Version 001 – Exam 3 – vanden bout – (51640) 1This print-out should have 31 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.Msci 21 1208001 10.0 pointsConsider the following electrode reactions:Fe3++ 1 e−→ Fe2+E0= +0.771 VI2+ 2 e−→ 2 I−E0= 0.535 VWhat would be E0cellfor the spontaneousreaction?1. +1.007 V2. +0.236 V3. −1.306 V4. −0.236 V5. +1.306 VChemPrin3e T12 59002 10.0 pointsIn the cell shown, A is a standard Zn2+| Znelectrode connected to a standard hydrogenelectrode (SHE).+ −VoltmeterSalt bri dgeH+APt(s)H2(g)If the voltmeter reading is +0.76 V, whichhalf-reaction occurs in the left-hand cell com-partment?1. Zn(s)→ Z n2+(aq) + 2 e−2. Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)3. Zn2+(aq) + 2 e−→ Zn(s)4. Zn2+(aq) + H2(g) → Zn(s) + 2 H+(aq)Cell Type 02003 10.0 pointsWhat is the cathode inMg(s) | Mg2+(aq) || Au+(aq) | Au(s)Mg2++ 2 e−→ Mg E◦red= −2.36Au++ e−→ Au E◦red= +1.69and what is the cell type?1. Au+(aq) | Au(s); a battery2. Mg(s) | Mg2+(aq); an electrolytic cell3. Mg(s) | Mg2+(aq); a battery4. Not enough informat ion is provided.5. Au+(aq) | Au(s); a n electro lytic cellEC Cell Nomenclature 006004 10.0 pointsIf the two half reacti ons below were used tomake an electrolytic cell, what species wouldbe consumed at the cathode?Half reaction E◦Cu2+(aq) + 2 e−−→ Cu(s) +0.34Fe3+(aq) + e−−→ 2 Fe2+(aq) +0.771. Fe2+(aq)2. Fe3+(aq)3. Cu(s)4. Cu2+(aq)ChemPrin3e T12 02005 10.0 pointsConsider the reactionS2O2−4(aq) → SO2−3(aq)in basic solution. How many electrons appearin the balanced half-reactio n?Version 001 – Exa m 3 – vanden bout – (51640) 21. 62. 33. 14. 25. 4ChemPrin3e T12 06006 10.0 pointsConsider the reactionZn(s) + OH−(aq) + H2O(ℓ) + NO−3(aq) →Zn(OH)2−4(aq) + NH3(g)If the coefficient of NO−3in the balancedequation is 1, how many electrons are trans-ferred in the reactio n?1. 42. 23. 64. 105. 8Msci 21 1204x007 10.0 pointsGiven the foll owing standard reduction po-tentialsFe3++ e−⇀↽Fe2++0.771 VCu2++ 2 e−⇀↽Cu +0.337 VSn2++ 2 e−⇀↽Sn −0.140 VWhich of the following species would be thestrongest oxidizing agent?1. Cu2+2. Cu3. Sn4. Fe2+5. Fe3+6. None of the species listed can act as anoxidizing agent.7. Sn2+ChemPrin3e T12 46008 10.0 pointsThe sta ndard voltage of the cellAg(s) | AgBr(s) | Br−(aq) || A g+(aq) | Ag(s)is +0.73 V at 25◦C. Calculate the equilibriumconstant for the cell reaction.1. 4.6 × 10−132. 5.1 × 10143. 2.0 × 10−154. 2.2 × 10125. 3.9 × 10−29LDE G from E Calc 001009 10.0 pointsWhat is ∆G◦for the reaction bel ow?Reaction E◦ClO−3+ 6 H+(aq)−→12Cl2(g) + 3 H2O(ℓ) +1.471. 194, 000 kJ · mol−12. −1, 418 kJ · mol−13. −709 kJ · mol−14. 194 kJ · mol−1Spontaneous Redox 002010 10.0 pointsConsider the half reactionsZn2++ 2 e−→ Zn(s) E◦= −0.76Ag1++ e−→ Ag (s) E◦= +0.80What will happen i f you mix a solutio n thatis 1M Zn(NO3)2, 1 M AgN O3and place intoVersion 001 – Exa m 3 – vanden bout – (51640) 3the solution a piece of solid Ag and a piece ofsolid Zn?1. nothing wi ll happen2. Ag meta l will form on the solid Ag3. Zn metal w ill form on t he solid Zn4. Zn metal w ill form on t he solid Ag5. Ag meta l will form on the solid ZnMlib 08 0086011 10.0 pointsA battery has two electrodes labeled anodeand cathode. Electro ns flow from the (anode,cathode) to the (anode, cathode) through theexternal circuit and (an oxidation, a reduc-tion) reaction occurs at the cathode.1. anode; cathode; reduction2. cathode; anode; reduction3. anode; cathode; oxidation4. cathode; anode; oxidationChemPrin3e T12 69012 10.0 pointsIf 8686 C of charge is passed through moltenmagnesium chloride, calculate the number ofmoles of Mg(ℓ) produced.1. 2.00 mol2. 0.0110 mol3. 0.0225 mol4. 0.0900 mol5. 0.0450 molCell Current013 10.0 pointsWhat is the average current generated in t heCu(s) | Cu2+(aq) || Fe3+(aq) | Fe(s)electrochemical cell if 50 g of Cu(s) ar e usedup i n a 24 hour period?Cu2++ 2 e−→ Cu E◦red= +0.22 VFe3++ 3 e−→ Fe E◦red= −0.04 V1. 13.00 amp2. 1.76 amp3. 2.64 amp4. 111.85 amp5. 42.17 ampLDE Nernst Equation Calc 001014 10.0 pointsA battery formed from the two half re-actions below di es (reaches equilibrium). If[Fe2+] was 0.24 M i n the dead bat tery, whatwould [Cd2+] be in the dead battery?Half reaction E◦Fe2+−→ Fe −0.44Cd2+−→ Cd −0.401. 120.3 M2. 5.4 M3. 0.0.0005 M4. 0.01 MConcentration Cell 001015 10.0 pointsConsider the following standard volta ic cellZn | Zn2+(1 M) || Cu2+(1 M) | CuWhat will happen if the concentration ofthe Cu2+side of t he cell is reduced to 10−2M?1. The voltage will decrease2. The voltage will stay the sameVersion 001 – Exa m 3 – vanden bout – (51640) 43. It depends on the value of E◦4. The voltage will increaseChemPrin3e T13 19016 10.0 pointsConsider t he concentration-time dependencegraph for two first-order reactions.[A]0Molar Co ncentrationof ReactantConcentration −→Time −→Which reacti on has the larger rate con-stant?1. the reaction represented by t he uppercurve2. Unable to determine3. the reaction represented by the lowercurveLDE Assigning Rate Expressions 003017 10.0 pointsConsider the reaction below. Which o f thefollowing is a correct expression of the rate?12S8(s) + 3 O2(g) −→ 2 SO3(g)1.2 · ∆[SO3]∆t2. −4 · ∆[S8]∆t3. −∆[SO3]2 · ∆t4.∆[O2]3 · ∆t5. −∆[O2]3 · ∆tChemPrin3e T13 04018 10.0 pointsThe rate of formation of NO2(g) in the reac-tion2 N2O5(g) → 4 NO2(g) + O2(g)is 5.78 (mol NO2)/L/s. What is the rate atwhich N2O5decomp oses?1. 2.89 (mol N2O5)/L/s2. 5.78 (mol N2O5)/L/s3. 0.723 (mol N2O5)/L/s4. 1.45 (mol N2O5)/L/s5. 11.6 (mol N2O5)/L/sChemPrin3e T13 18019 10.0 pointsConsider t he concentration-time dependencegraph for a first-order react ion.[A]0Molar Co ncentrationof ReactantABCConcentration −→Time −→At which point on the curve is the reactionfastest?1. The ra tes are the same at all point s.2. CVersion 001 – Exa m 3 – vanden bout – (51640) 53. A4. B5. A + t1/2ChemPrin3e T13 17020 10.0 pointsFor the reaction2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)the following data were collected.[NO(g)] [H2(g)] Rate, M/s0.10 0.10 0.00500.10 0.20 0.0100.10 0.30 0.0150.20 0.10 0.0200.20 0.20 0.040What is the rate law for this reaction?1. rate = k [H2] [NO]2. rate = k [H2] [NO]23. rate = k[H2][NO]24. rate = k[H2][NO]5. rate = k [H2]2[NO]6. None of theseChemPrin3e T13
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