1. Oxidation-Reduction Titrations PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 1

2. Introduction Oxidation-Reduction (Redox): Definitions & Terms. Oxidation Number Principles. Balancing Redox Equation by Half-Reaction Method. Electrochemical Cells and Electrode Potential. Oxidation Potential: Definition and Factors Affecting. Redox Titration Curves. Detection of End point in Redox Titrations. Standard Oxidizing Reagents and their Properties. Applications of Redox Titrations. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 2

3. Oxidation: It can be defined as loss of electrons or increase in oxygen content. Reduction: It can be defined as gain of electrons or increase of hydrogen content. Oxidizing agent: substance which get reduced. Reducing agent: substance which get oxidized. Both processes are combined and occur together so we combine them in one word as REDOX reaction. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 3

4. Oxidation-Reduction (Redox) Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ Fe 2+ —e Fe 3+ (Loss of electrons: Oxidation) Ce 4+ + e Ce 3+ (Gain of electrons: Reduction) In every redox reaction, both reduction and oxidation must occur. Substance that gives electrons is the reducing agent or reductant. Substance that accepts electrons is the oxidizing agent or oxidant. Reaction of ferrous ion with ceric ion Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half equation. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 4

5. Oxidation Number (O.N)  The O.N of a monatomic ion = its electrical charge.  The O.N of atoms in free un-combined elements = zero  The O.N of an element in a compound may be calculated by assigning the O.N to the remaining elements of the compound using the aforementioned basis and the following additional rules:  The O.N. for oxygen = –2 (in peroxides = –1).  The O.N. for hydrogen = +1 (in hydrides = —1).  The algebraic sum of the positive and negative O.N. of the atoms represented by the formula for the substance = zero.  The algebraic sum of the positive and negative O.N. of the atoms in a polyatomic ion = the charge of the ion. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 5

6. Oxidation Numbers of Some Substances SubstanceOxidation Numbers NaCl H 2 NH 3 H 2 O 2 LiH K 2 CrO 4 SO 4 2- KClO 3 Na = +1, Cl = —1 H = 0 N = —3, H = +1 H = +1, O = —1 Li = +1, H = —1 K = +1, Cr = +6, O = —2 O = —2, S = +6 K = +1, Cl = +5, O = —2 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 6

7. For manganese SpeciesMnMn 2+ Mn 3+ MnO 2 MnO 4 2  MnO 4  O.N.0+2+3+4+6+7 For nitrogen SpeciesNH 3 N2H4N2H4 NH 2 OHN2N2 N2ON2ONOHNO 2 HNO 3 O.N.–3–2–10+1+2+3+5 Oxidation states of manganese and nitrogen in different species PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 7

8. Balancing Redox Reactions using Half- Reaction Method Divide the equation into an oxidation half-reaction and a reduction half-reaction Balance these – Balance the elements other than H and O – Balance the O by adding H 2 O – Balance the H by adding H + – Balance the charge by adding e - Multiply each half-reaction by an integer such that the number of e - lost in one equals the number gained in the other Combine the half-reactions and cancel PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 8

9.  Balance each half reaction: MnO 4  + 5é Mn 2+ C 2 O 4 2  2 CO 2 + 2é 2 MnO 4  + 5 C 2 O 4 2  2 Mn 2+ + 10 CO 2 + 8 H 2 O  Balance oxygen atoms by adding water 2 MnO 4  + 5 C 2 O 4 2  + 16 H + 2 Mn 2+ + 10 CO 2 + 8 H 2 O  Balance hydrogen atoms by adding H +  Use the number of moles so as to make the electrons gained in one reaction equal those lost in the other one 2 MnO 4  + 5 C 2 O 4 2  2 Mn 2+ + 10 CO 2 MnO 4  + C 2 O 4 2  + H + Mn 2+ + CO 2 + H 2 O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 9

10. Electrochemical Cells Solution Pressure. The tendency of the metal to dissolve in a solution of its salt. Ionic Pressure. The tendency of the metal cations to deposit on its metal dipped into its solution.  Cu/Cu 2+ system: ionic pressure > solution pressure. Cu 2+ leaves the solution to deposit on Cu rod  Zn/Zn 2+ system: solution pressure > ionic pressure. Zn metal tends to dissolve forming Zn 2+ in solution. The potential difference between the metal rod (electrode) and the solution is known as electrode potential (E) Electrochemical cells consist of electrodes immersed in electrolyte solution and frequently connected by a salt bridge PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 10

11. anode//cathode Cu/CuSO 4 // ZnSO 4 /Zn PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 11

12. Nernest Equation for Electrode Potential (E) E t = electrode potential at temperature t. E  = standard electrode potential (constant depend on the system) R = gas constant T = absolute Temp. (t°C + 273) F = Faraday (96500 Coulombs) log e = ln (natural logarithm = 2.303 log) n= valency of the ion [M n+ ] = molar concentration of metal ions in solution E t = E o + log [M n+ ] RT nF E 25 °C = E o + log [M n+ ] 0.0591 n PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 12

13. E o is the electromotive force (emf) produced when a half cell (consisting of the elements immersed in a molar solution of its ions) is coupled with a standard hydrogen electrode (E  = zero). E 25 °C = E o + log [M n+ ] 0.0591 n Standard Electrode Potential (E) Standard Electrode Potential (E o ) SystemE° (volts)SystemE° (volts) Li / Li + –3.03Cd/Cd 2+ –0.40 K / K + –2.92Sn / Sn 2+ –0.13 Mg/Mg 2+ –2.37H 2 (pt) / H + 0.00 Al / Al 3+ –1.33Cu / Cu 2+ +0.34 Zn / Zn 2+ –0.76Hg / Hg 2+ +0.79 Fe / Fe 2+ –0.44Ag / Ag + +0.80 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 13

14. Measurement of the Electrode Potential By connecting to another electrode (galvanic cell), an electric current will then flow from the electrode having —ve potential to that having +ve potential (from Zn electrode to Cu electrode) The emf of the current can then be measured. The normal hydrogen electrode is used as a reference electrode. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 14

15. Normal Hydrogen Electrode (NHE) Consists of a piece of platinum foil coated with platinum black and immersed in a solution of 1 N HCl (with respect to H + ). H 2 gas (at 1 atm. Pressure) is passed. Platinum black layer absorbs a large amount of H 2 and can be considered as a bar of hydrogen, it also catalyses the half reaction: 2H + + 2e  H 2 Under these conditions: H 2 electrode potential = zero PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 15

16. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 16 Oxidation Potential When a platinum wire is immersed in a solution of redox couple like ferric/ferrous, an electron flow will occur on the surface of the wire leading to a potential difference between the wire and the solution of the redox couple which is called oxidation potential.

17. Standard Oxidation Potential (E) Standard Oxidation Potential (E o ) E 25 °C = E o + log [Oxidized] / [Reduced] 0.0591 n It is the e.m.f. produced when a half cell consisting of an inert electrode (as platinum), dipped in a solution of equal concentration of both the oxidized and reduced forms (such as Fe 3+ / Fe 2+ ), is connected with a NHE PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 17

18. +0.339 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 18

19. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 19

20. Oxidation Potentials: Electrochemical Series PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 20

21.  The potential of MnO 4  /Mn 2+ varies with the ratio [MnO 4  ]/[Mn 2+ ].  If ferrous is titrated with MnO 4  in presence of Cl , chloride will interfere by reaction with MnO 4  and gives higher results. Factors Affecting Oxidation Potential 1. Common Ion E 25 °C = E o + log [Oxid] /[Red] 0.0591 n Zimmermann’s Reagent (MnSO 4, H 3 PO 4 and H 2 SO 4 )  MnSO 4 has a common ion (Mn 2+ ) with the reductant that lowers the potential of MnO 4  /Mn 2+ system:  Phosphoric acid lowers the potential of Fe 3+ /Fe 2+ system by complexation with Fe 3+ as [Fe(PO 4 ) 2 ] 3 .  Sulphuric acid is used for acidification. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 21

22. 2. Effect of pH The oxidation potential of an oxidizing agent containing oxygen increases by increasing acidity and vice versa. E = E o + log 0.0591 5 MnO 4  /Mn 2+ [MnO 4  ][H + ] [Mn 2+ ] Potassium permanganate: MnO 4  + 8H + + 5e   Mn 2+ + 4H 2 O E = E o + log 0.0591 5 MnO 4  /Mn 2+ [MnO 4  ][H + ] 8 [Mn 2+ ] Potassium dichromate: Cr 2 O 7 2  + 14H + +6e   2Cr 3+ + 7H 2 O E = E o + log 0.0591 6 Cr 2 O 7 2  /Cr 3+ [Cr 2 O 7 2  ][H + ] 14 [Cr 3+ ] PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 22

23. E (I 2 /2I  ) system increases by the addition of HgCl 2 since it complexes with iodide ions. Hg 2+ + 4I   [HgI 4 ] 2  (low dissociation complex) 3. Effect of Complexing Agents E (Fe 3+ /Fe 2+ ) is reduced by the addition of F  or PO 4 3  due to the formation of the stable complexes [FeF 6 ] 3  and [Fe(PO 4 ) 2 ] 3  respectively. Thus, ferric ions, in presence of F  or PO 4 3  cannot oxidize iodide although E o (Fe 3+ /Fe 2+ ) = 0.77 while E o (I 2 /2I  ) = 0.54. Iodine: I 2 + 2e   2I  E = E o + log 0.0591 2 I 2 /I  [I 2 ] [I  ] 2 Ferric: Fe 3+ + e   Fe 2+ E = E o + log 0.0591 1 Fe 3+ /Fe 2+ [Fe 3+ ] [Fe 2+ ] PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 23

24. Addition of Zn 2+ salts which precipitates ferrocyanide: [Fe(CN) 6 ] 4- + Zn 2+  Zn 2 [Fe(CN) 6 ]  The oxidation potential of ferri/ferrocyanide system to oxidize iodide to iodine, although the oxidation potential of I 2 /2I - system is higher. 4. Effect of Precipitating Agents Ferricyanide: [Fe(CN) 6 ] 3  + e   [Fe(CN) 6 ] 4  E = E o + log 0.0591 1 Ferri/Ferro [[Fe(CN) 6 ] 3  ] [[Fe(CN) 6 ] 4  ] PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 24

25. In this reaction Cu 2+ oxidized I  although: E o Cu 2+ /Cu + = 0.16 and E o I 2 /2 I  = 0.54. Due to slight solubility of Cu 2 I 2, the concentration of Cu + is strongly decreased and the ratio Cu 2+ /Cu + is increased with a consequent increase of the potential of Cu 2+ /Cu + redox couple to about + 0.86 V, thus becoming able to oxidize iodide into iodine. ­ 4. Effect of Precipitating Agents Copper: 2 Cu 2+ + 4 I   2Cu 2 I 2 (ppt) + I 2 E = E o + log 0.0591 1 Cu 2+ /Cu + [Cu 2+ ] [Cu + ] PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 25

26. It is the plot of potential (E, volts) versus the volume (mL) of titrant. Redox Titration Curve Example: Titration of 100 ml 0.1 N Ferrous sulphate by 0.1 N ceric sulphate. Ce 4+ + Fe 2+ → Ce 3+ + Fe 3+ The change in potential during titration can be either measured or calculated using Nernest equation as follows: PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 26

27. At 0.00 mL of Ce +4 added, the potential is due to iron system: E = E o + (0.0592/n) log([Ox]/[Red])= 0.77 V No Ce +4 present; minimal, unknown [Fe +3 ]; thus, insufficient information to calculate E  After adding 10 mL Ce 4+ : E = 0.77 + 0.0591/1 log 10/90 = 0.71 V  After adding 50 mL Ce 4+ : E = 0.77 + 0.0591/1 log 50/50 = 0.77 V  After adding 90 mL Ce 4+ : E = 0.77 + 0.0591/1 log 90/10 = 0.82 V  After adding 99 mL Ce 4+ : E = 0.77 + 0.0591/1 log 99/1 = 0.88 V  A adding 99.9 mL Ce 4+ : E = 0.77 + 0.0591/1 log 99.9/0.1 = 0.94 V  After adding 100 mL Ce 4+ : the two potentials are identical PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 27

28. E = E O 1 + 0.0591/1 log [Fe 3+ ]/[Fe 2+ ]= 0.77 V E = E O 2 + 0.0591/1 log [Ce 4+ ]/[Ce 3+ ]= 1.45 V  After adding 100 mL Ce 4+ (end point): the two potentials are identical Summation of two equations: 2 E = E o 1 + E o 2 + 0.0591 log [Fe 3+ ][Ce 4+ ] [Fe 2+ ][Ce 3+ ] At the end point: Fe 2+ = Ce 4+, and Fe 3+ = Ce 3+ 2 E = E o 1 + E o 2 E = ( E o 1 + E o 2 ) / 2 E = 0.77 + 1.45 / 2 = 1.10 V PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 28

29. 204060801001201400 Potential (V) Ce 4+ (mL) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 29

30. Detection of End Point in Redox Titrations 1. Self Indicator (No Indicator) When the titrant solution is coloured (KMnO 4 ): KMnO 4 (violet) + Fe 2+ + H +  Mn 2+ (colourless) + Fe 3+. The disappearance of the violet colour of KMnO 4 is due to its reduction to the colourless Mn 2+. When all the reducing sample (Fe 2+ ) has been oxidized (equivalence point), the first drop excess of MnO 4  colours the solution a distinct pink. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 30

31. In Titration of Fe 2+ by Cr 2 O 7 2  Cr 2 O 7 2  + 3Fe 2+ + 14H +  2Cr 3+ + 3Fe 3+ + 7H 2 O The reaction proceeds until all Fe 2+ is converted into Fe 3+ Fe 2+ + Ferricyanide (indicator)  Ferrous ferricyanide (blue)]. Fe 2+ + [Fe(CN) 6 ] 3  → Fe 3 [Fe(CN) 6 ] 2-. The end point is reached when the drop fails to give a blue colouration with the indicator (on plate) Less accurate method and may lead to loss or contamination of sample. 2. External Indicator PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 31

32. 3. Internal Redox Indicator Redox indicators are compounds which have different colours in the oxidized and reduced forms. In ox + n e  = In red They change colour when the oxidation potential of the titrated solution reaches a definite value: E = E° + 0.0591/n log [In OX ]/[In red ] When [In ox ] = [In red ], E = E° Indicator colours may be detected when: [In ox ]/[In red ] = 1/10 or 10/1 hence, Indicator range: E = E° In ± 0.0591/n PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 32

33. Diphenylamine E < 0.73 V, colourless (red.). E > 0.79 V, bluish violet (ox.). E° = 0.76, n = 2. Range = 0.73 – 0.79 V. Ferroin indicator (1,10-phenanthroline-ferrous chelate). E° = 1.147, n = 1. Range = 1.088 – 1.206 V. E < 1.088 V, red (red.). E > 1.206 V, pale blue (ox.). PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 33

34. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 34

35. 4. Irreversible Redox Indicators In acid solutions, methyl orange is red. Addition of strong oxidants (Br 2 ) would destroy the indicator and thus it changes irreversibly to pale yellow colour Some highly coloured organic compounds that undergo irreversible oxidation or reduction Methyl Orange PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 35

36. Properties of Oxidizing Agents 1. Potassium permanganate (KMnO 4 ) 2. Potassium dichromate (K 2 Cr 2 O 7 ) 3. Iodine (I 2 ) 5. Bromate-bromide mixture 4. Potassium iodate (KIO 3 ) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 36

37. Very strong oxidizing agent, not a primary standard, self indicator. In acid medium: It can oxidize: oxalate, Fe 2+, Ferrocyanide, As 3+, H 2 O 2, and NO 2 . MnO 4  + 8H + + 5e  Mn 2+ + 4H 2 O In alkaline medium: MnO 4  + e  MnO 4 2  1. Potassium permanganate (KMnO 4 ) 2. Potassium dichromate (K 2 Cr 2 O 7 ) It is a primary standard (highly pure and stable). Used for determination of Fe 2+ (Cl  does not interfere); ferroin indicator. Cr 2 O 7 2  + 14H + +6e  2Cr 3+ + 7H 2 O In neutral medium: 4MnO 4  + 2H 2 O  MnO 2 + 4OH- + 3O 2 Unstable PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 37

38. Solubility of iodine in water is very small. Its aqueous solution has appreciable vapour pressure: Prepared in I  3. Iodine (I 2 ) I 2 + I  I 3  (triiodide ion) Iodine solution is standardized against a standard Na 2 S 2 O 3 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 38

39. Iodimetry: Iodimetry: Direct titration of reducing substances with iodine Iodometry: Iodometry: Back titration of oxidizing substances The oxidizing substance (Eº > + 0.54 V)is treated with excess iodide salt: 2MnO 4  + 10I  + 16H +  5I 2 + 2Mn 2+ + 8H 2 O Cr 2 O 7 2  + 6I - + 14H +  2Cr 3+ + 3I 2 + 7H 2 O The liberated Iodine is titrated with standard sodium thiosulphate (starch as indicator) The reducing substances (Eº < + 0.54 V) are directly titrated with iodine. Sn 2+ + I 2  Sn 4+ + 2I  2S 2 O 3 2  + I 2  S 4 O 6 2  + 2I  (Self indicator or starch as indicator) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 39

40. 4. Potassium iodate (KIO 3 ) It is strong oxidizing agent, highly pure, its solution is prepared by direct weighing. Andrew’s Reaction IO 3  + 5I  + 6H +  3I 2 + 3H 2 O (in 0.1 N HCl) Eq.W = MW/5 IO 3  + 2I  + 6H +  3I + + 3H 2 O Eq.W = MW/4 IO 3  + 2I 2 + 6H +  5I + + 3H 2 O (in 4-6 N HCl) Eq.W = MW/4 Determination of iodide with potassium iodate in 4-6 N HCl (chloroform as indicator) Starch can not be used. Potassium iodate prepared in molar PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 40

41. Upon acidification of bromate/bromide mixture, bromine is produced: BrO 3  + 5 Br  + 6 H +  3 Br 2 + 3 H 2 O Used for the determination of phenol and primary aromatic amines: dark OH + 3Br 2 OH Br Br Br + 3HBr 2,4,6-Tribromophenol Phenol 5. Bromate-bromide mixture The excess Br 2 is determined: Br 2 + 2I   I 2 + 2 Br  & I 2 + 2 Na 2 S 2 O 3  Na 2 S 4 O 6 + 2 I  Chloroform is added (dissolve TBP & indicator). Starch can be used PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 41

42. Applications of Redox Titrations 1.Determination of Free Metallic Elements Metallic iron Dissolved in FeCl 3 solution & the produced Fe 2+ is titrated with MnO 4  Fe + 2 FeCl 3  3 FeCl 2 (Zimmerman’s Reagent) 5 Fe 2+ + MnO 4  + 14 H +  5 Fe 3+ + Mn 2+ + 7 H 2 O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 42

43. 2. Determination of Halogen-Containing Compounds  Iodine in iodine tincture I 2 + 2 S 2 O 3 2   2 I  + S 4 O 6 2  Direct titration with thiosulphate  Bromine & chlorine I 2 + 2 S 2 O 3 2   2 I  + S 4 O 6 2  Back titration with thiosulphate, after treatment with excess KI. Cl 2 + 2I   l 2 + 2Cl  PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 43

44. 3. Determination of Peroxides Permanganatometrically. Direct titration with KMnO 4. 5 H 2 O 2 + 2 MnO 4  + 6 H +  5 O 2 + 2 Mn 2+ + 8 H 2 O Iodometry. Back titration with thiosulphate, after adding excess KI. H 2 O 2 + 2 I  + 2 H +  I 2 + 2 H 2 O  Hydrogen peroxide 4. Determination of Anions Oxalates and oxalic acid are strong reducing agents and can be titrated with standard KMnO 4 at 60 °C in the presence of dilute sulphuric acid. 5C 2 O 4 2- + 2MnO 4 - + 16H +  10CO 2 + 2Mn 2+ + 8H 2 O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 44