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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 t he 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
2
H
He
1.0
4.0
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.9
9.0
10.8
12.0
14.0
16.0
19.0
20.2
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
23.0
24.3
27.0
28.1
31.0
32.1
35.5
39.9
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.1
40.1
45.0
47.9
50.9
52.0
54.9
55.8
58.9
58.7
63.5
65.4
69.7
72.6
74.9
79.0
79.9
83.8
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.5
87.6
88.9
91.2
92.9
95.9
(98)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La *
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
137.3
138.9
178.5
180.9
183.9
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
64
65
66
67
68
69
70
71
87
88
89
104
105
106
107
108
109
Fr
Ra
Ac †
Unq
Unp
Unh
Uns
Uno
Une
(223)
226.0
227.0
(261)
(262)
(263)
(262)
(265)
(267)
58
59
60
61
62
63
*
†
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.1
140.9
144.2
(145)
150.4
152.0
157.3
158.9
162.5
164.9
167.3
168.9
173.0
175.0
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.0
(231)
238.0
(237)
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(259)
(260)
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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: Au 3+ + 3 e– → Au
E ° = 1.50V
Cl 2 + 2 e– → 2Cl–
E ° = 1.36V A
B
B . 3Cl2(g) + 2Au(s) → 6Cl–(aq) + 2Au+3(aq) C . Cl2(g) + Cl–(aq) → Au(s) + Au+3(aq)
D . 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.
C E
F Cl2 (1 atm)
D
Pt
A . 3Cl2(g) + 2Au+3(aq) → 6Cl–(aq) + 2Au(s)
4 . With switch C open, what is the initial reading on the voltmeter? Au
+3
Cl (l M )
Au (l M )
Figure 1
A . –2.80V B . –0.14V C . 0.14V D . 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 . Cl2(g), Cl–(aq) Au+3(aq), Au(s)
B . Au+3(aq) Au(s) Cl–(aq) Cl2(g)
C . Pt( s) Cl2(g) Cl–(aq) Au+3(aq) Au(s) D . 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.
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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.
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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
3
2
Pt
Pt
Pt
H2O NaOH
1
+2
Cu SO4 -2
Pt
SO 42– + 4H + + 2 e– → H 2SO 3 + H 2O
1 3 . Which of the following reactions will take place at Electrode 3? A. B. C. D.
A. B. C. D.
Figure 1
Cu 2+ + 2 e– → Cu Na+ + e– → Na O2 + 2H 2O + 4 e– → 4OH –
4 only 1 and 2 only 2 and 4 only 2, 3, and 4 only
Na+ +e– → Na 2H + + 2 e– → H 2 Na + H2O → 1/2H2 + NaOH 2O2– → O2 + 4 e–
1 4 . What is the ratio in the left-hand cell of the volume of hydrogen gas to that of oxygen gas?
H2O
Table 1 Half-reaction 2H + + 2 e– → H 2
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 . 0.39 grams B . 23.7 grams C . 47.4 grams D . 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 B 5.
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.
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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, w hich 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.
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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 E cell = E cathode – E anode, 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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