Cathode (Reduction) Half-ReactionE° (volts)Half-ReactionE° (volts)Formation Constants for Complex Ions at 25 oC.Chemistry 142 Name ___KEY______ Martin LarterExam 3 May 2009 Page 1 (24 points)Page 2 (26 points)Page 3 (26 points)Page 4 (23 points)Page 5 (14 points)Total (113 points)Percent (100 %)All work must be shown to receive credit. Give all answers to the correct number of significant figuresConstantsNA = 6.022 x 1023 mol-1h=6.626x10-34 J sec c=3.00x108 m sec-1R = 8.3145 J/(mol K) =0.08206 L atm/(mol K)1 C= 1 A⋅ sF = 96,485 C/molEquationsΔG = ΔH – TΔSΔ=hc/λΔGo = – RT Ln KeqΔG = –nFEEcell=Ecello−0.0257 vn lnQG = Go + RT ln QpH= -Log[H+] Ln [A] = Ln [A]o – kt [A] = [A]o e- ktt1/2 = Ln2/ k 1 = 1 + 2kt [A]2 [A]o2Kw=Ka*KbPT = P1 + P2 + P3 + .............. sg = kHPgx=-b ± (b 2 – 4ac) ½2aThe spectrochemical series:I- < Br- < S2- < SCN- < Cl- < NO3- < F- < OH- < C2O42- < H2O < NCS- < CH3CN < NH3 < en < bipy < phen < NO2- < PPh3 < CN- < COGrossmont CollegePeriodic Table IAVIIA NOBLEGASES1H1.008IIAIIIA IVA VA VIA1H1.0082He4.0023Li6.9414Be9.0125B10.816C12.017N14.018O16.009F19.0010Ne20.1811Na23.0012Mg24.30IIIB IVB VB VIB VIIB VIII VIII VIII IB IIB13Al27.0014Si28.0915P30.9716S32.0617Cl35.4518Ar39.9519K39.1020Ca40.0821Sc44.9622Ti47.9023V50.9424Cr52.0025Mn54.9426Fe55.8527Co58.9328Ni58.7029Cu63.5530Zn65.3831Ga69.7232Ge72.5933As74.9234Se78.9635Br79.9036Kr83.8037Rb85.4738Sr87.6239Y88.9140Zr91.2241Nb92.9142Mo95.9443Tc(99)44Ru101.145Rh102.946Pd106.447Ag107.948Cd112.449In114.850Sn118.751Sb121.852Te127.653I126.954Xe131.355Cs132.956Ba137.357La138.972Hf178.573Ta180.974W183.975Re186.276Os190.277Ir192.278Pt195.179Au197.080Hg200.681Tl204.482Pb207.283Bi209.084Po(209)85At(210)86Rn(222)87Fr(223)88Ra226.089Ac227.0104Rf(261)105Db(262)106Sg(263)107Bh(262)108Hs(265)109Mt(266)110??(269)58Ce140.159Pr140.960Nd144.261Pm(147)62Sm150.463Eu152.064Gd157.365Tb158.966Dy162.567Ho164.968Er167.369Tm168.970Yb173.071Lu175.090Th232.091Pa231.092U238.093Np(237)94Pu(244)95Am(243)96Cm(247)97Bk(247)98Cf(251)99Es(252)100Fm(257)101Md(258)102No(259)103Lr(260)Section 1: Multiple Choice (3 pts/question)1. Molten PbCl2 is subjected to electrolysis in order to form elemental lead and chlorine. Which of the following is true?A. Elemental chlorine gas is formed at the cathode and bubbles awayB. Elemental lead metal is formed and deposited at the anodeC. Electrons flow from the cathode to the anodeD. Chloride ions are the reducing agents in the reactionE. none of the abovePb2+(aq) + 2e-  Pb(s) -0.13 V Reduction, oxidizing agent, cathode2Cl-(aq)  Cl2(g) + 2e- - 1.36 V Oxidation, reducing agent, anode2. A metal can be protected from corrosion by making it: A. Either electrode in an electrochemical cell. B. The electrolyte in an electrochemical cell. C. The cathode in an electrochemical cell.D. The anode in an electrochemical cell. E. none of the above3. Use Eo value to calculate Ksp of Ni(OH)2 at 25oC (hint one of the electrodes is Ni (s))a) 2 x10-16 b) 6x1010 c) 30 d) 1 x10-9e) 4x10-13 Ni(OH)2 (s) + 2e-  Ni(s) + 2OH-(aq) -0.72 VNi(s)  Ni 2+ (aq) + 2e - 0.23 V Ni(OH)2  Ni2+(aq) + 2OH-(aq) -0.49 VΔG = –nFEmole- * 96,485 C/mol*(-0.49 V) = 9.46 x104 JLn Ksp = – ΔGo /RT = 9.46 x104 J / (8.3145 J/(mol K) *298 K) = -38.18Ksp = 2.6 x10-174. Rank Sn2+ (aq), I- (aq) and Cu (s) in order of strongest to weakest reducing agenta) I- (aq) > Sn2+ (aq) > Cu (s) b) Sn2+ (aq) > Cu (s) > I- (aq) c) Cu (s) > I- (aq) > Sn2+ (aq)d) I- (aq) > Cu (s) > Sn2+ (aq) e) Cu (s) > Sn2+ (aq) > I- (aq)Sn2+(aq)  Sn4+(aq) + 2e--0.15 VCu(s)  Cu2+ (aq) + 2e- -0.34 V2I-(aq)  I2(s) + 2e- -0.54 V5. The tetrahedral complex ion [Cu(H2O)4]2+ has magnetic properties that correspond to how many unpaired electrons?a) 4 b) 3 c) 2 d) 1 e) 0Cu2+ [Ar]3d96. A bidentate ligand alwaysA. Has bonds formed to two metal ionsB. Has two donar atoms C. Has a charge of 2+ or 2-.D. Forms complex ions with a charge of 2+ or 2-.E. Has medical uses.7. A particular complex ion is observed to have an orange red color at approximate wavelength of 630 nm. What is the value of the crystal field splitting energy for this complexa) 188.5 kJ/mol b) 171 kJ/mol c) 213.8 kJ/mold) 278.4 kJ/mol e) 249.3 kJ/molWavelength absorbed 480 nm ∆E = 6.626 x 10 -34 J*s (3.00 x 10 8 m/s) = 4.14 x 10-19 J/particle 480 x 10-9 m4.14 x 10-19 J/particle * 1kJ/1000 J * (6.022 x 1023 particles / 1mol) = 249 kJ/mol8. In the qualitative analysis schemes, initial separation of the metal cations into various groups is based on_______________a. Difference in acid/base properties of the metal cationsb. Amphoteric propertiesc. Difference in Ksp valuesd. Ability to form ammonia complexese. Hydrolysis of the metal cationSection 2: Short answer/ essay1. Consider a voltaic cell based on the following unbalanced reactionPb (s) + IO3- (aq)  Pb2+ (aq) + I2(s) a) Write the balanced oxidation and reduction half reactions under acidic conditions (label which is oxidation and which is reduction), the overall reaction equation and calculate Eocell. (9 pts)Oxidation: _________5*(Pb (s)  Pb 2+ (s) + 2 e -)_____________Reduction: ________10 e - 12 H + +2 IO3- (aq)  I2(s) + 6 H2O _________________________Overall Reaction: _____ 5 Pb (s) + 2 IO3- (aq) +12 H +  5 Pb 2+ (aq) + I2(s) + 6 H2O ____________b) Write the line notation for the cell __Pb(s)/Pb 2+ ║ IO3- ,H + / I2(s),Pt(s)___ (3 pts)c) Draw a fully labeled Free energy diagram for this reaction at 298 K. Indicate and clearly label on the diagram (i) a point where the reaction is spontaneous, and (ii) the equilibrium point. Explain your logic behind your graph (calculations are not necessary). (6 pts)Explain: EΔG = -d) Draw a picture of the cell. Indicate the composition of all parts of the cell, the direction of electron flow, and the direction of ion movements within each compartment. (8 pt)FreeEnergyReactants ProductsΔGo ReactantΔGo ProductsSpont RxnEquil2. Give the IUPAC name or formula, oxidation state, coordination number and electron configuration (8 pts)Name Formula Oxidation # Coordination # Electron Configamminecyanobis(ethylenediammine)colbalt(III)