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- ;  = -30.05916 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 1.110-36 , 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. 1.7487610-10 (3) [6 points] Calculate the solubility of AgCl (S) in 0.0100 mol kg-1 KCl without or with considering the activity coefficient (). 1.7487610-8 ; 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