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Formal potential analytical technique, P K MANI

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Formal potential analytical technique

Formal potential analytical technique

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  • 1. Formal potential Pabitra Kumar Mani , Assoc. Prof., Ph.D. ACSS, BCKV
  • 2. Standard potentials Eo are evaluated with full regard to activity effects and with all ions present in simple form: they are really limiting or ideal values and are rarely observed in a potentiometric measurement. In practice, the solutions may be quite concentrated and frequently contain other electrolytes; under these conditions the activities of the pertinent species are much smaller than the concentrations, and consequently the use of the latter may lead to unreliable conclusions. Also, the actual active species present (see example below) may differ from those to which the ideal standard potentials apply. For these reasons 'formal potentials' have been proposed to supplement standard potentials
  • 3. Formal Potential: The concept of formal potential was introduced by Swift and he defined it as the experimentally determined potential of a solution containing both the oxidised and reduced species each at a concentration of 1 formal together with other substances at a given concentration. 1F ≡ 1g formula weight per litre solution The std. Potential or E0 (normal potential) is a limiting or ideal value assuming that the reacting species are in their simple forms. Since in practice such ideal solutions are rarely found. The term formal potential has been proposed and found to be more practical one. [ Fe +3 ] Where the ionic sp. are aqueous forms 0.0591 E = E0 log [ Fe +2 ] 1 Fe+3 aq or Fe+3(H2O)6 ; Fe+2 aq or Fe+2 (H2O) simple form of Fe+3 or Fe+2
  • 4. In IM HCl soln, however part of Fe+3 exists as various sp.like FeCl+2, FeCl2+, FeCl4-2 (all in Feric state). Similarly Fe+2 as FeCl+, FeCl4-2 etc. Let, α be the fraction of total Fe+3 irons, β is the fraction of total Fe+2 iron existing in simple aqueous form 0 E = E 0 - 0.0591log α . CFe III .f Fe III aq β .CFe II .f Fe + 2aq α 0 = E - 0.0591 log - 0.0591log β 0/ = E - 0.0591 log C Fe III C Fe II f f Fe III aq E : normal potential i.e., std for ionic strength of the soln, where CFeIII and CFe+2 are the molar concentration of total Fe+3 and Fe+2 iron present in the soln.,f: activity coefficient − 0.0591 log Fe + 2 aq E 0/ = Formal C Fe III C potential Fe II
  • 5. Where Fe+3 forms a more stable complex in comparison to the Fe+2 by interaction with the anions or any comlexing reagent present in the soln. The value of the ratio α/β is less than 1 and in that case the value of Formal potential becomes more +ve which means the Fe+2 behaves as stronger reducing agent. PO4-3 is such an anion in presence of which Fe+2 is a stronger reducing agent. Medium E0/ 1M HClO4 -0.73 V 1M HCl -0.77V 1M H2SO4 -0.68V E0 : Fe+2/Fe+3 = -0.77V 0.5 M H3PO4 + 1M H2SO4 -0.61 V E0/ Diphenyl amine = -0.76V in (0.5M H3PO4 + 1M H2SO4) Minm 0.15 V difft of the redox potential value for making a titration accurate
  • 6. II III Fe(CN)6 ⇋ Fe(CN)6-3 + e4- 2I- = I2 + 2e- E0 = -0.36 V ; E0= -0.54 Volt I2 + 2Fe(CN)6-4 ⇋ 2I- +2Fe(CN)6-3 It can be seen that I2 should normally oxidise ferrocyanide to Ferricyanide. But in 1M HCl medium or 1M H2SO4 medium II III E0/ [(H2Fe(CN)6-2 /H2Fe(CN)6-1 ] = -0.71 V ferro 2[Fe(CN)6] α/β > 1 ferri -3 + 2I- ⇋ I2 + 2Fe(CN)6-4 Hence in 1M HCl medium Ferricyanide liberates I2 from I-. Similarly when Zn+2 is present in the soln it reacts with Ferrocyanide to form K2Zn3[Fe(CN)6]2 + +2 4- 2K + 3Zn + [Fe(CN)6] → K2Zn3[Fe(CN)6]2 And so in presence of Zn+2 sufficient K+Ferricyanide liberates I2 from KI and the rn can be used for indirect iodometric determination determination of Zn+2 in soln.
  • 7. Red Ox indicators InRed ⇋ InOx +ne- EIn [ InRed and InOx must have different colours ] In Ox 0.0591 = E In log [ In Red ] n 0 In order that indicator can be used in a redox titration the 0.0591 potential of the analyte soln must lies near 0 E Criteria for using a redox indicator In ± n 1. The indicator potential must be intermediate between that of the analyte and titrant system 2. At eq. point redox potential of the soln should change from 0.0591 to 0 0.0591 with sharp detectable E 0 In + E In − n n colour change 3.E0In or better Eo/In should differ by at least 0.15 V from the formal potential of the redox couple titrated
  • 8. Classification of RedOx Indicator (I) Metal Complexes [ MLx ]+p ⇋ [MLx ]+(p+n) InRed + ne- InOx Where L is a ligand, M= Metal , the two forms must have difft colour. e.g. [Fe-(Opn)3]+3 [Fe-(O-pn)3]+2 , Red colour pale blue less stable Ferroin Indicator E0In = -1.14 V E0/In (1MHCl) = -1.06 V As the red colour is more intense the colour change is actually observed at E0In value of -1.12 V
  • 9. Introducing various substitutes in the phenanthroline ring indicator potential value can be changed. Thus, with 4,7 dimethyl Ferroin , E0/In (1MHCl) = -0.88 V which is very suitable for titration of chromate or dichromate by Fe+2 soln. (Mohr’s titrant soln) when NO2- group is introduced RedOx potential value is more –ve -1.25V (Nitroferroin) II . Organic molecules forming a Redox Couple, Diphenyl amine (knop, 1924) in H2SO4 soln (DPA, base and insoluble in water)
  • 10. Diphenylbenzidine (colorless) Diphenylamine (colorless) → [O] Diphenylbenzidine violet Yellow product Red-Ox potential of In (0.5-1M H2SO4 + 0.5M H3PO4 ) medium, Eo/In = - 0.76 V Suitable for titration of Fe+2 by dichromate soln, in presence of H3PO4 Eo/In = - 0.61 V
  • 11. Due to low solubility of free DPA it is replaced by its sulfonic acid derivative. Diphenylamine 4-sulphonic acid The sodium or Ba salt of which is used as indicator in the form of 0.2% aq soln. The formal potential o/ E In = - 0.85 V (0.5M H2SO4) So, even without using H3PO4, DPA sulfonate indicator can be used safely in the titration of Fe+2 by dichromate. The violet colour Diphenyl benzidine is oxidised further to a yellow or reddish yellow product and hence in the reverse titration of chromate or dichromate by Mohr’s salt soln, DPA is not convenient, N-methyl DPA sulfonic acid is ecellent for reverse titration Eo/In = - 0.90 V (0.5M H2SO4) CH3 N-PHENYL ANTHRANILIC ACID is best for reverse titration
  • 12. Another group of Organic compounds are used as redox indicator of which “ Variamine blue” is a versatile indicator VARIAMINE BLUE In the pH range 1.5-6.3 it is colourless in reducing medium and blue in Oxidising medium. Eo/In = - 0.6 V (pH= 1.5) to 0.36 V (pH =6.3) It has been used in the titration using Ascorbic acid(Vitamin C) as titrant BTB can be used as a Red-Ox indicator in the titration of AsIII, NH3, I- etc. With ClO- hypochlorite soln, Deep blue (Reducing) – greenish yellow (Oxidising medium)
  • 13. Miscellaneous Methylene Blue EoIn = - 0.52 V Starch/ I- EoIn = - 0.53 V
  • 14. Permanganate titration   Oxidation with permanganate : Reduction of permanaganate KMnO4 Powerful oxidant that the most widely used. In strongly acidic solutions (1M H2SO4 or HCl, pH ≤ 1) MnO4– + 8H+ + 5e = Mn2 + + 4H2 O violet color Eo = 1.51 V colorless manganous KMnO4 is a self-indicator. In feebly acidic, neutral, or alkaline solutions MnO4– + 4H+ + 3e = MnO2 (s) + 2H2 O Eo = 1.695 V brown manganese dioxide solid In very strongly alkaline solution (2M NaOH) MnO4– + e = MnO42 – green manganate Eo = 0.558 V
  • 15. Standardization of KMnO4 solution Potassium permanganate is not primary standard, because traces of MnO2 are invariably present. Standardization by titration of sodium oxalate (primary standard) : 2KMnO4 + 5 Na2(COO)2 + 8H2SO4 = 2MnSO4 + K2SO4 + 5Na2SO4 + 10 CO2 + 8H2O 2KMnO4 mw 158.03 158.03 g / 5 ≡ 5 Na2(COO)2 ≡ 10 Equivalent mw 134.01 ≡ 134.01 g / 2 ≡ 1 Eq. 31.606 g ≡ 67.005 g 1N × 1000 ml ≡ 67.005 g x N × V ml ag x N = ( a g × 1N × 1000 ml) / (67.005 g × V ml)
  • 16. Preparation of 0.1 N potassium permanganate solution KMnO4 is not pure. Distilled water contains traces of organic reducing substances which react slowly with permanganate to form hydrous managnese dioxide. Manganesse dioxide promotes the autodecomposition of permanganate. 1) Dissolve about 3.2 g of KMnO4 (mw=158.04) in 1000ml of water, heat the solution to boiling, and keep slightly below the boiling point for 1 hr. Alternatively , allow the solution to stand at room temperature for 2 or 3 days. 2) Filter the liquid through a sintered-glass filter crucible to remove solid MnO2. 3) Transfer the filtrate to a clean stoppered bottle freed from grease with cleaning mixture. 4) Protect the solution from evaporation, dust, and reducing vapors, and keep it in the dark or in diffuse light. Preserve it in amber –coloured glass bottle. 5) Standardise from time to time. If in time managanese dioxide settles out, refilter the solution and restandardize it.
  • 17. Applications of permanganometry (1) H2O2 2KMnO4 + 5 H2O2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5O2 + 8H2O (2) NaNO2 2NaNO2 + H2SO4 = Na2SO4 + HNO2 2KMnO4 + 5 HNO2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5HNO3 + 3H2O (3) FeSO4 2KMnO4 + 510 FeSO4 + 8H2SO4 = 2MnSO4 + K2SO4 + 5Fe2(SO4)3 + 8H2O (4) CaO CaO + 2HCl = CaCl2 + H2O CaCl2 + H2C2O4 = CaC2O4 + 2HCl (excess oxalic acid) 2KMnO4 + 5 H2C2O4 + 3H2SO4 = 2MnSO4 + K2SO4 + 10CO2 + 8H2O (back tit) (5) Calcium gluconate [CH2OH(CHOH)4COO]2Ca + 2HCl = CaCl + 2CH2OH9CHOH)4COOH (NH4)2C2O4 + CaCl2 = CaC2O4 + 2 NH4Cl CaCl2 + H2SO4 = H2C2O4 + CaSO4
  • 18. The concentrations of the various species must be taken into consideration especially if combined redox and acid-base systems are involved. From the data, taken from Table 1.17 and I.l8 for example: one could draw the conclusion that arsenate ions will oxidize iodide: but the reaction cannot go in the opposite direction. This in fact is true only if the solution is strongly acid (pH ~ 0). The oxidation-reduction potential of the arsenate-arsenite system depends on the pH: At pH = 6 the potential of a solution containing arsenate and arsenite ions at equal concentrations decreases to +0'20 V. Under such circumstances therefore the opposite reaction will occur:
  • 19. Starch-Iodine complex Starch solution(05~ 1%) is not redox indicator. The active fraction of starch is amylose, a polymer of the sugar α-D-glucose ( 1,4 bond). The polymer exists as a coiled helix into which small molecules can fit. In the presence of starch and I–, iodine molecules form long chains of I5– ions that occupy the center of the amylose helix. ••••[I I I I I]– ••••[I I I I I]– •••• Visible absorption by the I5– chain bound within the helix gives rise to the characteristic starch-iodine color.
  • 20. Structure of the repeating unit of the sugar amylose. Schematic structure of the starch-iodine complex. The amylose chain forms a helix around I6 unit. View down the starch helix, showing iodine, inside the helix.
  • 21. Starch-Iodine Complex • • • Starch is the indicator of choice for those procedures involving iodine because it forms an intense blue complex with iodine. Starch is not a redox indicator; it responds specifically to the presence of I2, not to a change in redox potential. The active fraction of starch is amylose, a polymer of the sugar α-d-glucose. In the presence of starch, iodine forms I6 chains inside the amylose helix and the color turns dark blue
  • 22. 16-7 Methods Involving Iodine • Iodimetry: a reducing analyte is titrated directly with iodine (to produce I−). • iodometry, an oxidizing analyte is added to excess I− to produce iodine, which is then titrated with standard thiosulfate solution. • Iodine only dissolves slightly in water. Its solubility is enhanced by interacting with I- • A typical 0.05 M solution of I2 for titrations is prepared by dissolving 0.12 mol of KI plus 0.05 mol of I2 in 1 L of water. When we speak of using iodine as a titrant, we almost always mean that we are using a solution of I2 plus excess I−.
  • 23. Preparation and Standardization of Solutions • Acidic solutions of I3- are unstable because the excess I− is slowly oxidized by air: • In neutral solutions, oxidation is insignificant in the absence of heat, light, and metal ions. At pH ≳ 11, triiodide disproportionates to hypoiodous acid (HOI), iodate, and iodide. • An excellent way to prepare standard I3- is to add a weighed quantity of potassium iodate to a small excess of KI. Then add excess strong acid (giving pH ≈ 1) to produce I2 by quantitative reverse disproportionation:
  • 24. pabitramani@gmail.com http://www.bckv.edu.in