Oxidation Reduction Titrations

Oxidation Reduction Titrations

Oxidation-Reduction Titrations PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 1 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

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

(Gain of electrons: Reduction) In every redox reaction, both reduction and oxidation must occur. Substance that gives electrons is the reducing agent or reductant. Overall, the number of electrons in the oxidation half Substance that accepts electrons lost is the oxidizing agent or reaction must equal the number gained in the reduction oxidant. half equation.

PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 4 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

Oxidation Numbers of Some Substances Substance Oxidation Numbers NaCl H2 NH3 H2O2 LiH K2CrO4 SO42KClO3 Na = +1, Cl = 1 H = 0 N = 3, H = +1 H = +1, O = 1 Li = +1, H = 1 K = +1, Cr = +6,

O = 2, S = +6 K = +1, Cl = +5, PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 O = 2 O = 2 6 Oxidation states of manganese and nitrogen in different species For manganese Species Mn Mn2+ 0

+2 Species NH3 N2H4 O.N. 3 2 O.N. Mn3+ MnO2 +3 +4

MnO42 MnO4 +6 +7 For nitrogen NH2OH 1 N2 N2O NO HNO2 HNO3 0

+1 +2 +3 +5 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 7 Balancing Redox Reactions using HalfReaction 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 8 MnO4 + C2O42 + H+ Mn2+ + CO2 + H2O Balance each half reaction: MnO4 + 5 C2O42 Mn2+ 2 CO2 + 2 Use the number of moles so as to make the electrons gained in one reaction equal those lost in the other one 2 MnO4 + 5 C2O42 2 Mn2+ + 10 CO2 Balance oxygen atoms by

adding 2 MnO4water + 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 9 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 the metal 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) ANALYTICAL CHEMISTRY and the solutionPHARMACEUTICAL is known as electrode potential (E) 10 PHC 213 anode//cathode Cu/CuSO4 // ZnSO4 /Zn PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 11 Nernest Equation for Electrode Potential (E) RT

Et = E + n+ nF [Mat temperature ] Et = electrodelog potential t. o E = standard electrode potential (constant depend on the system) R = gas constant T = absolute Temp. (tC + 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 0.0591 E25 C = Eo + n n+ log [M ] PHARMACEUTICAL ANALYTICAL CHEMISTRY

PHC 213 12 o Standard Electrode Potential (E ) E25 C = Eo0.0591 + log [Mn+] n 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). System E (volts) System Li / Li+ K / K+ Mg/Mg2+ Al / Al3+ Zn / Zn2+

Fe / Fe2+ 3.03 Cd/Cd2+ 2.92 Sn / Sn2+ 2.37 H2 (pt) / H+ 1.33 Cu / Cu2+ 0.76 Hg / Hg2+ 0.44 AgCHEMISTRY / Ag+ PHARMACEUTICAL ANALYTICAL PHC 213 E (volts) 0.40 0.13 0.00 +0.34 +0.79

+0.80 13 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 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 15 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 16 o

Standard Oxidation Potential (E ) 0.0591 E25 C = E + n [Oxidized] / [Reduced] o log 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 17 +0.339 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

18 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 19 Oxidation Potentials: Electrochemical Series PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 20 Factors Affecting Oxidation Potential 1. Common Ion 0.0591 E25 C = E + n [Oxid] /[Red] 2+ o

log The potential of MnO4 /Mn varies with the ratio [MnO4]/[Mn2+]. If ferrous is titrated with MnO4 in presence of Cl , chloride will interfere by reaction with MnO4 and gives higher results. Zimmermanns 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. PHARMACEUTICAL ANALYTICAL CHEMISTRY Sulphuric acid is used for acidification. PHC 213 21 2. Effect of pH 0.0591 o = E + 5 [MnO4][H+]

EMnO4/Mn2+ 2+ [Mn ] log The oxidation potential of an oxidizing agent containing oxygen increases by increasing acidity and vice versa. Potassium permanganate: E MnO4/Mn2+ log Potassium dichromate: E Cr2O72/Cr3+ log MnO4 + 8H+ + 5e Mn2+ + 4H2O 0.0591o = E + 5 [MnO4][H+]8 [Mn2+] Cr2O72 + 14H+ +6e 2Cr3+ + 7H2O 0.0591

= Eo + 6 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 [Cr2O72][H+]14 [Cr3+] 22 3. Effect of Complexing Agents Iodine: E I2/I log I2 + 2e 2I [I2] 0.0591 = Eo + 2 [I]2

E (I2/2I) system increases by the addition of HgCl2 since it complexes with iodide ions. Hg2+ + 4I [HgI4]2 Ferric: Fe3+ + e Fe2+ (low dissociation complex) E Fe3+/Fe2+ log 0.0591 = Eo + 1 [Fe3+] [Fe2+] 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 although Eo(Fe3+/Fe2+) = 0.77 while Eo(I2/2I ) = 0.54. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 23 4. Effect of Precipitating Agents Ferricyanide: [Fe(CN)6]3 + e [Fe(CN)6]4 EFerri/Ferro log 0.0591 = Eo + 1 [[Fe(CN)6]3] [[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 24 4. Effect of Precipitating Agents Copper: 2 Cu2+ + 4 I 2Cu2I2 (ppt) + I2 E Cu2+/Cu+ log 0.0591 = Eo + 1 [Cu2+] [Cu+] 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 25 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 26 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 27 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: [Fe3+][Ce4+] + 0.0591

[Fe2+][Ce3+] 2 E = Eo1 + Eo2 log At the end point: Fe2+ = Ce4+ , and Fe3+ = Ce3+ 2 E = Eo1 + Eo2 E = ( Eo1 + Eo2 ) / 2 1.45 / 2 = 1.10 V PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 E = 0.77 + 28 Potential (V) 0 20 40 60 80 100120 140 Ce4+ (mL) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 29

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 30 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 31 PHC 213 3. Internal Redox Indicator Redox indicators are compounds which have different colours in the oxidized and reduced forms. Inox + n e = Inred 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 = EIn 0.0591/n PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213

32 Diphenylamine 2 H H H N N N diphenylbenzidine (colourless) + 2H+ + 2e diphenylamine (colourless) E = 0.76, n = 2.

E < 0.73 V Range = 0.73 0.79 V. H H + E < 0.73 V, colourless (red.). E > 0.79 V, bluish violet (ox.). + 2e N N + diphenylbenzidine (violet) E > 0.79 V

erroin 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.). N N Fe2+ N Fe3+ N 3 Ferroin (red) PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 + e

3 Ferrin (pale blue) 33 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 34 4. Irreversible Redox Indicators Some highly coloured organic compounds that undergo irreversible oxidation or reduction Methyl Orange (CH3)2N N SO3Na + Br2 + 2 H2O N Methyl orange

(CH3)2N NO + ON SO3Na + 4 HBr In acid solutions, methyl orange is red. Addition of strong oxidants (Br2) would destroy the indicator and thus it changesPHARMACEUTICAL irreversibly to pale yellow colour ANALYTICAL CHEMISTRY PHC 213 35 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 36 1. Potassium permanganate (KMnO4) Very strong oxidizing agent, not a primary standard, self indicator. In acid medium: MnO4 + 8H+ + 5e Mn2+ + 4H2O It can oxidize: oxalate, Fe2+, Ferrocyanide, As3+, H2O2, and NO2. MnO42 In alkaline medium: MnO4 + e n neutral medium:4MnO4 + 2H2O MnO2 + 4OH- + 3O2 Unstable 2. Potassium dichromate (K2Cr2O7)

It is a primary standard (highly pure and stable). Cr2O72 + 14H+ +6e 2Cr3+ + 7H2O Used for determination of Fe2+ (Cl does not interfere); ferroin indicator. PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 37 3. Iodine (I2) Solubility of iodine in water is very small. Its aqueous solution has appreciable vapour pressure: Prepared in I I2 + I I3 (triiodide ion) Iodine solution is standardized against a standard Na2S2O3 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 38

Iodimetry: Direct titration of reducing substances with iodine The reducing substances (E < + 0.54 V) are directly titrated with iodine. Sn2+ + I2 Sn4+ + 2I 2S2O32 + I2 S4O62 + 2I indicator or starch indicator) Iodometry:(Self Back titration ofasoxidizing substances The oxidizing substance (E > + 0.54 V)is treated with excess iodide salt: 2MnO4 + 10II + 16HH+ 5II2 + 2Mn2+ + 8H2O Cr2O72 + 6HI- + 14H+ 2Cr3+ + 3I2 + 7H2O The liberated Iodine is titrated with standard PHARMACEUTICAL ANALYTICAL CHEMISTRY sodium thiosulphate (starch as indicator) PHC 213

39 4. Potassium iodate (KIO3) It is strong oxidizing agent, highly pure, its solution is prepared by direct weighing. IO3 + 5II + 6H+ 3I2 + 3H2O (in 0.1 N HCl) Eq.W = MW/5I IO3 + 2I2 + 6H+ 5II+ + 3H2O (in 4-6 N HCl) Eq.W = MW/4 IO3 + 2I + 6H+ 3I+ + 3H2O Eq.W = MW/4 Andrews Reaction Determination of iodide with potassium iodate in 4-6 N HCl (chloroform as indicator) Starch can not be used. ANALYTICAL CHEMISTRY Potassium PHARMACEUTICAL iodate prepared in molar PHC 213 40

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: OH OH + 3Br2 dark Br Br + 3HBr Br Phenol 2,4,6-Tribromophenol 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 41 Applications of Redox Titrations 1. Determination of Free Metallic Elements Metallic iron Dissolved in FeCl3 solution & the produced Fe2+ is titrated with MnO4 Fe + 2 FeCl3 3 FeCl2 (Zimmermans Reagent) 5 Fe2+ + MnO4 + 14 H+ 5 Fe3+ + Mn2+ + 7 H2O PHARMACEUTICAL ANALYTICAL CHEMISTRY

PHC 213 42 2. Determination of Halogen-Containing Compounds 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 PHARMACEUTICAL ANALYTICAL CHEMISTRY PHC 213 43 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 + 2 H2O

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 44

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