MCAT Topical Tests

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GENERAL CHEMISTRY TOPICAL:

ElectrochemistryTest 1

Time: 21 Minutes*Number of Questions: 16

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

MCAT

2 as developed by

DIRECTIONS: Most of the questions in the followingtest are organized into groups, with a descriptivepassage preceding each group of questions. Study thepassage, then select the single best answer to eachquestion in the group. Some of the questions are notbased on a descriptive passage; you must also select thebest answer to these questions. If you are unsure of thebest answer, eliminate the choices that you know areincorrect, then select an answer from the choices thatremain. Indicate your selection by blackening thecorresponding circle on your answer sheet. A periodictable is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS

1H1.0

2He4.0

3Li6.9

4Be9.0

5B

10.8

6C

12.0

7N

14.0

8O

16.0

9F

19.0

10Ne20.2

11Na23.0

12Mg24.3

13Al

27.0

14Si

28.1

15P

31.0

16S

32.1

17Cl

35.5

18Ar39.9

19K

39.1

20Ca40.1

21Sc

45.0

22Ti

47.9

23V

50.9

24Cr52.0

25Mn54.9

26Fe

55.8

27Co58.9

28Ni

58.7

29Cu63.5

30Zn

65.4

31Ga69.7

32Ge72.6

33As

74.9

34Se

79.0

35Br

79.9

36Kr83.8

37Rb85.5

38Sr

87.6

39Y

88.9

40Zr

91.2

41Nb92.9

42Mo95.9

43Tc

(98)

44Ru

101.1

45Rh

102.9

46Pd

106.4

47Ag

107.9

48Cd

112.4

49In

114.8

50Sn

118.7

51Sb

121.8

52Te

127.6

53I

126.9

54Xe

131.355Cs

132.9

56Ba

137.3

57La *138.9

72Hf

178.5

73Ta

180.9

74W

183.9

75Re

186.2

76Os

190.2

77Ir

192.2

78Pt

195.1

79Au

197.0

80Hg

200.6

81Tl

204.4

82Pb

207.2

83Bi

209.0

84Po

(209)

85At

(210)

86Rn

(222)87Fr

(223)

88Ra

226.0

89Ac †227.0

104Unq(261)

105Unp(262)

106Unh(263)

107Uns

(262)

108Uno(265)

109Une(267)

*58Ce

140.1

59Pr

140.9

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

64Gd

157.3

65Tb

158.9

66Dy

162.5

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

71Lu

175.0

†90Th

232.0

91Pa

(231)

92U

238.0

93Np

(237)

94Pu

(244)

95Am

(243)

96Cm

(247)

97Bk

(247)

98Cf

(251)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

103Lr

(260)

GO ON TO THE NEXT PAGE.

Electrochemistry Test 1

KAPLAN 3

Passage I (Questions 1–6)

The components of a galvanic cell are shown inFigure 1. The left container contains a platinum electrodein a 1.0 M chloride solution; chlorine gas at oneatmosphere pressure is bubbled over the electrode. Theright container contains a gold electrode in a 1.0 M Au3+

solution. As can be seen in Figure 1, the circuit includes avoltmeter (A); a light bulb (B); two switches (C and E); asalt bridge (D); and a power source (F) that produces avoltage greater than that generated by the cell. At the startof 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

B

F

DCl2

(1 atm)

Cl(lM)Pt

Au(lM)

+3

Au

CE

A

Figure 1

1 . When switches C and E are open, what is the celldiagram 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)

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

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

B . The Pt electrode, because it is where oxidationoccurs.

C . The Au electrode, because it is where oxidationoccurs.

D . The Pt electrode, because it is where reductionoccurs.

4 . With switch C open, what is the initial reading onthe voltmeter?

A . –2.80VB . –0.14VC . 0.14VD . 2.80V

5 . According to Figure 1, in which direction are theelectrons flowing?

A . From right to leftB . From left to rightC . From the material being reduced to the material

being oxidizedD . From the cathode to the anode

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

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

the gold solution.

GO ON TO THE NEXT PAGE.

MCAT

4 as developed by

Questions 7 through 10 are NOTbased on a descriptive passage.

7 . Which of the following is true of an electrolytic cell?

A . An electric current causes an otherwisenonspontaneous 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.

8 . For a galvanic cell:

A . the cell potential is always positive.B . the products are less stable than the reactants.C . ΔG for the cell reaction is positive.D . the cell potential is always negative.

9 . A Faraday is:

A . the magnitude of the charge of 1 mole ofelectrons.

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.

1 0 . Which of the following would be classified as astrong electrolyte?

A . Benzoic acidB . WaterC . Hydrofluoric acidD . Potassium chloride

GO ON TO THE NEXT PAGE.

Electrochemistry Test 1

KAPLAN 5

Passage II (Questions 11–16)

Two electrolytic cells are set up in series andattached to a power source as shown in Figure 1. Thepower source produces 2.0 A at 0.5 V. Electrons areflowing as indicated by the arrows. Data from a table ofreduction potentials are shown in Table 1.

H2ONaOH

CuSO4

H2O

+2Pt

3

Pt

1

Pt

4

-2

Pt

2

PowerSource

Figure 1

Table 1Half-reaction E°(V)

2H+ + 2e– → H2 0.00Cu2+ + 2e– → Cu 0.34Na+ + e– → Na –2.71O2 + 2H2O + 4e– → 4OH– 0.40SO4

2– + 4H+ + 2e– → H2SO3 + H2O 0.20

1 1 . If the current is allowed to flow for 10 minutes, howmuch copper will be deposited on Electrode 1?(1F = 96,500 C/mole)

A . 0.39 gramsB . 23.7 gramsC . 47.4 gramsD . 63.5 grams

1 2 . At which electrode is oxygen gas liberated?

A . 4 onlyB . 1 and 2 onlyC . 2 and 4 onlyD . 2, 3, and 4 only

1 3 . Which of the following reactions will take place atElectrode 3?

A . Na+ +e– → NaB . 2H+ + 2e– → H2C . Na + H2O → 1/2H2 + NaOHD . 2O2– → O2 + 4e–

1 4 . What is the ratio in the left-hand cell of the volumeof hydrogen gas to that of oxygen gas?

A . 1:2B . 2:1C . 1:1D . 2:4

1 5 . Which electrodes are the cathodes?

A . 2 and 4B . 2 and 3C . 1 and 4D . 1 and 3

1 6 . What is the purpose of sodium hydroxide in the left-hand cell?

A . To protect the Pt from corrosion duringelectrolysis

B . To suppress the autoionization of water, whichin turn, facilitates hydrolysis

C . To act as a catalyst in the oxidation/reductionreaction

D . To serve as an electrolyte, which facilitatescurrent flow

END OF TEST

MCAT

6 as developed by

ANSWER KEY:1. C 6. C 11. A 16. D2. D 7. A 12. C3. B 8. A 13. B4. C 9. A 14. B5. B 10. D 15. D

Electrochemistry Test 1

KAPLAN 7

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 thenthe anode electrolyte solution. Next is a double vertical line; it represents the presence of a salt bridge, which separates theanode and the cathode compartments. To the right of the double lines appears the cathode electrolyte followed by a singlevertical 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 toknow that the cell potential must be greater than zero; (2) you need to know that the cell potential is equal to the cathodepotential minus the anode potential when they are written as reductions. All right, as I just said, in order for anelectrochemical reaction to be spontaneous, the cell potential must be greater than zero. For this question, this isaccomplished by subtracting the chlorine half-reaction from the gold half-reaction (you don’t have to actually do thesubtraction in order to answer this question. All you need to do is to notice that the gold potential is greater than the chlorinepotential). 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, sincechlorine gas--bubbling through the anode chamber--is a phase that must be included in the cell diagram, choice C is thecorrect choice.

2 . The correct answer to question 2 is answer choice D. When an electrochemical cell operatesspontaneously it is called a galvanic cell. There are a few things about galvanic cells that you should know: (1) the cellpotential 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 theplatinum electrode is the anode--this arrangement gives the necessary positive cell potential. Now, to determine the overallreaction you add the cathode reaction to the anode reaction. Remember to first make sure that each half-reaction has the samenumber of electrons; if they don’t, multiply one or both of them by an appropriate coefficient to balance the number ofelectrons.

Anyway, looking at the two half-reactions, you can see that the electrons are not balanced. In order to make theelectrons balance, the gold half-reaction must be multiplied by 2 and the chlorine half-reaction must be multiplied by 3. Allright, before you can add them together, you have to reverse the direction of the anode reaction since it is oxidation that occurshere and not reduction. So, reverse the direction of the chlorine half-reaction and add the two together. In doing this, theelectrons 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 answerthis question correctly. Granted, we have previously gone over these things (in questions 1 and 2), but since this is a topicalexam, I’ll go over them again. All right, to determine where--at which electrode--oxidation and reductions occurs you need toknow that for a spontaneous reaction, the overall cell potential must be greater than zero. Taking this into consideration, youneed to look at the half-reactions involved, and decide how to arrange them in such a way so as to get a positive overall cellpotential. For this reaction--as described earlier--the cell potential is positive when the gold half-reaction is the cathode--theplace 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 iswritten on the right in a cell diagram, that position is occupied by the cathode--the place where reduction takes place. ChoiceC is wrong because oxidation does not occur at the gold electrode--the cathode. Choice D is wrong because the platinumelectrode 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 electrochemicalreaction 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 twohalf-reaction presented in the passage, subtract them in such a way so as to get a number greater than zero. The only way thatyou’ll get a number greater than zero is if you subtract the reduction potential of chlorine--1.36 V--from the reductionpotential of gold--1.50V. Doing this gives choice C--0.14V. (If you had the case where no matter how you subtracted themyou 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--whereelectrons are lost--to the cathode--where electrons are gained. Therefore, according to Figure 1, the electrons are flowing fromthe left to the right: choice B. If you had trouble with this question, please review the explanations to questions 1 through4.

MCAT

8 as developed by

6 . Choice C is the answer to question 6. The light bulb will continue to glow until either the bulb burnsout--which is not one of the choices--or the flow of electrons stops. You should know an electrochemical cell has a cellpotential--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 electronswill stop flowing. This happens when equilibrium has been established (i.e., when the forward reaction rate equals the reversereaction rate, no net change in concentrations is taking place). At equilibrium, the concentration of no reactant or productgoes to zero, so choice A is wrong. Inert electrodes, such as platinum, do not dissolve in this environment. Answer B iswrong. 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, thosehaving a positive ΔG, can be driven to completion by passing an electric current through the solution; this process is knownas electrolysis and the cell is called an electrolytic cell. In an electrolytic cell, the anode is positively charged and the cathodeis negatively charged--the opposite of a galvanic cell. Just like a galvanic cell, oxidation occurs at the anode and reductionoccurs at the cathode. Answer choice A is correct because--as I just said--an electric current does indeed drive the reaction tocompletion. Answer choice B is wrong because oxidation--not reduction--occurs at the anode. Choice C is wrong because theelectrochemical reaction is NOT a spontaneous one. Choice D is wrong because the electrode at which electrons flow into iswhere 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 statedearlier, a spontaneous reaction has a negative delta G and a positive cell potential. Choice A--the cell potential of a galvaniccell 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 galvaniccell is negative, meaning that the products are more stable than the reactants. Choice D is wrong because the cell voltage of agalvanic 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 ofelectrons. The Faraday was named after the famous scientist Michael Faraday who was responsible for the discovery ofelectromagnetic induction, the laws of electrolysis, the discovery of benzene, and many other important discoveries. TheFaraday is used in the equation G = –nFE and in the Nernst equation. Choice C is wrong because this is the definition ofPlanck’s constant. Choice B is wrong because this is the definition of the electric dipole moment. Choice D is wrongbecause this is the definition of the i factor: a factor used in the discussion of the colligative properties. Again, choice A isthe correct answer.

1 0 . The correct answer to question 10 is choice D. An electrolyte is a substance that ionizes to yield anelectrically conducting solution. A strong electrolyte is one that ionizes completely or nearly completely, and a weakelectrolyte doesn’t ionize very much at all. Examples of strong electrolytes are NaCl, KCl, HCl, HBr, and HI. Examples ofweak 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 weakelectrolytes. Again, the correct answer is choice D.

Passage II (Questions 11–16)

1 1 . The correct answer to question 11 is answer choice A. In order to answer this question correctly youneed to know that the quantity of charge--Q --is equal to the current in Amps--i--times the time--t--in seconds. This will givean answer in coulombs. You now have to convert from coulombs into moles of electrons using the faraday constant: 96,500coulombs per mole of electrons, which is given to you in the question stem. Having moles of electrons, you need to convertto the number of moles of copper; this is done by looking at the second half-reaction in Table 1 and seeing that two moles ofelectrons 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 providedperiodic table. Let’s stop here for a minute and look at the answer choices, hoping that we can simplify things. Choice A isthe only answer choice less than one, all the others are quite a bit larger than one. Let’s take a look at our calculation: 2amps 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--times60--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 lessthan one, choice A is the correct answer.

Electrochemistry Test 1

KAPLAN 9

1 2 . The correct answer to question 12 is choice C. For this question, it’s now important that you knowwhat is going on. Two important points: the applied voltage potential is 0.5 volts and the voltages of cells connected inseries--which these are--are additive. So, the combined voltage of these two cells must be greater than –0.5 volts. (Youshould 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 fourthreaction 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 numberfour, oxygen gas is formed by the same reaction that occurs at electrode 2. So, using the familiar Ecell = Ecathode – Eanode, thecell 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 theoxidation of hydroxide ion, giving a total of minus 0.40 V. Adding the two cell voltages together, we get a total of minus0.46 V, so the applied voltage of 0.5 V is enough to drive the cell forward. So, after all that, oxygen is formed at electrodes2 and 4, making answer choice C the correct answer.

1 3 . The correct answer to question 13 is B. As was stated in the explanation to question 12, the reduction ofH+ to form hydrogen gas is occurring at electrode number 3. Choice A is wrong because the applied voltage, 0.5 V, is notenough to drive the reduction of sodium ion. Choice C is wrong because this is not even a half-reaction. Choice D is wrongbecause this reaction is an oxidation. Remember: if electrons are coming into an electrode a reduction is occurring, not anoxidation. Again, the correct answer is choice B.

1 4 . The correct answer to question 14 is choice B. In questions 12 and 13, we determined that reduction ofH+ to form hydrogen gas is occurring at electrode 3 and oxidation of hydroxide ion is occurring at electrode 4. What you needto do now is add the two half-reactions, remembering to balance the electrons of course. The reduction of H+ is the topreaction in Table 1, and hydroxide appears in the fourth reaction down, this reaction must be reversed before you add it to thehydrogen reaction. So, reversing the hydroxide equation, you can see that, in order to balance the electrons, the hydrogenequation 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.

1 5 . The correct answer to question 15 is answer choice D. If electrons are going into an electrode, reductionis occurring; the electrode at which reduction is occurring is the cathode. From the arrows in Figure 1, electrons are enteringelectrodes 1 and 3; these electrodes are the cathodes. Answer choice D is the correct answer.

1 6 . Answer choice D is correct. Pure water is not an electrolyte and will not conduct electricity, so an additionalsubstance must be added to serve as an electrolyte. The other statements are incorrect.