CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea. 1. The anti-cancer drug cis-platin is the complex: cis-[pt(nh ) (Cl) ]. In this complex, the

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1 CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea Sme useful cnstants: ln(10) = 2.303, R = J/ml@K, F = 96,00 cul/ml, 2.303RT/F = V at 2EC. Assume a temperature f 2EC unless tld therwise. Chse the ne best answer. Prgrammable calculatrs and telepathic amplifiers are nt permitted. 1. The anti-cancer drug cis-platin is the cmplex: cis-[pt(nh ) (Cl) ]. In this cmplex, the xidatin state f Pt is _+2_, the # f d electrns is _8_, and the crdinatin number is _4_. a. +2, 8, 4 b. +4, 6, 4 c. +2, 4, 6 d. +4, 10, 4 e. +2, 10, 4 2. Recently in Dr. Finklea s lab., an undergraduate synthesized the cmplex [C(en) ](Br) (where 3 3 en = ethylenediamine). What type f ismerism des this cmplex exhibit? a. cnstitutinal b. linkage c. inizatin d. enantimers en = ethylenediamine, which is a bidentate ligand. Octahedral cmplexes with identical bidentate ligands have nn-identical mirrr images. e. diastereismers The cmplex [Mn(H2O) 6] is high-spin. Hw many unpaired electrns are present? a Mn is a d case. 8 8 b c. 3 High-spin means that all f the electrns are unpaired. d. 4 e. 1

2 What is the cncentratin f Cu in a slutin cntaining 0.10 M [Cu(NH 3) 4 ]? K f = a M Cu + 4NH 3 º [Cu(NH 3) 4] 4 b M 0+x 0+4x 0.10 x M c M K f = = (0.10)/{(x)(4x) } d M 26x = 0.10/ e M x =.1 10 M. When phtgraphic film r paper is develped, the uncnverted AgBr is disslved in a fixer 2 bath cntaining thisulfate (S2O 3 ). What cncentratin f thisulfate is needed t disslve 1.0 mle f AgBr(s) in 1.0 L f water? 2 3 AgBr(s) + 2S2O 3 º [Ag(S2O 3) 2] + Br K = 16 a. 2.2 M x M b. 16 M K = 16 = {(1.0)(1.0)}/(x 2.0) 2 c. 0.2 M take square rt: 4.0 = 1.0/(x 2.0) d M x 2.0 = 1.0/4.0; x = 2.2 M e M 6. What treatment is used t precipitate Grup 4 metal catins in the Qualitative Analysis scheme? a. Add HCl until [Cl ] is abut 0.3 M. b. N cnditin will precipitate Grup 4 ins; they are sluble under all cnditins. c. Add NH 3 until the ph is 8, and saturate the slutin with H2S. d. Make sure the slutin is acidic (ph 1), and saturate the slutin with H2S. e. Add NH 3 and (NH 4) 2CO 3 t the slutin. 7. The famus smg mnster f Ls Angeles is tinted with N2O 4. Given the relevant free energies f frmatin, calculate the equilibrium cnstant fr the fllwing reactin: 2NO 2 º N2O4 )G kj/ml f a )G = (+97.8) 2(+1.3) = 4.8 kj = 4800 J b. 1.0 )G = 2.303RT lg(k) c lg(k) = (4800 J)/{(2.303)(8.314)(298)} = d K = 10 = 6.9 e

3 14 8. At rm temperature, the water dissciatin cnstant K w = What is )G fr the reactin: H2O(l) º H + OH? a kj 14 )G = (2.303)(8.314)(298)lg( ) = +79,900 J = kj b kj c kj d kj e kj 9. )G fr the fllwing reactin is +.6 kj. What is )G fr the reactin if [Ag ] = [Cl ] = M? AgCl(s) º Ag + Cl Wrk in Jules. a kj 2 10 Q = [Ag ][Cl ] = ( ) = b kj )G = )G RT lg(q) c kj 10 )G = (+,600) + (2.303)(8.314)(298) lg( ) d. +18 kj )G = (+,600) + (,800) = 200 J = 0.2 kj e. 0.2 kj Significant figures and units are imprtant here 10. Chse the false statement. a. During water electrlysis, hydrgen gas is prduced at the cathde. b. In cell ntatin, the cathde is n the right side. c. In a redx reactin, a reductant is xidized. d. When a mlecule is reduced, it lses electrns. (reductin - gain e ) e. In an xidatin half-reactin, electrns appear as a prduct. 11. What is the crrect cell ntatin fr the fllwing redx reactin? Cl 2(g) + 2Fe(s) º 2Cl + 2Fe Cl 2 is reduced 6 cathde; Fe(s) is xidized 6 ande a. Pt(s) * Cl 2(g), Fe(s) ** Cl, Fe * Pt(s) b. Pt(s) * Cl 2(g), Cl ** Fe * Fe(s) 2 c. Fe(s) * Fe ** Cl 2(g), Cl * Pt(s) d. Pt(s) * Cl 2(g), Cl ** Fe, Fe(s) * Pt(s) e. Cl 2(g) * Cl ** Fe * Fe(s) 3

4 12. )G fr the decmpsitin f hydrgen perxide (a redx reactin) is kj. What is )E fr this reactin? 2H2O 2 (l) º 2H2 O(l) + O 2(g) Data: H2O 2 + 2H + 2e º 2H2O O 2 (g) + 2H + 2e º H2O2 Wrk in Jules a V )G = nf)e ; n = 2 e transferred in cell reactin. b V 233,600 J = (2)(96,00))E c V )E = V d V e V 13. What is )E fr the fllwing cell? Hg(l) * Hg ** Zn * Zn(s) Data: 2 Hg + 2e º Hg(l) E = +0.8 V (ande) 2 Zn + 2e º Zn(s) E = 0.76 V (cathde) a V )E = (0.76 V) + (0.8 V) = 1.61 V b V c V d V e V 14. In the preceding electrchemical cell, identify the strnger xidant. a. 2+ Hg The cell reactin is: Hg(l) + Zn º Hg + Zn(s) b. Hg(l) The tw xidants are Zn and Hg. c. 2+ Zn The half-reactin with the mre psitive E has the strnger xidant. d. Zn(s) 1. What is )E fr the fllwing redx reactin if the cncentratins f all species in slutin are M? Pb(s) + 2Fe º Pb + 2Fe Data: 2 Pb + 2e º Pb(s) E = 0.13 V (ande) 3 2+ Fe + e º Fe E = V (cathde) a V )E = )E (2.303RT/nF) lg(q); n = 2 e transferred b V )E = (+0.77) + (+0.13) = V c V Q = {[Pb ][Fe ] }/[Fe ] = (0.010) /(0.010) = d V )E = (+0.90) (0.0916/2) lg(0.010) = (+0.90) (0.0916) e V )E = V 4

5 16. What is the ph in the left half cell f the fllwing cell if the cell ptential is +0.9 V? + 2+ Pt * H 2 (g) (P H2 = 1 atm), H (? M) ** Hg (1 M) * Hg(l) Data: 2 Hg + 2e º Hg(l) E = +0.8 V Ande: H 2(g) º 2H + 2e E = 0 V (by definitin; SHE) 2+ a. 3.4 Cell rxn: H 2(g) + Hg º 2H + Hg(l); n = 2 e transferred b )E = (+0.8) + (0) = +0.8 V c Q = [H ] /{P H2[Hg ]} = [H ] /{(1.0)(1.0)} = [H ] d )E = +0.9 = (+0.8) (0.0916/2) lg([h ] ) = (+0.8) + (0.0916/2)(2)pH e. 1.7 ph = {(+0.9) (+0.8)}/(0.0916) = 0.10/ = A greedy electrchemist cllects all the gld waste frm his clleagues and prceeds t electrdepsit the gld. The equipment and slutin nly permit a sustained current f A. What mass f gld is depsited after 1.00 hur? 3 Au + 3e º Au(s) n = 3e per Au(s) a g Q = i@t = (1.00 h)(60 min/hr)(60 s/min) = 3600 cul b g 3 Q/nF = (3600 cul)/{(2)(96,00)} = ml Au c g 3 ( ml Au)( g Au/ml) = 0.24 g Au d g e g 18. An aluminum can cntains 27. g f Al. Hw many culmbs f charge were required t prepare that much aluminum frm aluminum re? 3 Al + 3e º Al(s) n = 3e per Al(s) a. 96,00 cul (27 g Al)(1 ml Al/27.0 g) = 1.0 ml Al b cul Q/{(3)(96,00)} = 1.0 ml; Q = cul c cul d. 32,000 cul e cul 19. When the lead-acid battery is recharged, what happens? Chse the crrect statement. a. The cncentratin f sulfuric acid (as H and HSO 4 ) decreases. (npe; it increases) b. PbO 2 is cnverted t Pb at the ande. (xidatin ccurs at the ande) c. PbSO 4 is cnverted t Pb at the cathde.

6 20. Cnsider a redx reactin fr which )H is negative and )S is psitive. Which f the fllwing statements are true? I. )G is negative. II. )E is psitive. III. The equilibrium cnstant is greater than 1. True True True a. I nly )G = )H T )S b. I and III nly Since T is psitive, )G must be negative. c. All f the statements are true. )G = nf)e ; hence, )E is psitive. d. II nly )G = 2.303RT lg(k); hence, lg(k) > 0, e. Nne f the statements are true. and K > 1. 6