1. Oxidation-Reduction Titration Dr. A.K.M. Shafiqul Islam August 24, 2007

2. Oxidation Reduction Reaction Reactions of metals or any other organic compounds with oxygen to give oxides are labeled as oxidation. The removal of oxygen from metal oxides to give the metals in their elemental forms is labeled as reduction. In other words, oxidation is addition of oxygen and removal of hydrogen whereas reduction is addition of hydrogen and removal of oxygen.

3. Reactions of metals or any other organic compounds with oxygen to give oxides are labeled as oxidation. The removal of oxygen from metal oxides to give the metals in their elemental forms is labeled as reduction. In other words, oxidation is addition of oxygen and removal of hydrogen whereas reduction is addition of hydrogen and removal of oxygen. Oxidation Reduction Reaction

4. Chemical reaction Reduction-Oxidation reaction commonly known as RedOx reaction Reduction-Oxidation reaction commonly known as RedOx reaction Ox 1 + Red 2 ⇌ Red 1 + Ox 2 Ox 1 + Red 2 ⇌ Red 1 + Ox 2 Chemical reaction based oxidation and reduction reaction is known as RedOx reaction Chemical reaction based oxidation and reduction reaction is known as RedOx reaction Fe 2+ + Ce 4+ → Fe 3+ + Ce 3+ (1)

5. There are two kinds of electro chemical cells, galvanic or electrolytic. In galvanic cells, the chemical reaction occurs spontaneously to produce electrical energy. In a electrolytic cell, electrical energy is used to force the non spontaneous chemical reaction. Galvanic cells are of importance in our further discussion as we will be discussing the spontaneous chemical reaction to produce electrical energy. Electrochemical Cells

6. If a solution containing Fe 2+ is mixed with another solution containing Ce 4+, there will be a redox reaction situation due to their tendency of transfer electrons. If we consider that these two solution are kept in separate beaker and connected by salt bridge and a platinum wire that will become a galvanic cell. If we connect a voltmeter between two electrode, the potential difference of two electrode can be directly measured. The Fe 2+ is being oxidised at the platinum wire (the anode): Fe 2+ → Fe 3+ + e - The electron thus produced will flow through the wire to the other beaker where the Ce 4+ is reduced (at the cathode). Ce 4+ + e - → Ce 3+

8. The reaction involve electron transfer Fe 2+ → Fe 3+ + e - (2) Ce 4+ + e - → Ce 3+ (3) Equation (2) & (3) are called half reactions Equation (2) & (3) are called half reactions No half reaction can occur by itself No half reaction can occur by itself There must be an electron donner (reducing agent) and an electron accepter (oxidizing agent) There must be an electron donner (reducing agent) and an electron accepter (oxidizing agent) Fe 2+ is the reducing agent and Ce 3+ is the oxidizing agent Fe 2+ is the reducing agent and Ce 3+ is the oxidizing agent

9. Before we start the discussion of the oxidation reduction titration curve construction, we should understand the Nernst equation which was introduced by German scientist, Wlater Nernst in 1889. This equation express the relation between the potential of metal and metal ion and the concentration of the ion in the solution. Lets consider the following chemical reaction :- aA + bB ⇋ cC + dD The change in free energy is given by the equation ΔG = ΔG o + 2.3RT log [C]c x [D]d / [A]a x [B]b

10. From the relation ship of the free energy and cell potential, we can get ΔG = -nFE In a standard states, free energy will be ΔG o = -nFE o Hence, the above equation can be written as -nFE = -nFE o + 2.3RT log [C]c [D]d / [A]a [B]b

11. After dividing both side with –nF, we can get the expression as E = E o – 2.3 RT/ nF log [C]c [D]d / [A]a [B]b At 25ºC, the equation can be written as :- E = E o – 0.059/n log [C]c [D]d / [A]a [B]b This equation is known as the “Nernst equation” that correlate the electrode potential with the concentration of the ionic species.

12. Redox titration is monitored by observing the change of electrode potential. The titration curve is drawn by taking the value of this potential against the volume of the titrant added. Unlike other titration curve the p values are substituted by electrode potential values in the curve. The redox reaction is rapid and the system is always in equilibrium throughout the titration. The electrode potential of the two half reaction are always identical. If we consider the oxidation of Fe (II) with standard Ce (IV), then we can write the equation as follows :- Fe 2+ + Ce 4+ ⇋ Fe 3+ + Ce 3+ Redox Titration Curve

13. The electrode potential of the two half reaction will be always identical. For iron, the electrode potential will be E = Eº Fe – 0.059 log [Fe 2+ ] / [Fe 3+ ] For cerium the electrode potential will be E = Eº Ce – 0.059 log [Ce 3+ ] / [Ce 4+ ] Therefore, either of the electrode potential could be utilised to calculate the potential of the solution.

14. The utilisation of the either equation is based on the stage of titration. Prior to the equivalence point, the concentration of Fe(II) and Fe(III) are appreciable compare to Ce(IV) ion which is negligible because of the presence of large excess of Fe(II). Beyond the equivalence point, the concentration of Ce(IV) and Ce(III) is readily computed from the addition and the electrode potential for the Ce(IV) could be used. At equivalence point, the concentration of the oxidised and reduced forms of the two species are such that their attraction for electron are identical. At this point, the reactant species concentration and product species concentrations ratios are known and they are utilised to calculate the potential of the solution.

15. # 50.0 ml of 0.05M Fe 2+ is titrated with 0.1M Ce 4+ in a sulphuric acid media at all times. Calculate the potential of the inert electrode in the solution at various intervals in the titration and plot the titration curve. Use 0.68V as the formal potential of the Fe 2+ - Fe 3+ system in sulphuric acid and 1.44V for the Ce 3+ - Ce 4+ system. Initial step : After addition of 5.0 ml of Ce 4+ As because the Ce 4+ is too small, we are considering the iron (Fe) electrode potential to calculate the solution potential. [Fe 3+ ] = 5.0ml X 0.10M / (50.0 + 5.0) ml = 0.5 mmol / 55.0ml Similarly, [Fe 2+ ] = (50.0ml X 0.05M – 5.0 ml X 0.1M) / 55.0 ml = 2.0 mmol / 55.0 ml

16. Substituting the values to the standard electrode potential equation, we can get E = Eº Fe – 0.059 log [Fe 2+ ] / [Fe 3+ ] E = Eº – 0.059 log 2.0 / 0.5 E = 0.68 -0.036 = 0.64 V Step-2 : Equivalence point At equivalence point in the titration of Fe(II) and Ce(IV), the potential of the solution is controlled by both the half reaction. Eeq=E 0 Ce -0.059 log [Ce 3+ ] / [Ce 4+ ] Eeq = E 0 Fe - 0.059 log [Fe 2+ ] / [Fe 3+ ]

17. Adding the two expression, we can get 2Eeq = E 0 Ce + E 0 Fe - 0.059 log [Ce 3+ ] [Fe 2+ ]/ [Ce 4+ ][Fe 3+ ] At equivalence the concentration of Fe 3+ = Ce 3+ and Fe 2+ = Ce 4+ We can get, 2Eeq = E 0 Ce + E 0 Fe - 0.059 log [Ce 3+ ] [Ce 4+ ]/ [Ce 4+ ] [Ce 3+ ] Eeq = (E 0 Ce + E 0 Fe ) / 2 Eeq = (1.44 + 0.68) / 2 = 1.06 V

18. Step-3 : After addition of 25.1 ml Ce 4+ At this stage, the concentration of Fe(II) is very small and we can neglect the value and for convenience, we will utilise the Ce(IV) electrode potential to calculate the solution potential. [Ce 3+ ] = (25.0ml X 0.1M ) / (50.0 + 25.1) ml = 2.5 mmol / 75.1 ml [Ce 4+ ] = (0.1 ml X 0.1M) / 75.1 ml E = Eº Ce – 0.059 log [Ce 3+ ] / [Ce 4+ ]  E = 1.44 – 0.059 log 2.5 / 0.01 = 1.30 V

19. Redox Titrations Redox Titrations Equipment for obtaining a titration curve for a redox titration. Equipment for obtaining a titration curve for a redox titration.

20. RedOx Titration Curve

21. Redox Titrations The net balanced redox equation is the sum of the two half-cell reactions. It may be necessary to multiply one or both half-cells by some coefficient so that the same number of electrons are lost by the substance that is oxidized as are gained by the substance reduced.

22. Cu 0 (S) Cu 2+ (aq) + 2e - 2Ag + (aq) + 2e - 2Ag (S) Cu 0 (S) + 2Ag + (aq) + 2e- Cu 2+ (aq) + 2Ag (S) + 2e - Cu 0 (S) + 2Ag + (aq) Cu 2+ (aq) + 2Ag (S) Number of e - s involved in the overall reaction is 2 Balancing simple redox reactions

23. Fe +2 (aq) + MnO 4 - (aq) Mn +2 (aq) + Fe +3 (aq) Fe +2 (aq) Fe +3 (aq) + 1e - MnO 4 - (aq) Mn +2 (aq) Oxidizing half: Reducing half: Balancing atoms: MnO 4 - (aq)+ Mn +2 (aq) + 4H 2 O Balancing oxygens: Balancing complex redox reactions

24. MnO 4 - (aq)+8H + Mn +2 (aq) + 4H 2 O Balancing hydrogens: Oxidation numbers: Mn = +7, O = -2 Mn = +2 Balancing electrons: The left side of the equation has 5 less electrons than the right side MnO 4 - (aq)+8H + + 5e - Mn +2 (aq) + 4H 2 O Reducing Half Reaction happening in an acidic medium

25. Balancing complex redox reactions Final Balancing act: Making the number of electrons equal in both half reactions [Fe +2 (aq) Fe +3 (aq) + 1e - ]× 5 [MnO 4 - (aq)+8H + + 5e - Mn +2 (aq) + 4H 2 O]×1 5Fe +2 (aq) 5Fe +3 (aq) + 5e - MnO 4 - (aq)+8H + + 5e - Mn +2 (aq) + 4H 2 O 5Fe 2+ +MnO 4 - (aq)+8H + + 5e - 5Fe 3+ +Mn +2 (aq) + 4H 2 O + 5e -

26. Redox Titrations Many of the end point indicators used in redox titrations are substances that are either oxidized or reduced at given potentials. Many of the end point indicators used in redox titrations are substances that are either oxidized or reduced at given potentials. Much like acid-base indicators, the electrical potential at the end point oxidizes or reduces the indicator to one of the colored forms to signal the end point. Much like acid-base indicators, the electrical potential at the end point oxidizes or reduces the indicator to one of the colored forms to signal the end point.

27. RedOx Indicators If the titrant is highly colored, this color may be used to detect end point. If the titrant is highly colored, this color may be used to detect end point. 0.02 M potassium permanganate is deep purple. A dilute solution is pink. The product of its reduced form (Mn 2+ ) colorless. 0.02 M potassium permanganate is deep purple. A dilute solution is pink. The product of its reduced form (Mn 2+ ) colorless.

28. The unknown sample of iron contains, iron in Fe 2+ oxidation state. So we are basically doing a redox titration of Fe 2+ Vs KMnO 4 5Fe 2+ +MnO 4 - (aq)+8H + 5Fe 3+ +Mn +2 (aq) + 4H 2 O Titration of unknown sample of Iron Vs KMnO 4

29. 250mL V initial V final End point: Pale Permanent Pink color KMnO 4 V final - V inital = V used (in mL) Important requirement: The concentration of KMnO 4 should be known precisely.

30. Redox Indicators

31. Titrations Involving Iodine One of the most common redox titrations involve either using iodine (I 2 ) as a mild oxidizing agent or iodide (I - ) as a mild reducing agent. Iodine as oxidizing agent: I 2 + 2 e - = 2 I - When iodine is used as the titrant the method is known as iodimetry.

32. Titrations Involving Iodine I 2 is not very soluble in water (only about 1.3 x 10 -3 mol/L). It solubility is increased in the presence of excess iodide by the formation of the triiodide (I 3 - ) species, I 2 + I - = I 3 – I 2 + I - = I 3 – So it is really the triiodide species, though it will commonly be referred to as iodine that is involved in the chemical reactions.

33. Due to the difficulty is maintaining the concentration of I 2, (limited solubility in water, appreciable vapor pressure of the I 2.) iodometric methods are more commonly used. The amount of I 2 produced by the action of the oxidizing analyte on excess iodide is usually titrated with standardized sodium thiosulfate, Na 2 S 2 O 3. Due to the difficulty is maintaining the concentration of I 2, (limited solubility in water, appreciable vapor pressure of the I 2.) iodometric methods are more commonly used. The amount of I 2 produced by the action of the oxidizing analyte on excess iodide is usually titrated with standardized sodium thiosulfate, Na 2 S 2 O 3. Titrations Involving Iodine

34. Iodide as reducing agent: 2 I - = I 2 + 2 e - When iodine is produced by the addition of an oxidizing analyte to an excess of iodide, the method is as iodometry.

35. Titrations Involving Iodine Thiosulfate (S 2 O 3 -2 ) is commonly used in titration reactions involving iodine, both for iodimetric and iodometric methods. The iodimetric methods generally involve an excess of standard I 2 (as I 3 - ) followed by back titration with standard thiosulfate.

36. Titrations Involving Iodine In the iodometric methods, an excess of iodide is added to the sample of an oxidizing analyte and a stoichiometric amount of iodine (I 2 or I 3 - ) is produced. This iodine is titrated with a standard solution of thiosulfate. This reaction is shown on the following slide.

37. Titrations Involving Iodine

38. The reaction with iodine needs to occur in a solution whose pH < 9 to prevent side reactions which produce iodates (IO 3 - ). Generally acetic acid is added to the analyte mixture before titration to assure the proper pH. In some cases, appropriate pH buffers may also be added.

39. Titrations Involving Iodine The titrant solution of thiosulfate cannot be prepared directly. It is made to an approximate concentration and then standardized with a primary standard oxidizing agent by iodometry. The thiosulfate solution is unstable if the pH < 5, undergoing the following disproportionation reaction. S 2 O 3 -2 + 2 H + H 2 SO 3 + S

40. Titrations Involving Iodine This disproportionation reaction is prevented by using freshly boiled deionized water as the solvent and adding a small amount of NaOH. Although the thiosulfate needs to be stored in a basic solution, as mentioned earlier, its reaction as a reductant titrant needs to occur in an acid solution. This disproportionation reaction is prevented by using freshly boiled deionized water as the solvent and adding a small amount of NaOH. Although the thiosulfate needs to be stored in a basic solution, as mentioned earlier, its reaction as a reductant titrant needs to occur in an acid solution.

41. Chapter 16 – Iodine Methods The indicator for both iodimetric and iodometric titrations is a starch solution; in the presence of iodine, it shows a blue or purple color. The starch indicator should not be added when I 2 is in large excess because the desorption of I 2 from the starch molecule is not reversible. The indicator for both iodimetric and iodometric titrations is a starch solution; in the presence of iodine, it shows a blue or purple color. The starch indicator should not be added when I 2 is in large excess because the desorption of I 2 from the starch molecule is not reversible.

42. Starch Indicator The repeating amylose unit in the starch molecule. Starch is a polymer of amylose. The repeating amylose unit in the starch molecule. Starch is a polymer of amylose.

43. Starch Indicator The starch-iodine complex where the sugar chain forms a helix about I 6 units. The starch-iodine complex where the sugar chain forms a helix about I 6 units.

44. Iodimetric titrations: Titrations with standard iodine (actually I 3  ) Species analyzedOxidation reaction SO 2 SO 2 + H 2 O H 2 SO 3 H 2 SO 3 + H 2 O SO 2 4- + 4H + + 2e - H 2 SH 2 S S(s) + 2H + + 2e- Zn 2+, Cd 2+, M 2+ + H 2 S  MS(s) + 2H + Hg 2+, Pb 2+ MS(s) M 2+ + S + 2e -

45. Iodimetric titrations: Titrations with standard iodine (actually I 3  ) Species analyzedOxidation reaction Cysteine, glutathione,2RSH RSSR + 2H + + 2e - mercaptoethanol AldehydesH 2 CO + 3OH - HCO 2 - + 2H 2 O + 2e - Glucose (and other reducing sugar) O RCH + 3OH - HCO 2 - + 2H 2 O + 2e - Ascorbic acid See next slide (or vitamin C)

46. Ascorbic acid (or vitamin C)

47. Iodometric titrations: Titrations of iodine (actually I 3  ) produced by the analyte Species analyzedReaction HOClHOCl + H + + 3I - Cl - + I 3 - + H 2 O Br 2 Br 2 + 3 I - 2 Br - + I 3 - IO 3 - 2 IO 3 - + 16 I - + 12 H + 6 I 3 - + 6 H 2 O IO 4 - 2 IO 4 - + 22 I - + 16 H + 8 I 3 - + 8 H 2 O O 2 O 2 + 4 Mn(OH) 2 + 2 H 2 O 4 Mn(OH) 3 2 Mn(OH) 3 + 6 H + + 6 I - 2 Mn 2+ + 2 I 3 - + 6 H 2 O 2 Mn(OH) 3 + 6 H + + 6 I - 2 Mn 2+ + 2 I 3 - + 6 H 2 O H 2 O 2 H 2 O 2 + 3 I - + 2H + I 3 - + 2 H 2 O

48. Iodometric titrations: Titrations of iodine (actually I 3  ) produced by the analyte Species analyzedReaction O 3 O 3 + 3 I - + 2 H + O 2 + I 3 - + H 2 O NO 2 - 2 HNO 2 + 2 H + + 3 I - 2 NO + I 3 - + 2 H 2 O S 2 O 8 2- S 2 O 8 2- + 3 I - 2 SO 4 2- + I 3 - Cu 2+ 2 Cu 2+ + 5 I - 2 CuI(s) + I 3 - MnO 4 - 2 MnO 4 - + 16 H + + 15 I - 2 Mn 2+ + 5 I 3 - + 8 H 2 O MnO 2 MnO 2 (s) + 4 H + + 3 I - Mn 2+ + I 3 - + 2 H 2 O