 , MgSO 4 are known as 1-1, 2-1 and 2- 2 type electrolytes respectively. If we plot Λ m against C 1/2 , we obtain a straight line with intercept equal to Λ ° m and slope equal to ‘–A’. Λ m

58. The curve obtained for a strong electrolyte shows that there is only small increase in conductance with dilution Variation in Λ m with conc. for strong/weak electrolytes Reason : i ) A strong electrolyte is completely dissociated in solution and so the no. of ions remain constant. ii) At higher concentrations : the greater inter-ionic attractions retard the motion of ions and therefore the conductance falls with increasing concs . iii) With dilution : The ions are far apart and therefore the inter- ionic attractions decrease due to which the conductance increases with dilution and approaches a max. limiting value at infinite dilution. and consequently the number of ions in total volume of solution that contains 1 mol of electrolyte. Λ m

59. Variation in Λ m with conc. for strong electrolytes

60. Weak electrolytes In such cases Λ ° m increases steeply on dilution, specially near lower concentrations. Therefore, Λ ° m cannot be obtained by extrapolation of Λ m to zero concentration. Λ m i ) Weak electrolyte like acetic acid have lower degree of dissociation at higher concs . and hence for such electrolytes, the change in Λ m with dilution is due to increase in the degree of dissociation.

61. Example 3.6 : The molar conductivity of KCl solutions at different concentrations at 298 K are given below: c/mol L –1 Λm/S cm 2 mol –1 0.000198 148.61 0.000309 148.29 0.000521 147.81 0.000989 147.09 Show that a plot between Λ ° m and c 1/2 is a straight line. Determine the values of Λ ° m and A for KCl .

62. Kohlrausch law of independent migration of ions “The law states that limiting molar conductivity of a strong electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.” Λ ° m (NaCl) = λ Na + + λ Cl – Where λ Na + and λ Cl – are limiting molar conductivity of the sodium and chloride ions resp. Λ ° m = ν + λ + + ν – λ – Where, no. of cations = v + no. of anions = v – λ + and λ – are the limiting molar conductivities of the cation and anion resp.

63. Example Λ ° m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate Λo for Hac (CH3COOH) Solution: NaCl = Na + + Cl- HCl = H+ + Cl- NaAc = Na+ + CH3COO- To find for: CH3COOH = CH3COONa + HCl - NaCl

64. At infinite dilution a weak electrolyte dissociates completely , but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately. Therefore, Λ ° m for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions. At any concentration c, if α is the degree of dissociation Then α = Λ m / Λ ° m But for a weak electrolyte ka = c α2 1 – α = c Λ m 2 Λ ° m 2 ( 1 – Λ m / Λ ° m ) = c Λ m 2 Λ ° m ( Λ ° m – Λ m ) Application of Kohlrausch Law:

65. Example 3.9: The conductivity of 0.001028 mol L –1 acetic acid is 4.95 × 10–5 S cm –1 . Calculate its dissociation constant if Λ ° m for acetic acid is 390.5 S cm 2 mol –1 . 3.7 Why does the conductivity of a solution decrease with dilution? 3.8 Suggest a way to determine the Λ°m value of water. 3.9 The molar conductivity of 0.025 mol L –1 methanoic acid is 46.1 S cm 2 mol –1 . Calculate its degree of dissociation and dissociation constant. Given λ (H+)= 349.6 S cm 2 mol –1 and λ (HCOO – ) = 54.6 S cm 2 mol –1 Intext Questions

66. Faraday’s Laws of Electrolysis: First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation ). Quantitative Aspects of Electrolysis

67. Quantitative Aspects of Electrolysis Charge on one electron = 1.6021× 10 –19 C. Therefore, the charge on one mole of electrons = N A × 1.6021 × 10 –19 C = 6.02 × 10 23 × 1.6021 × 10 –19 C = 96487 C mol –1 = 96500 C This quantity of electricity is called Faraday (F) or Faraday’s constt . Ag + (aq) + e – → Ag(s) ( 1F or 96500C for 1mol or 108g of Ag) Al 3+ (aq) + 3e – → Al(s) ( 3F or 3 x 96500C for 1mol or 27g of Al)

68. Quantitative Aspect of First Law of Electrolysis W α Q W = Z Q, Z= M/ nF (where M= At.mass or molecular mass, n= charge on the metal, F=96500C) Q=I.t , (where I is the current passed (in A) and t is the time in sec) So, W= M.I.t nF Calculate the amount of charge in C and F for the following depositions: 2 mol of Mg 8g of Ca 1 mol of MnO 4- to Mn 2+

69. Q1. Calculate the amount of charge in C and F for the following depositions: 2 mol of Mg 8g of Ca 1 mol of MnO 4- to Mn 2+ Solutions: i ) 2 mol of Mg Mg 2+ (aq) + 2e – → Mg(s) (4F, 4x 96500C) 8g of Ca from molten CaSO 4 (Ca 2+ (aq) + 2e – → Ca(s) For 40g=== 2F, for 8g== 8 x 2/40 F, (y x 96500C) iii) 1 mol of MnO 4- to Mn 2+ 5e- + MnO 4- Mn +2 (5F, 5x 96500C)

70. Q 2: Find the time taken to deposit 1.27 g of Cu when current of 2A is passed through CuSO4 solution( cu= 63.5u). Solution: W = M.I.t nF 1.27 = 63.5 x 2 x t 2 x 96500 3.16 Three electrolytic cells A,B,C containing solutions of ZnSO 4 , AgNO 3 and CuSO 4 , respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? Solution: w1 = w2 = w3 E1 E2 E3 W Zn = W Ag E Zn E Ag

71. Example 3.10 : A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? Solution : t = 600 s Q = I × t = 1.5 A × 600 s = 900 C According to the reaction: Cu 2+ ( aq ) + 2e – = Cu(s) To deposit 1 mol or 63 g of Cu we require 2F or 2 × 96487 C For 900 C, the mass of Cu deposited = (63 g × 900 )/(2 × 96487 C ) = 0.2938 g.

72. Intext Questions 3.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? 3.11 Suggest a list of metals that are extracted electrolytically . 3.12 Consider the reaction: Cr 2 O 7 2– + 14H + + 6e – → 2Cr 3+ + 8H 2 O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr 2 O 7 2– ?

73. Products of Electrolysis Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., Pt or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. Some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these don’t seem to take place and extra potential (called overpotential ) has to be applied, which makes such process more difficult to occur.

74. Products of Electrolysis 1. Molten NaCl: the products of electrolysis are sodium metal and Cl 2 gas. At Cathode: Na+ (aq) + e – → Na (s) At Anode : Cl – (aq) → ½ Cl 2 (g) + e – 2. Aqueous NaCl : the products are NaOH, Cl 2 and H 2 . Cations formed are: H + , Na + At the cathode there is competition between the following reductions: Na+ (aq) + e – → Na (s) E o cell = – 2.71 V H+ (aq) + e – → ½ H 2 (g) E o cell = 0.00 V At the cathode the reaction with higher value of E o is preferred and, therefore, the reaction at the cathode during electrolysis is: H 2 O (l ) + e – → ½H 2 (g) + OH –

75. Anions formed are Cl – ions, OH- At the anode the following oxidation reactions are possible: Cl – (aq) → ½ Cl 2 (g) + e – E o cell = 1.36 V 2H 2 O (l )→ O 2 (g) + 4H + (aq) + 4e – E o cell = 1.23 V The reaction at anode with lower value of E o is preferred ( bcoz Eo is the reduction potential) and therefore, water should get oxidised in preference to Cl – , However, on account of overpotential of oxygen, first reaction is preferred . Thus, the net reactions may be summarised as: NaCl ( aq ) ⎯⎯⎯→ Na + ( aq ) + Cl – ( aq ) Cathode : H 2 O(l ) + e – → ½ H 2 (g) + OH – (aq) Anode : Cl – (aq) → ½ Cl 2 (g) + e– Net reaction : NaCl(aq) + H 2 O(l) → Na + (aq) + OH – (aq) + ½H 2 (g) + ½Cl 2 (g)

76. TIPS TO FIND OUT THE PRODUCTS: With Inert Electrodes: Reduction Potential of cations of Groups 1,2 and 13 < RP of H 2 O So, reduction of water to give H 2 gas will take place. 2H 2 O +2e- === H 2 + 2OH- Reduction Potential of cations of d- block metals > RP of H 2 O so, Metal ions get reduced to solid Metal For Eg Zn 2+ + 2e- === Zn (s) Oxidation Potential of F - , NO 3 - , SO 3 2- , SO 4 2- , S 2 O 3 2- < H 2 O So, H 2 O gets oxidized to O 2 2H 2 O === O 2 + 4H + + 4e- Oxidation Potential of Br - , Cl - , I - > H 2 O So, the ions get oxidized to corresponding gas 2Cl- === Cl 2 + 2e- 3.18 Predict the products of electrolysis in each of the following: ( i ) An aqueous solution of AgNO 3 with silver electrodes. (ii) An aqueous solution of AgNO 3 with platinum electrodes. (iii) A dilute solution of H 2 SO 4 with platinum electrodes. (iv) An aqueous solution of CuCl 2 with platinum electrodes.

77. TIPS TO FIND OUT THE PRODUCTS: With Active/ Metal Electrodes: For Eg. Products formed on electrolysis of CuSO 4 with Cu electrodes will be: At the Cathode, Cations will be: Cu 2+ , H + Cu 2+ + 2e- === Cu (s) At the Anode, Anions will be: SO 4 2- , OH - and also Cu metal electrode In this case the metal gets oxidized to ions Cu === Cu 2+ + 2e-

78. 3.18 Predict the products of electrolysis in each of the following: ( i ) An aqueous solution of AgNO 3 with silver electrodes. (ii) An aqueous solution of AgNO 3 with platinum electrodes. (iii) A dilute solution of H 2 SO 4 with platinum electrodes. (iv) An aqueous solution of CuCl 2 with platinum electrodes.

79. Batteries A battery is basically a galvanic cell where the chemical energy of the redox reaction is converted into electrical energy. There are mainly two types of batteries: Primary battery Secondary Battery 1. The reaction occurs only once and after use over a period of time battery becomes dead. 1. This after use can be recharged by passing current through it in the opposite direction. 2. It cannot be reused again. 2. It can be used again. 3. Eg. Dry cell, which is used in transistors and clocks. 3. eg. Lead storage battery, used in automobiles and invertors.

80. Primary Batteries A. Dry cell (known as Leclanche cell) 1. Construction: Anode: Zn container Cathode: C (graphite) rod surrounded by powdered MnO 2 and C. The space between the electrodes is filled by a moist paste of NH 4 Cl and ZnCl 2 2. Electrode Reactions : Anode: Zn(s) → Zn 2+ + 2e – Cathode: MnO 2 + NH 4 + + e – → MnO (OH) + NH 3 At cathode : Mn is r educed from the + 4 to +3 state. At Anode: Zn gets oxidized Ammonia produced in the reaction forms a complex with Zn 2+ to give [Zn(NH 3 ) 4 ] 2+ . The cell has a potential of nearly 1.5 V.

81. B. Mercury cell: 1. Construction: Anode: zinc – mercury amalgam Cathode: a paste of HgO and C Electrolyte : a paste of KOH and ZnO . Use: suitable for low current devices like hearing aids, watches, etc. 2. Electrode reactions : Anode: Zn(Hg) + 2OH – → ZnO(s) + H 2 O + 2e – Cathode: HgO + H 2 O + 2e – → Hg(l ) + 2OH – Overall reaction is represented by Zn(Hg) + HgO (s) → ZnO (s) + Hg(l ) The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.

82. 2. Cell reactions : when the battery is in use are: Anode: Pb(s) + SO 4 2– ( aq ) → PbSO 4 (s) + 2e – Cathode: PbO 2 (s) + SO 4 2– (aq) + 4H + (aq) + 2e – → PbSO 4 (s) + 2H 2 O (l) overall cell reaction consisting of cathode and anode reactions is: Pb(s)+PbO 2 (s)+2H 2 SO 4 (aq)→ 2PbSO 4 (s) + 2H 2 O(l) On charging the battery the reaction is reversed and PbSO 4 (s) on anode and cathode is converted into Pb and PbO 2 , respectively. A. lead storage battery 1. Construction: anode : lead Cathode : a grid of lead packed with PbO 2 Electrolyte : a 38% solution of H 2 SO 4 . Secondary Batteries:

83. B. N ickel-cadmium cell: which has longer life than the lead storage cell but more expensive to manufacture. The overall reaction during discharge is: Cd (s)+2Ni(OH) 3 (s) → CdO (s) +2Ni(OH) 2 (s) +H 2 O(l )

84. Fuel Cells Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells. H2-O2 fuel cell The cell was used for providing electrical power in the Apollo space programme. The water vapours produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions.

85. The electrode reactions are: Cathode: O 2 (g) + 2H 2 O(l ) + 4e – ⎯→ 4OH – (aq) Anode: 2H 2 (g) + 4OH – (aq) → 4H 2 O(l) + 4 e – Overall reaction being: 2H 2 (g) + O 2 (g) → 2 H 2 O(l ) Advantages of fuel cells: The fuel cell runs continuously as long as the reactants are supplied. Fuel cells produce electricity with an efficiency of about 70 % compared to thermal plants whose efficiency is about 40%. These are pollution free.

86. Corrosion In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron (commonly known as rusting ) occurs in presence of water and air. The chemistry of corrosion is considered as an electrochemical phenomenon.

87. At Anode: At a particular spot of iron, oxidation takes place and that spot behaves as anode and the reaction is : Anode: 2 Fe (s)→2 Fe 2+ + 4 e – E (Fe2+/Fe) = – 0.44 V At Cathode: Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+ ( which is believed to be available from H 2 CO 3 formed due to dissolution of carbon dioxide from air into water) This spot behaves as cathode with the reaction Cathode: O 2 (g) + 4 H + (aq) + 4 e – → 2 H 2 O (l ) E (H+/O2/H2O) = 1.23 V T he overall reaction being: 2Fe(s)+O 2 (g) + 4H + (aq) ⎯→ 2Fe 2+ (aq)+ 2H 2 O (l ) E o cell =1.67 V The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3. x H2O) and with further production of hydrogen ions. Corrosion is an electrochemical cell

88. Prevention of Corrosion One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done by covering the surface with paint or by some chemicals (e.g. bisphenol ). Another simple method is to cover the surface by other metals (Sn, Zn, etc.) that are inert or react to save the object . An electrochemical method is to provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which corrodes itself but saves the object.

89. NCERT EXERCISES 3.6 In the button cells the following reaction takes place : Zn(s) + Ag 2 O(s) + H 2 O(l) → Zn 2+ (aq) + 2Ag(s) + 2OH – (aq) Determine Δ r G o and E o for the reaction. 3.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. 3.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm –1 . Calculate its molar conductivity. 3.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10 –3 S cm –1 . 3.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Conc./M 0.001 0.010 0.020 0.050 0.100 10 2 × κ/ S m –1 1.237 11.85 23.15 55.53 106.74 Calculate Λ m for all concentrations and draw a plot between Λ m and c½. Find the value of Λ m . 3.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10 –5 S cm –1 . Calculate its molar conductivity and if Λ m for acetic acid is 390.5 S cm 2 mol –1 , what is its dissociation constant?

90. 3.12 How much charge is required for the following reductions: ( i ) 1 mol of Al 3+ to Al. (ii) 1 mol of Cu 2+ to Cu. (iii) 1 mol of MnO 4– to Mn 2+ . 3.13 How much electricity in terms of Faraday is required to produce ( i ) 20.0 g of Ca from molten CaCl 2 . (ii) 40.0 g of Al from molten Al 2 O 3 . 3.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H 2 O to O 2 . (ii) 1 mol of FeO to Fe 2 O 3 . 3.15 A solution of Ni(NO 3 ) 2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? 3.16 Three electrolytic cells A,B,C containing solutions of ZnSO 4 , AgNO 3 and CuSO 4 , respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

91. 3.17 Using the standard electrode potentials predict if the reaction between the following is feasible: ( i ) Fe 3+ ( aq ) and I – ( aq ) (ii) Ag + ( aq ) and Cu(s) (iii) Fe 3+ ( aq ) and Br – ( aq ) (iv) Ag(s) and Fe 3+ ( aq ) (v) Br 2 ( aq ) and Fe 2+ ( aq ). 3.18 Predict the products of electrolysis in each of the following: ( i ) An aqueous solution of AgNO 3 with silver electrodes. (ii) An aqueous solution of AgNO 3 with platinum electrodes. (iii) A dilute solution of H 2 SO 4 with platinum electrodes. (iv) An aqueous solution of CuCl 2 with platinum electrodes.