105bs10_warshel_e4_key

105bs10_warshel_e4_key - Chemistry 105bL 9:00 am. Section...

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Unformatted text preview: Chemistry 105bL 9:00 am. Section PLEASE PRINT YOUR NAME IN BLOCK LETTERS First Letter of Last Name Spring 2010 Name: EXAM #4 3 April 22, 2010 Last 4 Digits SID#: g, Professor Arieh Warshel Lab T.A.’s Name: Lab Day and Time: Question MaXimum T Score Grader 1 Please Sign Below: i Points 1 20 I certify that I have 2 L— 20 _| observed all the rules of _1 Academic Integrity while 3 20 taking this examination. 4 t 20 5 20 —‘ Signature TOTAL 100 INSTRUCTIONS: 1. There are 5 problems on 8 pages in this exam. Please check to see that you have a complete exam. 2. Please show all of your work to receive full credit. 3. You must use ink. Do not use pencil, erasable ink, or whiteout. 4. Note the equations, reduction potentials and periodic table on page 8. l. (20 points) /- (a) Draw a commercial battery (explain all components and write the relevant reaction in the L9 anode and cathode.) f , M e.» mill” M“ 2‘“ “a ‘7‘” *3 W: + 2H“ 01 i" 2c“ M KM aan/Ofi + 219% {Vi/1’9 04' w {Ml W .2” + 2W4”; 2H,, 0;, “(hf % Z7422, Mnyfi} f 2N”; 7”};3 (b) Draw a fuel cell (explain all components and write the relevant reaction in the anode and @ cathorilc). {a 2. (20 points) (a) What is 5° for this cell at 25°C? ZXLAfi" +¢a-> (b) Write the anode reaction. 9 grid?) “‘9 %%/?)+ Zc" (c) What will be the potential at 25°C of the above cell when the concentration of Ag+ is 1.0 M and that Oon2+ is 10‘10 M? 5 0,, @512- a; e g “’5 9- 42W)” 33/,Q+9,30==@ (d) What will be the concentration of Zn2+ that will lead to the condition 6 = 0 when [Ag*] = 1.0 M? @ __ ‘3’ 0-0:?2' [£th 0 ‘“ ’5’ "' W?“ 53’7"“ 1t LIZ 5‘2 [gsfls 5:? m t: 3. (20 points) _ _ (a) What will be the potential ofa Daniel cell with initial concentrations of 1.000 M CuSO4 and 1.000 M ZnSO4 in the cathode and anode, respectively, after 1999 M of OH” is added to the anode side. Note that KSp for Zn(OH)2 is 10‘”). 1+ C u. + 0 . 3 4:” £14” +- Wfl‘ua «211(2ng (3) .0120 I. 9?? a” W“:st .. 0 3‘74; 5': 44 lo 0.0005" 0 WWWM -- J3... ‘ZJRKNQL “0.3 27,” + WH.” tsf: /0 0 *0” 0.012] fx 2.x 2:- r z) [o :- [a.oa(+x)(2..x) ~(O.D¢I) 54X x: /.6x/o~¢ C%M?= aoooéé '3: 0'00/ 0 Toma, u» “" 2' Cu ’1'“ WWW—M‘”'“ — ' ’ ‘ a i ' T ' M “““““““““““ W ''' mm “W ,,,,,,,,,,,,,,,,,,,, ,s I./0 g :1 + 0,033? .1:th (b) Please Circle the correct answer. (i) The potential at the anode is negative. 5 (ii) A lead cell is very light. rue lg (iii) The Hall process is used in purifying water. True False (iv) Fuel cells use catalysts. (v) It’s more effective to do work by blowing air from True 3 high pressure balloon than by electricity. (vi)Al can be produced from distilled water by electrolysis. True (as; (vii) Al is a good reducing agent. @ False (viii) The absolute value of the standard reduction True False potential of H+ is zero. 4. (20 points) (3) Give the molecular formula (using symbols like K3[Cr(N)6]) for the following: (ii) pentaaquachlorocobalt (III) chloride C» CO S" a 7 661» @ (b) Give systematic names for the following: (i) [Crt‘NHmCuClLWmem (121‘) “lavian (ii) K4[M0(CN)2C14] . I (C) Is the following isomer optically active? @ . (d) Name a use of Cis Platinum. Answer: Cgéxm 0 *flafwe mm flea/W (6) Use a drawing and explain why EDTA binds metal ions more strongly than ' ' monodentate ligands. L L. L A\C«L L. / Z. L 5340 L'“L "KL (’ \ g) (L \ L /L. ....5 L/+\__”L_ LLJL kgflf .. Tflséiwfi'fl is fasifv‘u I?» M r4 {ésuz} M “ushfl’ 1‘5 "y ‘ a >9 ,2 $5, A57 04.44, a [‘5 Marc MlyafiVev £3 féS‘I ,1, W in W A! 6 5. (20 points) __ (a) How long will you have to run a current of 10.0 Amps (at 110V) through a solution of '= CuSO4 to obtain 34 grams of copper metal? Sljatx ngiflz’xfififéx gm" ‘5' 5 (WW I'M” loner/7 . :7 25m: 7’ 2.95)“ (b) Will you need more time or less time in order to obtain the same amount of copper metal when the potential is 220V. /,0%x/o¢s SW (c) What is the free energy of converting 1.0 g of Ag to Ag+ when paired with the hydrogen ' electrode? a g =2 ~ 0 .813 V A5»; me—e" [M 104.12" 9mm \ dw-“(l-Dg/lgxfifi “5'8”? 536:” +- 7/3”“ J” : +7.2wX/02’f Ow) Mark True or False for the following: ’7 7””) (i) E is equal to zero in every cell at equilibrium. if True? False X i . . ‘ - ' .x (i1) Nemst dlscovered the third law of thermodynamics. Clue/j False (iii) The effect of concentration on the emf is usually larger True than that of having different metals. -- (iv) Sn provides effective protection from corrosion. True @ (v) Electric cells can be used to determine Ksp. I False (vi) The electronic configuration of Cr+2 is [Ar]3d34s2. I {/F‘E kwyjfi Exam 4 Exam 4 VIII -, , j“, H II III IV V VI VII He ‘ l 008 4 003 V‘iiw ’ #f— “A 5 "6 fl”- m" 7 " i1 """"" _ ‘9’" "175‘ k Li Be B C N O F ' Ne ’ 0941 9012 1011 1201 1401 I60 1900 201:; ‘1‘1’ 12 M“ 473*” ‘ 14 15 ‘fiwfiflwflu 17" " ‘1'? fl Na M g Al Si P S Cl Ar 22 99 ’4 3| 2098 23 09 30 97 32 07 35 45 3'4 95 PETS """""" 30 ""21 ""22 23 23 25 W 20 27 28 29 30 31 32 33 34 35 1'?) ' K Ca Sc Ti V Cr Mn Fe Co Ni C u Zn Ga Ge As Se Br Kr 39 10 401011 44 909 47 33 50 94 51 990 54 9380 55 1147 5119332 51; 09 03 540 05 377 09 72 72 59 l 74 9210 73 90 79 90 111 110 37 ' 38 '39 1:0 " 41 42 43 44’ ""45 “30 I 47 48 F 49 '~50 51 "g 52 "t " 53 "T" 34 , Rb Sr Y Zr Nb M0 Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe ‘ 11547 8702 8891 9122 9291 9594 1991 1011 1029 10014 1079 1124 11418 11117 1213 1270 1269 1311 " "" "'5? 50 "57" 72 I 73 74 75 76 “T” 77 * 7s 79 “7‘ so 111' 82 83' ' a 84 " 35 31?" C 5 Ba La Hf Ta W Re 0 5 Ir Pt A u Hg Tl P b Bi Po At R n 1329 1373 1329 17135 1309 133115 11102 1902 1922 19509 L 1970 L 2000 2044 2072 2090 12091 1210) 1222 1 W137 ”””” "I """" 31?" "W 139 _ 104 _ 105 100 107 I 108 109 110 111 112 113"” ” 114 115 n” E M ' 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Uub Uut Uuq Uup 12231 220 o 227 0 J (2011 (262) 12031 12021 12051 12133) (2711 (2721 l """""""""""""""""""""""""""""""""" “38 59 667” 01 W02 03 _ 04 65 I 00 M W07 ‘ 08 7 "(Jim 76'“ I "71' 1 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 14019 1442 11451 1504 151,90 1573 15319 102 1049 1073 10119 1730 175 1 ’ ””” ” ' """""""" " "90 ‘91 ' 92 ' 93 ‘94 95 ’ 96 97 9s 1 " <59 ’ “100 101 """ " 10‘2"" Actinides Th Pa U Np Pu Am Cm Bk Cf ‘ Es Fm Md No Lr 232 0 231 0 2311 0 2370 (2441 (2431 12471 12471 (2511 l 1252) 12571 (258) (259) 1201 Additional Information g = 60 - £7; an "f g z 50 _ 0.059210% Q n 60 : 0.059210g10 KW 71 A0001): —96.5n8(v01t) AG = AGO + RT 1n(Q) AG" 9 -RT ln(K) AG“ 2 AH" — ms" R = 8.314 J 111011 K’1 = 0.08206 L atm mor‘ K" F = 96,485 C/mol Half reactions Fe+2 + 26‘ —> Fe(s) Cu+2 + 26‘ —-) Cu(s) Zn+2 + 213’ —> Zn(s) Ni+2 + 212’ —> Ni(s) Mg+2 + 26‘ —> Mg(s) Cd+2 + 26‘ —> Cd(s) 511*2 + 2e‘ —9 571(5) AP3 + 3e' ——9 Al(s) Ag'” +e' ——> Ag(s) 60 volts -O.44 0.34 -O.76 -0.23 -1.03 —0.40 -0.14 -1.66 0.80 m Exam 4 Exam 4 84 ...
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105bs10_warshel_e4_key - Chemistry 105bL 9:00 am. Section...

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