课程名称︰普通化学乙下 课程性质︰必修 课程教师︰李弘文 开课学院：理学院 开课系所︰心理系 考试日期（年月日）︰2009.05.06 考试时限（分钟）：110 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : I. True/False Section: (20%, 2 pts each) For a statement to be true, it must be always true. 1.F The outer s- and p-orbitals become closer in energy going down the group. 2.T In the Daniell cell, the anode is zinc metal and the cathode is copper metal. When the cell operates, the anode gets smaller and the cathode gets larger. 3.F The electron affinities of elements are always positive. 4.T Bromine (Br2) can be obtained from brines by oxidation of Br- by Cl2. 5.T In a working electrochemical cell (+ cell voltage), the electrons flow from the anode through the external circuit to the cathode. 6.F When KI(aq) is electrolyzed at a concentration of 1 M, the product at the anode is I2. 7.T A solution is prepared by mixing equal volumes of 0.40 M HF(aq) with 0.20 M KOH(aq). This solution is a buffer. 8.F Polarizability is a measure of the ease with which an electron cloud can be distorted and is greatest for electron-rich atoms like fluorine. 9.T The pH of 0.50 M HNO2(aq) is 1.8. Therefore, the pH of a solution that is 0.50 M HNO2(aq) and 0.10 M KNO2(aq) is greater than 1.8. 10.F When equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value. I. True/False Section: (20%, 2 pts each) For a statement to be true, it must be always true. 1.F The outer s- and p-orbitals become closer in energy going down the group. 2.T In the Daniell cell, the anode is zinc metal and the cathode is copper metal. When the cell operates, the anode gets smaller and the cathode gets larger. 3.F The electron affinities of elements are always positive. 4.T Bromine (Br2) can be obtained from brines by oxidation of Br- by Cl2. 5.T In a working electrochemical cell (+ cell voltage), the electrons flow from the anode through the external circuit to the cathode. 6.F When KI(aq) is electrolyzed at a concentration of 1 M, the product at the anode is I2. 7.T A solution is prepared by mixing equal volumes of 0.40 M HF(aq) with 0.20 M KOH(aq). This solution is a buffer. 8.F Polarizability is a measure of the ease with which an electron cloud can be distorted and is greatest for electron-rich atoms like fluorine. 9.T The pH of 0.50 M HNO2(aq) is 1.8. Therefore, the pH of a solution that is 0.50 M HNO2(aq) and 0.10 M KNO2(aq) is greater than 1.8. 10.F When equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value. 11.A Use the following: Pb(s) |PbSO4(s) |SO42-(aq, 0.60 M) || H+(aq, 0.70 M) |H2(g, 192.5 kPa) |Pt In this cell, if oEis 0.36 V at 25°C, what is the Nernst equation for the cell at this temperature? A) E = 0.36 - 0.012 85 ln[1.90/{(0.70)2(0.60)}] B) E = 0.36 + 0.012 85ln[1.90/{(0.70)2(0.60)}] C) E = 0.36 + 0.012 85 ln[192.5/{(0.70) (0.60)}] D) E = 0.36 - 0.025 69ln[l92.5/{(0.70)2(0.60)}] 12.A How many moles of O2(g) are produced by electrolysis of Na2SO4(aq) if 0.120 A is passed through the solution for 65.0 min? A) 0.00 121 mol B) 0.000 080 8 mol C) 0.002 42 mol D)0.00485 mol 13.D If a small amount of a strong base is added to buffer made up of a weak acid, HA, and the sodium salt of its conjugate base, NaA, the pH of the buffer solution does not change appreciably because A) the Ka of HA is changed. B) No reaction occurs. C) the strong base reacts with A-to give HA, which is a weak acid. D) the strong base reacts with HA to give A-, which is a weak base. 14.B What is the pH of an aqueous solution that is 0.10 M HCOOH (Ka =1.8 x 10-4) and 0.10M NaHCO2? A) 10.26 B) 3.74 C) 5.74 D) 2.38 15.A A buffer contains equal concentrations of NH3(aq) and NH4Cl(aq). What is the pH of the buffer? (Kb (NH3) = 1.8 x 10-5) A) 9.26 B) 4.74 C) 7.00 D) 13.00 16.C Write the proper cell diagram for the following reaction: 2AuCl(s) + H2(g)→ 2Au(s) + 2H+(aq) + 2Cl-(aq) A) Pt |Cl-(aq) |H+(aq) || H2(g) |AuCl(s) |Au(s) B) Au(s) |AuCl(s) |Cl-(aq) || H+(aq) |H2(g) |Pt C) Pt |H2(g) |H+(aq) || Cl-(aq) |AuCl(s) |Au(s) D) Pt |H2(g) |H+(aq) Cl-(aq) |Au(s) |Pt E) Au(s) |AuCl(s) |H+(aq) || Cl-(aq) |H2(g) |Pt 17.C Beryllium(Be) and aluminum(Al) have similar properties because A) both have similar electron affinities. B) both have almost identical ionization energies. C) both ions are highly polarizing due to their small size and high charge. D) both have virtually identical atomic radii. 18.D At the stoichiometric point in the titration of 0.130 M HCOOH(aq) with 0.130 M KOH(aq), A) the pH is 7.0. B) [HCOOH] = 0.0650 M. C) [HCO2-] = 0.130 M.D) the pH is greater than 7. 11.A Use the following: Pb(s) |PbSO4(s) |SO42-(aq, 0.60 M) || H+(aq, 0.70 M) |H2(g, 192.5 kPa) |Pt In this cell, if oEis 0.36 V at 25°C, what is the Nernst equation for the cell at this temperature? A) E = 0.36 - 0.012 85 ln[1.90/{(0.70)2(0.60)}] B) E = 0.36 + 0.012 85ln[1.90/{(0.70)2(0.60)}] C) E = 0.36 + 0.012 85 ln[192.5/{(0.70) (0.60)}] D) E = 0.36 - 0.025 69ln[l92.5/{(0.70)2(0.60)}] 12.A How many moles of O2(g) are produced by electrolysis of Na2SO4(aq) if 0.120 A is passed through the solution for 65.0 min? A) 0.00 121 mol B) 0.000 080 8 mol C) 0.002 42 mol D)0.00485 mol 13.D If a small amount of a strong base is added to buffer made up of a weak acid, HA, and the sodium salt of its conjugate base, NaA, the pH of the buffer solution does not change appreciably because A) the Ka of HA is changed. B) No reaction occurs. C) the strong base reacts with A-to give HA, which is a weak acid. D) the strong base reacts with HA to give A-, which is a weak base. 14.B What is the pH of an aqueous solution that is 0.10 M HCOOH (Ka =1.8 x 10-4) and 0.10M NaHCO2? A) 10.26 B) 3.74 C) 5.74 D) 2.38 15.A A buffer contains equal concentrations of NH3(aq) and NH4Cl(aq). What is the pH of the buffer? (Kb (NH3) = 1.8 x 10-5) A) 9.26 B) 4.74 C) 7.00 D) 13.00 16.C Write the proper cell diagram for the following reaction: 2AuCl(s) + H2(g)→ 2Au(s) + 2H+(aq) + 2Cl-(aq) A) Pt |Cl-(aq) |H+(aq) || H2(g) |AuCl(s) |Au(s) B) Au(s) |AuCl(s) |Cl-(aq) || H+(aq) |H2(g) |Pt C) Pt |H2(g) |H+(aq) || Cl-(aq) |AuCl(s) |Au(s) D) Pt |H2(g) |H+(aq) Cl-(aq) |Au(s) |Pt E) Au(s) |AuCl(s) |H+(aq) || Cl-(aq) |H2(g) |Pt 17.C Beryllium(Be) and aluminum(Al) have similar properties because A) both have similar electron affinities. B) both have almost identical ionization energies. C) both ions are highly polarizing due to their small size and high charge. D) both have virtually identical atomic radii. 18.D At the stoichiometric point in the titration of 0.130 M HCOOH(aq) with 0.130 M KOH(aq), A) the pH is 7.0. B) [HCOOH] = 0.0650 M. C) [HCO2-] = 0.130 M.D) the pH is greater than 7. 19.B Calculate the [H+] in an aqueous solution that is 0.0755 M HF and 0.100 M NaF. The value of Ka for HF is 3.5 x 10-4. A) 4.6 x 10-4 M B) 2.6 x 10-4 M C) 3.5 x 10-4 M D)0.176 M 20.A What is the equilibrium constant for the reaction HCN(aq) + OH-(aq)→CN-(aq) + H2O(l) A) KalKw B) Kb C) Kw/Ka D) Kw/Kb II. Integrated Questions (60%) 1. (8%) Complete and balance the following chemical reactions: (a). Al2O3(s) + OH-(aq)→ (b). H2S(g) + O2(g)→ (c). XeF6(s) + H2O(l)→ (d). H+ + MnO4- + Fe2+ → Mn2+ + Fe3+ + H2O 3. (12 pts) weak Acid-strong base titration Suppose that 25.00 mL of 0.200 M HCOOH(aq) is titrated with 0.300 M NaOH(aq). Ka=1.8x10-4 (a). calculate pH at half-point (b). calculate pH at stoichimetric point (c). sketch a titration curve. (a) (b) (c) (d) MnO4- + 5Fe2+ + 8H+ --> Mn2+ + 5Fe3+ + 4H2O (a) pH = pKa = 4 – log 1.8 = 3.74 (b) 25 x 0.2 = V x 0.3, V = 16.7 2HCOOHOOHHCOOHꄊ2. (20%) The galvanic cell below uses the following half-cells with their standard reduction potential specified. (a). Identify the half-reaction happening in the anode? Does the size of the electrode A increase or decrease? (b). Calculate the standard cell potential, Ecello ? (c). Calculate the Gibbs free energy of the whole reaction at 298 K. (d). Calculate the equilibrium constant at 298K. (e). Given the initial concentration of [Ag2+]= 12.0 M and [Ni2+]= 24.0 M, calculate the cell potential, Ecell? (a) Ni →Ni2++ 2e ; decrease- (b) Ecello = 0.80V –(-0.23)V= +1.03 V (c) GΔ= - 2 x 96500 x (1.03) = -198790 J (d) ΔGcell = -n*F*Ecell = -R*T*lnK -198790 = -8.314 x 298 x lnK lnK= 80.24 K=7.04 x 1034 (e) 1.03-0.0592/2 x log[24/122]= 1.05 (b) Ecello = +1.03 V (c)GΔ= -198790 J (d) 7.04 x 1034 (e) Ecell = 1.05 V --

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