6Electrochemistry test w solutions

GENERAL CHEMISTRY TOPICAL:
Electrochemistry
Test 1
Time: 21 Minutes*
Number of Questions: 16

* The timing restrictions for the science topical tests are optional. If
you are using this test for the sole purpose of content
reinforcement, you may want to disregard the time limit.


MCAT

DIRECTIONS: Most of the questions in the following
test are organized into groups, with a descriptive
passage preceding each group of questions. Study the
passage, then select the single best answer to each
question in the group. Some of the questions are not
based on a descriptive passage; you must also select the
best answer to these questions. If you are unsure of the
best answer, eliminate the choices that you know are
incorrect, then select an answer from the choices that
remain. Indicate your selection by blackening the
corresponding circle on your answer sheet. A periodic
table is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1
H
1.0

2
He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8
O
16.0

9
F

19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S
32.1

17
Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23
V
50.9

24
Cr

52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga
69.7

32
Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38
Sr
87.6

39
Y

88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd
106.4

47
Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53
I
126.9

54
Xe

131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re
186.2

76
Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82
Pb
207.2

83
Bi

209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Unq
(261)

105
Unp
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr
140.9

60
Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66
Dy
162.5

67
Ho

164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91
Pa
(231)

92
U

238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es
(252)

100
Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

GO ON TO THE NEXT PAGE.

2

as developed by


Electrochemistry Test 1
Passage I (Questions 1–6)

2 . When the cell described in the passage is operating
spontaneously, what is the overall chemical reaction?

The components of a galvanic cell are shown in

Figure 1. The left container contains a platinum electrode
in a 1.0 M chloride solution; chlorine gas at one
atmosphere pressure is bubbled over the electrode. The
right container contains a gold electrode in a 1.0 M Au3+
solution. As can be seen in Figure 1, the circuit includes a
voltmeter (A); a light bulb (B); two switches (C and E); a
salt bridge (D); and a power source (F) that produces a
voltage greater than that generated by the cell. At the start
of the experiment, the switches are open.
The relevant reduction potentials are the following:
Au3+ + 3e– → Au

E° = 1.50V

Cl2 + 2e– → 2Cl–

E° = 1.36V
A

B

C
E

F
Cl2
(1 atm)

D

Cl
(lM)

3Cl2(g) + 2Au+3(aq) → 6Cl–(aq) + 2Au(s)
3Cl2(g) + 2Au(s) → 6Cl–(aq) + 2Au+3(aq)
Cl2(g) + Cl–(aq) → Au(s) + Au+3(aq)
2Au+3(aq) + 6Cl–(aq) → 3Cl2 (g) + 2Au(s)

3 . When switch E is open, which electrode is the anode?
A . The Au electrode, because it is written on the
right.
B . The Pt electrode, because it is where oxidation
occurs.
C . The Au electrode, because it is where oxidation
occurs.
D . The Pt electrode, because it is where reduction
occurs.
4 . With switch C open, what is the initial reading on
the voltmeter?

Au

+3
Pt

A.
B.
C.
D.

Au
(lM)

Figure 1

A.
B.
C.
D.

–2.80V
–0.14V
0.14V
2.80V

5 . According to Figure 1, in which direction are the
electrons flowing?
A . From right to left
B . From left to right
C . From the material being reduced to the material
being oxidized
D . From the cathode to the anode

1 . When switches C and E are open, what is the cell
diagram for this galvanic cell?
A.
B.
C.
D.

Cl2(g), Cl–(aq)  Au+3(aq), Au(s)
Au+3(aq)  Au(s)  Cl–(aq)  Cl2(g)
Pt(s)  Cl2(g)  Cl–(aq)  Au+3(aq)  Au(s)
Pt(s)  Cl–(aq)  Au+3(aq)  Au(s)

6 . When switch C is closed, the light bulb will,
theoretically, continue to glow until:
A.
B.
C.
D.

all the Au+3 ions are consumed.
the Pt electrode dissolves.
the reaction has reached equilibrium.
the salt bridge allows chloride ion to mix with
the gold solution.

GO ON TO THE NEXT PAGE.

KAPLAN

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MCAT
Questions 7 through 10 are NOT
based on a descriptive passage.
7 . Which of the following is true of an electrolytic cell?
A . An electric current causes an otherwise

nonspontaneous chemical reaction to occur.
B . Reduction occurs at the anode.
C . A spontaneous electrochemical reaction produces
an electric current.
D . The electrode to which electrons flow is where
oxidation occurs.

1 0 . Which of the following would be classified as a
strong electrolyte?
A.
B.
C.
D.

Benzoic acid
Water
Hydrofluoric acid
Potassium chloride

8 . For a galvanic cell:
A.
B.
C.
D.

the cell potential is always positive.
the products are less stable than the reactants.
∆G for the cell reaction is positive.
the cell potential is always negative.

9 . A Faraday is:
A . the magnitude of the charge of 1 mole of
electrons.
B . the magnitude of the electric dipole.
C . a fundamental constant of nature equal to
6.63 × 10–34 J•s/photon.
D . a constant that accounts for the existence of ions
in solution.

GO ON TO THE NEXT PAGE.

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Electrochemistry Test 1
Passage II (Questions 11–16)

1 2 . At which electrode is oxygen gas liberated?

Two electrolytic cells are set up in series and
attached to a power source as shown in Figure 1. The
power source produces 2.0 A at 0.5 V. Electrons are
flowing as indicated by the arrows. Data from a table of
reduction potentials are shown in Table 1.

Power
Source

4

Pt

3

2

Pt

Pt

H2O
NaOH

1

+2

Cu
SO4 -2
H2O

Pt

4 only
1 and 2 only
2 and 4 only
2, 3, and 4 only

1 3 . Which of the following reactions will take place at
Electrode 3?
A.
B.
C.
D.

Na+ +e– → Na
2H+ + 2e– → H2
Na + H2O → 1/2H2 + NaOH
2O2– → O2 + 4e–

1 4 . What is the ratio in the left-hand cell of the volume
of hydrogen gas to that of oxygen gas?
A.
B.
C.
D.

Figure 1
Table 1
Half-reaction
2H+ + 2e– → H2
Cu2+ + 2e– → Cu
Na+ + e– → Na
O2 + 2H2O + 4e– → 4OH–
SO42– + 4H+ + 2e– → H2SO3 + H2O

A.
B.

C.
D.

E°(V)
0.00
0.34
–2.71
0.40
0.20

1:2
2:1
1:1
2:4

1 5 . Which electrodes are the cathodes?
A.
B.
C.
D.

2 and 4
2 and 3
1 and 4
1 and 3

1 6 . What is the purpose of sodium hydroxide in the lefthand cell?

1 1 . If the current is allowed to flow for 10 minutes, how
much copper will be deposited on Electrode 1?

(1F = 96,500 C/mole)
A.
B.
C.
D.

0.39 grams
23.7 grams
47.4 grams
63.5 grams

A . To protect the Pt from corrosion during
electrolysis
B . To suppress the autoionization of water, which
in turn, facilitates hydrolysis
C . To act as a catalyst in the oxidation/reduction
reaction
D . To serve as an electrolyte, which facilitates
current flow

END OF TEST

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MCAT
ANSWER KEY:
1.

C
2.
D
3.
B
4.
C
5.
B

6

6.
7.
8.
9.
10.

C
A
A
A
D

11.
12.
13.
14.
15.

A
C
B
B
D

16.

D

as developed by


Electrochemistry Test 1
ELECTROCHEMISTRY TEST 1 TRANSCRIPT
Passage I (Questions 1–6)
1.
The answer to question 1 is choice C. In a cell diagram--shorthand for the layout of an electrochemical cell-starting from the left, the anode appears first, followed by a single vertical line--which represents a phase boundary--and then
the anode electrolyte solution. Next is a double vertical line; it represents the presence of a salt bridge, which separates the
anode and the cathode compartments. To the right of the double lines appears the cathode electrolyte followed by a single
vertical line, and then at the far right, the cathode electrode is written.
So, all you need to do now is determine which compartment in Figure 1 is the cathode and which is the anode.
There are a few things that you must have committed to memory in order to answer this question correctly: (1) you need to
know that the cell potential must be greater than zero; (2) you need to know that the cell potential is equal to the cathode
potential minus the anode potential when they are written as reductions. All right, as I just said, in order for an
electrochemical reaction to be spontaneous, the cell potential must be greater than zero. For this question, this is
accomplished by subtracting the chlorine half-reaction from the gold half-reaction (you don’t have to actually do the
subtraction in order to answer this question. All you need to do is to notice that the gold potential is greater than the chlorine
potential). So, according to our rules for constructing cell diagrams, platinum is the anode, so it should appear on the far left.
Choices A and B can now be eliminated. Choices C and D are identical except that choice C has one more term--chlorine gas.

Looking at Figure 1 and the passage text, you can see that chlorine gas is bubbled through the anode chamber. Now, since
chlorine gas--bubbling through the anode chamber--is a phase that must be included in the cell diagram, choice C is the
correct choice.
2.
The correct answer to question 2 is answer choice D. When an electrochemical cell operates
spontaneously it is called a galvanic cell. There are a few things about galvanic cells that you should know: (1) the cell
potential is greater than zero; (2) reduction--the gain of electrons--occurs at the cathode; (3) oxidation--the loss of electrons-occurs at the anode. So, for this question, it was previously determined that the gold electrode is the cathode and that the
platinum electrode is the anode--this arrangement gives the necessary positive cell potential. Now, to determine the overall
reaction you add the cathode reaction to the anode reaction. Remember to first make sure that each half-reaction has the same
number of electrons; if they don’t, multiply one or both of them by an appropriate coefficient to balance the number of
electrons.
Anyway, looking at the two half-reactions, you can see that the electrons are not balanced. In order to make the
electrons balance, the gold half-reaction must be multiplied by 2 and the chlorine half-reaction must be multiplied by 3. All
right, before you can add them together, you have to reverse the direction of the anode reaction since it is oxidation that occurs
here and not reduction. So, reverse the direction of the chlorine half-reaction and add the two together. In doing this, the
electrons cancel, and you are left with choice D as the correct answer.
3.
The answer to question 3 is choice B. Determining at which electrodes oxidation and reduction occurs,
finding the cell potential, and assigning the proper labels to the electrodes--these are all things you need to know to answer
this question correctly. Granted, we have previously gone over these things (in questions 1 and 2), but since this is a topical
exam, I’ll go over them again. All right, to determine where--at which electrode--oxidation and reductions occurs you need to
know that for a spontaneous reaction, the overall cell potential must be greater than zero. Taking this into consideration, you
need to look at the half-reactions involved, and decide how to arrange them in such a way so as to get a positive overall cell
potential. For this reaction--as described earlier--the cell potential is positive when the gold half-reaction is the cathode--the
place where reduction occurs--and the platinum half-reaction is the anode--the place where oxidation occurs. This is choice B.
Taking a look at the other answer choices, choice A is wrong because even though it is true that the gold electrode is
written on the right in a cell diagram, that position is occupied by the cathode--the place where reduction takes place. Choice
C is wrong because oxidation does not occur at the gold electrode--the cathode. Choice D is wrong because the platinum
electrode is where oxidation takes place, not reduction. Again, choice B is correct.
4.

The answer to question 4 is C. Remember: a galvanic cell--which this is--is a spontaneous electrochemical
reaction that has a positive cell potential. Now, if the cell potential is positive, G is less than zero, by the relationship G
= –nFE. You should also remember that if G is less than zero, the reaction is spontaneous. Anyway, looking at the two
half-reaction presented in the passage, subtract them in such a way so as to get a number greater than zero. The only way that
you’ll get a number greater than zero is if you subtract the reduction potential of chlorine--1.36 V--from the reduction
potential of gold--1.50V. Doing this gives choice C--0.14V. (If you had the case where no matter how you subtracted them
you got a positive value, choose the value that is MORE positive.) Again the correct answer is choice C.
5.
The answer to question 5 is choice B. In a voltaic cell, the electrons always flow from the anode--where
electrons are lost--to the cathode--where electrons are gained. Therefore, according to Figure 1, the electrons are flowing from
the left to the right: choice B. If you had trouble with this question, please review the explanations to questions 1 through
4.

KAPLAN

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MCAT
6.
Choice C is the answer to question 6. The light bulb will continue to glow until either the bulb burns
out--which is not one of the choices--or the flow of electrons stops. You should know an electrochemical cell has a cell
potential--which you determined in question 4--because there is a flow of electrons. When the flow of electrons has stopped,
two important things happen: (1) the cell potential is zero; (2) the system is at equilibrium (recall the relationship G = nFE). So, choice C, the system has reached equilibrium, is the correct choice. When the potential reaches zero, the electrons
will stop flowing. This happens when equilibrium has been established (i.e., when the forward reaction rate equals the reverse
reaction rate, no net change in concentrations is taking place). At equilibrium, the concentration of no reactant or product
goes to zero, so choice A is wrong. Inert electrodes, such as platinum, do not dissolve in this environment. Answer B is
wrong. A salt bridge always stays intact, so choice D is wrong. Again, choice C is the correct response.
Discrete Questions
7.

The correct answer to question 7 is choice A. Electrochemical reactions that are nonspontaneous, those
having a positive ∆G, can be driven to completion by passing an electric current through the solution; this process is known
as electrolysis and the cell is called an electrolytic cell. In an electrolytic cell, the anode is positively charged and the cathode
is negatively charged--the opposite of a galvanic cell. Just like a galvanic cell, oxidation occurs at the anode and reduction
occurs at the cathode. Answer choice A is correct because--as I just said--an electric current does indeed drive the reaction to
completion. Answer choice B is wrong because oxidation--not reduction--occurs at the anode. Choice C is wrong because the
electrochemical reaction is NOT a spontaneous one. Choice D is wrong because the electrode at which electrons flow into is
where reduction occurs, not oxidation. Again, the correct answer is choice A.
8. Choice A is the correct answer for question 8. A spontaneous reaction occurs in a galvanic cell. As I stated
earlier, a spontaneous reaction has a negative delta G and a positive cell potential. Choice A--the cell potential of a galvanic
cell is positive--is the correct answer. Choice C is wrong because the delta G for a galvanic cell is negative, not positive-remember that G = —nFE. Choice B--the products are less stable than the reactants--is wrong because the G for a galvanic
cell is negative, meaning that the products are more stable than the reactants. Choice D is wrong because the cell voltage of a
galvanic cell is positive, not negative. Again the correct answer is choice A.
9.
Choice A is the correct answer for question 9: a Faraday is the magnitude of the charge of one mole of
electrons. The Faraday was named after the famous scientist Michael Faraday who was responsible for the discovery of
electromagnetic induction, the laws of electrolysis, the discovery of benzene, and many other important discoveries. The
Faraday is used in the equation G = –nFE and in the Nernst equation. Choice C is wrong because this is the definition of
Planck’s constant. Choice B is wrong because this is the definition of the electric dipole moment. Choice D is wrong
because this is the definition of the i factor: a factor used in the discussion of the colligative properties. Again, choice A is
the correct answer.
10.
The correct answer to question 10 is choice D. An electrolyte is a substance that ionizes to yield an
electrically conducting solution. A strong electrolyte is one that ionizes completely or nearly completely, and a weak
electrolyte doesn’t ionize very much at all. Examples of strong electrolytes are NaCl, KCl, HCl, HBr, and HI. Examples of
weak electrolytes are water, HF, acetic acid, benzoic acid, and ammonia. Choice D--KCl--is the only strong electrolyte and is,
therefore, the correct answer. Benzoic acid (choice A), water (choice B), and hydrofluoric acid (choice C) are all weak
electrolytes. Again, the correct answer is choice D.
Passage II (Questions 11–16)
11.

The correct answer to question 11 is answer choice A. In order to answer this question correctly you
need to know that the quantity of charge--Q --is equal to the current in Amps--i--times the time--t--in seconds. This will give
an answer in coulombs. You now have to convert from coulombs into moles of electrons using the faraday constant: 96,500
coulombs per mole of electrons, which is given to you in the question stem. Having moles of electrons, you need to convert
to the number of moles of copper; this is done by looking at the second half-reaction in Table 1 and seeing that two moles of
electrons give one mole of copper. You now need to divide your answer by two; this gives an answer in moles of copper.
This number must now be divided by the molecular weight of copper, which is 63.5 grams and is given in the provided
periodic table. Let’s stop here for a minute and look at the answer choices, hoping that we can simplify things. Choice A is
the only answer choice less than one, all the others are quite a bit larger than one. Let’s take a look at our calculation: 2
amps times 10 minutes times sixty seconds divided by 96,500 coulombs divided by 2 moles of electrons times 63.5 grams.
You should be able to quickly calculate that the numerator is about 72,000: 60--from the 60 seconds per minute term--times
60--the approximate atomic weight of copper--gives 3,600; 3,600 times 20--from 2 amps times 10 minutes--gives 72,000.
You can already see that the denominator is bigger than this--96,500 coulombs is down there. So, since the answer is less
than one, choice A is the correct answer.

8

as developed by


Electrochemistry Test 1
12.
The correct answer to question 12 is choice C. For this question, it’s now important that you know
what is going on. Two important points: the applied voltage potential is 0.5 volts and the voltages of cells connected in
series--which these are--are additive. So, the combined voltage of these two cells must be greater than –0.5 volts. (You
should know that the potential is negative because this is an electrolytic cell, not a galvanic one.) Starting from the left,
electrode number one has electrons coming into it, so a reduction is taking place. Copper is being reduced at this electrode.
At electrode number 2, oxidation must, therefore, be taking place; the oxidation that is occurring is the reverse of the fourth
reaction down in Table 1--oxygen gas is produced at this electrode. Because electrons are flowing into electrode number three,
a reduction is taking place--the reduction of H+ to form hydrogen gas--reaction one in Table 1. Finally, at electrode number

four, oxygen gas is formed by the same reaction that occurs at electrode 2. So, using the familiar Ecell = Ecathode – Eanode, the
cell potential of the right-hand cell is 0.34 V--from the reduction of copper--minus 0.40--from the oxidation of hydroxide ion,
giving a total of minus 0.06 V. For the left-hand cell, it’s 0 volts--from the reduction of H+--minus 0.40 V--again, from the
oxidation of hydroxide ion, giving a total of minus 0.40 V. Adding the two cell voltages together, we get a total of minus
0.46 V, so the applied voltage of 0.5 V is enough to drive the cell forward. So, after all that, oxygen is formed at electrodes
2 and 4, making answer choice C the correct answer.
13.
The correct answer to question 13 is B. As was stated in the explanation to question 12, the reduction of
H+ to form hydrogen gas is occurring at electrode number 3. Choice A is wrong because the applied voltage, 0.5 V, is not
enough to drive the reduction of sodium ion. Choice C is wrong because this is not even a half-reaction. Choice D is wrong
because this reaction is an oxidation. Remember: if electrons are coming into an electrode a reduction is occurring, not an
oxidation. Again, the correct answer is choice B.
14.
The correct answer to question 14 is choice B. In questions 12 and 13, we determined that reduction of
H+ to form hydrogen gas is occurring at electrode 3 and oxidation of hydroxide ion is occurring at electrode 4. What you need
to do now is add the two half-reactions, remembering to balance the electrons of course. The reduction of H+ is the top
reaction in Table 1, and hydroxide appears in the fourth reaction down, this reaction must be reversed before you add it to the
hydrogen reaction. So, reversing the hydroxide equation, you can see that, in order to balance the electrons, the hydrogen
equation must be multiplied by 2. Doing this and adding the two half-reactions gives 4H+ + 4OH− 2H2(g) + O2(g) +
2H2O. The ratio of hydrogen to oxygen is 2:1; answer choice B is the correct answer.
15.
The correct answer to question 15 is answer choice D. If electrons are going into an electrode, reduction
is occurring; the electrode at which reduction is occurring is the cathode. From the arrows in Figure 1, electrons are entering
electrodes 1 and 3; these electrodes are the cathodes. Answer choice D is the correct answer.
16.
Answer choice D is correct. Pure water is not an electrolyte and will not conduct electricity, so an additional
substance must be added to serve as an electrolyte. The other statements are incorrect.

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