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midterm1 - KEY J Last Name Lab Sec. # ; TA: Andreas...

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Unformatted text preview: KEY J Last Name Lab Sec. # ; TA: Andreas Toupadakis, Ph.D. CHEMISTRY 2C Section B Good luck © EXAM 1 Ins fruc flan: I CLOSED BOOK EXAM! No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back of the exam. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). This exam has 10 pages total. (1) Read each question carefully. (2) For Parts I and II, there is no partial credit given and only answers marked on this cover page will be graded. (3) The last 2 pages contain a periodic table and some u5efu| information. You may remove them for easy access. (4) If you finish early, RECHECK YOUR ANSWERS! U.C. Davis is an Honor Institution /28 Possible Points # 1—9 (4 points each) # 10-13 (7 points each) # 14 (8 points) #15 (10 points) #16 (18 points) Total Score ( 100 ) First Name ; Lab day/time: 1.abcd 2.ab©de [email protected] 4.abcd 5.abcd@ 6.bcde [email protected] 8.ab©de 9.acde 1-9 total points: 10.a®c d e 10-13 fatal points: (3 .fl/ Spring 2008 Multi le Choice (circle one) Name Spring 2008 Exam 1 (Page 2 of 10) Part I: Multiple Choice, Concepts (4 points each) Select the best answer and enter your choice on the cover sheet - No partial credit 1) 3) 4) The reaction taking place in a zinc-lead cell is: sz‘ + Zn —> Pb + an‘ E° = + 0.637 V Which of the following statements about this cell is TRUE? 0) The lead electrode is negatively charged. b) The shorthand notation is PbI sz’II an‘ I Zn c) The reaction will go in the reverse direction of the one indicated. d) E° > 0 means that starting with [sz’] = [an‘] = 1 M, when the cell is dead, it will be [Pb2*] > [Zn2*]. e) The zinc electrode is the anode. Choose the INCORRECT statement based on the following standard oxidation potentials. Mg —> Mg2+ + Ze' E° = +2.37 V Fe —> Fe2‘+ Ze' E° = +0.44 V Cu —> Cu2+ + 2e‘ E° = -0.34 V Zn —> Zn2‘+ 2e' E° = +0.76 V Sn —+ Snz‘ + 2e' E° = +0.14 V Ag —-> Ag‘+ e' E° = -0.80 V a) At standard concentrations, Fe will not displace an’from solution. b) Mg will displace an‘from solution. c) Fe will not displace H* from solution. d) Cu will not displace Snafrom solution. e) None of these. Consider the following reduction half—reactions and their respective standard reduction potentials at 25 °C. Cr2072'(aq) + 14H‘(aq) + 6e' —> 2Cr3‘(aq) + 7HZO Zn2*(aq) + Ze‘ —> Zn(s) E° = +1.33 V E° = - 0.763 V Which is the correct diagram of a working galvanic cell? a) Pt(s) | Cr3*(aq), Cr2072'(aq), H*(aq) II Zn2‘(aq) I Zn(s) b) Zn(s) I Zn2*(aq) II Cr2072'(aq), H‘(aq), Cr3*(aq) I Pt(s) c) Zn(s) | Zn2*(aq) ll H‘(aq), Cr3‘(aq) I Cr(s) d) Cr(s) I Cr3’(aq), H’(aq), Cr207z'(aq) II Zn(s) I Zn2‘(aq) e) Zn2’(aq) I Zn(s), H*(aq) II Cr2072‘(aq), Cr(s) I Cr3’(aq) Which element has the SMALEST number of unpaired electrons? a) Ni2+ b) Cu2+ c) Fe3+ d) V3’ e) Zn2+ Name Spring 2008 Exam 1 (Page 3 of 10) 5) The oxidation number of the metal atom in [Cr(NH3)6](N03)3 is: a) -1 b) —3 c) +2 d) +4 e) +3 6) List the elements: Os, Ag, Au, and Cd in order of INCREASING density. a) Cd<Ag<Au<Os b) Au<Ag<Cd<Os c) Ag<Cd<Os<Au d) Cd<Os <Au<Ag e) Au<Ag<Cd<Os 7) The correct name for [Co(NH3)4C|2]C| is: a) Dichlorotetraamminecobalt(III) chloride. b) Dichlorotetraamminecobaltate(III) chloride. c) Tetraamminedich|orocoba|t(III) chloride. d) Tetraamminedich|orocobaltate(III) chloride. e) Tetraamminedich|orocobalt(II) chloride. 8) Select the correct formula for tetraaquodichIorochromium(III) chloride. 0) [CF(H20)4CIZ]CI3 b) [CF‘(H20)4Cl2]C|2 C) [CP(H20)4C|2]CI d) [CP(H20)4]CI3 8) [CP(H20)4]CI2 9) The number of unpaired electrons in the [Cr(en)3,]2+ ion is: a) 1 b) 2 c) 3 d) 4 e) 5 Name Spring 2008 Exam 1 (Page 4 of 10) Part II: Multiple Choice, Short Calculations (7 points each) Select the best answer and enter your choice on the cover sheet — No partial credit 10. A resistance heater was wound around a 50 g metallic cylinder. A current of 0.65 A was passed through the heater for 24 5 while the measured voltage drop across the heater was 5.4 V. The temperature of the cylinder was 22.5 °C before the heating period and 29.8 °C at the end. If heat losses to the environment are neglected, what is the specific heat of the cylinder in cal g'1 K'l. 2.00 cal 9'1 K‘1 0.055 cal 9'1 K'1 5.0 cal 9'1 K'1 0.033 cal 9'1 K'1 0.099 col 9'1 K'1 09-09—53 t \APY’O‘xj input :I'V‘f :(O,6§M(§.Li (ills) :BL‘j :fiil- - :QOth ,7 Li l8 j’/(ul{/ L a ( VHV'LJYJ QbQOYl/Jfid —; (wuss) x(<g>€cif€c l/xeotl x AT 1c m5 'zCs’O 3\( CC) [(1Q.%~ 7,2-s\ l5] 11. How many hours is required for a current of 3.0 A to decompose electrolytically 18 g water? 2' 2;; mo w» m + 301 c. 4h d. 56h e. 18h 3 \ l9 sine We H'LO «we, we / “W” « Name Spring 2008 Exam 1 (Page 5 of 10) 12. What is the reduction potential of an electrode consisting of zinc metal in a solution in which the zinc ion concentration is 0.0100 M? a. Eredz-OBZZV 1+ b. Eredz—O.173V , — I ‘ ‘9 c. Ems-0.213 v 7’“ (Ml + 16 a ZV‘CQ EM 1 ' (“J6 V d. Ered=+0.189v K e. Ered=+0.390V LO M M 1+ _ ’ (Z 4,.vx (uq) [0.OlOO/Vl1 +16 "'">1V\CS\ Eved ‘1' - n. O ' C :: E ._ ._.—— 50 Q \(d ye.) h (j i M; (heme v) - 9:0ng 4323 i Aa Em ' (963:9 9_ " 0.0\ 00 «~— 134 The complex [Ti(HZO)6]3+ absorbs light of wavelength 510. nm. The value of the ligand field splitting (in kilojoules per mole) is: a. 235 kJ‘-mol'1 i) 3 Si 5 ; \.V b. 200 kJ-morI 0 WA“ > A L c. 275 kJ-mol'l ‘ . d. 335kJ-mol‘1 BE 3 k3 /w<>{ ions _ CLZ‘V”) e. 735 kJ-mol‘l ’ J; r . , pi [7:1 l/\~\I : t \ iéfimqw‘ __....._ (slaw < WM ) E. h 41 T i: O.Cé‘3 X\O w (OV\ / t wag ibn\ / 3. .‘ '1 (LI/VT (3.0’L’Lwoz Iqu M I {00st lo ,J, g " ""“" ‘ low _\ ifx 6 1% meme (000 j“ , W..-“ ,7 V v (‘4 ‘C — d} Name Spring 2008 Exam 1 (Page 6 of 10) ParT III: ParTial crediT may be given Please show all work for calculaTions 14. (8 poinTs) Is 1.0 M H*soluTion under hydrogen gas aT 1.0 aTm capable of oxidizing silver meTal in The presence of 1.0 M silver ion? (Yes or no? Circle The correcT answer below) afTer you SHOW YOUR WORK AND EXPLANATION FOR CREDIT. O I + - .. ~——> x;;fi\g5<.g\ ~—-> 2A3 (cm [1M1 + ’16 t :- 0.90 \/ T ’ _ 0 ill (0% LiMl T 1e "—9H1(3\[iokw’g E : (100 V (a ,1 + f » r y 1N: "4’ DATA, +1H(oqlL’lM1"’l(\5(uq\[lM3 +H1(3\ Uakw’X t :voféov Qpl: 2 \ :, \/<€c\-:’(.%%IO%9 /. R (3)6"(Lxu843 gaff <0) HA6 Y‘YW \AOS (mg bendean *0 kuke I yes or, v vv‘dev HA€§€ Comdih‘ovxs A 3(9\ (Dvnnult be OXKOi26d. 15. (10 poinTs) IT is known ThaT: e' + Co3+ —> Cc” wiTh E° = +1.82 V. Explain why Co3+ salTs are unsTable in waTer. HinT: Show ThaT The above poTenTial is sufficienT To oxidize waTer by wriTing The appropriaTe half reacTions, and overall reacTion. Find The value of The equilibrium consTanT and reIaTe H To The Thermodynamic sTabiliTy of Co3*salTs. m. + Q o —-’:> 1Hw<e\~"*01(3\+uH (oqwue E :- L223V 7>T 2* .0 V , vi) H46 1v HCCJ (uq\ ’5 L'CO (0q\ '3 +L82V 7_ 3+ . 1+7 .\ 0 ’ — -—#> Lycra (oxen +1HLO(€\"> HCO (Gun T Guy + L\\\ (om ‘1 Jr 0-59 V (\ x" 0.0697. .O§Cl'2 ) r: (e17? 1' [03 Ken 0- SC) ‘3 O keel V‘ L‘ ' \’<€q i9 §Q «govaoc W ‘ ' * . - - ‘ _ o ,_ 3 {be mm is p‘vochcufix) avomfiahve ice. \cc /§f Yew) (0 Com is Convevkt’d *9 (U (“A ‘ (w uwd pvo<~¥k0£€~l Name Spring 2008 Exam 1 (Page 7 of 10) 16. (18 points) A voltaic cell consists of two half-cells. One of the half—cells contains a platinum electrode surrounded by chromium (III) and dichromate ions. The other half-cell contains a platinum electrode surrounded by bromate ions and liquid bromine. Assume that the cell reaction, which produces a positive voltage, involves both chromium (III) and bromate ions. The cell is at 25 °C. Information for the bromate reduction half reaction is as follows: ZBrO3’(aq) + 12H+(aq) + 102‘ —> Br2(|) + 6HZO E°red = + 1.478 V (a) Write the anode half reaction, the cathode half reaction, and the overall equation for the cell. (b) Calculate E° for the cell. (c) Write the cell description in abbreviated notation. (d) For the redox reaction in (a), calculate Keq and [36° in KY. (e) Calculate the voltage of the cell when all ionic species except H+ are at 0.1500 M and the pH is at - 0.301. .- ' 2 6 8Y0") (0“) ‘l l4 “1000* SC\’ “LO? 2+ lQCY (can l’ (cums. (c) I Spring 2008 Exam 1 (Page 8 of 10) k\ (d) tr(e€€:_.____O'OSV\C’Z_/€on5keq .‘ 29“" 7 N » - 0 025032 Z l 2 LIA : ‘ , /< 7 ‘ ‘ . s _ V k- 30 00 “4 fl km;jjgw 3 “W? €(e/ae \ > : Woe Hgg C/w0€ €'-)<C-\S ; PK ‘2» << _______ W, 1‘ U l 2). ' x ~ q -3 x m m ' ‘ ‘47 "M, 7M7\‘\«_..__¢v_\g_w~‘wkmfi // Keq = 1.0x1o76 AG° = - 4.3x102 kJ (e) Ew/g/Z : (gaff - 949-22-— G V\ m ,w, m «030‘ «a CM? 21-0 r- - s v, 3 ” [(3% 3'0 [BYOJ] 6 C0. (€00) ‘0 (o. (goo) __ g ‘A‘ _ (O.i§00\s (ORG) H . xx“, l/N) ‘* r w N AC5 i\( ': 1.0% ‘0 :’ (QM : 043. 3:35“ Em (moxxom) 1 30 c) .__ x" a" \ I M. 7 I ‘ I" {Lira/1 Gage : + o. H V} “NV ! \ _Name Spring 2008 Exam 1 (Page 9 of 10) Periodic Table ___]__fi l l H r l "2794 mom 4902602 M7." 4 6 i; 9 m l I Be C U l~ NC _6 4i 'Jtllll’fi 1mm 14.00674 159994 myomm 29.1797 I4 IS l H Al ‘1 P S U 36.981538 285855 1037376] 31066 35.4527 35 3] 32 EE: 1‘ a: 3m gm ., l8 s Ar 2.} ‘ 0: Ma’, 393MB ' to :n 21 :2 11 24 B 1" t; t 21 So ' ‘i V ‘r Mn , ‘ hr vying“; ' I, 54‘9‘3 I v ‘ __ ' __. ' i ,, , ' _ ,, ,, ,, , L _ A _ ' ‘ L ' it] 41 42 43 u 46 47 4t 39 50 5] 52 53 54 Ar Nb {N 0 "l e Ru 1 P Ag Cd In t n Sb e I he 9| 23-! 92.90638 95 94 (93) ID“)? 102290550 “Um/ll [M36822 l [14“ “4118 “8310 [21.760 127.60 1213,9044? [3‘39 ' as . M ,. 72 7:1 74 75 '76 n n 79 l l l 'l a W Re ()5 Ir t Au 173.49 130.9416 1123.3»: 1mm 190.21 192211 195m: :95. 55 E j: :3 £323 : I V 956 304 107 Ills 109 1H "2 R1 [3h HS Mt l M (2:61] (2363] (2155] (272) Some useful equations and constants: PLEASE NOTE: Important values and equations required for calculations are given with the respective problem. The following may or‘ may not be of any use. Constants: R = 8.3145 J / mol K F = 96,485 C/ mol e' NA = 6.022 x 1023 R : 0.08206 L atm / mol K c = 2.998 x108 m / s h : 6.626 x10'34 J s lnm :109 m 1 cal = 4.18 J Equations: AG° : —nFE:,,, AG = —nFEce” AG° = AH° — TAS° AG = AG° + RT in Q ,0 a 0.0592 I, 0.0592 AC, : "RT anuq Eye/l : Ecell # cell : n log Keq AE : 17v 0 = v 1 Charge (Coulomb) = Current (Ampere) x Time (second) Energy (J) : Charge (Coulomb) X Voltage (V) Energy (J) = Current (A) x Time (s) x Voltage (V) Energy flow = (mass) x (specific heat) x (temperature change) Name Spring 2008 Exam 1 (Page 10 of 10) Reduction Half-Reaction (V) W‘t‘mngvr Flea) + 2 c’ —-—> 2 F (aq) 2.87 Weaker wivin Hzozmq) + 2 mag) + 2 e- -——> 2 H200) 1.78 reducing Wm Mn0;(aq) + 8 mag) + 5 e” -——> Mn2+(aq) + 4H20(1) 1.51 age“ 02(3) + 2 e- ——> 2 Cl'(aq) 1.36 Cr2072‘(aq) + 14 H+(aq) + 6 e" ——> 2 Cr3*(aq) + 7 HZOU) 1.33 0M + 4 H+(nq) + 4 e— ——-9 21-1200) 123 Br3(l) + 2 e‘ --> 2 Br‘(aq) 1.09 Agflaq) + e” ——-> Ag(s) 0.80 Fequ) + e" —-——> Fe2*(aq) 0.77 02(3) + 2 H+(aq) + 2e' ——> H202(m7) 0.70 I2(5) + 2 e” —> 2 l‘(aq) 0.54 02(g) + 2 H200) + 4 e" —-> 4 OH‘(aq) 0,40 Cu2+(aq) + 2 e“ ——> Cu(s) 0.34 Sn“(nq) + 2 e" -—v~> Sn2*(uq) 0.15 2H+(aq>f - I’b2+(uq) + 29' -——> Pb(s) ~0.13 Ni2+(aq) + 20‘ -—9 Ni(s) —O.26 Cd2+(uq) + 2 e“ ——> Cd(s) —0.4O F02*(aq) + 2 e‘ ——> Fe(s) -O.45 Zn2+(aq) + 2 e' ——> 211(5) —0.76 2 H200) + 29' -——> H2(g) + ZOE-{'(aq) —0.83 A13+(nq) + 3 e- —--9 Al(s) 4.66 Walk“, Mgzfiaq) + 2 c“ -—> Mg(s) —2.37 WWW“, midi/.ng Na+(”‘7) + 0‘ _’ Na(5) “271 rmimwgg agent I.i*(r1q) + e" —-—> Li(s) ~3.04 ...
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This note was uploaded on 10/27/2009 for the course CHE 2C taught by Professor Enderle during the Winter '08 term at UC Davis.

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midterm1 - KEY J Last Name Lab Sec. # ; TA: Andreas...

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