Chapter 17 - Practice Multiple Choice Test 1
(from Kotz 5^th Edition Chapter 21)

ANSWERS

The correct answer is:
ClO₄^- --> ClO₃^-

Explanation:
A reducing agent decreases the oxidation number of an atom in a compound or ion. A reducing agent provides electrons to the substance that is reduced.


Solution Reference:
Page 947, Oxidation-Reduction Reactions

1. Which of the following changes requires a reducing agent?
	a.	Cr2O7^2- --> CrO4^2-
	b.	ClO4^- --> ClO3^-
	c.	NH3 --> NH4^+
	d.	Al --> Al^3+
	e.	O^2- --> O2^2-

The correct answer is:
NO₃^-.

Explanation:
The oxidation number of the oxidizing agent is decreased.

Solution Reference:
Page 947, Oxidation-Reduction Reactions

2. In the following oxidation-reduction reaction 8H^+(aq) + 6Cl^-(aq) + Sn(s) + 4NO3^-(aq) --> SnCl6^2-(aq) + 4NO2(g) + 4H2O the oxidizing agent is
	a.	H^+.
	b.	Cl^-.
	c.	Sn.
	d.	NO3^-.
	e.	SnCl6^2-.

The correct answer is:
O₂; Fe and Cr

Explanation:
The oxidizing agent is reduced. Oxidation is an increase in oxidation number.

Solution Reference:
Page 947, Oxidation-Reduction Reactions

3. For the reaction: 4FeCr2O4 + 8Na2CO3 + 7O2 + 4H2SO4 --> 4Na2Cr2O7 + 2Fe2O3 + 4H2O + 8CO2 + 4Na2SO4 the oxidizing agent is _______ and the element(s) oxidized is (are) _______.
	a.	O2; Na2CO3
	b.	Na2CO3; O
	c.	O2; Fe and Cr
	d.	O2; CO2
	e.	O2; Cr, C

The correct answer is:
S^2-.

Explanation:
The sulfur in the sulfide ion is in its lowest oxidation so it cannot accept any more electrons, as it already has a noble gas configuration.

Solution Reference:
Page 206, Oxidation Numbers

4. A sulfur-containing species that cannot be reduced is
	a.	H2SO3.
	b.	SO4^2-.
	c.	S2O3^2-.
	d.	SO3^2-.
	e.	S^2-.

The correct answer is:
H₂O₂ --> OH^-

Explanation:

The oxidation number of oxygen changes from -1 to -2. This is reduction.

Solution Reference:
Page 206, Oxidation Numbers, and Page 947, Oxidation-Reduction Reactions

5. In which of the following half reactions is reduction occurring?
	a.	SO3 --> SO4^2-
	b.	SO2 --> HSO3^-
	c.	H2O2 --> OH^-
	d.	Mn^2+ --> Mn2O3
	e.	CrO4^2- --> Cr2O7^2-

The correct answer is:
1 and 4

Explanation:
There needs to be both a mass balance and a charge balance.

Solution Reference:
Page 948, Balancing Equations for Oxidation-Reduction Reactions

6. In the redox reaction 2Fe^3+ + 2Br^- --> Br2 + 2Fe^2+, which of the following are the correct half reactions? 1. 2Br^- --> Br2 + 2e 2. Br^- --> Br + e 3. 2Br^- + 2e --> Br2 4. Fe^3+ + e --> Fe^2+ 5. Fe^3+ --> Fe^2+ + e
	a.	1 and 4
	b.	2 and 5
	c.	2 and 4
	d.	3 and 4
	e.	3 and 5

The correct answer is:
four.

Explanation:
C₂H₅OH --> CH₃COOH

First, balance the oxygen atoms by adding one water to the left-hand side.
C₂H₅OH + H₂O --> CH₃COOH

Now balance the hydrogen atoms by adding H⁺ to the right-hand side.
C₂H₅OH + H₂O --> CH₃COOH + 4H⁺

Now balance the charge by adding electrons to the appropriate side.
C₂H₅OH + H₂O --> CH₃COOH + 4H⁺ + 4e

Solution Reference:
Page 948, Balancing Equations for Oxidation-Reduction Reactions

7. For the oxidation of ethanol to ethanoic acid (acetic acid) C2H5OH --> CH3COOH (not balanced) the number of electrons that must be added to the right side is
	a.	zero.
	b.	one.
	c.	two.
	d.	three.
	e.	four.

The correct answer is:
4 OH^- on the left side and 4 electrons on the right.

Explanation:
ClO^- + 4 OH^- --> ClO₃^- + 4e + 2H₂O

Solution Reference:
Page 948, Balancing Equations for Oxidation-Reduction Reactions

8. When the half-reaction involving the conversion of ClO^- to ClO3^- in basic solution is balanced, there are
	a.	4 OH^- + 4e on the left-hand side.
	b.	4 OH^- on the left side and 4 electrons on the right.
	c.	2 OH^- on the left side and 2 electrons on the right.
	d.	2 OH^- on the left side and 2 electrons on the left.
	e.	1 OH^- on the left side and 2 electrons on the right.

The correct answer is:
20.

Explanation:

8H₂O + 4MnO₄^- + 3ClO₂^- + 12 OH^- --> 4MnO₂ + 3ClO₄^- + 16 OH^- + 6H₂O

2H₂O + 4MnO₄^- + 3ClO₂^- --> 4MnO₂ + 3ClO₄^- + 4OH^-

Solution Reference:
Page 948, Balancing Equations for Oxidation-Reduction Reactions

9. The following reaction occurs in basic solution: ___ H2O + ___ MnO4^- + ___ ClO2^- --> ___ MnO2 + ___ ClO4^- + ___ OH^- Balance the equation above and then add together all of the coefficients in the balanced equation. The sum of these coefficients is
	a.	15.
	b.	20.
	c.	24.
	d.	25.
	e.	30.

The correct answer is:
from the anode to the cathode.

Explanation:
Oxidation occurs at the anode and reduction at the cathode, so electrons must flow from the anode to the cathode.

Solution Reference:
Page 958, Chemical Change Leading to an Electric Current

10. In a voltaic cell, the electron flow is always from
	a.	the positive electrode to the negative electrode.
	b.	through the salt bridge.
	c.	from the oxidizing agent through the salt bridge to the reducing agent.
	d.	from the reducing agent through the salt bridge to the oxidizing agent.
	e.	from the anode to the cathode.

The correct answer is:
-625 kJ

Explanation:


Solution Reference:
Page 962, E° and DG°

11. Calculate the Gibbs free energy change for the reaction below when the initial concentrations of Cr^3+ and Cu^2+ are 1.00 M and the cell potential is 1.08 volts.
	a.	-1450 kJ
	b.	-625 kJ
	c.	-232 kJ
	d.	-208 kJ
	e.	+208 kJ

The correct answer is:
Fe + Cu²⁺ --> Fe²⁺ + Cu

Explanation:
Iron is more reactive than Cu (Fe is a stronger reducing agent).

Solution Reference:
Page 964, Calculating the Potential E° of an Electrochemical Cell

12. Iron reacts with hydrochloric acid to produce hydrogen and iron(II) ions, but copper does not react with hydrochloric acid to produce hydrogen. From this information we can predict that the following reaction would be spontaneous.
	a.	Cu + Fe^2+ --> Fe + Cu^2+
	b.	2Fe^2+ + Cu^2+ --> 2Fe^3+ + Cu
	c.	Fe + Cu^2+ --> Fe^2+ + Cu
	d.	2Cu^+ --> Cu + Cu^2+
	e.	Fe^2+ + Cu^2+ --> Fe^3+ + Cu^+

The correct answer is:
remain unchanged.

Explanation:
The cell potential of a complete reaction is independent of any value assigned to the reference half-cell. The potential is related to the difference in energy of reactants and the products.

Solution Reference:
Page 964, Calculating the Potential E° of an Electrochemical Cell

13. If the value of -0.50 V were assigned to the standard hydrogen half-cell instead of the currently accepted value of zero, the cell potential of a given electrochemical cell would
	a.	increase by 0.50 V.
	b.	decrease by 0.50 V.
	c.	increase by 1.0 V.
	d.	decrease by 1.0 V.
	e.	remain unchanged.

The correct answer is:
0.61 volts

Explanation:


Solution Reference:
Page 964, Calculating the Potential E? of an Electrochemical Cell

14. Given the standard reduction potentials

Cr^3+ + 3e --> Cr	-0.74 volts
Pb^2+ + 2e --> Pb	-0.13 volts

What is the standard cell potential, E?, for the reaction
2Cr + 3Pb²⁺ --> 2Cr³⁺ + 3Pb

a.

-0.87 volts

b.

1.09 volts

c.

0.61 volts

d.

-1.87 volts

e.

none of these

The correct answer is:
Mn.

Explanation:
Only three reducing agents, Br^-, Sn, and Mn, are among the five responses.
Mn --> Mn²⁺ + 2e      E? = 1.03
The strongest reducing agent has the largest positive half-cell potential.

Solution Reference:
Page 968, Using Standard Potentials

15. The strongest reducing agent of this series is
	a.	Mn^2+.
	b.	Br2.
	c.	Br^-.
	d.	Sn.
	e.	Mn.

The correct answer is:
Cu⁺.

Explanation:
In this problem only two oxidizing agents, Cu⁺ and Mn²⁺, are given.

This reaction has a positive cell potential, so it will go as it is written.
This reaction has a negative cell potential, so it will not go as it is written.

Solution Reference:
Page 968, Using Standard Potentials

16. The element or ion that will oxidize mercury but will not oxidize bromide is
	a.	Mn.
	b.	Cu^+.
	c.	Mn^2+.
	d.	Ag.
	e.	Cu.

The correct answer is:
H⁺

Explanation:
There are three oxidizing agents to consider (H⁺, Al³⁺, and Pb²⁺).

There is a more positive potential for the H⁺, H₂ half-cell. It is helpful for a student to memorize an abbreviated electromotive series.

Solution Reference:
Page 968, Using Standard Potentials

17. If all of the species below are in their standard states, which is the strongest oxidizing agent?
	a.	H^+
	b.	K
	c.	F^-
	d.	Al^3+
	e.	Pb^2+

The correct answer is:
Electrons will flow from Tl through the external circuit to Al.

Explanation:
Electrons will flow from the anode (Al) to the cathode (Tl).

Solution Reference:
Page 968, Using Standard Potentials

18. Given the following standard electrode potentials: If the concentrations of ions are all 1.00 M and an electrochemical cell is constructed from Al, Al^3+, Tl^+, and Tl, which of the following statements is incorrect?
	a.	Electrons will flow from Tl through the external circuit to Al.
	b.	If the standard hydrogen half-cell were substituted for the Tl half-cell, the voltage would be 1.66 V.
	c.	The cell voltage would be 1.32 V.
	d.	The negative terminal of the voltmeter should be attached to Al for a positive reading.
	e.	The following electrode reaction will occur spontaneously: Tl^+(aq) + e --> Tl

The correct answer is:
6

Explanation:

There is a 6-electron change in the balanced half-reactions.

Solution Reference:
Page 973, The Nernst Equation

19. For cell reaction 3Hg + Cr2O7^2- + 14H^+ --> 2Cr^3+ + 3Hg^2+ + 7H2O, the standard cell potential is 0.48 V. To determine the cell potential at nonstandard conditions, the value that should be used for n in the Nernst equation is
	a.	1
	b.	2
	c.	3
	d.	4
	e.	6

The correct answer is:
1 only

Explanation:
By increasing the pH we are lowering the H+ concentration. More students usually select response d than a. Increasing the amount of Hg does affect the potential because we haven't changed the concentration of the Hg, as it is a pure phase.

Solution Reference:
Page 973, The Nernst Equation

20. For cell reaction 3Hg + Cr2O7^2- + 14H^+ --> 2Cr^3+ + 3Hg^2+ + 7H2O, the standard cell potential is 0.48 V. Which change(s) will result in an increase in the cell potential? 1. Increasing the dichromate ion concentration 2. Increasing the pH 3. Increasing the mercury ion concentration 4. Increasing the amount of mercury
	a.	1 only
	b.	2 only
	c.	3 only
	d.	1 and 4 only
	e.	2 and 4 only

The correct answer is:
the [Zn²⁺] was larger than [Ag⁺].

Explanation:
Zn + 2Ag⁺ --> Zn²⁺ + 2Ag

If [Ag⁺] = [Zn²⁺] = 1.0 M, then E = E° = 1.56 V. If Zn²⁺ is larger than Ag⁺, then E < E°.

Solution Reference:
Page 973, The Nernst Equation

21. For a cell consisting of Zn, Zn^2+, Ag^+, and Ag, the standard cell potential is 1.56 V. A cell using these reagents was made and the observed potential was 1.35 V. A possible explanation for the observed voltage is
	a.	there was 0.0100 mole of Zn^2+ and 0.200 mole of Ag^+.
	b.	the [Zn^2+] was larger than [Ag^+].
	c.	the Ag electrode was twice as large as the Zn electrode.
	d.	the volume of the Zn^2+ solution was larger than the volume of the Ag^+ solution.
	e.	the volume of the Ag^+ solution was larger than the volume of the Zn^2+ solution.

The correct answer is:
0.56

Explanation:


Solution Reference:
Page 989, Counting Electrons

22. A current of 10 amperes is passed through molten magnesium chloride for 3.0 hours. How many moles of magnesium metal could be produced by this electrolysis (one Faraday = 96,500 coulombs)?
	a.	1.1
	b.	2.2
	c.	0.56
	d.	0.37
	e.	0.22

The correct answer is:
Cl₂ is produced at the anode.

Explanation:
Oxidation occurs at the anode. Chlorine is produced commercially by electrolysis.

Solution Reference:
Page 992, Chlorine and Sodium Hydroxide

23. When a sodium chloride solution is electrolyzed using platinum electrodes
	a.	Cl2 is produced at the anode.
	b.	O2 is produced at the anode.
	c.	O2 is produced at the cathode.
	d.	H2 is produced at the anode.
	e.	Na is produced at the cathode.

The correct answer is:
3 only

Explanation:
Aluminum is the only element listed that is a stronger reducing agent than iron. The iron in a "tin can" rusts faster because the tin facilitates the oxidation of the iron.

Solution Reference:
Page 581, Corrosion

24. Protection of iron from corrosion can be accomplished by making an electrical contact between iron and certain other metals. Metal(s) that would provide protection is (are) 1. Hg. 2. Sn. 3. Al. 4. Ag.
	a.	1 only
	b.	3 only
	c.	1 and 2 only
	d.	2 and 4 only
	e.	3 and 4 only

The correct answer is:
the sodium ions and chloride ions carry a current through the solution.

Explanation:
Many students select response b. Chloride ions can only function as a reducing agent. The sodium chloride acts as an electrolyte. Cars rust more rapidly in areas where roads are salted during the winter. Why?

Solution Reference:
Page 981, Corrosion

25. A piece of iron half immersed in a sodium chloride solution will corrode more rapidly than a piece of iron half immersed in pure water because
	a.	the sodium ions oxidize the iron atoms.
	b.	the chloride ions oxidize the iron atoms.
	c.	the chloride ions form a precipitate with iron.
	d.	the chloride ions increase the pH of the solution.
	e.	the sodium ions and chloride ions carry a current through the solution.