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4.1 Redox Titrations By Dr. P. B. Thakur

Name: Dr. Pramod B. Thakur Class: T.Y.B.Sc Subject: Analytical Chemistry Title of Topic: Redox Titrations. T.Y.B.Sc Analytical Chemistry Paper IV- USCH504, UNIT- IV. 4.1 Redox Titrations By Dr. P. B. Thakur. Redox Titrations. General Introduction Theory of Redox indicator

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4.1 Redox Titrations By Dr. P. B. Thakur

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  1. Name: Dr. Pramod B. ThakurClass: T.Y.B.ScSubject: Analytical Chemistry Title of Topic: Redox Titrations

  2. T.Y.B.Sc Analytical Chemistry Paper IV- USCH504, UNIT- IV 4.1 Redox Titrations By Dr. P. B. Thakur

  3. Redox Titrations • General Introduction • Theory of Redox indicator • Criteria for choosing an indicator for redox titration • Use of diphenyl amine and ferroin as redox indicators • Construction of Titration curve in the case of • Fe(II) Vs Ce(IV) • Fe(II) Vs dichromate ( Cr2O7-2 )

  4. General Introduction: • Redox = Reduction + Oxidation • Reduction = Addition of electron Mz+ + n e- M(z-n)+ example: Ce4+ + 1 e- Ce3+ • Oxidation = Removal of electron Mz+ - n e- M(z+n)+ example: Fe2+ - 1 e- Fe3+ • Redox titration involve the titration reaction where transfer of electron between the reactant (titrand) and titrant takes place.

  5. Criteria for reaction in redox titration • Reaction should be rapid • Reaction should go to the completion • Reaction should be stoichiometric • Reaction should provide easy detection of end point

  6. General reaction Oxi1 + Red2 Red1 + Oxi2 Oxi1 & Oxi2 = Oxidizing agent Red1 & Red2 = Reducing agent example: Fe+2 + Ce+4 Fe+3 + Ce+3 Oxi1 + Red2 Red1 + Oxi2 Number of electrons donated and number of electrons accepted should be equal Since two Redox couple are present, the electrode potential of each redox couple must be identical which will be the electrode potential of the system Esystem = E Fe+2 / Fe+3 = E Ce+3 / Ce+4 Esystem is the potential which is referred as the electrode potential of the system

  7. Redox Indicator: • Electrode potential is the parameter which changes during the course of titration and magnitude of the change is maximum in the vicinity of the equivalence point. • So indicator used for redox titration must change color due to sudden change in the electrode potential of the system at the equivalence point. • Such indicator are known as Redox Indicator. Reducing agent Oxidizing agent

  8. Theory of Redox Indicator: • Certain organic compounds are capable to undergo oxidation-reduction reactions and the color of the oxidized form of indicator in solution is significantly different than that of reduced form

  9. If In = any redox indicator, then In (ox) + ne- In (Red) By using Nearst equation, the electrode potential of the indicator can be given as, EIn = EoIn - 0.05916 X log [ In(Red) ]………eq 1 n [ In(ox) ] EoIn is the standard electrode potential of indicator which is constant for given redox indicator. n = number of electrons So, EIn will depend on the ratio of [ In(Red) ] [ In(ox) ]

  10. So, Colour of the indicator in the solution will depend on the ratio of [ In(Red) ] [ In(ox) ] as colour is depend on potential EIn Because when EIn changes, the ratio of the reduced form of indicator to the oxidized form of indicator will change.

  11. The colour of reduced form will dominant when [ In(Red) ] ≥ 10 or [ In(ox) ] ≤ 10 [ In(ox) ] [ In(Red) ] The colour of Oxidized form will dominant when [ In(ox) ] ≥ 10 or [ In(Red) ] ≤ 10 [ In(Red) ] [ In(Ox) ] By substituting these in eq 1, we get EIn = EoIn ± 0.05916 n

  12. Criteria for choosing an indicator for redox titration EIn = EoIn ± 0.05916 n From the above equation, it is clear that the potential at which a colour transition will occur depend on the standard potential of the indicator system. Hence the transition potential of the redox indicator must be equal to or nearly equal to the equivalence point potential of the system to give accurate end point at equivalence point.

  13. Oxidation-Reduction Indicators

  14. Use of organic molecule as redox indicators • The redox indicators are organic molecule that undergo structural changes upon being oxidized or reduced. • We will consider the two examples 1. Use of diphenyl amine as redox indicators 2. Use of ferroin as redox indicators

  15. Use of diphenyl amine as redox indicators • Diphenyl is the first indicator used in the redox titration by Knop in 1924 for the titration of Fe(II) with K2Cr2O7. • In the presence of a strong oxidizing agent, diphenyl amine undergoes a series of reaction as follows:

  16. The main indicator reaction is the transfer of Diphenyl benzidine to Diphenyl benzidine violet (Purple). • The transition potential is 0.76 v at 250C for diphenyl amine indicator. • In titration of Fe(II) & K2Cr2O7 the equivalence poin potential is 1.17 V. • Thetransition potential of indicator andequivalence point potential should be equal or nearly equal to give colour change at the equivalence point. • The transition potential of diphenyl amine indicator is less than that of equivalence point potential, so colour change will occur before the equivalence point which will introduce error in measurement.

  17. To avoid such error, mixture of sulphuric acid and phosphoric acid is used. • Sulphuric acid form the diphenyl benzidine sulphonate salt which increases the solubility of the indicator in aqueous medium due to which transition potential of the indictor increases to 0.85 v from previous 0.76 v • Phosphoric acid forms stable colourless ferric phosphate complex which considerably decrease the equivalence potential value. • So, the equivalence point potential is now closer to the transition potential of indicator due to which colour change will takes place close to the equivalence point.

  18. Use of Ferroin as redox indicators • Ferroin is organic compound known as 1,10 phenonthrolines or orthophenanthrolines. • It is capable of forming intensely coloued, highly stable complexwith Fe(II).

  19. The complex is called Ferroin and represented as (Phen)3.Fe+2 • It can be shown as below • The indicator reaction involves, (Phen)3.Fe+3 + e- = (Phen)3.Fe+2 (Pale Blue)(Red) EoRed = 1.06 v

  20. Transition potential of ferroin indicator is 1.06 v • In the titration of Fe(II) Vs Ce(IV), the equivalence point potential is 1.105 v • As the transition potential of ferroin indicator is very close to equivalence point potential of Fe(II) and Ce(IV) system, the ferroin is the most suitable indicator for this sysyem which gives colour change at exactly equivalence point

  21. Construction of Titration curve • For Redox titration, titration curve is the plot of electrode potential of the system Versus volume of the titrant (reduing or oxidizing agent). • Electrode potential of the system can be obtained by setting up electrochemical cell or by calculating theoretically using nearst equation.

  22. Redox titrations reaction • Redox titrations reactions in general classified in two main group • Titraton where H3O+ or OH- do not participate directly in the redox reaction. eg. Fe(II) Vs Ce(IV) ii) Titraton where H3O+ or OH- participate directly in the redox reaction. eg. Fe(II) Vs KMnO4 Titration curve for each of this system will be discussed further

  23. Titration curve of Fe(II) Vs Ce(IV) • In this redox titration, acid solution of Fe(II) is titrated with standard solution of Ce(IV). • Consider the titration of 10 cm3 of 0.1 M Fe+2 with 0.1 M Ce+4 solution in the presence of H2SO4. • The titration reaction correspond to this can be given as Fe+2 + Ce+4 Fe+3 + Ce+3 • The electrode potential of the system at each stage of the titration can be calculated by using Nearst equation as below,

  24. At the beginning of the titration: • At the beginning of the titration, No Ce+4 is added to Fe+2 solution, so solution will contain only Fe+2 ions. • Electrode potential is determined from Fe+2/ Fe+3 redox couple • At the beginning no Fe+3 will be present so at this stage the electrode potential is not significant. Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  25. Before the equivalence point: 1. When 2 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 2 cm3 0.1 M Ce+4 has been added to 10 cm3, 0.1 M Fe+2 solution, then 2 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be [Fe+3] = 2 X 0.1 12 and 8 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 8 x 0.1 12 [Fe+2] = 8 = 4 [Fe+3] 2 Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  26. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 4 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [4] = 0.771- 0.05916 X 0.6021 Esystem = 0.735 V

  27. 2. When 5 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 5 cm3 0.1 M Ce+4 has been added to 10 cm3, 0.1 M Fe+2 solution, then 5 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be [Fe+3] = 5 X 0.1 15 and 5 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 5 x 0.1 15 [Fe+2] = 1 [Fe+3] Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  28. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 1 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [1] = 0.771- 0.05916 X 0 Esystem = 0.771 V

  29. 3. When 9.9 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 9.9 cm3 0.1 M Ce+4 has been added to 10 cm3, 0.1 M Fe+2 solution, then 9.9 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be [Fe+3] = 9.9 X 0.1 19.9 and 0.1 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 0.1 x 0.1 19.9 [Fe+2] = 0.1 = 0.0101 [Fe+3] 9.9 Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  30. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 0.0101 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [0.0101] = 0.771- 0.05916 X [-1.9956] Esystem = 0.889 V

  31. 3. At equivalence point When 10 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 10 cm3 0.1 M Ce+4 has been added to 10 cm3, 0.1 M Fe+2 solution, then all Fe+2 will oxidized in to Fe+3. Similarly all Ce+4 will reduced in to Ce+3. So, at equivalence point, two redox couple are present are as follows, Fe+2 -Fe+3 and Ce+4 -Ce+3 Hence, the potential of the system will be the sum of the potential of this two redox couple Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  32. By using Nearst equation, electrode potential of the two redox couple system can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] + EoPt/Ce+3- Ce+4- 0.05916 X log [Ce+3] n [Ce+4] = 2 equivalence point potential 2 E eq. Point potential = EoPt/Fe+2- Fe+3+ EoPt/Ce+3- Ce+4 – 0.05916 X log [Fe+2] [Ce+3] [Fe+3] [Ce+4] At equivalence point, [Fe+3] = [Ce+3] and [Fe+2] = [Ce+4] [Fe+2] [Ce+3] = 1 [Fe+3] [Ce+4]

  33. EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 EoPt/Ce+3- Ce+4= Standard electrode potential = 1.44 Substituting these values in above equation, we get 2 E eq. Point potential = 0.771 + 1.44 – 0.05916 X log [1] = 0.771 + 1.44 E eq. Point potential = 2.211 = 1.105 v 2 E eq. Point potential = 1.105 v

  34. 4. After the equivalence point When 10.1 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 10.1 cm3 0.1 M Ce+4 has been added then 0.1 ml Ce+4 will be in excess. At this stage, only Ce+4 -Ce+3 redox couple is responsible for electrode potential and it calculated as Esystemt =EoPt/Ce+3- Ce+4- 0.05916 X log [Ce+3] n [Ce+4] Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  35. EoPt/Ce+3- Ce+4 =1.44 v [Ce+4] = 0.1 X 0.1 20.1 and [Ce+3] = 10 X 0.1 20.1 [Ce+3] = 10 = 100 [Ce+4] 0.1 Esystem =EoPt/Ce+3- Ce+4- 0.05916 X log [Ce+3] n [Ce+4] = 1.44-0.05916 X log [100] Esystem = 1.322 v

  36. 4. After the equivalence point When 11 cm3 0.1 M Ce+4 is added Fe+2 + Ce+4 Fe+3 + Ce+3 When 11 cm3 0.1 M Ce+4 has been added then 1 ml Ce+4 will be in excess. At this stage, only Ce+4 -Ce+3 redox couple is responsible for electrode potential and it calculated as Esystemt =EoPt/Ce+3- Ce+4- 0.05916 X log [Ce+3] n [Ce+4] Ce+4 Titrant 10 cm3 of 0.1 M Fe+2

  37. EoPt/Ce+3- Ce+4 =1.44 v [Ce+4] = 1 X 0.1 21 and [Ce+3] = 10 X 0.1 21 [Ce+3] = 10 = 10 [Ce+4] 1 Esystem =EoPt/Ce+3- Ce+4- 0.05916 X log [Ce+3] n [Ce+4] = 1.44-0.05916 X log [10] Esystem = 1.381 v

  38. Titration curve of Fe(II) Vs Ce(IV) titration Esystem Volume of 0.1 M Ce(IV) added

  39. Titration curve of Fe(II) Vs Cr2O7-2 • In this redox titration, acid solution of Fe(II) is titrated with standard solution of K2Cr2O7-2. • Consider the titration of 10 cm3 of 0.1 M Fe+2 against with 0.0167 M K2Cr2O7-2 at pH 1. • The titration reaction correspond to this can be given as 6Fe+2+Cr2O7-2 +14H3O+ 6Fe+3+2Cr+3 + 21H2O • The electrode potential of the system at each stage of the titration can be calculated by using Nearst equation as below,

  40. At the beginning of the titration: • At the beginning of the titration, No Cr2O7-2 is added to Fe+2 solution, so solution will contain only Fe+2 ions. • Electrode potential is determined from Fe+2/ Fe+3 redox couple • At the beginning no Fe+3 will be present so at this stage the electrode potential is not significant. 10 cm3 of 0.1 M Fe+2

  41. Before the equivalence point: 1. When 2 cm3 0.0167 M Cr2O7-2 is added 6Fe+2+Cr2O7-2 +14H3O+ 6Fe+3 + 2Cr+3 + 21H2O When 2 cm3 0.0167 M Cr2O7-2 has been added to 10 cm3, 0.1 M Fe+2 solution, then 2 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be, [Fe+3] = 2 X 0.1 12 and 8 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 8 x 0.1 [Fe+2] = 4 12 [Fe+3] Cr2O7-2 Titrant 10 cm3 of 0.1 M Fe+2

  42. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 4 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [4] = 0.771- 0.05916 X 0.6021 Esystem = 0.735 V

  43. 2. When 5 cm3 0.0167 M Cr2O7-2 is added 6Fe+2 + Cr2O7-2 +14H3O+ 6Fe+3 + 2Cr+3 + 21H2O When 5 cm3 0.0167 M Cr2O7-2 has been added to 10 cm3, 0.1 M Fe+2 solution, then 5 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be [Fe+3] = 5 X 0.1 15 and 5 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 5 x 0.1 15 [Fe+2] = 1 [Fe+3] Cr2O7-2 Titrant 10 cm3 of 0.1 M Fe+2

  44. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 1 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [1] = 0.771- 0.05916 X 0 Esystem = 0.771 V

  45. 3. When 9.9 cm3 0.0167 M Cr2O7-2 is added 6Fe+2 + Cr2O7-2 +14H3O+ 6Fe+3 + 2Cr+3 + 21H2O When 9.9 cm3 0.0167 M Cr2O7-2 has been added to 10 cm3, 0.1 M Fe+2 solution, then 9.9 cm3 of Fe+2 will oxidized in to Fe+3 So, resulting concentration of Fe+3 will be, [Fe+3] = 9.9 X 0.1 19.9 and 0.1 cm3 of Fe+2 will remain unreacted, So, concentration of Fe+2 will be [ Fe+2] = 0.1 x 0.1 19.9 [Fe+2] = 0.1 = 0.0101 [Fe+3] 9.9 Cr2O7-2 Titrant 10 cm3 of 0.1 M Fe+2

  46. By using Nearst equation, electrode potential can be given as Esystem = EoPt/Fe+2- Fe+3- 0.05916 X log [Fe+2] n [Fe+3] EoPt/Fe+2- Fe+3 = Standard electrode potential = 0.771 V n = no of electrons involved = 01 [Fe+2] = 0.0101 [Fe+3] So we get, Esystem = 0.771- 0.05916 X log [0.0101] = 0.771- 0.05916 X [-1.9956] Esystem = 0.889 V

  47. 3. At equivalence point When 10 cm3 0.0167 M Cr2O7-2 is added 6Fe+2 + Cr2O7-2 +14H3O+ 6Fe+3 + 2Cr+3 + 21H2O When 10 cm3 0.0167 M Cr2O7-2 has been added to 10 cm3, 0.1 M Fe+2 solution, then all Fe+2 will oxidized in to Fe+3. Similarly all Cr2O7-2 will reduced in to Cr+3. So, at equivalence point, two redox couple are present are as follows, Fe+2 -Fe+3 and Cr+3 -Cr2O7-2 Hence, the potential of the system will be the sum of the potential of this two redox couple Cr2O7-2 Titrant 10 cm3 of 0.1 M Fe+2

  48. The respective Half reactions are Fe+3 + e- Fe+2 and Cr2O7-2 + 14 H3O+ +6e- 2Cr+3 + 21 H2O Potential of Fe+2- Fe+3 redox couple is given as, Eeq = EoFe+2- Fe+3- 0.05916 X log [Fe+2].....eq 1 1 [Fe+3] Potential of Cr+3- Cr2O7-2 redox couple is given as, Eeq = EoCr+3- Cr2O7-2- 0.05916 X log [Cr+3]2 6 [Cr2O7-2][H3O+]14

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