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PHYSICAL CHEMISTRY By: Shailendra Kumar

1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 42. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna

Page No.: 1

1. Balance the following equation and determine the sum of the coefficients of the reactants.

___ SnO2– + ___ Bi3

+ + ___ OH– → ___ SnO32– + ___ Bi + ___ H2O

a. 5 b. 10

c. 6 d. 8

e. 9


___Cu+ ___ NO3– + ___ H+ → ___ Cu2

+ + ___ NO2 + ___ H2O

a. 5 b. 7

c. 6 d. 8

e. 9


___ HSnO2– ___ CrO4

2– + ___ H2O → HsnO3– + ___CrO2

– + ___ OH–

a. 5 b. 7

c. 6 d. 4

e. 8

4. Balance the following equation and indicate the coefficients of MnO4 and OH–.

___ MnO4– + ___ S2

+ ___ H2O → ___MnO2 + ___ S + ___ OH–

a. 2, 6 b. 3, 8

c. 4, 8 d. 2, 8

e. 3, 6

5. Balance the following redox equation and indicate the coefficents of NH3 and H2O.

___ Al + ___ NO3– + ___ OH– ___H2O → ___Al(OH)4

– + ___ NH3

a. 2, 10 b. 2, 12

c. 4, 18 d. 3, 18

e. 3, 14

6. Balance the following redox equation and indicate the coefficents of MnO42– and MnO2.

___ MnO42– + ___ H2O → ___ MnO4

– + ___ OH– + ___ NH3

a. 5, 3 b. 4, 3

c. 6, 3 d. 3, 1

e. 3, 2

PHYSICAL CHEMISTRY

SCIENCE TUTORIALS; Opp. Khuda Baksh Library, Ashok Rajpath, PatnaPIN POINT STUDY CIRCLE; House No. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna

Topic: Redox Reactions

Classes at: -

by: SHAILENDRA KR.Meq. Approach



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7. Balance the following reaction in acidic solution and determine the coefficient of H+ and its location(right or left side) in the equation.

__ HNO2 + __ HClO2 → __ N2O + __ ClO3–

a. 6, right b. 2, right

c. 4, left d. 2, left

e. 4, right


__ NO3– + __ MnO4

2– → __ NO + __MnO4–

a. 4, right b. 6, left

c. 6, right d. 2, right

e. 4, left


__ NO3– + __ MnO4

2– → __ NO + MnO4–

a. 4, right b. 4, left

c. 6, right d. 2, left

e. 3, right

10. Balance the following reaction in alkaline solution and determine the coefficient of OH– and itslocation (right or left side) in the equation.

__ Se + __ P → __ PH3 + __ Se

a. 6, right b. 8, right

c. 4, left d. 6, left

e. 4, right


__ P + __ Te → __ H2PO2– + __ Te2

–

a. 4, left b. 4, right

c. 6, right d. 2, left

e. 6, left


__ Al + __ NO3– → __ Al (OH)4

– + __ NH3

a. 5, left b. 4, right

c. 2, left d. 2, right

e. 4, left

13. How many electrons are transferred in the following reaction?

S2O82– (aq) + 2 H2O (l) → H2O2 (aq) + 2 SO2– (aq) + 2 H+ (aq)



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a. 1 b. 4

c. 5 d. 3

e. 2


2 Fe 3+ (aq) + 3 Cu (s) → 3 Cu2

+ (aq) + 2 Fe (s)

a. 1 b. 3

c. 4 d. 2

d. 6


3 MnO4– + 5 Cr + + 24 H+ → 3 Mn2

+ + 5 Cr3+ + 12 H2O

a. 1 b. 4

c. 5 d. 3

e. 2

16. How many minutes are required to deliver 3.21 x 106 coulombs using a current of 5.0 x 102 .Aused in the commercial production of chlorine?

a. 5.3 x 104 b. 8.3

c. 6.4 x 102 d. 6.4 x 103

e. 1.1 x 102

17. How many coulombs are provided by a current of 0.010 mA in the calculator battery that canoperate for 1000 hours?

a. 10 b. 1.0

c. 0.010 d. 1.0 x 103

e. 0.10

18. For how many minutes must a current of 1.22 amp be provided to deliver 512 coulombs?

a. 625 b. 10.4

c. 420 d. 4.00

e. 7.00

19. Passage of a current for 318 seconds through a silver coulometer results in the deposition of0.640 g of silver. What is the amperage?

a. 1.71 b. 1.98

c. 1.80 d. 1.65

e. 1.89


a. 1.31 b. 1.00

c. 1.22 d. 1.07

e. 1.14



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a. 1.21 b. 0.89

c. 1.09 d. 0.93

e. 1.14

22. A AgNO3 solution is electrolyzed using a current of 2.50 A. How many grams of Ag plate out in60.0 min? (Atomic weight : Ag = 107.868)

a. 12.7 b. 7.80

c. 9.20 d. 11.8

e. 10.1

23. How many millgrams silver is deposited if an electric current of 0.150 A flows through a silvernitrate solution for 20.0 min? (Atomic weight : Ag = 107.9)

a. 50.0 b. 403

c. 302 d. 100

e. 201

24. An automobile battery is charged at 5.00 A for exactly 5 hours. What mass of PbO2 is formed?

(Atomic weights: Pb = 207.2, O = 16.00)

2 PbSO4 (s) + 2 H2O (l) → Pb (s) + PbO2 (s) + 2 H2SO4 (aq)

a. 82.2 b. 222

c. 161 d. 55.5

e. 112

25. Electrolysis can be used to determine atomic masses. A current of 0.500 A deposits 0.213 g of acertain metal in 30.0 minutes. Calculate the atomic mass of the metal if n = 4.

a. 65.4 b. 54.9

c. 91.2 d. 112.4

e. 47.9


a. 45.0 b. 88.9

c. 69.7 d. 138.9

e. 27.0


a. 122 b. 177

c. 106 d. 152

e. 63.6

28. How many seconds are needed to produce 1.06 g of thallium from a thallium (I) solution using1.53 amp? (Atomic weight : Ti = 204.4)



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a. 327 b. 179

c. 286 d. 224

e. 146

29. How many minutes will it take to plate out 3.00 g of Cd from a Cd(NO3)2 solution using a currentof 3.00 A? (Atomic weight: Cd = 112.41)

a. 26.2 b. 17.3

c. 25.2 d. 14.3

e. 19.8

30. How many seconds are needed to produce 0.109 g of zirconium from a zirconium (IV) solutionusing k1.45 amp? (Atomic weight: Zr = 91.2)

a. 318 b. 476

c. 332 d. 506

e. 210

31. If 0.872 g of Ag is deposited on the cathode of a silver coulometer, and the process takes 15.0min, what was the rate of the current (A)? (Atomic weight: Ag = 107.868)

a. 0.867 b. 0.624

c. 13.0 d. 1.05

e. 8.67

32. If 0.123 g of Al is plated out from Al3+ and the process takes 15.0 min, what was the rate of current

(A) ? (Atomic weight: Al = 26.98)

a. 1.47 b. 1.64

c. 0.812 d. 0.490

e. 1.21

33. Calculate the current (mA) required to deposit 0.28 g of platinum metal in 5.0 hours from asolution of PtCl6

2–. (Atomic weight: Pt = 195.09)

a. 31 b. 62

c. 310 d. 95

e. 12

34. How many Faradays are required to reduce 0.25 g of Nb (V) to the metal?

(Atomic weight: Nb = 92.9)

a. 1.3 x 10–3 b. 7.8 x 10–3

c. 1.3 x 10–2 d. 2.7 x 10–2

e. 2.7 x 10–3

35. How many Faradays are required to reduce 0.10 g of Zr (IV) to the metal? (At Wt.: Zr = 91.2)

a. 4.4 x 10–3 b. 7.8 x 10–3

c. 1.1 x 10–3 d. 3.9 x 10–3

e. 8.6 x 10–4

36. How many Faradays are required to reduce 0.20 g of Eu3+ to the metal? (At Wt.: Eu = 151.96)



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a. 8.6 x 10–4 b. 1.3 x 10–3

c. 3.9 x 10–3 d. 7.8 x 10–3

e. 4.3 x 10–4

37. A 200.0 mL sample of 0.250 M Cu2+ is electrolyzed for 20.0 minutes with a current of 1.50 A. What

is the remaining [Cu2+]? (Atomic weight : Cu = 63.5)

a. 0.185 b. 0.156

c. 0.171 d. 0.203

e. 0.221

38. If 0.500 L of a 0.600 M SnSO4 solution is electrolyzed for a period of 30.0 min using a current of4.60 A. If inert electroduces are used, what is the final concentration of Sn2

+ remaining in thesolution?

a. 0.430 b. 0.342

c. 0.389 d. 0.514

e. 0.544

39. A 250.0 mL sample of 0.200 M Cr3+ is electrolyzed for 45.0 minutes with a current of 1.40 A. What

is the remaining [Cr3+]? (Atomic weight : Cr = 52.00)

a. 0.162 b. 0.043

c. 0113 d. 0.148

e. 0.175

40. An aqueous solution of a complex ion of a cadminium is electrolyxed for 30.0 minutes with acurrent of 7.5 A to deposit 7.86 g of cadmium. What is the charge of cadmium in the complex ion?

(Atomic weight : Cd = 112.41)

a. 1 b. 4

c. 2 d. 3

e. 5

41. An electrolysis of a oxytellurium complex ion using 1.20 A for 30 min produces 0.714 g of tellurium.What is the charge of tellurium in the material? (Atomic weight: Te = 127.4)

a. 6 b. 2

c. 4 d. 1

e. 3

42. An aqueous solution of a platinum salt is electrolyzed for 1.00 hour using a currenta of 1.50 A toproduce 2.73 g of metallic platinum. What is the charge on the ion? (Atomic weight: = 195.1)

a. 1 b. 3

c. 4 d. 2

e. 6

43. In the electrolysis of a CuSO4 solution, how many grams of Cu are plated out on the cathode inthe time that it takes to liberate 5.00 liter of O2 (g), measured at STP, at the anode?

(Atomic weights: Cu = 63.546; O = 16.00)



Page No.: 7

a. 4.32 b. 7.09

c. 28.3 d. 35.5

e. 14.2

44. In the electrolysis of molten MgCl2, how many liters of Cl2 (g), measured at STP, are produced inthe same time that it takes to plate out 10.0 g of Mg? (Atomic weights : Mg = 24.30; Cl = 35.45)

a. 7.58 b. 5.78

c. 9.28 c. 10.9

e. 4.64

45. How many grams of Ni are deposited in the electrolysis of a solution of NiSO4 in the same timethat it takes to deposit 0.575 g of Ag in a silver coulometer arranged in series with the NiSO4 cell?

(Atomic weights: Ni = 58.69; Ag = 107.868)

a. 0.124 b. 0.312

c. 0.156 d. 0.202

e. 0.272

46. Ammonium perchlorate, NH4ClO4, used in the solid fuel in the booster rockets on the spaceshuttle, is prepared from sodium perchlorate, NaClO4, which is produced commercially by theelectrolysis of a hot, stirred solution of sodium chloride. How many faradays are required toproduce 1.00 kg of sodium perchlorate? (Atomic weights: Na = 22.99, Cl = 35.45, O = 16.00)

NaCl + 4 H2O → NaClO4 + 4 H2

a. 18.3 b. 65.3

c. 8.16 d. 31.6

e. 40.3

47. In the commercial preparation of aluminum, aluminum oxide, Al2O3, is electrolyzed at 1000°C.How many coulombs of electricity are required to give 5.12 kg of aluminum? Assume the cathodefollowing reaction. (Atomic weight: Al = 26.98)

Al3+ + 3 e– → Al

a. 3.21 x 107 b. 5.49 x 107

c. 5.49 x 104 d. 1.82 x 107

e. 1.82 x 107

48. A layer of chromium metal 0.23 mm thick is to be plated on an auto bumper with a total area of0.32 m2 from a solution containing CrO4

2–? What current flow A) is required for this electroplate ifthe bunper is to be plated in 60 s? The density of chromium metal is 7.20 g/cm3.

(Atomic weight: Cr = 52.00)

a. 1.6 x 103 b. 9.8 x 104

c. 1.6 x 104 d. 3.3 x 104

e. 4.9 x 103

49. Electrolysis of molten NaCl in a Downs cell occurs at 7.0 volts and at a current of 4.0 x 104 amps.How many grams of Na (s) can be produced in one day assuming 100% efficiency?(Atomic weights: Na = 22.9, Cl = 35.45)



Page No.: 8

a. 8.2 x 105 b. 5.0 x 104

c. 1.0 x 105 d. 8.2 x 104

e. 4.1 x 105

50. Magnesium is produced commerical by the electrolysis of molten MgCl2 in cell with capacities inmetric tons (1.00 x 103 kg). How many metric tons of maganesium can be produced per dayusing a current of 1.00 x 105 A?

(Atomic weight : Mg = 24.31)

a. 1.09 b. 1.32

c. 1.65 d. 2.18

e. 2.56

51. Borateem is a peroxyborate bleache. Sodium peroxyborate, NaBO3, can be prepared by electrolyticoxidation of borax (Na2B4O7) solutions. How many grams of NaBO3 can be prepared in 24.0 h ifthe current is 20.0 A ? (Atomic weights : Na = 22.99, B = 10.87, O = 16.00)

Na2B4O7 (aq) + 10 NaOH (aq) → 4NaBO3 (aq) + 5 H2O (l) + 8 Na+ (aq) + 8 e-

a. 733 b. 884

c. 642 d. 1.24 x 103

e. 1.46 x 103

52. In the Hall process, aluminum is produced by the electrolysis of molten Al2O3 by the followingreactions. How many seconds would it take to produce enough aluminum by the Hall process tomake case of aluminum soft-drink cans (24 cans) if each can uses 5.00 g of Al, a current of50.000 amp is employed, and the efficiency of the cell is 90.0%?

anode : C + 2 O2– → CO2 + 4 e–

Cathode : 3 e– + Al3+ → Al

a. 33.3 b. 28.6

c. 30.4 d. 25.7

e. 22.6

53. Chlorine, Cl2, is produced comercially by the electrolysis of aqueous sodium chloride. The anodereaction is given below. How many minutes will it take to produce 1.18 kg of chlorine if the currentis 5.00 x 102 A ? (Atomic weight : Cl = 35.45)

2 Cl– (aq) → Cl2 (g) + 2 e–

a. 214 b. 107

c. 54.0 d. 152

e. 75.0

54. How many hours are required to plate chromium (density = 7.1 g/cm3) from a solution of Cr3+ on

a steel object with a surface area of 0.10 m2 and form a 0.1 mmthick coating using a current of1.50 A ? (Atomic weight : Cr = 52.00)

a. 24 b. 73

c. 48 d. 12

e. 89

55. Given the listed standard electrode potential, what is E for the cell:



Page No.: 9

[S2O32– (aq) + 2 OH– (aq) + O2 (g) → 2 SO3

2– (aq) + H2O (l)]

2 S2O32– (aq) + 3 H2O (l) + 4 e– → S2O3

2– + 6 OH– E = –0.580 V

O2 (g) + 2 H2O (l) + 4 e– → 4 OH– (aq) E = +0.401 V

a. -0.981 b. +0.981

c. +0.490 d. +0.179

e. -0.179


[Ag+ (aq) + Fe2+ (aq) → Ag (s) + Fe3

+ (aq)] ?

Fe2+ (aq) + 1 e– → Fe2

+ (aq) E = +0.77 V

Ag+ (aq) + 1 e– → Ag (s) E = +0.80 V

a. -0.03 b. +0.03

c. -1.57 d. +0.74

e. +1.57


[2 VO2+ (aq) + 4 H+ (aq) + Ni (s) → 2 VO2

+ (aq) + H2O (l) + Ni2+ (aq)] ?

Ni2+ (aq) + 2 e– → Ni (s) E = -0.250 V

VO2+ (aq) + 2 H+ (aq) + e– → VO2

+ (aq) + H2O (l) E = +1.00 V

a. -2.25 b. +1.25

c. +0.75 d. +1.75

e. -0.50

58. Based on this cell notation for a spontaneous reaction:

Ag (s) | AgCl (s) | Cl– (aq) || Br– (aq) | Br2 (l) | (C)

a. electrons flow in the external circuit toward the Ag electrode

b. C is the cathode c. none of the above are correct

d. Br2 becomes reduced e. AgCl becomes reduced

59. Based on this cell notation for a spontaneous reaction, at the cathode:

(Pt) | H2 (g) | H+ (aq) || Cl– (aq) | AgCl (s) | Ag (s)

a. H2 becomes oxcidized b. H+ becomes reduced

c. AgCl becomes reduced d. Ag becomes oxidized

60. Based on this cell notation for a spontaneous reaction:

Mg (s) | Mg2+ (aq) || Sn2+ (aq) | Sn (s)

a. Sn becomes oxidized b. none of the above are correct

c. Mg2+ becomes reduced d. Mg is the cathode

e. electrons flow in the external circuit toward the Mg electrode

61. From the listed standard electrode potentials, what is E° for the cell:

Cr (s) | Cr3+ (aq) || Cd2

+ (aq) | Cd (s)

Cr3+ (aq) + 3 e– → Cr (s) E° = –0.74 V

Cd2+ (aq) + 2 e– → Cd (s) E° = -0.44 V



Page No.: 10

a. +1.80 b. +0.30

c. +1.34 d. –1.18

e. -0.16


(C) | Fe2+ (aq) || Fe3

+ (aq) || Ag+ (aq) | Ag (s)

Fe3+ (aq) + 1 e– → Fe2

+ (aq) E° = +0.77 V

Ag+ (aq) + 1 e– → Ag (s) E° = +0.80 V

a. -1.57 b. +0.03

c. +1.57 d. +0.74

e. -0.03


Bi (s) | BiO+ (aq) || MnO4+ (aq) || Mn2

+ (aq) | (C)

BiO+ (aq) + 2 H+ + 3 e– → Bi (s) + H2O E° = +0.32 V

MnO4+ (aq) + 8 H+ + 5 e– → Mn2

+ (aq) + 4 H2O E° = +1.52 V

a. +1.83 b. +1.19

c. -1.83 d. -1.19

e. +1.42

64. The E° = 0.59 V for the following reaction. The standard electrode potential for N2 as written is -0.23 V. What is the standard electrode potential for the reduction of Fe(CN)6

3–.

N2H5+ (aq) + 4 Fe(CN)6

3– (aq) º N2 (g) + 5 H+ (aq) + 4Fe(CN)64- (aq)

N2 (g) + 5 H+ (aq) + 4 e- → N2H5+ (aq)

Fe(CN)63+ (aq) + e- → Fe(CN)6

4- (aq)

a. -0.82 b. +0.36

c. -0.36 d. +0.82

e. -0.41

65. The E° for the following cell is +0.34 V. Using E° = -1.00 V for the In (OH)3/In couple , calculate E°for the SbO2

–/Sb half-reaction.

In (s) | In (OH)3 (s) || SbO2– (aq) | Sb (s)

a. +0.66 b. -0.66

c. +1.34 d. -1.34

e. +0.82

66. The E° for the following cell is +2.777 V. Using E° = +0.799 V for the Ag+/Ag couple , calculate E°for the V3+/V half-reaction.

V (s) | V3+ (aq) || Ag+ (aq) | Ag (s)

a. +1.798 b. -1.798

c. -3.576 d. +1.693

e. +3.576



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67. The E° for the following cell is +0.46 V. Using E° = +1.06 V for the Br2/Br– couple , calculate E° forthe MnO4

–/MN2+ half-reaction.

(C) | Br– | Br2 (l) || MnO4– (aq) || Mn2+ (aq) | (C)

a. -1.51 b. +1.51

c. +0.76 d. +0.66

e. -0.66

68. Consider the following half-cell reactions and associated standard half-cell potentials anddetermine which species is the best reducing agent.

PO43- (aq) + 2H2O (l) + 2 e– → HPO3

2- + 3 OH– (aq) E° = -1.05 V

PbO2 (s) + H2O (l) + 2 e– → PbO (s) + 2 OH– (aq) E° = +0.28 V

IO3– (aq) + 2 H2O (l) + 4 e– → IO– (aq) + 4 OH– (aq) E° = +0.56 V

a. PO3– b. IO–

c. IO3– d. HPO3

2–

e. PbO

69. Consider the following half-cell reactions and associated standard half-cell potentials anddetermine which species is the best oxidizing agent.

S2O62– (aq) + 4 H+ (aq) + 2 e– → 2 H2SO3 (aq) E° = +0.60 V

Fe3+ (aq) + e– → Fe2

+ (aq) E° = +0.771 V

VO2+ (aq) + 2 H+ (aq) + e– → VO2

+ (aq) + H2O (l) E° = +1.00 V

N2O (aq) + 2 H+ (aq) + 2 e– → N2 (g) + H2O (l) E° = +1.77 V

a. S2O62– b. N2

c. Fe3+ d. N2O

e. VO2+

70. Consider the following half-cell reactions and associated standard half-cell potentials anddetermine which species is the best oxidizing agent.

AuBr4– (aq) + 3 e– → Au (s) + 4 Br– (aq) E° = -0.858 V

Eu3+ (aq) + e– → Eu2

+ (aq) E° = -0.43 V

Sn2+ (aq) + 2 e– → Sn (s) E° = -0.14 V

IO– (aq) + H2O (l) + 2 e– → I– (aq) + 2 OH– (aq) E° = +0.49 V

a. Br– b. IO– Eu3+

c. Sn2+ d. AuBr4

–

71. Based on the following information arrange four metals, A, B, C, and D in order of increasingability to act as reducing agents.

I. only A, C and D react with 1 M HCl to give H2 (g).

II. When A is added to solutions of the other metal ions, metallic B and C are formedbut not D.

a. C < B < A < D b. B < C < D < A

c. B < D < C < A d. B < C < A < D

e. C < B < D < A



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I. only A, B and C react with 1 M HCl to give H2 (g).

II. When C is added to solutions of the other metal ions, metallic B and D are formed

III. Metal C does not reduce An+

a. A > C > D > B b. B > D > A > C

c. C > A > B > D d. A > C > B > D

e. C > A > D > B


I. A, B and D do not react with 1 M HCl to give H2 (g).

II. When A is added to solutions of the other metal ions, no metallic products are obtained

III. Metal D does not reduct with Bn+

a. B > C > D > A b. B > C > A > D

c. B < D < C < A d. C > B > D > A

e. C > B > A > D

74. The value of E° at 25°C for the following reaction is 0.030 V. Calculate the equilibrium constant.

Ag+ (aq) + Fe2+ (aq) º Ag (s) + Fe3

+ (aq)

a. 6.4 b. 0.32

c. 3.2 d. 0.64

e. 10

75. The value of E° at 25°C for the following reaction is 0.98 V. Calculate the equilibrium constant.

S2O32– (aq) + 2 OH– (aq) + O2 (g) º 2SO3

2– (aq) + H2O (l)

a. 3.6 x 1016 b. 2.6 x 1033

c. 1.6 x 1066 d. 1.3 x 1033

e. 8.0 x 1065

76. The E° at 25°C for the following reaction is 0.22 V. Calculate the equilibrium constant at 25°C.

H2 (g) + 2 AgCl (s) º 2 Ag (s) + 2 HCl (aq)

a. 5.2 x 106 b. 5.2 x 103

c. 2.7 x 107 d. 5.2 x 108

e. 2.7 x 103

77. Calculate the equilibrium constant for the following reaction using the standard electrode potentials.

N2H5+ (aq) + 4 Fe (CN)6

3– (aq) º N2 (g) + 5 H+ (aq) + 4 Fe(CN)64– (aq)

N2 (g) + 5 H+ + 4 e– → N2H5+ (aq) E° = -0.23 V

e (CN)63– (aq) + 4 e– → Fe(CN)6

4– (aq) E° = +0.36 V

a. 7.3 x 1039 b. 8.5 x 1029

c. 7.3 x 1025 d. 9.2 x 1097

e. 8.5 x 1019



Page No.: 13

78. Calculate the equilibrium constant for the following reaction using the standard electrode potentials.

2 Cr3+ (aq) + 3 Ni (s) º 2 Cr (s) + 3 Ni2

+ (aq)

Cr3+ (aq) + 3 e– → Cr (s) E° = -0.913 V

Ni2+ (aq) + 2 e– → Ni (s) E° = -0.250 V

a. 1.6 x 1067 b. 1.2 x 1051

c. 1.2 x 1039 d. 7.5 x 1022

e. 1.2 x 1014

79. Given the following standard electrode potentials, calculate the Ksp for Ag2CrO4.

Ag2CrO4 (s) + 2 e––→ 2 Ag (s) + CrO42– (aq) E° = +0.446 V

Ag+ (aq) + 1 e– → Ag (s) E° = +0.799 V

a. 4.0 x 10-10 b. 2.4 x 10-13

c. 1.2 x 10-12 d. 5.3 x 10-9

e. 3.1 x 10-11

80. Given the following standard electrode potentials, calculate the Ksp for AgBr.

AgBr (s) + 1 e––→ Ag (s) +Br– (aq) E° = +0.071 V

Ag+ (aq) + 1 e– → Ag (s) E° = +0.799 V

a. 6.2 x 10-15 b. 1.8 x 10-10

c. 5.0 x 10-13 d. 8.6 x 10-17

e. 4.1 x 10-11

81. Given the following standard electrode potentials, calculate the Ksp for PbS.

PbS + 2 e––→ Pb (s) +S2– (aq) E° = – 0.930 V

Pb2+ (aq) + 2 e– → Pb (s) E° = – 0.126 V

a. 4.3 x 10-25 b. 5.4 x 10-26

c. 6.9 x 10-28 d. 1.2 x 10-22

e. 3.1 x 10-23

82. The equilibrium constant for the following general reaction is 10. Calculate E° for the cell.

X2 (s) + Y2+ (aq) → X2

2+ (aq) + Y (s)

a. +0.09 b. +0.12

c. +0.03 d. -0.03

e. -0.12

83. The equilibrium constant for the following general reaction is 2.5 x 10-45. Calculate E° for the cell.

2 X2 (s) + 3 Y2+ (aq) → 2 X 2

3+ (aq) + 3 Y (s)

a. -0.26 b. -0.32

c. +0.03 d. +0.44

e. +0.32

physical chemistry by: shailendra kumar physical chemistry · physical chemistry by: shailendra...

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