Pharmaceutical Analytical Chemistry - 3 - Redox Reactions Lecture Notes PDF

Pharmaceutical Analytical Chemistry- 3 Redox Reaction  Oxidation reduction reactions (redox) Oxidation reduction reaction are those reaction that involve oxidation and reduction. Definitions: An old definitions of oxidation and reduction: Oxidation: It was defined as , combinat...

Pharmaceutical Analytical Chemistry- 3 Redox Reaction  Oxidation reduction reactions (redox) Oxidation reduction reaction are those reaction that involve oxidation and reduction. Definitions: An old definitions of oxidation and reduction: Oxidation: It was defined as , combination of substance with oxygen or loss of hydrogen. Fe0 + O FeO 2 FeO + O Fe2O3 Reduction: It was defined as , combination of substance with hydrogen or loss of oxygen. C + 2 H2 CH4 But sometimes oxidation or reduction involve no oxygen or hydrogen transfer, e.g. SnCl2 + 2 FeCl3 2 FeCl2 + SnCl4 reduction oxidation Fe3+ is reduced to Fe2+, its positive charge is decreased through gaining of electrons. While, Sn2+ is oxidized to Sn4+, its positive charge is increased through loss of electrons. The modern definitions of oxidation and reduction is: Oxidation It was defined as , loss of electrons (increasing +ve charge or decreasing –ve charge). Reduction: It was defined as , gaining of electrons (decreasing +ve charge or increasing –ve charge). Oxidizing agent (oxidant) It is that compound which gain electrons and is reduced (known as conjugated reductant). n e + Oxidant Conjugated reductant Reducing agent (reductant) It is that compound which lose electrons and is oxidized (known as conjugated oxidant). Reductant Conjugated oxidant + n e ▪ Each oxidant has a conjugate reductant and each reductant has a conjugate oxidant, the stronger the oxidant, the weaker is its conjugate reductant and vice versa.  ▪ During reaction between oxidant of an element and reductant of another element, oxidant is converted to its conjugated reductant and the reductant is converted to its conjugated oxidant and also two redox system is established as in the following example Cr2 O72-/2 Cr3+ and Fe3+/Fe2+ systems are established. Oxidant 1 + Reductant2 Conjugated reductant1 + Conjugated oxidant2 Cr2O72- + 6 Fe2+ + 14 H+ 2Cr3+ + 6 Fe3+ + 7H2O ▪ If oxidant has great tendency to gain electrons, its conjugated reductant is not easy to give electrons and vice versa. Therefore each system has a predominate character, either giving or gaining electrons. According to the specific character of each system , it becomes oxidizing or reducing system. The following table contains the most famous oxidants and reductants: Name Symbol Potassium permangenate KMnO4 Oxidants Potassium dichromate K2Cr2O7 Potassium bromate KBrO3 Potassium iodate KIO3 Ferrous sulphate FeSO4 Metallic iron Fe Sodium thiosulphate Na2S2O3 Reductants Sodium arsenite Na3ASO3 Oxalic acid H2C2O4 Oxalate C2O42-  Equivalent weight of oxidant and reductant 1-Old definitions The weight of the substance that react with or contain 1.008 gm of available hydrogen or 8.000 gm of available O2. By available is meant being used in oxidation or reduction. acid medium 2KMnO4 K2O + 2MnO + 5O Equivalent weight of KMnO4 = mol.wt. of KMnO4/5 alkaline medium K2O + 2MnO2 + 3O 2KMnO4 Equivalent weight of KMnO4 = mol.wt. of KMnO4/3 K2Cr2O7 K2O + Cr2O3 +3O Equivalent weight = mol.wt. of K2Cr2O7/6.  2- Modern definitions of equivalent weight oxidant and reductant a- Equivalent weight = mol.wt. /No of electrons gained or lost by the substance - 2+ + MnO4 + 5 Fe + 8H 5Fe3+ + Mn2+ + 4H2O Equivalent weight of KMnO4 = mol.wt. of KMnO4/5 b- Equivalent weight is defined as: Equivalent weight = mol.wt. /change in oxidation number of the element Change in oxidation number of Mn is from +7 to +2. Eq.wt = KMnO4/5 Oxidation potential (E25 ºC) ▪ If an inert electrode (e.g platinum) is immersed in a solution containing a redox system, electromotive force (EMF) is produced which known as oxidation potential. ▪ If oxidant has great tendency to gain electrons, the system is described as oxidizing system and has a positive potential (i.e. platinum wire will acquire positive charge). On the other hand, if the reductant has great tendency to donate electrons, the system is described as reducing system and has a negative potential (i.e. platinum wire will acquire negative charge). ▪ The value of oxidation potential can not be measured directly, but measured against reference electrode e.g. normal hydrogen electrode (N.H.E). ▪ From the previously mentioned, redox reaction is transfer of electrons from reductant to oxidant and consequently system of +ve potential will oxidize that of –ve potential or that of less positive potential. N.B. 1. Number of electrons gained by oxidant or lost by reductant must be equal, as electrons can not exist in solution. 2. Electrons transfer from reductant to oxidant until equilibrium is established will depend on the relative concentration and capacity of oxidant and reductant to gain or donate electrons. Nernest equation for calculation of oxidation potential RT [ox] Et = E0 + loge nF [red] Et = oxidation potential at tº C. E0 = standard oxidation potential. R = gas constant = 8.314 T = absolute temperature (273 + t ºC). n = number of electrons gained by oxidant or lost by reductant. F = faraday (Unit of quantity of electricity). loge = natural logarithm, that is to base 2.718. [ox] or [red] = molar concentration of oxidant and reductant. Nernst equation can be simplified by introducing known values of R and F and converting the natural logarithm (loge) to the base 10 by multiplication by 2.303. 0.0001982 T [ox] Et = E0 + log n [red] At a temperature of 25 ºC, it becomes o 0.0591 [ox] E25 C= E0 + log n [red] ▪ From Nernst equation , one can conclude that, the oxidation potential of a redox system will vary according to variation in the ratio of [ox]/ [red] at constant temperature and acidity. ▪ So, E25ºC of a redox system may increased or decreased with the increase and decrease in the concentration of oxidant and reductant. ▪ For evaluation of the magnitude of each redox system , oxidation potential is measured in presence of equimolar concentration of oxidant and reductant. Applying in Nernst equation, when the ratio of [ox]/[red] = 1, so, log 1 = zero. o 0.0591 [ox] E25 C= E0 + log n [red] E25ºC = E0 Standard oxidation potential (E0) It is EMF (or potential) produced when there is equilibrium between equimolar concentration of oxidant and reductant. The magnitude of the standard oxidation potential is the measure of oxidizing or reducing strength of a redox system. Electrode potential If metal or (non-metal) electrode is immersed in solution of its ions, EMF is produced which is known as electrode potential, which is due to transfer of electrons between element and its ions. Transfer of electrons between element and its ions, depends on the nature of element which can be represented by the following equation (in case of metal): 0 Solution pressure M Mn+ + n e Ionic pressure According to the nature of the metal, it may has: Tendency to loose electrons and converted to its ions, i.e. metal has high solution pressure, e.g.. Znº, Feº, Coº, Niº Tendency of metal ion to accept electrons and converted to its metal, i.e. metal has high ionic pressure, e.g. Cuº, Hgº, Agº, Ptº. Examples: a- Assume a metal of high solution pressure (e.g. Znº) is immersed in solution of its ions (e.g. Zn2+), the following equilibrium will be established: 0 2+ Zn Zn + 2 e The electrode potential developed in this case has a negative sign (i.e. the sign of potential is similar to the charge on metal electrode).  b- Assume a metal of high ionic pressure (e.g. Cuº) is immersed in solution of its ions (e.g. Cu2+), the following equilibrium will be established: Cu 0 Cu2+ + 2 e The electrode potential developed in this case has a positive sign. On addition of Znº to a solution of Cu2+ salt the following equilibriums are established: Zn 0 Zn2+ + 2 e 2 e + Cu 2+ Cu0 0 Zn + Cu 2+ Zn2+ + Cu0  mV e- → e- → e- → Zn0 Cu0 Zn2+ Cu2+ Zn0 → Zn2+ + 2e- Cu2+ + 2e- → Cu0 Znº displace Cu2+ in solution. i.e. M/Mn+ of negative potential displace that of positive potential. Calculation of electrode potential Electrode potential could be calculated from Nernst equation of electrode potential: a- In case of metal/ metal ion system (Mº/Mn+) E25 o 0.0591 n+ C= E0 + log [M ] n E 25 ºC is the electrode potential at 25 ºC. E0 is standard electrode potential (constant). n is the number of electrons gained by Mn+ or lost from Mº. [Mn+ ] is the molar concentration of metal ions.  b- In case of non-metal/ non-metal ion system (Mº/Mn-) 0.0591 n- E25 o C= E0 - log [M ] n E 25 ºC is the electrode potential at 25 ºC. E0 is standard electrode potential (constant). n is the number of electrons gained by non-metal or lost from non-metal ion. [Mn- ] is the molar concentration of non-metal ions. So, electrode potential is related to [Mn+] or [Mn-]. In case of Mº/Mn+ system, electrode potential is directly proportional to [Mn+] and is inversely proportional to [Mn-] in Mº/Mn-. Standard electrode potential (E0) ▪ When element is immersed in a molar solution of its ions, electrode potential produced is known as standard electrode potential. ▪ It is quantitative measure of readiness to loose electron in case of metal or gaining of electrons in case of non-metal to give their ions. ▪ According to standard electrode potential Mº/Mn+ and Mº/Mn- systems are arranged in what is called electrochemical series. ▪ Mº/Mn+ of negative potential displace that of positive potential, Mº/Mn- of positive potential displace that of less positive potential. Examples: Zn0/Zn2+ (E0= -0.76 volts) # Hg0/Hg2+ (E0= + 0.79 volts) Zn + Hg0 2+ Zn2+ + Hg0 0 2+ 0 2+ Fe /Fe (E0= - 0.44 volts) # Cu /Cu (E0= + 0.34 volts) 0 Fe + Cu 2+ Fe2+ + Cu0 - - Cl2/2 Cl (E0= + 1.36 volts) # I2/ 2 I (E0= + 0.54 volts) Cl2 + 2 I - 2 Cl - + I2 Measurement of standard oxidation potential or standard electrode potential: This can be achieved by connecting with reference electrode normal hydrogen electrode (N.H.E.) which has potential equals zero as represented by the following diagram: ▪ Notes ▪ During redox reactions, that system of + ve potential oxidizes that of –ve potential or that of less + ve one. ▪ For quantitative redox reaction, the difference in standard oxidation potential between two systems involved in the reaction must be at least 0.4 volt. ▪ If the difference is less than 0.4 volt, the reaction is reversible (not quantitative). ▪ In this case we have to increase the difference to be 0.4 volt to shift the reaction forward. This can be achieved by increasing oxidation potential of the system entering the reaction by its oxidant or decreasing oxidation potential system entering the reaction by its reductant. This can be achieved by increasing or decreasing the ratio of [ox]/[red] for that system.  Factors affecting oxidation potential 1- Common ion effect ▪ The oxidation potential of MnO4-/ Mn2+ system, varies with the ratio of MnO4-/ Mn2+. If Fe2+ as ferric chloride is titrated with KMnO4. unless the oxidation potential of MnO4-/ Mn2+ is reduced Cl- will be attacked by KMnO4, leading to higher result. ▪ However, decreasing oxidation potential of MnO4-/ Mn2+ decreases the difference between oxidation potential of MnO4-/ Mn2+ and also oxidation potential of Fe3+/Fe2+. ▪ Therefore oxidation potential of Fe3+/Fe2+ must also be decreased, to facilitate the quantitative determination of Fe2+ as ferric chloride with KMnO4. This can be achieved by addition of Zimmermann’s reagent: MnO4-/ Mn2+ (E = + 1.52 V). Cl2/Cl- (E = + 1.36 V). Zimmermann’s reagent composition 1. MnSO4 (50 gm): reduce E MnO4-/ Mn2+ by common ion effect. 2. H3PO4 (100 ml): reduce E Fe3+/Fe2+ through formation [Fe (PO4)2]3- complex. 3. Conc. H2SO4 (100 ml cooled): used for acidification. 2- Complexing agent During titration of Fe2+ with I2, the reaction is reversible as the difference between oxidation potential of Fe3+/Fe2+ (E = + 0.77 v) and I2/I- (E = + 0.535 v) is less than 0.4 volt. - 3+ 2+ 2 Fe + I 2 I + 2 Fe 2 To increase oxidation potential of I2/I- 2 I- 2 e + I2 0.0591 [I2] E - = E0 + log I2/2I n [I - ]2 Addition of HgCl2 forms HgI42- complex, decreasing [I-] leading to increase of [I2] /[I-] thus oxidation potential of I2/I- is increased. To decrease oxidation potential of Fe3+/Fe2+ Fe3+ + e Fe2+ 3+ 0.0591 [Fe ] E = E0 + log 3+ 2+ 2+ Fe /Fe n [Fe ] ▪ Addition of F- or PO43- form complex with Fe3+ as [FeF6]3- or [Fe (PO4)2]3-, thus the value of [Fe3+] /[ Fe2+] will be decreased and hence the oxidation potential of Fe3+/Fe2+. ▪ So, to facilitate the quantitative oxidation of Fe2+ to Fe3+, HgCl2, F- or PO43- must be added. 3- Precipitating agent ▪ The presence of precipitating agent will affect the oxidation potential of a redox system through precipitation of reduced or oxidized form. ▪ e.g. the addition of Zn2+ salt to [Fe (CN)6]3- / [Fe (CN)6]4- , will precipitate [Fe (CN)6]4- , thus the value of [Fe (CN)6]3- / [Fe (CN)6]4- will be increased and consequently the oxidation potential of [Fe (CN)6]3- / [Fe (CN)6]4- will be increased. 4- Effect of hydrogen ion concentration Hydrogen ion concentration has strong effect on the oxidation potential of oxidizing agents containing oxygen, where the oxidation potential increases by increasing acidity and decreases by decreasing it. Example: 1- KMnO4 reacts as oxidant at different pH values, where oxidation potential of KMnO4 varies with variation in hydrogen ion concentration: a- KMnO4 in acidic medium (MnO4-/Mn2+): 5 e + MnO4- + 8 H+ Mn2+ + 4 H2O E = E0 + 0.0591 [MnO4- ] [H+]8 log n [Mn2+] b- KMnO4 in slightly alkaline medium (MnO4-/MnO2) 3 e + MnO4- + 4 H+ MnO2+ 2 H2O E = E0 + 0.0591 [MnO4- ] [H+]4 log n [MnO2] 2- K2Cr2O7 in acidic medium (Cr2O72-/2 Cr3+) Cr2O72- + 6 e + 14 H + 2 Cr3+ + 7 H2O E = E0 + 0.0591 [Cr2O72- ] [H+]14 log n [Cr3+]2 3- In case of the reaction between ASO43-/ASO33- (E = + 0.56 volt) system and I2/ 2 I- (E = + 0.54 volt) ▪ The reaction is reversible as the difference between oxidation potential of the two system is less than 0.4. ▪ To shift the reaction (a) forward , oxidation potential of ASO43-/ASO33- should be increased by increasing hydrogen ion concentration by addition of conc. HCl (conc. H2SO4 or HNO3 can not be used as both of them can oxidize I- to I2). ▪ To shift the reaction (b) forward , oxidation potential of ASO43-/ASO33- should be decreased by decreasing hydrogen ion concentration by rendering the medium alkaline with sodium bicarbonate. a ASO43- - +2I +2H + ASO33- + I2 + H2O b  Redox titration curves These curves show the change in potential during the progress of titrations. Redox titration curve is the plot of potential against ml of titrant. E.g. titration of 100 ml of 0.1 N FeSO4 with standard 0.1 N Ce (SO4)2. Fe 2+ + Ce4+ Fe3+ + Ce3+ cerric cerrous The change in potential during titration can be calculated from Nernst equation. A- During titration: 1- Assume we add 10 ml of 0.1 N Ce4+ salts, the ratio of [Fe3+]/[Fe2+] will be 10/90 E 0.0591 [Fe3+] Fe3+ /Fe 2+ = E0 + log n [Fe2+ ] 0.0591 [Fe3+] E 3+ 2+ = E0 + log Fe /Fe n [Fe2+ ] 10 = E0 + 0.0591 log n 90 10 = 0.77 + 0.0591 log n 90 = 0.69 volt 2- After addition of 50 ml of 0.1N Ce4+ salts, the ratio of [Fe3+]/[Fe2+] will be 50/50, E= 0.77 volt. 3- After addition of 90 ml of 0.1N Ce4+ salts, the ratio of [Fe3+]/[Fe2+] will be 90/10, E= 0.81 volt. 4- After addition of 99 ml of 0.1N Ce4+ salts, the ratio of [Fe3+]/[Fe2+] will be 99/1, E= 0.87 volt. 5- After addition of 99.9 ml of 0.1N Ce4+ salts, the ratio of [Fe3+]/[Fe2+] will be 99.9/ 0.1, E= 0.93 volt. B- At end point: The potential of the two system are equal, moreover, [Fe2+]= [Ce4+], [Fe3+]= [Ce3+]. 3+ 0.0591 [Fe ] E = E01 + log (1) 2+ n [Fe ] 0.0591 [Ce4+] E = E02 + log (2) n [Ce3+ ] If the two equation are added : 0.0591 [Fe3+] [Ce4+] 2E = E01 + E02 + log n [Fe2+ ] [Ce3+ ] 2 E = E01 + E02 E = 1.10  C- After addition of excess cerric sulphate: 1- Assume we add 0.1 ml of 0.1 N Ce4+ salts excess, the ratio of [Ce4+]/[Ce3+] will be 0.1/100 4+ [Ce ] E = E0 + 0.0591 log 2 n [Ce3+ ] 0.1 = E02 + 0.0591 log n 100 = 1.27 volt 2- After addition of 1 ml of 0.1N Ce4+ salts excess, , E= 1.33 volt. 3- After addition of 10 ml of 0.1N Ce4+ salts, E= 1.39 volt. It is clear that, there is a sudden change (inflection) in E at the end point. Inflection in potential at the end point depends on: 1. Difference in the standard potentials of the two oxidation-reduction systems that are involved. 2. Complexing agent or precipitating agent. 3. Dilution. 4. The ratio of oxidant/ reductant of the two systems.  Detection of the end point 1- No indicator method ▪This is well illustrated by potassium permanganate, one drop of which will impart a visible pink coloration to several hundred milliliters of solution, even in the presence of slightly colored ions, such as iron(III). ▪The colors of cerium(IV) sulfate and of iodine solutions have also been employed in the detection of end points, but the color change is not so marked as for potassium permanganate. ▪This method has the drawback that an excess of oxidizing agent is always present at the end point. For work of the highest accuracy, blank test must be done. 2- External indicator method It depend on carrying out spot test with indicator outside the flask by removing one drop of titrated solution and test for complete reaction by the presence of the first excess of the titrant or complete absence of the sample. E.g. during titration of Fe2+ with Cr2O72-: ▪ The first excess of the titrant is tested for using diphenyl carbazide, which oxidized to the red diphenyl carbazone by the first excess of dichromate. ▪ The complete absence Fe2+ sample is test for using ferricyanide [Fe (CN)6]3-, which give blue ppt. with Fe2+, when all Fe2+ is oxidized to Fe3+ will give brown color of Fe [Fe (CN)6] indicating the end point. 2- 2+ + Cr2O7 + 6 Fe + 14 H 6 Fe3+ + 2Cr3+ + 6 H2O  Also titration of Zn2+ with K4[Fe(CN)6] solution: where uranyl acetate or uranyl nitrate solution is used as external indicator. The end point is obtained when a drop of the solution just imparts a brown color to the indicator. 3- Internal indicator method 1. They are highly colored substances that may be reversibly oxidized or reduced and exhibits different colors in the oxidized and reduced forms. 2. Indicator potential must be intermediate between sample potential and titrant potential. 3. Each redox indicator changes color over a certain potential range. This potential range must be small so that the indicator error in titration should be small as possible. 4. Indicator must undergo reversible oxidation reduction. In OX + n e Inred 5. Color change of the redox indicator may be detected when the ratio of [In OX]/ [ Inred] not more than 10/1 or 1/10 for the reduced form, hence: 0.0591 1 10 E = E0 + log or n 10 1 E = E0 -- + 0.0591 n Examples of internal redox indicator: a- Diphenylamine ▪ It is redox indicator (E0 = 0.76 volt and n =2), using the above equation, the range of diphenylamine is 0.73 - 0.79 V. ▪ At potential below 0.73 V , the color of the reduced form appear (colorless). At potential 0.79 V or more the color of the oxidized form predominates (blue violet). ▪ Between 0.73 - 0.79 V, the color of the solution change gradually from colorless to blue-violet. It is used as sodium diphenylaminesulphonate or a solution of diphenylbenzidine in concentrated sulfuric acid. ▪ Diphenylamine is used during titration of Fe2+ with Cr2O72-, E of Fe3+/ Fe2+ = 0.77 volt, E of Cr2O72-/ 2 Cr3+ = 1.36 volt. ▪ However, E of diphenylamine = 0.76 volt is not intermediate between the potential of the two system, therefore, we have to add PO43- or F- to lower E of Fe3+/ Fe2+ for rendering E of the indicator intermediate.  HN Oxid NH NH + 2 H+ + 2 e Red Diphenylamine I Diphenylamine II (colorless) (colorless) (diphenylbenzidine) Red Oxid N N + 2 H+ + 2 e Diphenylamine III (violet) However, N-phenylanthranilic acid which has E0 = 1.08 volt, therefore it can be used during titration of Fe2+ with Cr2O72- without the addition of PO43- or F-. b- 1, 10 orthophenanthroline (Ferroin) ▪ One of the best oxidation-reduction indicators is the 1,10- phenanthroline-iron (II) complex. ▪ The base 1,10-phenanthroline combines readily in solution with iron (II) salts in the molecular ratio 3 base: 1 iron (II) ion forming the intensely red l,10-phenanthroline-iron(II) complex ion. ▪ with strong oxidizing agents, the iron(III) complex ion is formed, which has a pale blue color. ▪ The standard redox potential is 1.06 volts. It is used during titration of Fe2+ with Ce4+ salt, E of Fe3+/ Fe2+ = 0.77 volt, E of Ce4+/Ce3+ = 1.45, so the indicator has intermediate potential between the two reacting system. ▪ The only disadvantage of this indicator is some what expensive , so , it can be replaced by the less costly N- phenylanthranilic acid or α, ά dipyridyl. ▪ They form similar complex with ferrous and ferric iron showing similar indicator properties. 3+ 2+ N N N oxid Fe Fe 3 2+ + Fe red N N N 3 3 Ferroin Ferrin intense red bale blue  General application Any element exist in more than one valence (oxidation no.) can be determined through redox reactions. Substances to be determined by redox reactions must fulfill the following requirement: 1. The solution of the substance to be analyzed must be free from other reducing or oxidizing agents which may interfere during the reaction, therefore some separation are essential before titration. 2. The substance to be analyzed must be in appropriate oxidation state, e.g. iron must be present as either Fe2+ or Fe3+. 3. To adjust the oxidation state of the substance to be analyzed, some reducing agents (pre-reductant) or oxidizing agent (pre-oxidant) are used, which must be completely removed before titration. Examples of pre-reductants and pre-oxidants: Pre-reductants: - Sulfite (SO32-): Excess is removed by boiling. - Stannous Chloride (SnCl2): Excess is removed by precipitation with HgCl2. - Metallic zinc (Znº): Excess is removed by filtration. Pre-oxidants: - Persulphate (S2O82-): Excess is removed by boiling. - Hydrogen peroxide (H2O2): Excess is removed by boiling in acid medium. - Iodate (IO4-): Excess is removed by precipitation with Hg2+. - Chlorate (ClO3-): Excess is removed by boiling with HCl. - Nitric acid (HNO3): excess is removed by evaporation.  I- Oxidation with potassium permanganate KMnO4 is valuable and powerful oxidizing agent. Properties of KMnO4 : 1- Purity: KMnO4 is impure substances as it contain MnO2. 2- Stability: KMnO4 is decomposed on standing according to the following equation: This is reaction is catalyzed by acids, light, MnO2 therefore, we have to avoid the preparation of KMnO4 in acidic solution, in addition we have to filter KMnO4 solution to remove MnO2 and store KMnO4 solution in dark colored bottle for protection from light. So, from the previously mentioned KMnO4 is not primary standard titrant and therefore must be standardized. 3- Preparation: 1. Dissolve potassium permanganate in distilled water. 2. Boil gently for 15-30 minutes (to destroy any oxidisable impurities in water) and allow the solution to cool to the laboratory temperature. 3. Filter the solution through non-reducing filter (e.g. a funnel containing a plug of purified glass wool, or through a sintered-glass ). 4. Then the resulting filtrate should be standardized against primary standard. 4- Standardization: a- Standardization against iron wire: Dissolve iron wire in sulfuric acid where equivalent amount of FeSO4is produced. Titrate FeSO4 solution against prepared KMnO4 solution till first pink. b- Standardization against sodium oxalate: Acidified solution of sodium oxalate solution with dilute sulphuric acid was heated to 70 ºC, and titrated with the permanganate solution slowly and with constant stirring until the first permanent faint pink color was obtained. this reaction is auto-catalyzed by the obtained Mn2+. 70 0C 10 CO2 + 2 Mn2+ + 8 H2O 2 MnO4- + 5 C2O4 2- + + 16 H c- Standardization against arsenic(III) oxide: This procedure, which utilizes arsenic (III) oxide as a primary standard and potassium iodide or potassium iodate as a catalyst for the reaction. Dissolve As2O3 in NaOH, then acidify with HCl, add a drop of KI to catalyze the reaction, then titrate the solution with the prepared KMnO4 solution till first pink persists for 30 seconds. As2O3 + 4 OH - 2 HAsO32- + H2O 5 H3AsO3 + 2 MnO4- + 6 H+ 5 H3AsO4 + 2 Mn2+ + 3H2O 5- Interference with chloride (Cl-) Chloride ion interfere during titration with KMnO4 solution (e.g. titration of Fe2+ as FeCl2). To avoid this interference we have to add Zimmernann’s reagent. 6- In acid medium: KMnO4 is reduced to Mn2+: Substance to be titrated Oxidation product Oxalate CO2 Fe2+ Fe3+ [Fe (CN)6]4- [Fe (CN)6]3- As3+ As5+ Sb3+ Sb5+ H2O2 O2 SO2/SO32- SO42- NO2- NO3- ( In acid medium KMnO4 may be reduced to Mn3+ , however this occur in the presence of fluoride or PO43- to form stable complex with Mn3+. 7- In alkaline medium: KMnO4 is reduced to MnO2 ( in moderately alkaline medium) and reduced to MnO42- ( manganate) this occur in strongly alkaline medium and in presence of Ba2+ to form insoluble barium manganate.  Examples of samples can be determined using KMnO4 as titrant I- Determination of iron compounds 1- Determination of Fe2+ salt Acidified solution of Fe2+ salt is titrated against standard solution of KMnO4 , till first pink color. 2 MnO4- 2+ + + 5 Fe + 8 H 5 Fe3+ + 2 Mn2+ + 4H2O in case of FeCl2 , we have to add Zimmermann’s reagent To avoid interference of chloride. 2- Determination of Fe3+ salt ▪ Ferric ion can be determined by direct titration with standard reducing agent such as titanous chloride solution using methylene blue as indicator which at end point reduced to leuco-compound (colorless) by the first excess of titrant. Also, SCN- can be used as indicator. Fe3+ + Ti3+ Fe 2+ + Ti4+ ▪ Fe3+ salt can be determined after reduction to Fe2+ using pre-reductant, then equivalent amount of Fe2+ salt is titrated against standard solution of KMnO4. a- Reduction of Fe3+ to Fe2+ with SnCl2 ▪ Concentrated stannous chloride solution is added drop wise with continuous stirring to hot solution of ferric salt (in 5-6 N HCl) until disappearance of the yellow color of Fe3+ ions. ▪ Excess reducing agent is removed by adding saturated HgCl2 solution were Hg2Cl2 is precipitated as a silky white PPT. ▪ If too much SnCl2 is added , heavy ppt. of Hg2Cl2 is produced and another sample must be taken. 2 FeCl3 + SnCl2 2 FeCl2 + SnCl4 SnCl2+ 2 HgCl2 SnCl 4 + Hg2Cl2 b- Reduction with Zn metal and sulphuric acid ▪ Complete reduction is tested by SCN- (spot test) , after complete reduction , excess hydrogen is expelled by heating and excess Zn is removed by filtration through glass wool. After quantitative reduction of ferric to ferrous, the resulting ferrous is titrated with standard KMnO4 till first pink. ▪ Reduction is done in flask fitted with a bunzene valve to avoid atmospheric oxidation of ferrous to ferric. 0 Zn + 2 H + Zn2+ + 2[H] 3+ 2+ + Fe + [H] Fe +H c- Reduction with amalgamated Zn ▪ Amalgamated zinc is an excellent reducing agent for ferric ion. ▪ Zinc reacts rather slowly with acids, but upon treatment with a dilute solution of a mercury (II) salt, Zn metal is covered with a thin layer of mercury; the amalgamated Zn reacts quite readily. ▪ Reduction with amalgamated zinc is usually carried out in reductor known as Jones reductor, this done by passing acidic solution of ferric ion through column packed with amalgamated Znº. 0 Zn + HgCl2 ZnCl2 + Hg0 3+ 0 2+ 2+ 2 Fe + Zn 2 Fe + Zn Jones reductor N.B. ▪ From the previously mentioned , reducing substance such as Zn powder or Fe metal can be determined by adding known excess Fe3+ salt where equivalent amount of Fe2+ is produced which is then titrated with standard solution of KMnO4. Zinc oxide and iron oxide do not interfere. ▪ Oxidizing substances capable of oxidizing Fe2+ to Fe3+ are allowed to react with known excess of standard ferrous sulphate solution. The residual Fe2+ can be titrated with standard solution of KMnO4. e.g. persulphate (S2O82-), chlorate (ClO3-) and manganese dioxide (MnO2). 2- 3+ S2O82- + 2 Fe2+ 2 SO4 + 2 Fe - 3+ ClO3- + 6 Fe2+ + 6 H+ Cl + 6 Fe + 3H2O 2+ MnO2 + 2 Fe + 4 H + Mn2+ + 2 Fe3+ + 2 H2O 3- Determination of Fe3+/Fe2+ mixture 1. For Fe2+: Titrate the acidified mixture with standard solution of KMnO4. 2. For Total : reduce Fe3+ to Fe2+ by SnCl2, titrate the total mixture in the form of ferrous with standard solution of KMnO4 in presence of Zimmermann’s reagent. By difference Fe3+ can be calculated. 4- Determination of metallic iron Iron metal can be treated with CuSO4 or HgCl2, where Feº displaces Cu2+ or Hg2+ producing equivalent amount of Fe2+ which can be titrated with standard solution of KMnO4 in presence of Zimmermann’s reagent, if HgCl2. Fe0 + CuSO4 Cu0 + FeSO4 Fe/Fe2+ = - 0.44 Cu/Cu2+ = + 0.34 Fe0 + HgCl2 Hg0 + FeSO FeCl2 4 Hg/Hg2+ = + 0.79 Also, iron can be dissolved in H2SO4 producing an equivalent amount of FeSO4 which can be titrated with standard solution of KMnO4. 5- Determination of Feº/Fe2O3 mixture 1. For Feº: sample mixture is added to hot solution of CuSO4 where Feº displace Cu2+producing equivalent amount of FeSO4, filter and titrate FeSO4 with standard solution of KMnO4. 2. For Fe2O3: Dissolve the residue in H2SO4, where Fe2O3 is converted to Fe2(SO4)3 which, is reduced to FeSO4 and titrate FeSO4 with standard solution of KMnO4. 6- Determination of ferrocyanide [Fe (CN)6]4- Acidified solution of ferrocyanide is determined by oxidation with KMnO4 till the first yellowish brown color according to the following equation: MnO4- + 5 [Fe(CN)6]4-+ 8 H+ 5 [Fe(CN)6]3- + Mn2+ + 4H2O 7- Determination of ferricyanide ([Fe (CN)6]3-) First ferricyanide is reduced to ferrocyanide with sodium peroxide (excess of sodium peroxide is removed by boiling), then acidify the solution and titrate with standard KMnO4 solution. 2 [Fe(CN)6]3- + Na2O2 [Fe(CN)6]4- + 2 Na+ + O2 MnO4- + 5 [Fe(CN)6]4-+ 8 H+ 5 [Fe(CN)6]3- + Mn2+ + 4H2O 8- Determination of ferricyanide / ferrocyanide mixture 1. For [Fe (CN)6]4- : titrate acidified mixture with with standard KMnO4 solution. 2. Total: Reduce [Fe (CN)6]3- to [Fe (CN)6]4- using sodium peroxide (excess of sodium peroxide is removed by boiling), then acidify the solution and titrate with standard KMnO4 solution. By difference [Fe (CN)6]3- can be calculated. 9- Determination of Hydrogen peroxide ▪ Hydrogen peroxide is usually encountered in the form of an aqueous solution containing about 6 %, 12 % or 30 % hydrogen peroxide, and referred to as '20-volume', '40-volume', and 100-volume' hydrogen peroxide respectively; Thus 1 mL of '100-volume' hydrogen peroxide will yield 100 mL of oxygen measured at standard temperature and pressure. ▪ The following reaction occurs when potassium permanganate solution is added to hydrogen peroxide solution acidified with dilute sulphuric acid: ▪ The equation shows that in this reaction H2O2 act as a reducing agent and is oxidized to oxygen. H2O2 + 2 MnO4- +6H + 5 O2 + 2 Mn++ + 8 H2O + H2O2 - e O2 + 2 H 10- Determination of Oxalates a- Soluble oxalates: Oxalic acid and oxalates are strong reducing agent , can be titrated with standard KMnO4 solution in presence of dilute sulphuric acid at 70 ºC to catalyze the reaction till first pink. The reaction is slow at the beginning but once small amount of Mn2+ is formed the reaction become very rapid. 0 70 C 5C2O4 2- +2 MnO4 - + + 16 H 10 CO2 + 2 Mn++ + 8 H2O Oxalate or 0 70 C 5 H2C2O4 + 2 MnO4- + 6 H+ 10 CO2 + 2 Mn++ + 8 H2O Oxalic acid b- Cations forming insoluble oxalates: Substances which form PPT with oxalates (e.g. Pb2+, Ca2+, Ba2+, Sr2+, Mg2+) can be determined by precipitation them quantitatively with oxalic acid and either: 1- The washed ppt. is transferred quantitatively to a beaker, dilute H2SO4 is added and titration is done standard KMnO4 solution at 70 ºC. 2- The residual oxalic acid in the filtrate and washing is back titrated with standard KMnO4 solution at 70 ºC. The method is used for the determination of calcium salts and litharge (PbO) which is dissolved in glacial acetic acid and diluted with water and then precipitated as oxalate. Oxidizing substances capable of oxidizing oxalates are determined by treating them with measured excess of oxalate and the excess oxalate is back titrated with standard KMnO4 solution at 70 ºC. e.g. persulphate (S2O82-), chlorate (ClO3-), nitrate (NO3-) and manganese dioxide (MnO2). Care should be taken to avoid strong heating as oxalic acid decomposed by heating ClO3- + 3C2O42-+ 6H + Cl- + 6 CO2 + 3H2O 2 NO3- + 3C2O42-+ 8 H+ 2 NO + 6 CO2 + 4H2O Ag + H2S2O8 + H2C2O4 2 H2SO4+ 2 CO2 MnO2 + C2O42-+ 4H + Mn2+ + 2 CO2+ 2 H2O  II- Oxidation with potassium dichromate Potassium dichromate is powerful oxidizing agent as potassium permanganate but it has several advantages over potassium permanganate: 1- It can be obtained in pure form. 2- It is stable up to its fusion point, and it is therefore an excellent primary standard. Standard solutions of exactly known concentration can be prepared by weighing out the pure dry salt and dissolving it in the proper volume of water. Furthermore, the aqueous solutions are stable if protected from evaporation. 3- Potassium dichromate is used only in acid solution, and is reduced rapidly at the ordinary temperature to a green chromium (III) salt. 4- It can not oxidize hydrochloric acid, potassium dichromate is therefore of particular value in the determination of ferrous ion. It can not serve as self indicator in this determination, as the orange dichromate is reduced to the green chromium ion, but end point can be detected using diphenylamine as indicator.  III- Oxidation with Ceric (Ce4+) ▪ Cerium (IV) sulphate is a powerful oxidizing agent. It can be used only in acidic solution, to avoid precipitation of [CeO2. x H2O] due to hydrolysis of cerric salt in neutral or alkaline medium. ▪ Oxidation potential of Ce4+/Ce3+ system depends on the acid used, it has highest potential in the presence of HClO4. ▪ Chloride does not interfere with Ce4+ because of the slow rate of oxidation of Cl-. ▪ Ammonium hexanitro cerate (NH4)2Ce(NO3)6 is primary standard chemical due to its high purity and stability. ▪ The solution has intense yellow color and in hot solutions which are not too dilute the end point may be detected without an indicator Applications: Determination of Fe2+ salt ▪ Acidified solution of Fe2+ salt is titrated with standard ceric sulphate in presence of 1,10 orthophenanthrolene or phenylanthranilic acid. ▪ Also, methyl red can be used as irreversible redox indicator, in this case titration with cerric sulphate is continued until the red color of the indicator disappears. ▪ During titration of Fe2+ salt with standard ceric sulphate, ceric sulphate can not be used as self indicator due to the yellow color of the produced Fe3+ salt. Fe 2+ + Ce4+ Fe3+ + Ce3+ cerric cerrous