20 Applications of Oxidation/Reduction Titrations

20 Applications of Oxidation/Reduction Titrations 20A AUXILIARY OXIDIZING AND REDUCING REAGENTS 20A1 Auxiliary Reducing Reagents Table 201 Uses of the Walden Reductor and the Jones Reductor Jones Walden Ag(s) + Cl AgCl(s) + e Zn(Hg)(s) Zn 2+ + Hg + 2e Fe 3+ + e Fe 2+ Fe 3+ + e Fe 2+ Cu 2+ + e Cu + Cu 2+ + 2e Cu(s) H 2 MoO 4 + 2H + + e + MoO2 + 2H 2 O H 2 MoO 4 + 6H + + 3e Mo 3+ + 4H 2 O UO 2+ 2 + 4H + + 2e U 4+ + 2H 2 O UO 2+ 2 + 4H + + 2e U 4+ + 2H 2 O UO 2+ 2 + 4H + + 3e U 3+ + 2H 2 O V() + 4 + 2H + + e VO 2+ + 3H 2 O V() + 4 + 4H + + 3e V 2+ + 4H 2 O TiO 2+ not reduced TiO 2+ + 2H + + e Ti 3+ + 4H 2 O Cr 3+ not reduced Cr 3+ + e Cr 2+ 20A2 Auxiliary Oxidizing Reagents Sodium Bismuthate: Mn(II) MnO 4 NaBiO 3 (s) + 4H + + 2e BiO + + Na + + 2H 2 O Ammonium Peroxydisulfate, ammonium persulfate, (NH 4 ) 2 S 2 O 8 in acidic soln: Cr(III) dichromate Ce(III) Ce(IV) Mn(II) permanganate S 2 O 2 8 + 2e 2 2SO 4 Eº = 2.01 V The oxidations are catalyzed by traces of silver ion. The excess reagent is readily decomposed by a brief period of boiling: S 2 O 2 8 + 2H 2 O 4SO 2 4 + O 2 + 4H + Sodium Peroxide and Hydrogen Peroxide H 2 O 2 + 2H + + 2e 2H 2 O Eº = 1.78 V boiling: H 2 O 2 2H 2 O + 2O 2 (g) 20B APPLING STANDARD REDUCING AGENTS 20B1 Iron (II) Solutions iron(ii) ammonium sulfate, Fe(NH 4 ) 2 (SO 4 ) 2.6H 2 O (Mohr's salt) iron(ii) ethylenediamine sulfate, FeC 2 H 4 (NH 3 ) 2 (SO 4 ) 2.4H 2 O (Oesper's salt) Fig. 201 A Jones reductor. 137

Airoxidation of iron (II) takes place rapidly in neutral solutions but is inhibited in the presence of acids, with the most stable preparations being about 0.5 M in H 2 SO 4. oxidizing agents excess of standard Fe(II) standard soln of pot. dichromate or Ce(IV). Application: organic peroxides, hydroxylamine, Cr(VI), Ce(IV), Mo(VI), nitrate, chlorate, perchlorate and numerous other oxidants. 20B2 Sodium Thiosulfate iodine thiosulfate 2 2 2S 2 O 3 S 4 O 6 + 2e excess KI analyte/slightly acidic solution iodine standard solution of Na 2 S 2 O 3 ex: determination of sod. Hypochlorite in bleaches Ocl + 2I + 2H + Cl + I 2 + H 2 O (unmeasured excess KI) 2 I 2 + 2S 2 O 3 2I 2 + S 4 O 6 Detecting End Points in Iodine/Thiosulfate Titrations 1. disappearance of the iodine color 5 10 6 M I2 discernible color 2. starch indicator deep blue color Starch undergoes decomposition in solution with high I 2 concentration. In titration of excess I 2 with Na 2 S 2 O 3, addition of the indicator must be deferred until most of the (b) I 2 has been reduced. Fig. 202 Thousands of glucose molecules polymerize to form huge molecules of β amylose as shown in (a). Molecules of βamylose tend to assume a helical structure. The iodine species I 3 as shown in (b) is incorporated into the amylose helix. The Stability of Sodium Thiosulfate Solutions decompose: S 2 O 2 3 + H + HSO 3 + S(s) ph, microorganisms, concentration of the solution, presence of Cu(II) ion and exposure to sunlight. Standardizing Thiosulfate Solutions primary standard: pot. iodate/excess KI (pot. dichromate, pot. bromate, pot. hydrogen iodate, pot. ferricyanide and metallic copper)/ excess KI. IO 3 + 5I + 6H + 3I 2 + 2H 2 O 1 mol IO 2 3 = 3 mol I 2 = 6 mol S 2 O 3 thiosulfate 138

Ex. 201 A solution of Sod. thiosulfate was standardized by dissolving 0.1210 g KIO 3 (214.00 g/mol) in water, adding a large excess of KI, and acidifying with HCl. The liberated I 2 required 41.64 ml of the thiosulfate soln to decolorize the blue starch/iodine complex. Calculate the molarity of the Na 2 S 2 O 3. amount Na C Na 2S2O3 2 S 2 O 3 = 0.1210g KIO 3.3925 mmol = = 0.08147 M 41.64 ml 3 1mol 0.21400 g or 6 = 3.3925 mmol 121mg 3 2 214 mg/mmol = 0.08147M 41.64 ml Tab 202 Applications of Sodium Thiosulfate as Reductant Analyte HalfReaction Special Condition IO 4 IO 4 + 8H + + 7e ½ I 2 + 4H 2 O Acid solution IO 4 + 2H + + 2e IO 3 + H 2 O Neutral solution IO 3 IO 3 + 6H + + 5e ½ I 2 + 3H 2 O Strong acid BrO 3, ClO 3 XO 3 + 6H + + 6e X + 3H 2 O Strong acid Br 2, Cl 2 X 2 + 2I I 2 + 2X NO 2 HNO 2 + H + + e NO(g)+ H 2 O Cu 2+ Cu 2+ + I + e CuI(s) O 2 O 2 + 4Mn() 2 (s) + 2H 2 O 4Mn() 3 (s) Basic solution Mn() 3 (s) + 3H + + e Mn 2+ + 3H 2 O Acidic solution O 3 O 3 (g) + 2H + + 2e O 2 (g) + H 2 O Organic peroxide RO + 2H + + 2e R + H 2 O 20C APPLYING STANDARD OXIDING AGENTS Table 203 Some common oxidants used as standard solutions Reagent and Reduction Standard Standardized Formula product Potential, V with Indicator* Stability# KMnO 4 Mn 2+ 1.51 Na 2 C 2 O 4, Fe, As 2 O 3 MnO 4 (b) KBrO 3 Br 1.44 KBrO 3 (1) (a) Ce(IV), Ce 4+ Ce 3+ 1.44 Na 2 C 2 O 4, Fe, As 2 O 3 (2) (a) K 2 Cr 2 O 7 Cr 3+ 1.33 K 2 Cr 2 O 7, Fe (3) (a) I 2 I 0.536 BaS 2 O 3 H 2 O, Na 2 S 2 O 3 starch (c) * (1) αnaphthoflavone; (2) 1,10phenanthroline iron(ii) complex (ferroin); (3) diphenylamine sulfonic acid. # (a) Indefinitely stable; (b) moderately stable, requires periodic standardization; (c) somewhat unstable, require frequent standardization. Eº' in H 2 SO 4. 139

20C1 The Strong OxidantsPotassium Permanganate and Cerium(IV) MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O Eº = 1.51 V (in 0.1 M strong acid) Ce 4+ + e Ce 3+ Eº = 1.44 V (1M H 2 SO 4 ) = 1.70 V (1M HClO 4 ) = 1.61 V (1M HNO 3 ) Comparison of the Two Reagents Ce 4+ MnO 4 in sulfuric acid stable not oxidize Cl decompose slowly oxidize Cl HCl soln of analyte can be used cannot be used primarystandardgrade salt available selfindicator no color of MnO 4 cost (1L 0.02 M soln) $ 2.20 (4.40) $ 0.08 in < 0.1 M strong acid tendency to form ppt Detecting the End Points indicators: KMnO 4 solution intense purple color diphenylamine sulfonic acid 1, 10phenanthroline complex of Fe(II) 2MnO 4 + 3Mn 2+ + H 2 O 5MnO 2 (s) + 4H + K = 10 47 equilibrium [MnO 4 ] rate: slow end point fades only gradually over 30s. in Ce(IV) titration: indicator: Fe(II) complex of 1,10phenanthroline or one of its substitute derivatives (Table 203) C 12 H 8 N 2 + Fe 2+ Fe(C 12 H 8 N 2 ) 2+ 3+ 3 Fe(C 12 H 8 N 2 ) 3 + e Ferrous complex (Ferroin) red Ferric complex (Ferriin) weak blue The Preparation and Stability of Standard Solutions KMnO 4 soln: not entirely stable 4MnO 4 + 2H 2 O 4MnO 2 (s) + 3O 2 (g)+ 4 decomposition reaction is slow catalyzed by light, heat, acids, bases, Mn(II) and MnO 2. 140

Ex. 202 Described how you would prepare 2.0 L of an approximately 0.010 M soln of KMnO 4 (158.03 g/mol). KMnO 4 needed = 2.0 L 0.010 M 158.03 g/mol = 3.16 g Dissolve about 3.2 g of KMnO 4 in a little water. After solution is complete add water to bring the volume to about 2.0 L. Heat the solution to boiling for a brief period, and let stand until it is cool. Filter through a glassfiltering crucible and stored in a clean dark bottle. Analytically Useful Cerium(IV) Compounds Name Formula Molar Mass Cerium(IV) ammonium nitrate Ce(NO 3 ) 4 2NH 4 NO 3 548.2 Cerium(IV) ammonium sulfate Ce(SO 4 ) 2 2(NH 4 ) 2 SO 4 2H 2 O 632.6 Cerium(IV) hydroxide Ce() 4 208.1 Cerium(IV) hydrogen sulfate Ce(HSO 4 ) 4 528.4 Primary Standards Sodium Oxalate. 2MnO 4 + 5H 2 C 2 O 4 + 6H + Mn(II) as a catalyst (autocatalysis) 2Ce 4+ + H 2 C 2 O 4 2Ce 3+ + 2H + + 2CO 2 2Mn 2+ + 10CO 2 (g) + 8H 2 O Ex. 203 You wish to standardize the soln in Ex.202 against pure Na 2 C 2 O 4 (134.00 g/mol). If you want to use between 30 and 45 ml of the reagent for the standardization, what range of masses of the primary standard should you weigh out? for a 30mL titration: amount KMnO 4 = 30 ml 0.010 M = 0.30 mmol mass Na 2 C 2 O 4 = 0.30 mmol 5/2 0.134 = 0.101 g for a 45mL titration: mass Na 2 C 2 O 4 = 45 0.010 5/2 0.134 = 0.151 g Ex. 204 A 0.1278g sample of primarystandard Na 2 C 2 O 4 required exactly 33.31 ml of the KMnO 4 solution in Ex. 202 to reach the end point. What was the molarity of the KMnO 4 reagent? amount Na 2 C 2 O 4 = 0.1278 g 1 mmol/0.134 g = 0.95373 mmol CKMnO 4 = 0.95373 mmol (2/5) (1/33.31 = 0.01145 M 141

Using Potassium Permanganate and Cerium(IV) Solutions: Table 205 Ex. 205 Aqueous solution containing approximately 3% (w/w) H 2 O 2 are sold in drug stores as a disinfectant. propose a method for determining the peroxide content of such a preparation using the standard soln described in Exs.203 and 4. Assume that you wish to use between 35 and 45 ml of the reagent for a titration. 5H 2 O 2 + 2MnO 4 + 6H + 5O 2 + 2Mn 2+ + 8H 2 O 35 45 ml reagent: amount KMnO 4 = (35 ~ 45) ml 0.01145 M = 0.401~0.515 mmol amount H 2 O 2 = (0.401 ~ 0.515) mmol (5/2) = 1.00~1.29 mmol mass sample = (1.00 ~ 1.29) 0.03401 (100/3) = 1.1 ~1.5 g Thus we could weigh out from 1.1 to 1.5 g samples. These should be diluted to perhaps 75 to 100 ml with water and made slightly acidic with dilute H 2 SO 4 before titration. 20C2 Potassium Dichromate Cr 2 O 7 2 + 14H + + 6e 2Cr 3+ + 7H 2 O Eº = 1.33 V orange green in 1 M HCl or H 2 SO 4 Eº' = 1.0 ~ 1.1 V Advantages: stable, can be boiled without decomposition and do not react with HCl, primarystandard reagent is available and at a modest cost. Disadvantage: lower electrode potential and the slowness reaction. Preparing Dichromate Solutions reagentgrade K 2 Cr 2 O 7 dried at 150 to 200 before being weighed indicator: diphenylamine sulfonic acid, violet (oxidized) colorless (reduced) Applying Potassium Dichromate Solutions 1. titration of Fe(II): in moderate conc. of HCl Cr 2 O 2 7 + 6Fe 2+ + 14H + 2Cr 3+ + 6Fe 3+ + 7H 2 O 2. indirect determination of oxidizing agents (nitrate, chlorate, permanganate, dichromate and organic peroxides): analyte/acidic solution + measured excess Fe(II) backtitrated excess Fe(II) 142

Ex. 206 A 5.00mL sample of brandy was diluted to 1.000 L in a volumetric flask. The ethanol(c 2 H 5 ) in a 25.00mL aliquot of the diluted soln was distilled into 50.00 ml of 0.02000 M K 2 Cr 2 O 7, and oxidized to acidic acid with heating. 3C 2 H 5 + 2Cr 2 O 2 7 + 16H + 4Cr 3+ + 3CH 3 CO + 11H 2 O After cooling, 20.00 ml of 0.1253 M Fe 2+ were pipetted into the flask. The excess Fe 2+ was then titrated with 7.46 ml of the standard K 2 Cr 2 O 7 to a diphenylamine sulfonic acid end point. Calculate the percent (w/v) C 2 H 5 (46.07 g/mol) in the brandy. amount K 2 Cr 2 O 7 = (50.00 + 7.46) ml 0.02000 = 1.1492 mmol K 2 Cr 2 O 7 consumed by Fe 2+ = 20.00 0.1253 1/6 = 0.41767 mmol K 2 Cr 2 O 7 consumed by C 2 H 5 = 1.1492 0.41767 = 0.73153 mmol mass C 2 H 5 = 0.73153 (3/2) 0.04607 = 0.050552 g percent C 2 H 5 = 0.050552/(5.00 25.00/1000) 100 % = 40.44 % 20C3 Iodine weak oxidizing agents: determination of strong reductants I 3 + 2e 3I Eº = 0.536 V advantages: selectivity, sensitive and reversible indicator disadvantage: lack stability Properties of Iodine Solutions I 2 (s) + I I 3 K = 7.1 10 2 lack stability: volatility of iodine, slowly attacks most organic materials, airoxidation of iodide ion ( conc.). 4I + O 2 (g) + 4H + 2I 2 + 2H 2 O Standardizing and Appling Iodine Solutions Standardization: anhydrous Na thiosulfate or Ba thiosulfate Iodimetry: (direct method) I 2 reducing agents (ex: thiosulfate or arsenites) Iodometry: (indirect method) oxidizing agents + excess KI I 2 thiosulfate Indicator: 1. Starch indicator solution I 2 + I I 3 + starch I 3 starch complex (bluepurple) 2. CCl 4, HCCl 3, CS 2 I 2 in CCl 4, HCCl 3, CS 2 violet color 143

Preparation of 0.1 N Iodine solution 12.7 g I 2 + 40 g KI/20 ml H 2 O adding H 2 O to 1 L Standardization Primary standard: Arsenic (III) oxide, As 2 O 3 As 2 O 3 + 6Na 2Na 3 AsO 3 + 3H 2 O Na 3 AsO 3 + 3HCl H 3 AsO 3 + 3NaCl H 3 AsO 3 + H 2 O + I 2 H 3 AsO 4 + 2HI ph : H 3 AsO 3 + H 2 O + I 2 H 3 AsO 4 + 2H + + 2I ph : I 2 + 2 IO + I + H 2 O 3IO IO 3 + 2I strong oxidizing agent 滴 定 中 加 入 NaHCO 3 [ph: 7~8] Na 3 AsO 3 + I 2 + 2NaHCO 3 Na 3 AsO 4 + 2NaI + 2CO 2 + H 2 O mg As2O3 Calculation: N of Iodine = 197.8 / 4 ml Iodine (1) Direct Iodimetric Titration a. H 3 AsO 3 + H 2 O + I 2 H 3 AsO 4 + 2H + + 2I b. Assay of Ascorbic Acid (Vit C) O HO O (enediol) CH CH 2 O O CH CH 2 + I 2 O O (αdiketone) (2) Residual Titration (I 2 Na 2 S 2 O 3 ) NaHSO 3 + I 2 +H 2 O NaHSO 4 + 2HI I 2 + 2 Na 2 S 2 O 3 2NaI + Na 2 S 4 O 6 (3) Iodometry: Sample + KI (excess) I2 Na2S2O3 + 2I + 2H+ (+2) (1e) (+1) a. CuSO 4 + 4KI 2CuI + I 2 + 2K 2 SO 4 I 2 + 2Na 2 S 2 O 3 2NaI + Na 2 S 4 O 6 144

b. Assay of sodium hypochlorite solution (NaOCl) NaOCl + H + HOCl (+1) (2e) (1) HOCl + 2KI + HOAc I 2 + KCl + KOAc + H 2 O I 2 + 2Na 2 S 2 O 3 2NaI + Na 2 S 4 O 6 Table 206 Some Applications of Iodine Solutions Analyte HalfReaction As H 3 ASO 3 + H 2 O H 3 AsO 4 + 2H + + 2e Sb H 3 SbO 3 + H 2 O H 3 SbO 4 + 2H + + 2e Sn Sn 2+ Sn 4+ + 2e H 2 S H 2 S S(s) + 2H + + 2e SO 2 SO 2 3 + H 2 O SO 2 4 + 2H + + 2e 2 S 2 O 3 2 2 S 2 O 3 S 4 O 2 6 + 2e N 2 H 4 N 2 H 4 N 2 (g) + 4H + + 4e Ascorbic acid C 6 H 8 O 6 C 6 H 6 O 6 + 2H + + 2e ** Dichloroindophenol Titration Determination of Ascorbic acid Preparation O N Cl HO NH Cl Cl Cl Blue in basic sol'n colorless Pink in acid sol'n End point: pink (selfindicator) Vit C titration in metaphosphoric acid and acetic acid sol'n 20C4 Potassium Bromate as a Source of Bromine Primarystandard KBrO 3 is available, stable standard 0.1 N Bromine sol'n (Koppeschaar's sol'n): (3 g KBrO 3 + 15 g KBr)/1 L H 2 O Assay of sample: aniline, phenol, salicylic acid, resorcinol etc. BrO 3 + 5Br + 6H + 3Br 2 + 3H 2 O standard soln excess 1 mol KBrO 3 = 3 mol Br 2 2I + Br 2 I 2 + 2Br 2 I 2 + 2S 2 O 3 S 4 O 2 6 +2I (excess KI) 145

Br Br + 3 Br 2 + 3 H + + 3Br Br (ppt) soluble in CHCl 3 CO Br Br + 3Br 2 + HBr + CO 2 Br Substitution Reactions halogen substitution: replacement of H in an aromatic ring by a halogen. determination of aromatic compound that contain strong orthoparadirecting groups, particularly amines and phenols. Ex: 1. Determination of 8hydroxyquinoline N + 2Br 2 2. Determination of aluminum Al 3+ N ph 49 Br Br + 2HBr + ( OC H N) ( S ) + 3 + 3HOC H N Al H 9 6 hot 4M HCl 3+ ( OC H N) ( ) 3HOC H N + Al S 9 6 3 9 6 Al 9 H6N + 6Br2 3HOC9H4NBr + 6HBr 3HOC 2 Ex. 207 A 0.2891g sample of an antibiotic powder containing sulfanilamide was dissolved in HCl and the solution diluted to 100.0 ml. A 20.00mL aliquot was transferred to a stoppered flask and 25.00 ml of 0.01767 M KBrO 3 added. About 10 g of KBr was added to form Br 2, which brominated the sulfanilamide in the sample. After 10 min, an excess of KI was added and the liberated iodine titrated with 12.92 ml of 0.1215 M sodium thiosulfate. The reaction are BrO 3 + 5Br + 6H + 3Br 2 + 3H 2 O 9 6 3 NH 2 NH 2 Br Br + 2Br 2 + 2H + + 2Br SO 2 NH 2 SO 2 NH 2 146

Br 2 + 2I 2Br + I 2 (excess KI) I 2 + 2S 2 O 3 2 S 4 O 6 2 + 2I Calculate the % NH 2 C 6 H 4 SO 2 NH 2 (172.21 g/mol) in the powder. total amount Br 2 = 25.00 ml 0.01767 M 3 = 1.32525 mmol amount excess Br 2 = amount I 2 = 12.92 ml 0.1215 M (1/2) = 0.78489 mmol The amount of Br 2 consumed by the sample = 1.32525 0.78489 = 0.54036 mmol mass analyte = 0.54036 (1/2) 0.17221 = 0.046528 g 0.046528 % analyte = 100% = 80.47% 20.00mL 0.2891 100mL 172.21 (0.01767 3 2 25 0.1215 12.92) 5 or 4 100% = 80.47% 289.1 Addition Reactions H C C H H H + Br 2 H C C Br Br 20C5 Determining Water with the Karl Fischer Reagent H Karl Fischer Reagent: I 2, SO 2, organic base such as pyridine (C 5 H 5 N) or imidazole /CH 3 or lowmolecularmass alcohol In aprotic solvent: I 2 + SO 2 + 2H 2 O 2HI + H 2 SO 4 2 mol H 2 O 1 mol I 2 Classical chemistry: use anhydrous methanol as solvent, and excess pyridine C 5 H 5 N I 2 + C 5 H 5 N SO 2 + C 5 H 5 N + H 2 O 2C 5 H 5 N HI + C 5 H 5 N SO 3 C 5 H 5 N + SO 3 + CH 3 C 5 H 5 N(H)SO 4 CH 3 1 mol H 2 O 1 mol I 2, 1 mol SO 2, 3 mol C 5 H 5 N Pyridinefree chemistry Replaced by other amines: imidazole (1) Solvolysis: 2R + SO 2 RSO + 3 + R 2 (2) Buffering: B + RSO 3 + R + 2 BH + SO 3 R + R (3) Redox: B I 2 + BH + SO 3 R + B + H 2 O BH + SO 4 R + 2BH + I 1 mol H 2 O 1 mol I 2 Interfering reactions 147