UNIT OF STUDY: PHYSICAL CHEMICAL ANALYSIS AND TREATMENT OF EXPERIMENTAL DATA

Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam HUE’S UNIVERSITY HUE’S UNIVERSITY OF EDUCATION CHEMISTRY DEPARTMENT -?&@ - ESSAY UNIT OF STUDY: PHYSICAL CHEMICAL ANALYSIS AND TREATMENT OF EXPERIMENTAL DATA TOPIC: EXERCISES ABOUT COULOMETRIC METHODS Instruction Lecturer: Student: Assoc Prof Ngo Van Tu Nguyen Thi Thanh Tam Class : Chemistry 3E Student’s code: 16S2011055 School year: 2018-2019 Hue, May 2019 Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam ACKNOWLEDGEMENT A completed study would not be done without any assistance Therefore, the author who conducted this essay gratefully gives acknowledgement to their support and motivation during the time doing this essay: “EXERCISES ABOUT COULOMETRIC METHODS” On my honor, I would first like to express my endless thanks and respectfulness to my supervisor Associate Professor Dr Ngo Van Tu His kindly support and continuous advices went through the process of studying, researching and completing my topic I would also like to sincerely thank the teachers in the Faculty of Chemistry at Hue University’s College of Education for their help and guidance during the past time I also acknowledge my thankfulness to all fellow students Without their encouragement and enthusiastic support, my essay would have been impossible to be done effectively Although I have constantly tried to improve the essay, it is inevitable to have mistakes due to limited knowledge and English ability I am looking forward to receiving comments from you and others so that my knowledge in this field could be improved I sincerely thank you all Hue, May 2019 Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam TABLE OF CONTENTS PART 1: BEGINNING I Reasons for choosing this topic .1 II Research Purposes .1 III Research tasks IV Objects of study V Methods for study PART 2: CONTENTS Exercise 1: Exercise 2: Exercise 3: Exercise 4: Exercise 5: Exercise 6: Exercise 7: Exercise 8: 10 Exercise 9: 10 Exercise 10: 11 Exercise 11: 12 Exercise 12: 13 Exercise 13: 14 Exercise 14: 16 Exercise 15: 18 Exercise 16: 19 Exercise 17; 20 Exercise 18: 21 Exercise 19: 22 Exercise 20: 23 PART 3: CONCLUSIONS .26 REFERENCE MATERIALS .27 Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam PART 1: BEGINNING I Reasons for choosing this topic Coulometry is based on an exhaustive electrolysis of the analyte By exhaustive we mean that the analyte is completely oxidized or reduced at the working electrode or that it reacts completely with a reagent generated at the working electrode There are two forms of coulometry: controlled-potential coulometry, in which we apply a constant potential to the electrochemical cell, and controlled-current coulometry, in which we pass a constant current through the electrochemical cell For this reason, Coulometric method is a useful quantitative method of electrochemical analysis methods in modern analytical method Therefore, I have carried out the topic: " EXERCISES ABOUT COULOMETRIC METHODS” with the desire to provide a useful reference to teachers and students This topic also helps students, students to learn, reference exercises or in reference books of foreign countries Hopefully this topic is a useful document for readers who are interested in studying chemistry Thank you sincerely II Research Purposes - Promoting my ability to explore, learn myself From that, building good exercises for students - Synthesizing exercise-system to improve knowledge and skill solving it for students at the university - Learn more deeply about analytical exercises, especially electrolytic analysis methods - Approaching the method of solving electrochemical analysis exercises of some countries in the world III Research tasks Search and synthesized chemically some exercises electrochemical analysis by electrolysis of our country and other countries IV Objects of study Exercises about electrochemical methods of analysis : Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam Electrolysis (electrogravimetric and coulometric methods) V Methods for study Based on the knowledge learned, conduct searches, research, analyze, compare and synthesize sources: textbooks of our country and other countries, the examination questions, etc Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam PART 2: CONTENTS Exercise 1: To analyze a brass alloy, a 0.442-g sample is dissolved in acid and diluted to volume in a 500-mL volumetric flask Electrolysis of a 10.00-mL sample at –0.3 V versus a SCE reduces Cu2+ to Cu, requiring a total charge of 16.11 C Adjusting the potential to –0.6 V versus a SCE and completing the electrolysis requires 0.442 C to reduce Pb2+ to Pb Report the %w/w Cu and Pb in the alloy Solution The reduction of Cu2+ to Cu requires two electrons per mole of Cu (n = 2) We calculate the moles and the grams of Cu in the portion of sample being analyzed Q 16.11C  8.348 10 mol Cu  nF mol e 96487C  mol Cu mol e  63.55g Cu 8.348 10 mol Cu   5.30110 g Cu mol Cu N Cu  This is the Cu from a 10.00 mL portion of a 500.0 mL sample; thus, the %w/w copper in the original sample of brass is 500.0 mL 5.30110 g Cu  10.00 mL 100 60.0%w / w Cu 0.442 g sample For lead, we follow the same process, thus Q 0.422 C  2.19 10 mol Pb  nF mol e 96487C  mol Pb mol e  207.2 g Pb 2.19 10  mol Pb   4.53 10  g Pb mol Pb 500.0 mL 4.53 10  g Pb  10.00 mL 100 5.12%w / w Pb 0.442 g sample N Pb  Exercise 2: Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam The purity of a sample of Na2S2O3 was determined by a coulometric redox titration using I- as a mediator, and I3- as the “ titrant” A sample weighing 0.1342 g is transferred to a 100-mL volumetric flask and diluted to volume with distilled water A 10.00-mL portion is transferred to an electrochemical cell along with 25 mL of 1M KI , 75 mL of a pH 7.0 phosphate buffer, and several drops of a starch indicator solution Electrolysis at a constant current of 36.45 mA required 221.8 s to reach the starch indicator end point Determine the purity of the sample Solution: 2  The coulometric titration of S 2O3 with I is : S2O32 ( aq )  I 3 (aq )  S4O62 (aq)  3I  (aq ) 2 2 2 Oxidizing S 2O3 to S 4O6 requires one electron per S 2O3 ( n=1) Solving for the grams of Na2S2O3 gives mNa2 S2O3  ItANa2 S2O3 nF  0.03645 221.8 158.1 0.01325 g 196487 This represents the amount of Na2S2O3 in a 10.00-mL portion of a 100-mL of the sample, thus 0.1325 g of Na2S2O3 is present in the original sample The purity of the sample, therefore, is % mNa2 S2O3  0.1325 g Na2 S 2O3 100 98.73% 0.1342 g  sample 2 Note that the calculation is worked as if is S 2O3 oxidized directly at the working electrode instead of in solution Exercise 3:   Suppose we wish to electrolyze I to I3 in a 0.10 M KI solution containing 3.0  ×10-5 M I3 at pH 10.00 with fixed at 1.00 bar Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam (a) Find the cell voltage if no current is flowing  (b) Then suppose that electrolysis increases [ I3 ]s to 3.0 ×10-4 M, but other concentrations are unaffected Suppose that the cell resistance is 2.0 Ω, the current is 63 mA, the cathode overpotential is 0.382 V, and the anode overpotential is 0.025 V What voltage is needed to drive the reaction? Solution: (a) The open-circuit voltage is E(cathode)- E(anode): Cathode : 2H 2O + 2e  H (g) + 2OH  E o  0.828 V Anode : E o 0.535 V I3  2e  3I  0.05916 log( PH2 [OH  ]2 ) 2 0.05916  0.828  log[   1.00   1.0  10    =0.591 V   E(cathode)  0.828   [I  ]3  0.05916 E  anode  = 0.535  log     [I3 ]   [0.10]3  0.05916 0.535  log  =0.490 V 5   [3.0 10 ]  E= E  cathode   E  anode  = 1.081 V We would have to apply 1.081 V to force the reaction to occur  (b) Now E(cathode) is unchanged but E(anode) changes because [I3 ]s is different  from [I3 ] in bulk solution  [I  ]3  0.05916 E  anode   0.535  log     [I3 ]   [0.10]3  0.05916 0.535  log  =0.520 V 4   [3.0 10 ]  E E  cathode   E  anode   IR  overpotentials  0.591 V  0.520 V  (2.0 )(0.063A)  0.382 V  1.644 V Instead of 1.081 V, we need to apply 1.644 V to drive the reaction Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam Exercise 4: Prepare 0.125 g of an alloy sample with 100 ml of H 2SO4 to dissolve the sample completely Here iron is transferred into solution in the form of Fe 2+ Transfer the whole solution to the electrolyte flask Add 50,00 ml of Ce (III) 0,0400M solution Electrolysis with current intensity I = 0.0500A using Pt electrode The titration stop is achieved after 511.2 seconds Calculate the% Fe content in the alloy Solution: When electrolysis occurs oxidation of Fe2+ at anot: Fe2+  Fe3+ + e At the end of electrolysis, Ce3+ oxidation will occur : Ce3+  Ce4+ + e Ce4+ produces Fe2+ oxidation: Ce4+ + Fe2+  Ce3+ + Fe3+ And when Fe2+ is completely finished, a small amount of Ce 4+ electrolyte will soar and we can end electrolysis AIt 55.85  0.050 511.2 0.01479 gam n 96500 196500 0.01479 % Fe  100 12.17% 0.1215 mFe  Exercise 5: To determine the antimony in Sb2O3 form in the analytical sample, we weigh 15 grams of sample and acidify with HCl to obtain the solution In addition to the solution, obtain an excess of NaBr solution and then quantify antimony by titration with the constant current I, knowing that when conducting electrolysis with a flow of 2.4125 mA for a period of 12 minutes a Write the reaction occurs Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam b Calculate Sb2O3 content in ppm units Given that: MSb2O3 291.5 Solution: The reaction occurs when sample dissolution: Sb 2O3  12HCl  2H 3[SbCl6 ]  3H 2O (1) [SbCl6 ]3  [SbCl ]  2e Anode : Br  Cathode :  Br2  2e 2H   2e  H The reaction occurs when titration of electric quantity: [SbCl6 ]3  Br2  [SbCl6 ]  2Br  (2) Applying Faraday's law, we have: m Br2 AIt 160 2.4125 10  12 60   0.00144(g) nF 96500 From (1) and (2) we have the mass of Sb2O3 in the sample: mSb2O3  So: 0.00144 291.5 0.00131175(g) 160 ppmSb2O3 0.00131175 10  87.45 ppm 15 Exercise 6: Electrolysis of Co2+ 0.100M and Cd2+ 0.0500M solutions a Calculate Co2+ concentration when Cd2+ starts electrolysis b Calculation of cathode potential to reduce Co2+ ion concentration to 1.0 10-6 M For Cd2+ to start electrolysis: Cd2+ + 2e  Cd o Given that: E oCo2 /Co =  0.280V; E Cd =  0.402V 2 /Cd Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam Electrodes reaction in cathode: Co2+ + 2e  Co Applying the law of Faraday we have: m AIt nFm 96500 5.00  t  16355.93s 4.54 h nF AI 59 1 Thus, to precipitate 0.500 grams of Co, electrolysis must be carried out with the current intensity of 1A in a time of 4.54 hours b Calculate the electrolysis time to precipitate 0.602 grams of Co 3O4 onto the cathode At the cathode: 3Co2+ + 4H2O + 6e & t Co3O4& + 4H2& nFm 96500 0.602  14463s 4h AI 241 1 Thus, to precipitate 0.602g Co3O4, it is required to conduct electrolysis with current intensity 1A for hours Exercise 12: Is a quantitative separation of Cu2+ and Pb2+ by electrolytic deposition feasible in principle? If so, what range of cathode potential (vs SCE) can be used? Assume that solution is initially 0.1000 M in each ion and that quantitative removal of an ion is realized when only part in 10,000 remains underposited In Appendix 5, we find Cu 2  2e   Cu(s) E o 0.337V Pb 2  2e  & E o  0.126V Pb(s) Solution It is apparent that copper will begin to deposit before lead because the electrode potential for Cu2+ is large than that for Pb 2+ Let us first calculate the cathode potential required to reduced the Cu2+ concentration to 10-4 of its original concentration ( that is, to 1.00×10-5 M) Substituting into the Nernst equation, we obtian 14 Lecturer: Assoc Prof Ngo Van Tu E 0.337  Student: Nguyen Thi Thanh Tam 0.059 log 0.189V 1.00 10 Similarly, we can derive the cathode potential at which lead begins to deposit: E  0.126  0.059 log  0.156V 0.100 Therefore, if the cathode potential is maintained between 0.189 V and -0.156 V (vs SHE), a quantitative separation should in theory occur To convert these potentials to potentials relative to a staturated calomel electrode, we treat the reference electrode as the anode and write Ecell = Ecathode  ESCE  0.189  0.244=  0.055V and Ecell =  0.156  0.244=  0.400V Therefore, the cathode potential should be kept between -0.055 V and -0.400V versus the SCE Exercise 13: Conduct electrolysis of 100 ml of a mixture of Cu(NO3)2 0.15 M and 0.1 M AgNO3 (inert electrode) with a current of 1A Calculate the mass of substances obtained in anode and cathode when the electrolysis time is: a) 16 minutes seconds; b) 32 minutes 10 seconds; c) 64 minutes 20 seconds Solution: According to the article we have: Electrode reactions: At the cathode: Ag  + e → Ag Cu 2 + 2e → Cu 15 Lecturer: Assoc Prof Ngo Van Tu At the anode: 2H2O  Student: Nguyen Thi Thanh Tam O2& + 4H+ + 4e  The electrolysis time of Ag is: m AIt nFm F 96500  t n Ag   0.01  965s 16m5s nF AI I 2 Time to Cu electrolysis: 2F 96500 t n Cu   0.015  2895s 48m15s I Thus, after 16 minutes seconds + 48 minutes 15 seconds = 64 minutes 20 seconds,  2 both Cu and Ag are electrolyzed a) After 16 minutes seconds At the cathode: Cu 2  has not been electrolyzed; Ag electrolysis At that time: n Ag n Ag 0.01mol  m Ag 0.01 108 1.08g At the anode: m O2  AIt 32 1 (16.60  5)  0.08g nF 96500 b) After 32 minutes 10 seconds:  At the cathode: Ag has been electrolyzed:  m Ag  1.08g 2 Electrolytic time Cu is: t = 32 minutes10 seconds - 16 minutes 5seconds = 16minutes seconds m Cu  AIt 64 1 (16 60  5)  0.32g nF 96500 16 Lecturer: Assoc Prof Ngo Van Tu Student: Nguyen Thi Thanh Tam At the anode: m O2  AIt 32 1 (32.60  5)  0.16g nF 96500 c) After 64 minutes 20 seconds:  2 Ag At the cathode: and Cu are all electrolyzed  m Ag  1.08g n Cu n Cu  0.015mol  mCu 0.015 64 0.96g At the anode: m O2  AIt 32 1 (34.60  5)  0.32g nF 96500 Exercise 14: Electrolyte of NiSO4 0.100M solution with pH = 2.00 using Pt electrode a The cathode potentiality needed to have Ni precipitate in the cathode b Calculate the voltage needed to have the first electrolysis process c Calculating the voltage that must be applied to [Ni2+] is equal to 1.0 10-4 M Electrolytic flask resistance: R = 3.15; I = 1.10A Solution: a At cathode: Ni2+ +2e  Ni To have Ni precipitate in cathode, the cathode E c  E Ni2 / Ni Which 17 ... electrochemical analysis exercises of some countries in the world III Research tasks Search and synthesized chemically some exercises electrochemical analysis by electrolysis of our country and other... Conduct electrolysis of 100 ml of a mixture of Cu(NO3)2 0.15 M and 0.1 M AgNO3 (inert electrode) with a current of 1A Calculate the mass of substances obtained in anode and cathode when the electrolysis... flask and diluted to volume with distilled water A 10.00-mL portion is transferred to an electrochemical cell along with 25 mL of 1M KI , 75 mL of a pH 7.0 phosphate buffer, and several drops of