Shandong University Physical Chemistry(2)Quarter Exam-2 2019-2020 Aoademmio Year Fall Semester 号国國五六[七九团十分分人 分■人 2020-2021 Academic Year. Fall Semester IL Fill the blanks with the best answers (4 points for each blank, totally The Second Quarter Exam of Physical Chemistry(2) [2-1] At 298.2 K, 0.005 mol kg CuSO4 solution obey Lewis empirical law, then the 舞分[人 L. Choose the best answer for the following questions using a cirele(4 points for 12-2] For cell Pu s)IHi(g, p)IH SO.(0. 1 mol kgr)Oa(g,pe))Pu(s), its electromotive force 密 each question, totally 20 points) is 1. V, when 2 mol electrons are transferred, A, He--28606 k mol, then energy 2 be constants, which conditions should be met electrochemical capacitance of Oz is 3350.7 mAh g. (Al, b, c, d; (B)a, b, e, d, e, (c)b, e, d; ( D)b, e, d, e. [2-3] The electrode reaction of Fe Fe O4- is 3Fe+80H--Fe:O4+4l a-strong electrolyte, b--infinite dilution, c-at constant temperature; d-in certain solvent; e-with certain counter ion. [I-2] Which of the following factor is not the reason for the high mean activity coefficient of [2-4]When we use Puc as catalyst for hydrogen evolution instead of Fe for electrolysis of 封 LiCI solution at high concentration. a NaCl solution and get 22.4 L (STP) hydrogen at a current density of I A dm-2, the electric energy saved is 94.57 k. (The a of Tafel equation for Pt and Fe is 0.07 and 0.56 while b is the same) 1.018-405X 10-(/C-25). When this cell discharge at 298.2 K, the reversible heat [2-5] The overpotential of hydrogen on Cu electrode is 0.23 V. If we want to perform (Ore)sent off by the cell is electrorefining of Cu using pure Cu as cathode and 0.2 mol kg-CuSO4 with 0.05 mol kg- (A)One>0, (B)One Ean (B)Ec< Ean,(C)Eear- Ean: (D)uncertain. 第
2019-2020 Academic Year Fall Semester 题号 一 二 三 四 五 六 七 八 九 十 总分 总分人 得分 2020-2021 Academic Year, Fall Semester The Second Quarter Exam of Physical Chemistry (2) ---------------------------------------------------------------------------------------------- 得分 阅卷人 I. Choose the best answer for the following questions using a circle (4 points for each question, totally 20 points) [1-1] For a given ion B, if we want its molar conductivity (B,m) and ionic mobility (UB) to be constants, which conditions should be met? (A)a, b, c, d; (B) a, b, c, d, e; (C) b, c, d; (D) b, e, d, e. a—strong electrolyte; b--infinite dilution; c—at constant temperature; d—in certain solvent; e—with certain counter ion. [1-2] Which of the following factor is not the reason for the high mean activity coefficient of LiCl solution at high concentration. (A) Severe hydration of Li+ ; (B) Strong repulsion between ions; (C) Decreased permittivity of water; (D) increased solution viscosity. [1-3] Fora reversible cell, its temperature dependence of electromotive forces is E/V = 1.018 – 4.05×10-5 (t/ oC-25). When this cell discharge at 298.2 K, the reversible heat (Qre) sent off by the cell is (A) Qre > 0; (B) Qre Ean; (B) Ecat < Ean; (C) Ecat = Ean; (D) uncertain. 得分 阅卷人 II. Fill the blanks with the best answers (4 points for each blank, totally 24 points) [2-1] At 298.2 K, 0.005 mol kg-1 CuSO4 solution obey Lewis empirical law, then the activity of this solution is ___6.64×10-6____. [2-2] For cell Pt(s)H2(g, p)H2SO4(0.01 mol kg-1 )O2(g, p)Pt(s), its electromotive force is 1.229 V, when 2 mol electrons are transferred, rH m = -286.06 kJ mol-1 , then energy conversion efficient from chemical to electric is __82.92%___. The theoretical electrochemical capacitance of O2 is ___3350.7_mAh g-1 . [2-3] The electrode reaction of FeFe3O4OHis _3Fe + 8OH- ⎯→Fe3O4 + 4H2O + 8e-__. [2-4] When we use Pt/C as catalyst for hydrogen evolution instead of Fe for electrolysis of a NaCl solution and get 22.4 L (STP) hydrogen at a current density of 1 A dm-2 , the electric energy saved is __94.57_ kJ. (The a of Tafel equation for Pt and Fe is 0.07 and 0.56 while b is the same) . [2-5] The overpotential of hydrogen on Cu electrode is 0.23 V. If we want to perform electrorefining of Cu using pure Cu as cathode and 0.2 mol kg-1 CuSO4 with 0.05 mol kg-1 H2SO4 as electrolyte without evolution of hydrogen. The potential of Cu cathode should keep higher than ____-0.39_ V. (without considering activity coefficient) Shandong University Physical Chemistry (2) Quarter Exam-2 College Major Grade Student No. Name … … … … … … … … … … … … … … 密 … … … … … … … … … … 封 … … … … … … … … … … 线 … … … … … … … … … … … … … 第 1 页 共 2 页
Shandong University Physical Chemistry(2)Quarter Exam-2 2020-2021 Academic Year Fall Semester 舞分[舞 For Zn-plated iron: anode Zn-Zn2++2e, cathode: 2H+++2e=H Ill. Answer the following questions: (Totally 34 points) For Sn-plated iron: anode Fe-Fe+ 2e [3-1][20 points] The following figure is the Pourbaix diagram of iron-water system at 298.2 (2)14 points] For hydrogen evolution on Zn, Fe, Sn, the Tafel equation can be written as nV-a+ blg(JA dm). The a in the Tafel equation is 1.24 V, 0.70 V, and 1.20 V respectively, while b is 0. 118 V/dec. If the potential of a Sn-plated iron is-0.20 V, then the density ( for hydrogen evolution which is related to the corrosion rate of iron is 0-12 adm-2 (3)[4 points] The active dissolution of metals is usually much faster, so the revolution of 中狲豁 hydrogen is the r.d. s of the corrosion. Then, for the Sn-plated iron, the corrosion rate of (slower or faster) than that of iron without Sn plating in the same conditions 餐分 (I)(12 points)Write the electrode reaction corresponding to line(c),(h), and (m)and give the pH-dependence of electrode potential for these lines. IV Calculation( totally 14 points): (c)Fe2++3H2O→Feo+3H+e;g=0-3×0.05916pH 4-1][14 points] According to the data of 298.2 K listed in the following table, (h)Fe(OH+HO→→FeOH3+H+c,mo-0.059l6pH Electrodes H/H? Cl/AeCIAg Cr/CH m)Fe(OH+H2O→→FeO2+5H+3e;g(5/3)×0.05916pH 0.000000 (2)(4 points) The solubility product of Fe(OH)s is 1. 1x10-, define the exact pH of line(a).I(1)(4 points]Design a reversible cell which can be used to measure the activity product of AgCl(Ka, Agcl) (3)[4 points] Is iron more stable in acidic or alkaline medium? Use the Pourbaix diagram to Ag(s)lAg*l lcr-lAgCI(s)lAg(s) give explanation to your judgement.[ Hint: line g corresponds to hydrogen evolution in (2)(4 points] Calculate the Kap. alkaline solution] it is in the iron stable region (3)[6 points] Calculate the solubility of AgCI(S)in 0.0100 mol kg-KCl without or with 3-2][14 points] considering the activity coefficient(x).1.74876x10-: S(0.01+S)=a (1)[8 points] For Zn-plated iron and Sn-plated iron, if the plating layer is broken, write the anodic and cathodic reaction of micro-cell in weak acidic media 第2页共2页
2020-2021 Academic Year Fall Semester 得分 阅卷人 III. Answer the following questions: (Totally 34 points) [3-1] [20 points] The following figure is the Pourbaix diagram of iron-water system at 298.2 K. (1) (12 points) Write the electrode reaction corresponding to line (c), (h), and (m) and give the pH-dependence of electrode potential for these lines. (c) Fe2+ + 3H2O ⎯→ Fe(OH)3 + 3H+ + e- ; = -30.05916 pH (h) Fe(OH)2 + H2O ⎯→ Fe(OH)3 + H+ + e- ; = -0.05916 pH (m) Fe(OH)3 +H2O ⎯→ FeO4 2- + 5H+ + 3e - ; = -(5/3)0.05916 pH (2) [4 points] The solubility product of Fe(OH)3 is 1.110-36 , define the exact pH of line (a). 2.014 (3) [4 points] Is iron more stable in acidic or alkaline medium? Use the Pourbaix diagram to give explanation to your judgement. [Hint: line g corresponds to hydrogen evolution in alkaline solution] it is in the iron stable region [3-2] [14 points] (1) [8 points] For Zn-plated iron and Sn-plated iron, if the plating layer is broken, write the anodic and cathodic reaction of micro-cell in weak acidic media. For Zn-plated iron: anode___Zn⎯→Zn2+ + 2e-____; cathode:__2H++2e- ⎯→ H2____ For Sn-plated iron: anode___Fe_⎯→Fe2+ + 2e-____; cathode:__ 2H++2e- ⎯→ H2____ (2) [4 points] For hydrogen evolution on Zn, Fe, Sn, the Tafel equation can be written as /V = a + blg(J/A dm-2 ). The a in the Tafel equation is 1.24 V, 0.70 V, and 1.20 V respectively, while b is 0.118 V/dec. If the potential of a Sn-plated iron is -0.20 V, then the current density (J) for hydrogen evolution which is related to the corrosion rate of iron is _1.38×10-12_A dm-2 . (3) [4 points] The active dissolution of metals is usually much faster, so the revolution of hydrogen is the r.d.s of the corrosion. Then, for the Sn-plated iron, the corrosion rate of iron is ______ (slower or faster) than that of iron without Sn plating in the same conditions. 得分 阅卷人 IV. Calculation (totally 14 points): [4-1] [14 points] According to the data of 298.2 K listed in the following table, Electrodes H+ /H2 Cl - /AgCl/Ag Ag + /Ag Cl - /Cl2 /V 0.000000 0.22216 0.79940 1.35793 (1) [4 points] Design a reversible cell which can be used to measure the activity product of AgCl (Kap, AgCl). Ag(s)Ag+Cl-AgCl(s)Ag(s) (2) [4 points] Calculate the Kap, AgCl. 1.7487610-10 (3) [6 points] Calculate the solubility of AgCl (S) in 0.0100 mol kg-1 KCl without or with considering the activity coefficient (). 1.7487610-8 ; 2 (0.01 ) K ap S S + = Shandong University Physical Chemistry (2) Quarter Exam-2 学院 专业 级 学号 姓名 … … … … … … … … … … … … … … 密 … … … … … … … … … … 封 … … … … … … … … … … 线 … … … … … … … … … … … … … 第 2 页 共 2 页
Shandong University_ Phvsical Chemistry (2)Quarter Exam-2 2018-2019 Academic Year Fall Semester 14-28 points At 298 K, the resistance of a conductance cell filled with 0. I mol kgKCI and 0.002414 mol kg acetic acid(HAc)is 23.78 Q2 and 3942 Q2, respectively. Th conductivity of o I mol kg-KCI is 1. 289 S mrat 298K. [the limiting molar conductivity of HAc is 0.03907S m2mor-I ( Calculate the degree of ionization of HAc at this concentration, REAR 3. s dm? mol-l a=Am/1°m=0.0824 (2)Calculate the ionization equilibrium constant of HAc. 封
201 8 -201 9 Academic Year Fall Semester [ 4 -2] [8 points] At 298 K, the resistance of a conductance cell filled with 0.1 mol kg - 1 KCl and 0.002414 mol kg - 1 acetic acid (HAc) is 23.78 and 3942 , respectively. The conductivity of 0.1 mol kg - 1 KCl is 1.289 S m - 1 at 298 K. [the limiting molar conductivity of HAc is 0.03907 S m 2 mol - 1 ] (1) Calculate the degree of ionization of HAc at this concentration ; s s x x R R = ; =7.776 10 - 3 S m - 1 m = 3.2211 S d m 2 mol - 1 = m/ m = 0.0824 (2) Calculate the ionization equilibrium constant of HAc. 2 5 1.7884 10 1c K − = = − Shandong University Physical Chemistry (2) Quarter Exam - 2 学院 专业 级 学号 姓名 ……………………………………密…………………………封…………………………线………………………………… 第 2 页 共 2 页
