Question 1

Choose the INCORRECT statement.

Accepted Answer

A)  A double vertical line separates half cells in a cell diagram. 
B)  A single vertical line separates phases in a cell diagram. 
C)  A voltaic cell is a galvanic cell. 
D)  Voltaic cells produce an electron flow. 
E)  Electrolytic cells are cells where electron flow is caused by spontaneous reactions. 
A)  A double vertical line separates half cells in a cell diagram. 
B)  A single vertical line separates phases in a cell diagram. 
C)  A voltaic cell is a galvanic cell. 
D)  Voltaic cells produce an electron flow. 
E)  Electrolytic cells are cells where electron flow is caused by spontaneous reactions.

Question 2

Using the following reduction half-reactions and their respective standard reduction potentials at 25°C, draw a diagram of a galvanic cell. 
Cr2O72-(aq) + 14 H+(aq) + 6e- → 2 Cr3+(aq) + 7 H2O   &emsp; E° = +1.33 V
Zn2+(aq) + 2 e- → Zn(s)  &emsp;  E° = -0.763 V

Accepted Answer

A)  Zn2+(aq) ∣ Zn(s), H+(aq) || Cr2O72-(aq), Cr(s) ∣ Cr3+(aq) 
B)  Zn(s) ∣ Zn2+(aq) || H+(aq), Cr3+(aq) ∣ Cr(s) 
C)  Cr(s) ∣ Cr3+(aq), H+(aq), Cr2O72-(aq) || Zn(S) ∣ Zn2+(aq) 
D)  Zn(s) ∣ Zn2+(aq) || Cr2O72-(aq), H+(aq), Cr3+(aq) ∣ Pt(s) 
E)  Pt(s) ∣ Cr3+(aq), Cr2O72-(aq), H+(aq) || Zn2+(aq) ∣ Zn(s) 
A)  Zn2+(aq) ∣ Zn(s), H+(aq) || Cr2O72-(aq), Cr(s) ∣ Cr3+(aq) 
B)  Zn(s) ∣ Zn2+(aq) || H+(aq), Cr3+(aq) ∣ Cr(s) 
C)  Cr(s) ∣ Cr3+(aq), H+(aq), Cr2O72-(aq) || Zn(S) ∣ Zn2+(aq) 
D)  Zn(s) ∣ Zn2+(aq) || Cr2O72-(aq), H+(aq), Cr3+(aq) ∣ Pt(s) 
E)  Pt(s) ∣ Cr3+(aq), Cr2O72-(aq), H+(aq) || Zn2+(aq) ∣ Zn(s)

Question 3

Glass electrodes are used for direct measurement of:

Accepted Answer

A)  pH 
B)  pKa 
C)  Ksp 
D)  Q 
E)  pKw 
A)  pH 
B)  pKa 
C)  Ksp 
D)  Q 
E)  pKw

Question 4

Determine E°cell at 25 °C for the following reaction: Pb2+(aq) + Cu(s) → Cu2+(aq) + Pb(s)
The half-reactions are:
Pb2+(aq) + 2 e- → Pb(s) E° = -0.125 V
Cu2+(aq) + 2 e- → Cu(s) E° = +0.337 V

Accepted Answer

A)  0.212 V 
B)  -0.462 V 
C)  0.462 V 
D)  -0.212 V 
E)  0.424 V 
A)  0.212 V 
B)  -0.462 V 
C)  0.462 V 
D)  -0.212 V 
E)  0.424 V

Question 5

Choose the INCORRECT statement.

Accepted Answer

A)  One mole of hydrogen ions will have a positive charge of one Faraday. 
B)  One Faraday of charge will release 35.5 g of Cl2 in the electrolysis of a chloride solution. 
C)  One Faraday of charge will plate out two moles of Cu from a solution of CuSO4(aq). 
D)  One mole of electrons has a charge of l Faraday. 
E)  One mole of hydrogen gas requires 2 Faradays. 
A)  One mole of hydrogen ions will have a positive charge of one Faraday. 
B)  One Faraday of charge will release 35.5 g of Cl2 in the electrolysis of a chloride solution. 
C)  One Faraday of charge will plate out two moles of Cu from a solution of CuSO4(aq). 
D)  One mole of electrons has a charge of l Faraday. 
E)  One mole of hydrogen gas requires 2 Faradays.

Question 6

Will magnesium metal displace Al3+ ion from an aqueous solution? Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V
Al3+(aq) + 3 e- → Al(s) E° = -1.676 V

Accepted Answer

A)  No, since E°cell is negative. 
B)  Yes, since E°cell is negative. 
C)  No, the reverse reaction is spontaneous. 
D)  Yes, since E°cell is positive. 
E)  No, the system is at equilibrium. 
A)  No, since E°cell is negative. 
B)  Yes, since E°cell is negative. 
C)  No, the reverse reaction is spontaneous. 
D)  Yes, since E°cell is positive. 
E)  No, the system is at equilibrium.

Question 7

A constant current of 15.0 A is passed through an electrolytic cell for 60.0 min. How many Coulombs of charge are passed through the cell?

Accepted Answer

A)  0.00933 
B)  0.560 
C)  0.00560 
D)  900 
E)  15 
A)  0.00933 
B)  0.560 
C)  0.00560 
D)  900 
E)  15

Question 8

When Ecell < 0, the reaction is spontaneous.

Accepted Answer

A) True 
 B)False

Question 9

A solution of AuCl3 is electrolyzed by passing 85.0 A of current for 17.6 min. What mass of gold plates out?

Accepted Answer

A)  61.1 g 
B)  1.02 g 
C)  183 g 
D)  0.483 g 
E)  290 g 
A)  61.1 g 
B)  1.02 g 
C)  183 g 
D)  0.483 g 
E)  290 g

Question 10

A copper electrode weighs 35.42 g before the electrolysis of a copper (II) sulfate solution and 36.69 g after the electrolysis has run for 20.0 s. What was the amperage of the current used?

Accepted Answer

A)  2.12 A 
B)  0.0335 A 
C)  48.2 A 
D)  96.4 A 
E)  193 A 
A)  2.12 A 
B)  0.0335 A 
C)  48.2 A 
D)  96.4 A 
E)  193 A

Question 11

What is the cell diagram for the spontaneous cell involving the Cl2 ∣ Cl- (1.358 V) and the Sn2+ ∣ Sn (-0.137V) half cells?

Accepted Answer

A)  Pt ∣ Cl2(g) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s) 
B)  Sn(S) ∣ Sn2+(aq) || Cl-(aq) ∣ Cl2(g) ∣ Pt 
C)  Cl2(g) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s) 
D)  Sn(S) || Sn2+(aq) || Cl-(aq) ∣ Cl2(g) 
E)  Sn(S) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s) 
A)  Pt ∣ Cl2(g) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s) 
B)  Sn(S) ∣ Sn2+(aq) || Cl-(aq) ∣ Cl2(g) ∣ Pt 
C)  Cl2(g) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s) 
D)  Sn(S) || Sn2+(aq) || Cl-(aq) ∣ Cl2(g) 
E)  Sn(S) ∣ Cl-(aq) || Sn2+(aq) ∣ Sn(s)

Question 12

A solution of Al(NO3)3 is electrolyzed by passing 144 A of current for 102 min. What mass of aluminum plates out?

Accepted Answer

A)  27.4 g 
B)  760 g 
C)  82.1 g 
D)  2.3 g 
E)  75.3 g 
A)  27.4 g 
B)  760 g 
C)  82.1 g 
D)  2.3 g 
E)  75.3 g

Question 13

Primary batteries' cell reactions cannot be reversed.

Accepted Answer

A) True 
 B)False

Question 14

Choose the INCORRECT statement.

Accepted Answer

A)  Corrosion of metals is an oxidation reduction process. 
B)  Rust is a form of corrosion. 
C)  Metals can be protected by cathodic protection. 
D)  Cathodic protection is attaching a more active metal to the protected metal. 
E)  The active metal is called a sacrificial cathode. 
A)  Corrosion of metals is an oxidation reduction process. 
B)  Rust is a form of corrosion. 
C)  Metals can be protected by cathodic protection. 
D)  Cathodic protection is attaching a more active metal to the protected metal. 
E)  The active metal is called a sacrificial cathode.

Question 15

In a galvanic cell, oxidation occurs at the cathode.

Accepted Answer

A) True 
 B)False

Question 16

Determine E°cell for the reaction: Pb2+ + Zn → Pb + Zn2+. The half reactions are: Pb2+(aq) + 2 e- → Pb(s)  &emsp;  E° = -0.125 V
Zn2+(aq) + 2 e- → Zn(s)   &emsp; E° = -0.763 V

Accepted Answer

A)  0.638 V 
B)  -0.638 V 
C)  0.888 V 
D)  -1.276 V 
E)  -0.888 V 
A)  0.638 V 
B)  -0.638 V 
C)  0.888 V 
D)  -1.276 V 
E)  -0.888 V

Question 17

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI2(S) → Pb(s) + 2 I-(aq) E° = -0.365 V
2 e- + Pb2+(aq) → Pb(s) E° = -0.126 V

Accepted Answer

A)  8 × 10-9 
B)  9 × 10-5 
C)  2 × 10-17 
D)  5 × 10-13 
E)  6 × 10-5 
A)  8 × 10-9 
B)  9 × 10-5 
C)  2 × 10-17 
D)  5 × 10-13 
E)  6 × 10-5

Question 18

What is E° of the spontaneous cell made from Ag+/Ag (0.80 V) and Cl2/Cl- (1.36 V) half cells?

Accepted Answer

A)  2.16 V 
B)  -2.16 V 
C)  -0.56 V 
D)  0.56 V 
E)  1.36 V 
A)  2.16 V 
B)  -2.16 V 
C)  -0.56 V 
D)  0.56 V 
E)  1.36 V

Question 19

Choose the INCORRECT statement.

Accepted Answer

A)  The work available in a cell = -zFEcell. 
B)  ΔG° = -zFE°cell 
C)  When Ecell < 0, the reaction is spontaneous. 
D)  When Ecell = 0, the reaction is at equilibrium. 
E)  Reversing a cell reaction changes the sign of Ecell. 
A)  The work available in a cell = -zFEcell. 
B)  ΔG° = -zFE°cell 
C)  When Ecell < 0, the reaction is spontaneous. 
D)  When Ecell = 0, the reaction is at equilibrium. 
E)  Reversing a cell reaction changes the sign of Ecell.

Question 20

Determine Ecell at 25°C for the following reaction at 25°C: Al ∣ Al3+(0.435 M) || Sn2+(2.12 × 10-3 M) ∣ Sn
3 Sn2+ + 2 Al(s) → 2 Al3+ + 3 Sn(s)
Al3+ + 3 e- → Al(s)   &emsp; E° = -1.676 V
Sn2+ + 2 e- → Sn(s)   &emsp; E° = -0.137 V

Accepted Answer

A)  1.227 V 
B)  1.611 V 
C)  1.562 V 
D)  1.487 V 
E)  1.467 V 
A)  1.227 V 
B)  1.611 V 
C)  1.562 V 
D)  1.487 V 
E)  1.467 V

Exam 20: Electrochemistry

Choose the INCORRECT statement.

Using the following reduction half-reactions and their respective standard reduction potentials at 25°C, draw a diagram of a galvanic cell. Cr2O72-(aq) + 14 H+(aq) + 6e- → 2 Cr3+(aq) + 7 H2O E° = +1.33 V Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

Glass electrodes are used for direct measurement of:

Determine E°cell at 25 °C for the following reaction: Pb2+(aq) + Cu(s) → Cu2+(aq) + Pb(s) The half-reactions are: Pb2+(aq) + 2 e- → Pb(s) E° = -0.125 V Cu2+(aq) + 2 e- → Cu(s) E° = +0.337 V

Choose the INCORRECT statement.

Will magnesium metal displace Al3+ ion from an aqueous solution? Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Al3+(aq) + 3 e- → Al(s) E° = -1.676 V

A constant current of 15.0 A is passed through an electrolytic cell for 60.0 min. How many Coulombs of charge are passed through the cell?

When Ecell < 0, the reaction is spontaneous.

A solution of AuCl3 is electrolyzed by passing 85.0 A of current for 17.6 min. What mass of gold plates out?

A copper electrode weighs 35.42 g before the electrolysis of a copper (II) sulfate solution and 36.69 g after the electrolysis has run for 20.0 s. What was the amperage of the current used?

What is the cell diagram for the spontaneous cell involving the Cl2 ∣ Cl- (1.358 V) and the Sn2+ ∣ Sn (-0.137V) half cells?

A solution of Al(NO3)3 is electrolyzed by passing 144 A of current for 102 min. What mass of aluminum plates out?

Primary batteries' cell reactions cannot be reversed.

Choose the INCORRECT statement.

In a galvanic cell, oxidation occurs at the cathode.

Determine E°cell for the reaction: Pb2+ + Zn → Pb + Zn2+. The half reactions are: Pb2+(aq) + 2 e- → Pb(s) E° = -0.125 V Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI2(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb2+(aq) → Pb(s) E° = -0.126 V

What is E° of the spontaneous cell made from Ag+/Ag (0.80 V) and Cl2/Cl- (1.36 V) half cells?

Choose the INCORRECT statement.

Determine Ecell at 25°C for the following reaction at 25°C: Al ∣ Al3+(0.435 M) || Sn2+(2.12 × 10-3 M) ∣ Sn 3 Sn2+ + 2 Al(s) → 2 Al3+ + 3 Sn(s) Al3+ + 3 e- → Al(s) E° = -1.676 V Sn2+ + 2 e- → Sn(s) E° = -0.137 V

Matter - Its Properties and Measurement

Atoms and the Atomic Theory

Chemical Compounds

Chemical Reactions

Introduction to Reactions in Aqueous Solutions

Gases

Thermochemistry

Electrons in Atoms

The Periodic Table and Some Atomic Properties

Chemical Bonding I: Basic Concepts

Chemical Bonding II: Additional Aspects

Intermolecular Forces: Liquids and Solids

Solutions and Their Physical Properties

Chemical Kinetics

Principles of Chemical Equilibrium

Acids and Bases

Additional Aspects of Acid-Base Equilibria

Solubility and Complex-Ion Equilibria

Spontaneous Change: Entropy and Free Energy

Main Group Elements I: Groups 1, 2, 13, and 14

Main Group Elements II: Groups 18, 17, 16, 15, and Hydrogen

The Transition Elements

Complex Ions and Coordination Compounds

Nuclear Chemistry 1-41

Structures of Organic Compounds

Reactions of Organic Compounds

Chemistry of the Living State

Filters

Using the following reduction half-reactions and their respective standard reduction potentials at 25°C, draw a diagram of a galvanic cell. Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6e- → 2 Cr³⁺(aq) + 7 H₂O E° = +1.33 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

A solution of AuCl₃ is electrolyzed by passing 85.0 A of current for 17.6 min. What mass of gold plates out?

What is the cell diagram for the spontaneous cell involving the Cl₂ ∣ Cl- (1.358 V) and the Sn²⁺ ∣ Sn (-0.137V) half cells?

A solution of Al(NO₃)₃ is electrolyzed by passing 144 A of current for 102 min. What mass of aluminum plates out?

Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

What is E° of the spontaneous cell made from Ag⁺/Ag (0.80 V) and Cl₂/Cl- (1.36 V) half cells?

Determine Ecell at 25°C for the following reaction at 25°C: Al ∣ Al³⁺(0.435 M) || Sn²⁺(2.12 × 10^-3 M) ∣ Sn 3 Sn²⁺ + 2 Al(s) → 2 Al³⁺ + 3 Sn(s) Al³⁺ + 3 e- → Al(s) E° = -1.676 V Sn²⁺ + 2 e- → Sn(s) E° = -0.137 V