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Introduction to Oxidation-Reduction Titrations in Pharmaceutical Analytical Chemistry PHC 213

Learn about redox reactions, oxidation numbers, balancing redox equations, electrode potential, and applications of redox titrations in pharmaceutical analytical chemistry. Understand the principles behind oxidation and reduction, detecting end points, and more.

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Introduction to Oxidation-Reduction Titrations in Pharmaceutical Analytical Chemistry PHC 213

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  1. Oxidation-Reduction Titrations PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  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

  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

  4. Fe2+ —e Fe3+ (Loss of electrons: Oxidation) Ce4+ + e Ce3+(Gain of electrons: Reduction) Oxidation-Reduction (Redox) Reaction of ferrous ion with ceric ion Fe2+ + Ce4+ Fe3+ + Ce3+ 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. 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

  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

  6. OxidationNumbers of Some Substances Substance Oxidation Numbers NaCl H2 NH3 H2O2 LiH K2CrO4 SO42- KClO3 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

  7. For nitrogen Species NH3 N2H4 NH2OH N2 N2O NO HNO2 HNO3 O.N. –3 –2 –1 0 +1 +2 +3 +5 Oxidation states of manganese and nitrogen in different species For manganese Species Mn Mn2+ Mn3+ MnO2 MnO42 MnO4  O.N. 0 +2 +3 +4 +6 +7 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  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 H2O • 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

  9. MnO4 + C2O42 + H+Mn2+ + CO2 + H2O • Balance each half reaction: MnO4 + 5é Mn2+ C2O42 2 CO2 + 2é • Use the number of moles so as to make the electrons gained in one reaction equal those lost in the other one 2 MnO4 + 5C2O42 2 Mn2+ + 10 CO2 • Balance oxygen atoms by adding water 2 MnO4 + 5 C2O42 2 Mn2+ + 10 CO2 + 8 H2O • Balance hydrogen atoms by adding H+ 2 MnO4 + 5 C2O42 + 16 H+2 Mn2+ + 10 CO2 + 8 H2O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  10. Electrochemical Cells Electrochemical cells consist of electrodes immersed in electrolyte solution and frequently connected by a salt bridge Solution Pressure.The tendency of the metal to dissolve in a solution of its salt. Ionic Pressure.The tendency of themetal cations to deposit on its metal dipped into its solution. • Cu/Cu2+ system: ionic pressure > solution pressure. Cu2+ leaves the solution to deposit on Cu rod • Zn/Zn2+ system: solution pressure > ionic pressure. Zn metal tends to dissolve forming Zn2+ in solution. The potential difference between the metal rod (electrode) and the solution is known as electrode potential (E) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  11. anode//cathode Cu/CuSO4 // ZnSO4 /Zn PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  12. Et = Eo + log [Mn+] RT nF E25 °C = Eo + log [Mn+] 0.0591 n Nernest Equation for Electrode Potential (E) Et = 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) loge = ln (natural logarithm = 2.303 log) n= valency of the ion [Mn+] = molar concentration of metal ions in solution PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  13. System E° (volts) System E° (volts) Li / Li+ –3.03 Cd/Cd2+ –0.40 K / K+ –2.92 Sn / Sn2+ –0.13 E25 °C = Eo + log [Mn+] Mg/Mg2+ –2.37 H2 (pt) / H+ 0.00 Al / Al3+ –1.33 Cu / Cu2+ +0.34 0.0591 Zn / Zn2+ –0.76 Hg / Hg2+ +0.79 n Fe / Fe2+ –0.44 Ag / Ag+ +0.80 Standard Electrode Potential (Eo) Eo 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). PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  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

  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+). H2 gas (at 1 atm. Pressure) is passed. Platinum black layer absorbs a large amount of H2 and can be considered as a bar of hydrogen, it also catalyses the half reaction: 2H+ + 2e  H2 Under these conditions: H2 electrode potential = zero 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. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  17. E25 °C = Eo + log [Oxidized] / [Reduced] 0.0591 n Standard Oxidation Potential (Eo) 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 Fe3+ / Fe2+), is connected with a NHE PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  18. +0.339 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  19. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

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

  21. E25 °C = Eo + log [Oxid] /[Red] 0.0591 n Factors Affecting Oxidation Potential 1. Common Ion • The potential of MnO4/Mn2+ varies with the ratio [MnO4]/[Mn2+]. • If ferrous is titrated with MnO4 in presence of Cl , chloridewill interfere by reaction with MnO4 and gives higher results. Zimmermann’s Reagent (MnSO4, H3PO4 and H2SO4) • MnSO4 has a common ion(Mn2+) with the reductant that lowers the potential of MnO4/Mn2+ system: • Phosphoric acid lowers the potential of Fe3+/Fe2+ system by complexation with Fe3+ as [Fe(PO4)2]3. • Sulphuric acid is used for acidification. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  22. 2. Effect of pH E = Eo + log [MnO4][H+] 0.0591 MnO4/Mn2+ 5 [Mn2+] E = Eo + log [MnO4][H+]8 0.0591 MnO4/Mn2+ 5 [Mn2+] Potassium dichromate: Cr2O72 + 14H+ +6e 2Cr3+ + 7H2O E = Eo + log [Cr2O72][H+]14 0.0591 Cr2O72/Cr3+ 6 [Cr3+] The oxidation potential of an oxidizing agent containing oxygen increases by increasing acidity and vice versa. Potassium permanganate: MnO4 + 8H+ + 5e Mn2+ + 4H2O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  23. E = Eo + log [I2] 0.0591 I2/I 2 [I]2 E = Eo + log [Fe3+] 0.0591 Ferric: Fe3+ + eFe2+ Fe3+/Fe2+ 1 [Fe2+] 3. Effect of Complexing Agents Iodine: I2 + 2e2I • E (I2/2I) system increases by the addition of HgCl2 since it complexes with iodide ions. • Hg2+ + 4I [HgI4]2 (low dissociation complex) • E (Fe3+/Fe2+) is reduced by the addition of F or PO43 due to the formation of the stable complexes [FeF6]3 and [Fe(PO4)2]3 respectively. Thus, ferric ions, in presence of F or PO43 cannot oxidize iodide althoughEo(Fe3+/Fe2+) = 0.77whileEo(I2/2I ) = 0.54. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  24. E = Eo + log [[Fe(CN)6]3] 0.0591 Ferri/Ferro 1 [[Fe(CN)6]4] 4. Effect of Precipitating Agents Ferricyanide: [Fe(CN)6]3 + e[Fe(CN)6]4 Addition of Zn2+ salts which precipitates ferrocyanide: [Fe(CN)6]4- + Zn2+ Zn2 [Fe(CN)6] The oxidation potential of ferri/ferrocyanide system to oxidize iodide to iodine, although the oxidation potential of I2/2I- system is higher. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  25. E = Eo + log [Cu2+] 0.0591 Cu2+/Cu+ 1 [Cu+] 4. Effect of Precipitating Agents Copper: 2 Cu2+ + 4 I 2Cu2I2 (ppt) + I2 In this reaction Cu2+ oxidized I although: EoCu2+/Cu+ = 0.16 and EoI2/2 I = 0.54. Due to slight solubility of Cu2I2,the concentration of Cu+ is strongly decreased and the ratio Cu2+/Cu+is increased with a consequent increase of the potential of Cu2+/Cu+redox couple to about + 0.86 V, thus becoming able to oxidize iodide into iodine. ­ PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  26. Redox Titration Curve It is the plot of potential (E, volts) versus the volume (mL) of titrant. Example: Titration of 100 ml 0.1 N Ferrous sulphate by 0.1 N ceric sulphate. Ce4+ + Fe2+ → Ce3+ + Fe3+ The change in potential during titration can be either measured or calculated using Nernest equation as follows: PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  27. At 0.00 mL of Ce+4 added, the potential is due to iron system: E = Eo + (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 Ce4+: E = 0.77 + 0.0591/1 log 10/90 = 0.71 V • After adding 50 mL Ce4+: E = 0.77 + 0.0591/1 log 50/50 = 0.77 V • After adding 90 mL Ce4+: E = 0.77 + 0.0591/1 log 90/10 = 0.82 V • After adding 99 mL Ce4+: E = 0.77 + 0.0591/1 log 99/1 = 0.88 V • A adding 99.9 mL Ce4+: E = 0.77 + 0.0591/1 log 99.9/0.1 = 0.94 V • After adding 100 mL Ce4+: the two potentials are identical PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  28. 2 E = Eo1 + Eo2 + 0.0591 log [Fe3+][Ce4+] [Fe2+][Ce3+] • After adding 100 mL Ce4+ (end point): the two potentials are identical E = EO1 + 0.0591/1 log [Fe3+]/[Fe2+]= 0.77 V E = EO2 + 0.0591/1 log [Ce4+]/[Ce3+]= 1.45 V • Summation of two equations: • At the end point: Fe2+ = Ce4+ , and Fe3+ = Ce3+ 2 E = Eo1 + Eo2 E = ( Eo1 + Eo2 ) / 2E = 0.77 + 1.45 / 2 = 1.10 V PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  29. Potential (V) 0 20 40 60 80 100 120 140 Ce4+ (mL) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  30. Detection of End Point in Redox Titrations 1. Self Indicator (No Indicator) When the titrant solution is coloured (KMnO4): KMnO4 (violet) + Fe2+ + H+ Mn2+ (colourless) + Fe3+. The disappearance of the violet colour of KMnO4 is due to its reduction to the colourless Mn2+. When all the reducing sample (Fe2+) has been oxidized (equivalence point), the first drop excess of MnO4 colours the solution a distinct pink. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  31. 2. External Indicator • In Titration of Fe2+ by Cr2O72 • Cr2O72 + 3Fe2+ + 14H+ 2Cr3+ + 3Fe3+ + 7H2O • The reaction proceeds until all Fe2+ is converted into Fe3+ • Fe2+ + Ferricyanide (indicator)  Ferrous ferricyanide (blue)]. • Fe 2+ + [Fe(CN)6]3→ Fe3[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. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

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

  33. 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.). Diphenylamine E° = 0.76, n = 2. Range = 0.73 – 0.79 V. E < 0.73 V, colourless (red.). E > 0.79 V, bluish violet (ox.). PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  34. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  35. 4. Irreversible Redox Indicators Some highly coloured organic compounds that undergo irreversible oxidation or reduction Methyl Orange In acid solutions, methyl orange is red. Addition of strong oxidants (Br2) would destroy the indicator and thus it changes irreversibly to pale yellow colour PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  36. Properties of Oxidizing Agents 1. Potassium permanganate (KMnO4) 2. Potassium dichromate(K2Cr2O7) 3. Iodine (I2) 4. Potassium iodate (KIO3) 5. Bromate-bromide mixture PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  37. MnO4 + 8H+ + 5e Mn2+ + 4H2O 4MnO4 + 2H2O  MnO2+ 4OH- + 3O2 Unstable MnO4 + e MnO42 2. Potassium dichromate(K2Cr2O7) It is a primary standard (highly pure and stable). Used for determination of Fe2+ (Cl does not interfere); ferroin indicator. Cr2O72 + 14H+ +6e 2Cr3+ + 7H2O 1. Potassium permanganate (KMnO4) Very strong oxidizing agent, not a primary standard, self indicator. In acid medium: It can oxidize: oxalate, Fe2+, Ferrocyanide, As3+, H2O2, and NO2. In alkaline medium: In neutral medium: PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  38. I2 + I I3 (triiodide ion) 3. Iodine (I2) Iodine solution is standardized against a standard Na2S2O3 Solubility of iodine in water is very small. Its aqueous solution has appreciable vapour pressure: Prepared in I PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  39. Iodimetry: Direct titration of reducing substances with iodine The reducing substances (Eº < + 0.54 V) are directly titrated with iodine. Sn2+ + I2 Sn4+ + 2I 2S2O32 + I2  S4O62 + 2I (Self indicator or starch as indicator) Iodometry:Back titration of oxidizing substances The oxidizing substance (Eº > + 0.54 V)is treated with excess iodide salt: 2MnO4 + 10I + 16H+5I2 + 2Mn2+ + 8H2O Cr2O72 + 6I- + 14H+  2Cr3+ + 3I2 + 7H2O The liberated Iodine is titrated with standard sodium thiosulphate (starch as indicator) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  40. IO3 + 5I+ 6H+ 3I2 + 3H2O (in 0.1 N HCl) Eq.W = MW/5 IO3 + 2I2 + 6H+ 5I+ + 3H2O (in 4-6 N HCl) Eq.W = MW/4 IO3 + 2I + 6H+ 3I+ + 3H2O Eq.W = MW/4 4. Potassium iodate (KIO3) It is strong oxidizing agent, highly pure, its solution is prepared by direct weighing. Andrew’s Reaction 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

  41. O H O H B r B r d a r k + 3 B r 2 + 3 H B r B r P h e n o l 2 , 4 , 6 - T r i b r o m o p h e n o l 5. Bromate-bromide mixture Upon acidification of bromate/bromide mixture, bromine is produced: BrO3 + 5 Br + 6 H+ 3 Br2 + 3 H2O Used for the determination of phenol and primary aromatic amines: The excess Br2 is determined: Br2 + 2I  I2 + 2 Br &I2 + 2 Na2S2O3  Na2S4O6 + 2 I Chloroform is added (dissolve TBP & indicator). Starch can be used PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  42. Applications of Redox Titrations Determination of Free Metallic Elements Metallic iron Dissolved in FeCl3 solution & the produced Fe2+ is titrated with MnO4 Fe + 2 FeCl33 FeCl2(Zimmerman’s Reagent) 5 Fe2+ + MnO4 + 14 H+ 5 Fe3+ + Mn2++ 7 H2O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  43. Iodine in iodine tincture Direct titration with thiosulphate I2 + 2 S2O32 2 I + S4 O62 • Bromine & chlorine Back titration with thiosulphate, after treatment with excess KI. Cl2 + 2Il2 + 2Cl I2 + 2 S2O32 2 I + S4 O62 2. Determination of Halogen-Containing Compounds PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

  44. 3. Determination of Peroxides • Hydrogen peroxide Permanganatometrically. Direct titration with KMnO4. 5 H2O2 + 2 MnO4 + 6 H+ 5 O2+ 2 Mn2++ 8 H2O Iodometry.Back titration with thiosulphate, after adding excess KI. H2O2 + 2 I + 2 H+ I2+ 2H2O 4. Determination of Anions Oxalates and oxalic acid are strong reducing agents and can be titrated with standard KMnO4 at 60 °C in the presence of dilute sulphuric acid. 5C2O42- + 2MnO4- + 16H+ 10CO2 + 2Mn2+ + 8H2O PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

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