Chapter 17 - Practice Multiple Choice Test 2
(from Kotz 5^th Edition Chapter 20)

ANSWERS

The correct answer is:
N₂H₄ --> N₂

Explanation:
An oxidizing agent increases an oxidation number of an atom in a compound. An oxidation agent removes electrons from the substance that is oxidized.


Solution Reference:
Page 947, Oxidation-Reduction Reactions

1. Which of the following changes requires an oxidizing agent?
	a.	NH3 --> NH4^+
	b.	N2H4 --> N2
	c.	NO3^- --> N2O
	d.	NO2^- --> N2O3
	e.	NO2 --> N2O4

The correct answer is:
C₂O₄^2-.

Explanation:
The reducing agent is oxidized.


Solution Reference:
Page 947, Oxidation-Reduction Reactions

2. In the balanced equation 2MnO4^- + 5C2O4^2- + 16H^+ --> 2Mn^2+ + 10 CO2 + 8H2O the reducing agent is
	a.	MnO4^-.
	b.	C2O4^2-.
	c.	H^+.
	d.	Mn^2+.
	e.	CO2.

The correct answer is:
FeCr₂O₄; Fe₂O₃ and Na₂Cr₂O₇

Explanation:
The reducing agent is oxidized.

Solution Reference:
Page 947, Oxidation-Reduction Reactions

3. For the reaction: 4FeCr2O4 + 8Na2CO3 + 7O2 + 4H2SO4 --> 4Na2Cr2O7 + 2Fe2O3 + 4H2O + 8CO2 + 4Na2SO4 The reducing agent is ______ and the oxidation product(s) is (are) ______.
	a.	Na2CO3; CO2 and Na2Cr2O7
	b.	FeCr2O4; Fe2O3 and Na2Cr2O7
	c.	FeCr2O4; Fe2O3 and H2O
	d.	FeCr2O4; Fe2O3, Na2Cr2O7 and H2O
	e.	Na2CO3; CO2 and H2O

The correct answer is:
Zn(OH)₄^2-

Explanation:

Let y represent the oxidation number of Zn in Zn(OH)₄^2-.
y + 4 (-2) + 4 (+1) = -2
y = +2
or
y + 4 OH^- = -2
y + 4 (-1) = -2; y = +2

Solution Reference:
Page 206, Oxidation Numbers

4. In which of the following species does the underlined element have an oxidation number of +2?
	a.	Zn(OH)4^2-
	b.	CrO2Cl2
	c.	HNO2
	d.	PH4^+
	e.	NO2

The correct answer is:
Fe³⁺ --> Fe²⁺ + e

Explanation:
The half-reaction in e does not have a charge balance.
+3 is not equal to +2 - 1

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

5. All of the following half-reactions are balanced EXCEPT
	a.	2H2O + 2e --> H2 + 2OH^-
	b.	NO3^- + 4H^+ + 3e --> NO + 2H2O
	c.	2Ta + 5H2O --> Ta2O5 + 10H^+ + 10e
	d.	H3PO3 + H2O --> H3PO4 + 2H^+ + 2e
	e.	Fe^3+ --> Fe^2+ + e

The correct answer is:
10

Explanation:
To balance this half-reaction, we need only to balance the charges. The left-hand side of the equation has zero charge, therefore we get 0 = -2 + 12 + y; y = -10 or 10 electrons.

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

6. I2 + 6H2O --> 2IO3^- + 12H^+ + ______ e How many electrons are needed to balance the half-reaction above?
	a.	2
	b.	6
	c.	8
	d.	10
	e.	12

The correct answer is:
6H⁺ + 6e on the right-hand side.

Explanation:
C₃H₆O₂ + 2H₂O --> C₃H₄O₄ + 6H⁺ + 6e

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

7. Glycidol can be oxidized in acid solution to malonic acid. When the half-reaction C3H6O2 --> C3H4O4 is balanced, there are
	a.	2H^+ + 2e on the left-hand side.
	b.	2H^+ + 4e on the right-hand side.
	c.	4H^+ + 4e on the right-hand side.
	d.	4H^+ + 4e on the left-hand side.
	e.	6H^+ + 6e on the right-hand side.

The correct answer is:
4

Explanation:


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

8. Balance the equation below. Choose from the numbers given as answers the one that is the coefficient of the underlined substance in the balanced ionic reaction: Zn + NO3^- --> Zn(OH)4^2- + NH3 (basic solution)
	a.	1
	b.	2
	c.	4
	d.	6
	e.	8

The correct answer is:
two on the right.

Explanation:
3(SO₃^2- + 2OH^- --> SO₄^2- + H₂O + 2e)
2(2H₂O + MnO₄^- + 3e --> MnO₂ + 4OH^-)

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

9. For the reaction between permanganate ion and sulfite ion in basic solution, the unbalanced equation is MnO4^- + SO3^2- --> MnO2 + SO4^2-. When this equation is balanced using the smallest whole number coefficients possible, the number of OH^- ions is
	a.	two on the right.
	b.	two on the left.
	c.	three on the right.
	d.	four on the right.
	e.	four on the left.

The correct answer is:
both c and d

Explanation:
Oxidation occurs at the anode. Oxidation is an increase in oxidation number. Oxidation is a loss of electrons.
Cu --> Cu²⁺ + 2e
Zn --> Zn²⁺ + 2e

We can see from the two equations above that there is both a loss of electrons and an increase in oxidation numbers.

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

10. Which of the following processes could occur at the anode of an electrochemical cell?
	a.	Cu^2+ + 2e --> Cu
	b.	Zn^2+ + 2e --> Zn
	c.	Cu --> Cu^2+ + 2e
	d.	Zn --> Zn^2+ + 2e
	e.	both c and d

The correct answer is:
-807 kJ

Explanation:


Solution Reference:
Page 962, E° and DG°

11. Calculate the standard Gibbs free energy change for the reaction: 2Fe + 3Cl2 --> 2Fe^3+ + 6Cl^- given the following electrode potentials:
	a.	765 kJ
	b.	-269 kJ
	c.	-404 kJ
	d.	-765 kJ
	e.	-807 kJ

The correct answer is:
0.15 V.

Explanation:
2H⁺ + 2e --> H₂ at 1.00 M H⁺ and 1.00 atm H₂ is the reference half-cell and it has been assigned the potential of 0.00 volt.

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

12. A cell with a potential of 0.15 V has the cell reaction 2Cu^2+(aq) + H2 --> 2Cu^+(aq) + 2H^+(aq) If the concentrations of ions were 1.0 molar and the pressure of H2 were 1.0 atmosphere, then E° for the half reaction Cu^2+(aq) + e --> Cu^+(aq) would be
	a.	-0.075 V.
	b.	-0.15 V.
	c.	0.075 V.
	d.	0.15 V.
	e.	0.30 V.

The correct answer is:
decrease by 1.00 V and remain unchanged.

Explanation:
The potential between two half-cells is independent of any cell convention.

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

13. If the value of -1.00 volts was assigned to the standard hydrogen half-cell instead of the currently accepted value of zero 2H^+(aq) + 2e --> H2 E? = -1.00 V then all reduction half-cell potentials and all cell potentials would respectively
	a.	decrease by 1.00 V and decrease by 1.00 V.
	b.	decrease by 1.00 V and remain unchanged.
	c.	decrease by 1.00 V and increase by 1.00 V.
	d.	increase by 1.00 V and increase by 1.00 V.
	e.	increase by 1.00 V and decrease by 1.00 V.

The correct answer is:
1.08 V

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 V
Cu^2+ + 2e --> Cu	+0.34 V

what is the standard cell potential for the reaction
2Cr + 3Cu²⁺ --> 3Cu + 2Cr³⁺?

a.

-1.08 V

b.

-0.40 V

c.

0.40 V

d.

1.08 V

e.

2.50 V

The correct answer is:
Sn²⁺.

Explanation:
The oxidizing agents are Al³⁺ and Sn²⁺.


Solution Reference:
Page 968, Using Standard Potentials

15. An element or ion that will oxidize iron but will not oxidize mercury is
	a.	Al^3+.
	b.	Sn^2+.
	c.	Br^-.
	d.	Cu.
	e.	Al.

The correct answer is:
Al.

Explanation:
The reducing agents are H₂, Br^-, and Al.
Al --> Al³⁺ + 3e      1.67 V
The half-reaction written with the species functioning as a reducing agent that has the largest positive potential is the strongest reducing agent.

Solution Reference:
Page 968, Using Standard Potentials

16. The strongest reducing agent of this series is
	a.	H2.
	b.	Br2.
	c.	Br^-.
	d.	Al.
	e.	Al^3+.

The correct answer is:
Cu.

Explanation:
Fe, Sn, and Al will all displace H₂ from water.
Cu will not displace H₂.
A reaction of HNO₃ acid with copper is represented by the equation below.
3Cu + 8H⁺ + 2NO₃^- --> 3Cu²⁺ + 2NO + 4H₂O

Solution Reference:
Page 968, Using Standard Potentials

17. An element that would dissolve in HNO3 without the evolution of H2 would be
	a.	Fe.
	b.	Sn.
	c.	Cu.
	d.	Al.

The correct answer is:
the size of the anode.

Explanation:
A small 12-volt motorcycle battery has the same potential as a large 12-volt car battery.

Solution Reference:
Page 973, The Nernst Equation

18. The voltage produced in the reaction Fe + Cu^2+(aq) --> Cu + Fe^2+(aq) is independent of
	a.	the metal used as the anode.
	b.	the concentration of Fe^2+.
	c.	the concentration of Cu^2+.
	d.	the temperature.
	e.	the size of the anode.

The correct answer is:
E = E°.

Explanation:


Solution Reference:
Page 973, The Nernst Equation

19. From a consideration of the Nernst equation when the concentration of the reactants and products are all 1.0 molar then
	a.	E = 1.0.
	b.	E = E°.
	c.	E = 0.
	d.	E° = 0.
	e.	Kc = Q.

The correct answer is:
DG° is positive and K is less than 1.

Explanation:
If E° is negative then DG° is positive since DG° = -nFE°. If E° is negative, then K must be less than one since
.

Solution Reference:
Page 962, E° and DG°, and Page 975, E° and the Equilibrium Constant

20. For a certain oxidation-reduction reaction, E° is negative. This means that
	a.	DG° is negative and K is less than 1.
	b.	DG° is negative and K is greater than 1.
	c.	DG° is zero and K is greater than 1.
	d.	DG° is positive and K is greater than 1.
	e.	DG° is positive and K is less than 1.

The correct answer is:
decreasing the Mg²⁺ concentration.

Explanation:
Mg + 2Ag⁺ <--> Mg²⁺ + 2Ag

By applying LeChatelier's principle, we can predict the effect that a change in concentration will have on the cell potential (or use the Nernst equation).
a. Decrease because Ag⁺ will precipitate as AgCl.
b. This will shift the equilibrium to the left, thus decreasing the cell potential.
c. No effect.
d. No effect.

Solution Reference:
Page 973, The Nernst Equation

21. A cell consists of a magnesium electrode immersed in a solution of magnesium chloride and a silver electrode immersed in a solution of silver nitrate. The two half-cells are connected by means of a salt bridge. It is possible to increase the voltage of the cell by
	a.	addition of sodium chloride to both half-cells.
	b.	increasing the Mg^2+ concentration and decreasing the Ag^+ concentration.
	c.	increasing the size of the Mg electrode and decreasing the size of the Ag electrode.
	d.	decreasing the size of the Mg electrode and increasing the size of the Ag electrode.
	e.	decreasing the Mg^2+ concentration.

The correct answer is:
decrease by 31.8 g.

Explanation:

Many students select answer e. The first step in solving many chemistry problems is to write a balanced equation.

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

22. A copper electrode in CuSO4 solution is connected by a wire to a silver electrode in a AgNO3 solution. The two solutions are also joined by a salt bridge Current is allowed to flow through the cell until the mass of the silver electrode has changed by 108 g. During this time the mass of the copper electrode will
	a.	decrease by 108 g.
	b.	increase by 108 g.
	c.	increase by 31.8 g.
	d.	decrease by 31.8 g.
	e.	decrease by 63.5 g.

The correct answer is:
0.100 Faraday.

Explanation:


Solution Reference:
Page 989, Counting Electrons

23. A steady current of 10.0 amp flowing for 965 sec. corresponds to
	a.	9650 electrons.
	b.	0.100 Faraday.
	c.	0.100 volt.
	d.	0.100 coulomb.
	e.	None of these.

       Question 24 of 25

The correct answer is:
O₂ and H₂.

Explanation:
The Li₂SO₄ is an electrolyte that was added to help transport the current in order to facilitate the electrolysis of water.

Oxidation occurs at the anode (oxygen) and reduction occurs at the cathode (hydrogen).

Solution Reference:
Page 985, Electrolysis

24. When a dilute aqueous solution of Li2SO4 is electrolyzed, the products formed at the anode and cathode, respectively, are
	a.	O2 and H2.
	b.	H2 and O2.
	c.	O2 and H^+.
	d.	H2 and OH^-.
	e.	O2 and Li.

The correct answer is:
PbO₂ is formed at the anode during charging.

Solution Reference:
Page 976, Batteries and Fuel Cells

25. The following reaction takes place in a lead storage battery. PbO2(s) + Pb(s) + 2H2SO4(aq) --> 2PbSO4(s) + 2H2O(l) Which state is true?
	a.	The concentration of H2SO4 increases as the battery discharges.
	b.	Pb is formed at the anode during discharging.
	c.	PbO2 is formed at the anode during charging.
	d.	The mass of Pb decreases during charging.
	e.	The mass of PbSO4 remains constant during charging and discharging.