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Oxidation-Reduction Titrations Dr M.Afroz Bakht. Introduction Oxidation-Reduction (Redox): Definitions & Terms. Oxidation Number Principles. Balancing.

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Presentation on theme: "Oxidation-Reduction Titrations Dr M.Afroz Bakht. Introduction Oxidation-Reduction (Redox): Definitions & Terms. Oxidation Number Principles. Balancing."— Presentation transcript:

1 Oxidation-Reduction Titrations Dr M.Afroz Bakht

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.

3 Oxidation-Reduction (Redox) 3 Fe 2+ + Cl 2 2Fe Cl  3 Fe 2+ — 2e 2 Fe 3+ (Loss of electrons: Oxidation) Cl e 2 Cl - (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 chlorine gas Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half equation.

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 special cases= —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.

5 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

6 For manganese SpeciesMnMn 2+ Mn 3+ MnO 2 MnO 4 2  MnO 4  O.N For nitrogen SpeciesNH 3 N2H4N2H4 NH 2 OHN2N2 N2ON2ONOHNO 2 HNO 3 O.N.–3–2– Oxidation states of manganese and nitrogen in different species

7 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

8 Al + HCl = AlCl 3 + H Oxidation: ( Al = Al e - )  2 = 6 e - Reduction: ( 2 H + + 2e = H 2 )  3 = 6 e - Redox: 2 Al + 6 H + = 2 Al H 2 Balancing Redox Reactions Using the Half- Reaction Method

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 CO H 2 O  Balance oxygen atoms by adding water 2 MnO 4  + 5 C 2 O 4 2  + 16 H + 2 Mn CO 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 CO 2 MnO 4  + C 2 O 4 2  + H + Mn 2+ + CO 2 + H 2 O

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

11 anode//cathode Cu/CuSO 4 // ZnSO 4 /Zn

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 = 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+ ] n

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+ ] 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 Al / Al 3+ –1.33Cu / Cu Zn / Zn 2+ –0.76Hg / Hg Fe / Fe 2+ –0.44Ag / Ag

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.

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

16 Standard Oxidation Potential (E) Standard Oxidation Potential (E o ) E 25 °C = E o + log [Oxidized] / [Reduced] 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

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20 Oxidation Potentials: Electrochemical Series

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] 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.

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 MnO 4  /Mn 2+ [MnO 4  ][H + ] [Mn 2+ ] Potassium permanganate: MnO 4  + 8H + + 5e   Mn H 2 O E = E o + log MnO 4  /Mn 2+ [MnO 4  ][H + ] 8 [Mn 2+ ] Potassium dichromate: Cr 2 O 7 2  + 14H + +6e   2Cr H 2 O E = E o + log Cr 2 O 7 2  /Cr 3+ [Cr 2 O 7 2  ][H + ] 14 [Cr 3+ ]

23 E (I 2 /2I  ) system increases by the addition of HgCl 2 since it complexes with iodide ions. Hg I   [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  ) = Iodine: I 2 + 2e   2I  E = E o + log I 2 /I  [I 2 ] [I  ] 2 Ferric: Fe 3+ + e   Fe 2+ E = E o + log Fe 3+ /Fe 2+ [Fe 3+ ] [Fe 2+ ]

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 Ferri/Ferro [[Fe(CN) 6 ] 3  ] [[Fe(CN) 6 ] 4  ]

25 In this reaction Cu 2+ oxidized I  although: E o Cu 2+ /Cu + = 0.16 and E o I 2 /2 I  = 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 V, thus becoming able to oxidize iodide into iodine. ­ 4. Effect of Precipitating Agents Copper: 2 Cu I   2Cu 2 I 2 (ppt) + I 2 E = E o + log Cu 2+ /Cu + [Cu 2+ ] [Cu + ]

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:

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 = /1 log 10/90 = 0.71 V  After adding 50 mL Ce 4+ : E = /1 log 50/50 = 0.77 V  After adding 90 mL Ce 4+ : E = /1 log 90/10 = 0.82 V  After adding 99 mL Ce 4+ : E = /1 log 99/1 = 0.88 V  A adding 99.9 mL Ce 4+ : E = /1 log 99.9/0.1 = 0.94 V  After adding 100 mL Ce 4+ : the two potentials are identical

28 E = E O /1 log [Fe 3+ ]/[Fe 2+ ]= 0.77 V E = E O /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 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 = / 2 = 1.10 V

29 Potential (V) Ce 4+ (mL)

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.

31 In Titration of Fe 2+ by Cr 2 O 7 2  Cr 2 O 7 2  + 3Fe H +  2Cr Fe H 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

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° /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 ± /n

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 = – V. E < V, red (red.). E > V, pale blue (ox.).

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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

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 )

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 H 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 H 2 O In neutral medium: 4MnO 4  + 2H 2 O  MnO 2 + 4OH- + 3O 2 Unstable

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

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

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

41 Upon acidification of bromate/bromide mixture, bromine is produced: BrO 3  + 5 Br  + 6 H +  3 Br 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 Br  & I Na 2 S 2 O 3  Na 2 S 4 O I  Chloroform is added (dissolve TBP & indicator). Starch can be used

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 H 2 O

43 2. Determination of Halogen-Containing Compounds  Iodine in iodine tincture I S 2 O 3 2   2 I  + S 4 O 6 2  Direct titration with thiosulphate  Bromine & chlorine I 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 

44 3. Determination of Peroxides Permanganatometrically. Direct titration with KMnO 4. 5 H 2 O MnO 4  + 6 H +  5 O Mn H 2 O Iodometry. Back titration with thiosulphate, after adding excess KI. H 2 O I  + 2 H +  I 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 MnO H +  10CO 2 + 2Mn H 2 O


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