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by: Carmela Kilback


Carmela Kilback
GPA 3.92


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Class Notes
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This 5 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 321 at University of Washington taught by Staff in Fall. Since its upload, it has received 25 views. For similar materials see /class/192590/chem-321-university-of-washington in Chemistry at University of Washington.




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Date Created: 09/09/15
Experiment 6 Manganese Note s Preparation I Dry WSO4HZO for 30 minutes in the 125 C oven I This experiment is highly compatible with the Fluoride experiment for multiplexing your time PreLab Calculations I Standard addition calculations can be confusing at rst You may wish to plot Absorbance vs Added Mn mgmL for your calibration curve Experiment I Weight 020 7 022 g K104 I Make sure that all of the K104 is dissolved before proceeding I Crush larger particles swirl the ask periodically until no K104 particles are visible I Allow 5 7 10 minutes more time swirling occasionally I Total dissolution time will be at least 15 minutes I A correct blank solution is vital for standard addition methods I If you seen a brown tinge in your KlVlIlO4 solutions some irreversible reduction has occurred You will need to reprepare the solution Possible causes are dirty glassware residual soap or overcooking I You may divide this experiment into two parts I Day 1 Boil the unknown and standard solutions in K104 let asks cool and cover with para lm I Day 2 Finish the experiment Review Question Reminder This review question is not required 7 it is designed to extend your knowledge about the experiment at hand Other common applications of standard addition methods include spiking a 5 7 100 mL sample with a very small volume 10 7 100 uL ofa concentrated standard These methods rely on having a negligible dilution due to the added standard Design an experiment where you would analyze trace Mn levels by standard addition Assume 100 ppb 7 100 ppm Mn with 10 mL of sample available Assume that you can reproducibly deliver 10 ML of spike solution What concentration of Mn do you need in this solution to analyze a sample of 1 ppm Mn 10 and 100 ppm How precise does the spectrometer need to be PreLab Calculation The answer is include units 144 145 gt6 CHAPTER 14 FUNDAMENTALS OF ELECTROCHEMXSTY second 51mm polenliul volts mensums me W11 Lhal can be done by or mm In dunoezcl1mdnmb olcharge as it moves from um Pain ID number a 111502177 33 x1049 Clclcclmn 62413063 x 10W ulecmmsc b F 96451309 China a 1 cuulombss Evury mo of 02 accqu 4 mm or e39 15 m0107dzy 6411101 waxy 74m 11H mul e39s 115 05 x b 1 PowerE 5m w115 V 435 A The tasting human uses 16 umuu us much current us me refrigemlur c Powcr E 11V7L5A 79w a I 63900 V 3mm 300x103Cs 211 x 103 n 300 x10393 Cs 1 r 9549 X m Clmnl5022x 10 1 2 mm b P E 600 V 301 103 A Leona2W 1 so x 10 21s 137 x c 300mm 1500s 337x 1o9e1ecuuns 550x105 mol 1 E W W100 WZ00x 103 m 447v a 1 20 s 21 Oxidaux 37 x 10 59 15 3 963 x10quot Ie s 07 25207 402 25 ucml c 100 g 9103 I 11212 gmul 392 111111131520 1392 mm a Euglx 0393 mol 9649 x 10 Clam 86 C d Currant A coulomsz 861 350 s 143 A a Oxidation numbns ofreaclnnts N in Nam c1 in c103 Al 3 7 H Onda ml numbers ofpmducLs N c1 A1 n 1 3 0 Formula weight of macIams 6FW NH4CIO4 INFW A1 97475 Heal released per yarn 9 334 UI974 75 g 9376 Ug M Fundamentals of Electrochemistry 1 17 147 148 149 a 1410 V 1411 1412 1413 In a galvanic cell two halfreactions are physically separated from each other At the anode an oxidation reaction generates electrons that can ow through the electric circuit to reach the cathode where a r duction reaction occurs The favorable free energy change for the net reaction provides the driving force for electrons to ow through the circuit There must be a connector such as a salt bridge between the two halfcells to allow ions to ow to maintain electroneutrality a Fes l FeOs l KOHaq l Ag20s l Ags oxidation Fes 20Hquot FeOs H20 2e reduction Ag20 H20 2equot x39 2Ags ZOH39 b Pbs l PbSO4s l K2804aq l I H2804aq IPbSO4s I Pb02s I Pbs oxidation Pbs so a PbSO4s 2e39 reduction PbOz 4H so 39 2e PbSO4s 2H20 b oxidation Fe2 v Fe3 e reduction CrzOg39 14H 6e39 v 2Cr3 7H20 c Cr2027 6Fe2 14H 2Cr3 6Fe3 7H20 1a oxidation Zns quotlt Zn2 2e reduction C120 2e39 F 2Cl39 b One mol of C12 requires 2 mol of equot 39 Moles of C12 consumed in 100 hr mol of e39hr g 100 x 103 glam x 104 Cmo13 600 shr 187 mol of 12 132 kg C12 is strongest because it has the most positive reduction potential Become stronger Cr2027 MnO 103 Unchanged C12 Fe3 a Since it is harder to reduce FeIII to FeII in the presence of CN39 FeIII is stabilized more than FeII Chapmr 14 1 Since 11 is mm In Educ F3011 to FeII m the prcmnce of phenanlhmline 11111 1s stabilized mute Lhnn F2011 1 4 1 4 Equot mcaaurad Al equilibnum E gees down a zero Equot is a constant Ihat does 1101 14 15 a 2115 Zn2n1 M I Cu2o1 M I Cus a Cns ghthalf em 011 25 h f cel Zn 21239 2 2115 E 0339 rmln g 0762 7ng 1101 V 5111 the vohage is positive electrons are transferred from Zn 117 Cu The ma munch is 0112 2115 Cum 2112 b Sine Cu 10115 are consumed m me right halfrcell Z1121onx must migrale fro e an halfcell 11110 the sill budge 11 help balms charge Ihope you 11k Our 2111 because that is what your bady will mks up 005915 F111 3 H113 LOO760 7 Wm i 70359 v 1416 a E 0233 log 1 E 70238 m lug 1417 1 Pls l 13121 I mamq 010 M II A1N033aq 0010 M A1s b gmhalfcell A134 3e39 A1s 1677V f1 halfcell Huang a 2131 1078 V 15 716777W10gw3101 1715 V mm 01011 mm 11gb halfrce 1511 hunted 13 10737 1 EVE717164711372 elnclmns ow from he rightrhand electrode 0 Lb lefthm e 31 pontaneons reaction 15 2 354 v Since the vohage is negauvc 1 1mm Reduction occurs 11 he 1237mm excumdc 2320 A10 313139 A13 c 143 mL 0113 446 g 0279 mo1 011m 120 gofAl 0445 11101 of Al The macnon rcquircs 32 11101 of Br for evcry mol 01 AL The 1112 will be m1 up 151 I Fundamentals of Electrochemistry r i l 39 1 19 1418 1419 1420 1421 1422 1 0231 mL of Brg 0721 g of Br2 451 x 10 3 mol Brg 902x10393 mole 870 C Work Eq 150870 131 k e 1 PR 100 x10394120 x103 289 x 104 A 299 x 109 mol 67s 997 x 1010 mol Als 269 x 108 gs a right halfcell E 0222 0059 16 log Cl39 02812 V left halfcell E 0350 log F 2 V02903 V E E E 02812 02908 0572 V b Pb2 KspfoerF2F392 36x 1080102 36x 106M Ag KSp for AgClCl39 18x 1010010 18x 109M right halfcell E 0799 0059 16 log A1g 02812 V 0059 16 1 left halfcell E 0126 7 log Pb2 02870 V E E E 02812 02870 0568 V The agreement between the two calculations is reasonable 1n 108314510 J mol391 K39129815 K 10 RT 005916 nF1n Q n 96485309 x 104 c moll g Q n logQ IfT 0 C 27315 K 1an i0 i g logQ WhenT 37 C 31015 K l gln Q 9991 log Q 11 0010 00 091439 0798 3 E Ag Ag 005 1610g 727276012 0010 ooo898 gt E Ag1Ag 07993V Br39 H20 AG 1F1584 AG 1F1098 AG A6 AG 2FE Balanced reaction HOBr 2639 H HOBr 9 gBrz Brz gt Br39 HOBr gt Brquot 1F1584 1F1098 E3 2F 1 1341V 2X 2639 e 2Xs E1E X3 2e39 2 X 39 E3E 3X X3 2Xs E E E Whenever gt B then will be greater than 0 and disproportionation will be spontaneous


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