Electrochemistry and Nuclear Chemistry

Which one of the following is NOT a characteristic of the anodic electrode in an electrochemical cell? Question 8 Answer Choices A. It is the site of reduction. B. Anions from the salt bridge migrate toward it. C. It is the location where free electrons are formed. D. It is often consumed during the course of the reaction.

A. "It is the site of reduction" is NOT a characteristic of the anode in an electrochemical cell. The anode is the site of oxidation, not reduction.

A ß- particle is emitted by a lithium nucleus. The resulting nucleus will be an isotope of: Question 10 Answer Choices A. beryllium. B. boron. C. lithium. D. helium.

A. A β- particle is emitted by a lithium nucleus. The resulting nucleus will be an isotope of beryllium. The result of beta decay is the transformation of a neutron to a proton and an expelled electron (β- particle). Therefore, the atomic number increases by one. So, if the original atom was lithium (Z = 3), the daughter atom will be beryllium (Z = 4).

Based on the following data: Li+(aq) + e- Li(s) E° = -2.71 V Na+(aq) + e- Na(s) E° = -3.05 V which one of the following must be true? Question 14 Answer Choices A. Sodium metal is more easily oxidized than lithium metal. B. Lithium metal is more easily oxidized than sodium metal. C. Lithium metal is more easily reduced than sodium metal. D. Both lithium and sodium ions are spontaneously reduced by H2

A. Based on the following data sodium is more easily oxidized than lithium. Li+(aq) + e- Li(s) E° = -2.71 V Na+(aq) + e- Na(s) E° = -3.05 V Since the reduction potentials for Li+(aq) and Na+(aq) are both negative, neither process is spontaneous, so "both sodium and lithium are spontaneously reduced" is wrong. Flipping each of the reactions listed in the table reveals that Na(s) is more easily oxidized than Li(s), since 3.05 V > 2.71 V. Neither ion is spontaneously reduced by H2, since the oxidation of H2 is generally defined as 0, and the overall redox equations would have a negative voltage. The redox potentials given do not give any insight into the further reduction of Na or Li metal.

A drained battery is placed onto its charger causing its normal oxidation-reduction reaction to proceed in reverse. Which of the following best describes the changes in the cell? Question 18 Answer Choices A. Both electrodes maintain the same charge following reversal of the reaction. B. Electron flow occurs from the cathode to the anode. C. The reverse reaction occurs spontaneously. D. The cathode becomes the site of oxidation.

A. Both electrodes maintain the same charge following reversal of the reaction upon placing the battery on its charger. When a discharged battery is attached to a voltage source, the reaction proceeds in reverse. This results in the reversal of the anode and cathode (the cathode is still the site of reduction) and the flow of electrons from the new anode to the new cathode. This reversal of the normally spontaneous reaction of a battery requires an attached voltage source, so it would be nonspontaneous, or an electrolytic cell. In an electrolytic cell, the cathode has a negative charge while the anode has a positive charge. Given that the electrodes changed from anode to cathode (and cathode to anode), the electrodes maintain the same charge following reversal of the cell. This answer can always be arrived at using process of elimination as the remaining answer choices are not true for any electrolytic cell.

n an electrolytic cell containing molten MgCl2(l): Question 20 Answer Choices A. Mg2+ is reduced at the cathode, and Cl- is oxidized at the anode B. Mg2+ is oxidized at the anode, and Cl- is reduced at the cathode. C. Mg2+ is reduced at the anode, and Cl- is oxidized at the cathode. D. Mg2+ is oxidized at the cathode, and Cl- is reduced at the anode.

A. In an electrolytic cell containing molten MgCl2(l) Mg2+ is reduced at the cathode, and Cl- is oxidized at the anode. The anode is always the site of oxidation, and the cathode is always the site of reduction, so choices "Mg2+ is reduced at the anode, and Cl- is oxidized at the cathode" and "Mg2+ is oxidized at the cathode, and Cl- is reduced at the anode" are eliminated. Since Mg2+ will not lose any more electrons and Cl- will not gain any more electrons, the choice "Mg2+ is oxidized at the anode, and Cl- is reduced at the cathode" is wrong. The answer is "Mg2+ is reduced at the cathode, and Cl- is oxidized at the anode".

In an electrolytic cell, the electrochemical reaction is: Question 19 Answer Choices A. nonspontaneous, with oxidation occurring at anode. B. nonspontaneous, with oxidation occurring at the cathode. C. spontaneous, with oxidation occurring at the anode. D. spontaneous, with oxidation occurring at the cathode.

A. In an electrolytic cell, the electrochemical reaction is nonspontaneous, with oxidation occurring at anode. Electrolytic cells involve nonspontaneous chemical reactions, so "spontaneous, with oxidation occurring at the anode" and "spontaneous, with oxidation occurring at the cathode" are eliminated. Since the anode is the site of oxidation (and the cathode the site of reduction) in any electrochemical cell, the answer is "nonspontaneous, with oxidation occurring at anode".

In the construction of a "lemon battery", a galvanized nail coated in zinc and a penny are inserted on opposite ends of a lemon. Each metal is then attached to a voltmeter by a wire and a reading of -0.85 V is obtained. Which of the following best describes this reading? Question 6 Answer Choices A. The reaction is spontaneous but the leads from the voltmeter to the anode and cathode have been reversed. B. The reaction is nonspontaneous and will proceed without a power source. C. The reaction is spontaneous but requires a salt bridge to complete the circuit. D. The reaction is nonspontaneous and requires a power source to proceed.

A. In the construction of a "lemon battery", a galvanized nail coated in zinc and a penny are inserted on opposite ends of a lemon. Each metal is then attached to a voltmeter by a wire and a reading of -0.85 V is obtained. This is because the reaction is spontaneous but the leads from the voltmeter to the anode and cathode have been reversed. The "lemon battery" must proceed via a spontaneous reaction or there would be no flow of current. An incomplete circuit would result in no electron flow. Therefore obtaining a negative reading must be due to a reversal of the leads.

What is the oxidation state of iron in FeO42-? Question 20 Answer Choices A. +6 B. +2 C. +3 D. +8

A. The oxidation state of iron in FeO42- is +6. Oxygen maintains an oxidation state of -2, so Fe must be +6 since the molecule is an anion with charge -2.

Given the reactions and standard reduction potentials below, which expression best represents the Eocell of the spontaneous reaction between Al and Sn+2? 2 Al(s) + 3 Sn+2(aq) → 2 Al+3(aq) + 3 Sn+4(aq) Al+3(aq) + 3e- → Al(s) Eo = -1.66 V Sn+2(aq) + 2e- → Sn+4(aq) Eo = +0.15 V Question 1 Answer Choices A. (0.15 V) - (-1.66 V) B. 3(0.15 V) - 2(-1.66 V) C. (-1.66 V) - (0.15 V) D. 2(-1.66 V) - 3(0.15 V)

A. (0.15 V) - (-1.66 V)

Which of the following is LEAST likely to disrupt the function of a single-chamber electrolytic cell designed for electroplating silver? Question 11 Answer Choices A. Severing the salt bridge B. Removing the voltage source C. Exhausting silver cation D. Drop in reaction vessel fluid level below the anode

A. Electrolytic cells do not require two separate reaction vessels: this eliminates the need for a salt bridge, making choice A the best answer. Electroplating is a nonspontaneous process, so without a voltage source the reaction cannot occur (eliminating choice B). With no starting material (Ag+), the reaction cannot occur (eliminate choice C), and with no contact of the solution to the source of electrons, there will be no closed circuit so electrons cannot flow (eliminate choice D).

Given the titration curve below, what is the pKa of the most acidic proton in maleic acid (C4H4O4)? A. 1.9 B. 4.0 C. 6.1 D. 8.5

A. Half-equivalence points on titration curves are the point where the measured pH is equal to the pKa for a given proton and are located at the middle of the buffering region (the flat line) of the curve. During a titration of an acid, the most acidic protons are lost first, thus we are being asked to identify the first half-equivalence point on the above titration curve. The first pKa of maleic acid is 1.9 (choice A is correct).

Given one mole of each of the following substances in one liter of water, which would result in a solution with the highest pH? Question 31 Answer Choices A. HCN (pKa = 9.2) B. HC5H3N4O3 (pKa = 3.9) C. HC3H5O2 (pKa = 4.9) D. HCHO2 (pKa = 3.7)

A. Here we are given four solutions with identical molarities and are asked to find the solution with the highest pH. (Remember that pH decreases as [H+] increases.) As all solutions have an equal concentration, the weakest acid (the one that dissociates the least) will result in the highest pH. We can find the weakest acid by finding the compound with the highest pKa (answer A is correct). Had Ka been provided in place of pKa, we would have looked for the lowest Ka.

In the net reaction of the pyruvate dehydrogenase complex displayed below, NAD+ functions as a(n): pyruvate + HSCoA + NAD+ → acetyl-CoA + NADH + H+ + CO2 Question 16 Answer Choices A. oxidizing agent B. reducing agent. C. catalyst. D. enzyme.

A. NAD+ forms an additional bond to hydrogen to form NADH, a reduction reaction, indicating that NAD+ functions as an oxidizing agent (choice A is correct and choice B is not). While the pyruvate dehydrogenase complex does rely upon enzymatic activity, enzymes are peptides and NAD+ is a nucleotide derivative, eliminating choice D. It is also not a catalyst since it is used up in the reaction, making choice C incorrect.

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) If a galvanic cell with the overall equation shown above was connected to a voltage source and forced to run in reverse, which of the following would be the cathode? Question 28 Answer Choices A. Cu(s) B. Ag+(aq) C. Cu2+(aq) D. Ag(s)

A. Neither of the ions in solution can serve as the cathode, so choices B and C can be eliminated. The cathode is the site of reduction (red cat). In the reverse reaction, silver is being oxidized to silver ion while copper ion is being reduced to copper solid. This makes Cu(s) the cathode (choice A is correct) and Ag(s) the anode.

In a galvanic cell, the half reaction with the higher reduction potential: Question 2 Answer Choices A. will be the reduction and take place at the cathode. B. will be the oxidation and take place at the cathode. C. will be the oxidation and take place at the anode. D. will be the reduction and take place at the anode.

A. will be the reduction and take place at the cathode.

According to the reduction potentials given above, Mn3+ is a much stronger oxidizing agent than Fe3+. However, one would expect that Fe3+ with its larger nuclear charge and slightly smaller size should gain electrons more easily than Mn3+. Which one of the following best accounts for this behavior? Question 21 Answer Choices A. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration. Correct Answer (Blank) B. Fe3+ has a higher ionization energy than Mn3+, so it attracts electrons more strongly than Mn3+. C. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the less stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a 3d5 configuration to a 3d6 configuration, which is more stable since it is closer to a filled 3d10 configuration. D. Mn3+ is more electropositive than Fe3+, so it should attract electrons more strongly.

According to the reduction potentials given above, Mn3+ is a much stronger oxidizing agent than Fe3+. However, one would expect that Fe3+ with its larger nuclear charge and slightly smaller size should gain electrons more easily than Mn3+. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration best accounts for this behavior. The statement "Mn3+ is more electropositive than Fe3+, so it should attract electrons more strongly" is not correct because an electropositive species should not have a greater affinity for electrons than an electronegative one. And even if the statement "Fe3+ has a higher ionization energy than Mn3+, so it attracts electrons more strongly than Mn3+" were a true statement, it doesn't answer the question ("Why does Mn3+ attract electrons more strongly than Fe3+ does?"). The statement "the Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration" is correct, and the statement "the Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the less stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a 3d5 configuration to a 3d6 configuration, which is more stable since it is closer to a filled 3d10 configuration" is false: The 3d subshell can hold 10 electrons (2 electrons in each of the five equivalent d orbitals), and stability is associated with half-filled (and filled) sets of equivalent orbitals. Therefore a 3d5 electron configuration (that is, a half-filled 3d subshell) is more stable than a 3d4 or a 3d6 configuration.

Which of the following are advantageous properties for a titrant to have in a redox titration? I. The net electrochemical reaction must have positive electric potential. II. The titrant must be able to be standardized or exist in a pure solid form. III. The titrant must have different colors for its reduced and oxidized forms. Question 1 Answer Choices A. II and III only B. I and II only C. I only D. II only

B. Advantageous properties for a titrant in a redox titration are that the net electrochemical reaction must have positive electric potential, and the titrant must be able to be standardized or exist in a pure solid form. This is a Roman numeral question, so assess each numbered item as true or false, and eliminate answer choices with each decision. The reaction must be spontaneous for the titration to occur, meaning ΔG must be negative and Eo must be positive. Item I is therefore true, and the "II only" and "II and III only" choices are therefore incorrect. All titrants for all titrations need to have a known concentration with a high degree of certainty. Thus a solution must be able to be made from a dry, pure sample of known mass and volume, or a solution must be able to be standardized with another compound which does have a known concentration. Item II is true as well, so the "I only" choice is incorrect. While there is no reason to even read Item III from a strategy perspective, although having a titrant that has two different colors for its oxidized and reduced forms within the titration is helpful, it is not required. All redox titrations have the advantage that by measuring the potential of the solution during the titration, one can determine the equivalence point of the titration by using graphical methods of analysis.

Al3+ + 3e- Al E° = -1.67 Au3+ + 3e- Au E° = +1.50 Based on the half-reactions listed above, which of the following has the strongest activity? Question 3 Answer Choices A. Al3+ as an oxidizing agent B. Al as a reducing agent C. Au3+ as a reducing agent D. Au as a reducing agent

B. Al3+ + 3e- Al E° = -1.67 Au3+ + 3e- Au E° = +1.50 Based on the half-reactions listed above, Al as a reducing agent has the strongest activity. Strong reducing agents readily lose electrons and are easily oxidized. Likewise, strong oxidizing agents readily gain electrons and are easily reduced. In the half reactions shown, Al3+ gains electrons (is reduced), but the reduction potential for the reaction is negative, indicating it is not spontaneous. Therefore Al3+ is not a good oxidizing agent. Since Au3+ is shown spontaneously gaining electrons (positive E°), not losing them, it cannot be a reducing agent. Taking the reverse of the reactions given, the oxidation potential for Al becomes positive, indicating that Al easily loses electrons. Since it is oxidized easily, it is a good reducing agent, making this choice the best answer. Since the oxidation potential for Au is negative, gold does not lose electrons easily, making it a poor reducing agent.

Carbon-14 has a half-life of 5700 years. Approximately what percent of a carbon-14 sample will remain after 2000 years? Question 11 Answer Choices A. 22% B. 78% C. 52% D. 35%

B. Carbon-14 has a half-life of 5700 years. Approximately 78% of a carbon-14 sample will remain after 2000 years. Since it takes 5700 years for 50% of a 14C sample to decay, if we wait only 2000 years, much less than 50% will have decayed, leaving much more than 50% remaining. Therefore, 78% is the best choice.

Consider the following reduction reactions and their standard potentials: Cu2+ + 2e- ? Cu(s)E° = +0.337 V Zn2+ + 2e- ? Zn(s)E° = -0.763 V If a piece of Zn metal is dropped into a CuSO4(aq) solution, and a piece of Cu metal is dropped into a ZnSO4(aq) solution, then: Question 5 Answer Choices A. the zinc metal will begin to dissolve in the CuSO4 solution since Zn(s) is a stronger oxidizing agent than Cu(s). B. the zinc metal will begin to dissolve in CuSO4 solution since Zn(s) is a stronger reducing agent than Cu(s). C. the copper metal will begin to dissolve in the ZnSO4 solution since Cu(s) is a stronger oxidizing agent than Zn(s). D. the copper metal will begin to dissolve in the ZnSO4 solution since Zn(s) is a stronger reducing agent than Cu(s).

B. Consider the following reduction reactions and their standard potentials: Cu2+ + 2e- → Cu(s) E° = +0.337 V Zn2+ + 2e- → Zn(s) E° = -0.763 V If a piece of Zn metal is dropped into a CuSO4(aq) solution, and a piece of Cu metal is dropped into a ZnSO4(aq) solution, then the zinc metal will begin to dissolve in CuSO4 solution since Zn(s) is a stronger reducing agent than Cu(s). The voltages given for these redox half-reactions indicate that Cu2+ likes to be reduced, but Zn2+ doesn't. Therefore, Zn metal will be oxidized in CuSO4 solution—so Zn metal is the reducing agent—and Cu would precipitate out. The net ionic equation is Zn(s) + Cu2+ → Cu(s) + Zn2+.

Which of the following species is the oxidizing agent in the following redox reaction? Zn + Cu2+ Zn2+ + Cu Question 18 Answer Choices A. Zn B. Cu2+ C. Cu D. Zn2+

B. Cu2+ is the oxidizing agent in the following redox reaction. Zn + Cu2+ Zn2+ + Cu The oxidizing and reducing agents are always reactants, not products, in a redox reaction. This immediately eliminates choices Zn2+ and Cu. The oxidizing agent is the species that gets reduced, that is, the one that gains electrons; this is Cu2+.

Given that a parent and a daughter nucleus are isotopes of the same element, the ratio of alpha to beta decays which produced the daughter must be which of the following? Question 12 Answer Choices A. 2 to 1 B. 1 to 2 C. 1 to 1 D. 2 to 3

B. Given that a parent and a daughter nucleus are isotopes of the same element, the ratio of alpha to beta decays which produced the daughter must be 1 to 2. Alpha decay results in the loss of two protons, while ordinary beta decay (that is, β- decay) results in the gain of one proton. Since isotopes of the same element have the same number of protons, the number of protons lost by alpha decay must equal the number gained by beta decay. Therefore, twice as many beta decays as alpha decays must occur, so the ratio of alpha to beta decays is 1:2.

H2 is removed from an electrochemical cell in which metallic magnesium is oxidized, and H+ is reduced. At STP the volume of gas is measured to be 11 L. If the reaction has proceeded for 2 hours at constant current, what was the value, in amps, of the current? (F = 96,500 C/mol) Question 7 Answer Choices A. 19.9 A B. 13.4 A C. 3.2 A D. 6.5 A

B. H2 is removed from an electrochemical cell in which metallic magnesium is oxidized, and H+ is reduced. At STP the volume of gas is measured to be 11 L. If the reaction has proceeded for 2 hours at constant current, the value of the current is 13.4 A. (F = 96,500 C/mol) At STP 1 mole of gas occupies 22.4 L. Thus, 11 L accounts for half a mole of H2. Since 2 electrons are required to produce each molecule of H2, the production of 0.5 moles of gas requires a total of 1 mole of e-. Current can then be obtained:

Given the following reaction occurring in a galvanic cell, what is the mass of hydroxide produced when the cell is run for 20 minutes at 3 A? (Faraday's constant = 96,500 C/mol e-) ClO4-(aq) + H2O(l) + Sn2+(aq) → ClO3-(aq) + 2 OH-(aq) + Sn4+(aq) Question 13 Answer Choices A. 0.34 grams B. 0.63 grams C. 0.17 grams D. 0.41 grams

B. The mass of hydroxide produced when the cell is run for 20 minutes at 3 A is 0.63 grams. Using unit analysis, we can convert from time the current flows into mass of hydroxide as follows: The molar ratio for electrons can most easily be determined by looking at the Sn half-reaction: Sn2+ → Sn4+ + 2 e- Since 2 mol e- are needed to oxidize 1 mol Sn2+ to 1 mol Sn4+, 2 mol e- are transferred in the overall balanced redox reaction as well. Therefore, there is a 2 mol OH- to 2 mol e- ratio.

The oxidation states of sulfur in H2SO4 and H2SO3 are, respectively: Question 13 Answer Choices A. +2 and +4. B. +6 and +4. C. +4 and +6. D. +4 and +2.

B. The oxidation states of sulfur in H2SO4 and H2SO3 are, respectively +6 and +4. In these molecules, the oxidation number of each H is +1, and the oxidation number of each O is -2. Since the sum of all the oxidation numbers in any neutral molecule must be zero, if we let x denote the oxidation number of S in H2SO4, then we must have 2(+1) + x + 4(-2) = 0, which implies x = +6. The answer must therefore be +6 and +4.

Which of the following statements about stable chemical isotopes is FALSE? Question 2 Answer Choices A. Isotopes can be separated by both chemical and physical means. B. Changing the number of neutrons has little effect on nuclear reactivity. C. Many stable isotopes are naturally occurring. D. Theoretical rate constants of decay are very small.

B. The statement about stable chemical isotopes that is FALSE is "changing the number of neutrons has little effect on nuclear reactivity." Changing the number of neutrons directly affects nuclear stability, particularly for lighter elements. The remaining choices are true statements. Isotopes can be separated by physical (centrifugation) and chemical means (variable reaction rate). Rate constants of decay for a stable isotope would likely be very small leading to a long half-life of decay. Stable isotopes are those that are most likely to be naturally occurring.

The concentration of Cu2+ in a sample can be determined by titration with iodide, as described in the equation below. 2 Cu2+ + 4 I− → 2 CuI + I2 If an endpoint, determined by the iodide-starch test, is achieved after titration with 2.2 g of KI, how many moles of electrons have been transferred? Question 6 Answer Choices A. 0.001 moles B. 0.007 moles C. 0.014 moles D. 0.026 moles

B. 2.2 g of KI (MW = 166 g/mol) represents: 2.2/166 = 22/16 x (10−2) = 3/2 x (10−2) ≈ 0.015 moles I− According to the stoichiometry of the reaction, 2 of the 4 total I− ions are oxidized to I0, while 2 remain I−, meaning ~0.007 moles of electrons have been transferred.

Addition of which of the following to Cd(s) would result in no electrochemical reaction between the two? Fe2+ + 2 e− → Fe(s) E° = -0.44 V Cd2+ + 2 − → Cd() E° = -0.40 V Sn2+ + 2 e− → Sn() E° = -0.14 V Ag+ + e− → Ag() E° = 0.799 V Br2(l) + 2 e− → 2 Br− E° = 1.078 V Question 1 Answer Choices A. Sn2+(aq) B. Fe2+(aq) C. Ag+(aq) D. Br2(l)

B. Addition of Fe2+ to Cd(s) will result in a total cell potential of -0.44 V + 0.40 V = -0.04 V which, given its negative sign, is a non-spontaneous reaction at standard conditions. Doing the same calculation for the other materials in the remainder of the choices yields a positive voltage, and a spontaneous reaction.

An unknown quantity of KOH was added to a liter of water in order to prepare a stock solution. The pH of the solution was measured to be 10.0. Assuming there were no other compounds in the solution, what quantity of KOH was added to the water? Question 23 Answer Choices A. 10-2 moles B. 10-4 moles C. 10-10 moles D. 10-14 moles

B. Given the final pH was measured at 10.0, the concentration of hydronium ion is 10-10 M and we can calculate the quantity of hydroxide in solution to be 10-4 M (given Kw = 10-14). A pure water solution would already contain 10-7M OH-, but as this number is so much smaller than the final concentration (10-4M), the quantity of KOH added was likely very close to 10-4 moles (answer B is correct).

Two moles of propanoic acid, CH3CH2CO2H (pKa = 4.9), are added to a liter of water. What is the pH once equilibrium has been reached? Question 32 Answer Choices A. 0.47 B. 2.3 C. 4.6 D. 7.7

B. Here we are asked to find the pH of a solution containing a weak acid (HA). To do this, we can use an ICE table to find the solution (note: we round the pKa to 5.0). Since pH = -log[H+], and log 10A = A, we can estimate that the pH of the solution is between 2 and 3 because the exponent is between -2 and -3.

Addition of which of the following compounds to a 1 M solution of H2C2O4 (pKa = 1.2) would best help maintain the pH of a solution at 1.5? Question 26 Answer Choices A. H2C2O4 B. HC2O4- C. HCl D. NaCl

B. If the goal is to maintain pH, we are attempting to create a buffer. Buffers are composed of conjugate pairs (weak acids and/or bases). Addition of more of the same acid would have no impact on the buffering potential of this solution (eliminate choice A). Addition of a different and strong acid would drop the pH dramatically, so eliminate choice C. Sodium chloride would not affect the pH (eliminate choice D). The answer should be the conjugate base of oxalic acid (H2C2O4), which is hydrogen oxalate (HC2O4-), or choice B.

What is the final pH after 30 milliliters of 3 x 10-3 M HBr is mixed with 20 milliliters of 4 x 10-3 M NaOH? Question 4 Answer Choices A. 2.5 B. 3.7 C. 7.0 D. 11.5

B. In the reaction of a strong acid and a strong base, protons generated by the acid react with hydroxide ions generated by the base and any remaining protons or hydroxide ions are then responsible for the final pH of the solution. Here we have (0.03 L)(3 x 10-3 M) = 9 x 10-5 mol H+ and (0.02 L)(4 x 10-3 M) = 8 x 10-5 mol OH- resulting in 1 x 10-5 mol of H+ remaining. This means the solution should be acidic, so we can eliminate choices C and D. The final proton concentration is 1 x 10-5 mol/0.05 L = 0.2 x 10-3 = 2 x 10-4M H+ resulting in a pH of 3.7 (choice B is correct).

If it takes 31.8 years for a sample of 60Co to decay to 1.5% of its original mass, what is the half-life of this isotope? Question 25 Answer Choices A. 6.4 years B. 5.3 years C. 3.2 years D. 0.5 years

B. It is best to approach half-life questions by determining how many half-lives have passed in the context of the problem. The ratio of total decay time to half-life is equal to the number of half-lives. Therefore, given the total decay time of 31.8 years, if we can determine how many half-lives have passed we can calculate the half-life. Finding the number of half-lives is straightforward based on percentages by applying the definition of half-life: the time it takes for half of a radioactive sample to decay. after 1 half-life, 50% of original sample remains after 2 half-lives, 25% of original sample remains after 3 half-lives, ~12% of original sample remains after 4 half-lives, ~6% of original sample remains after 5 half-lives, ~3% of original sample remains after 6 half-lives, ~1.5% of original sample remains Since 6 half-lives have passed in 31.8 years, the half-life must be 31.8/6, or 5.3 years.

An intern in a chemistry laboratory decided to measure the voltage of a galvanic cell and, while constructing the cell, selected a lead chloride salt bridge. Unfortunately the measured voltage of the cell was dramatically lower than anticipated and an unexpected solid began forming in one of the beakers. What caused this deviation from the expected voltage? Question 17 Answer Choices A. Side reactions due to the oxidation of lead B. Side reactions due to the reduction of lead C. Side reactions due to the oxidation of chloride D. Side reactions due to the reduction of chloride

B. Salt bridges are ideally composed of inert ions which serve to balance charge in a galvanic cell (or two-chamber electrolytic cell). If a salt bridge is composed instead of lead chloride, the lead cation may take part in side reactions. The cation is most likely to accept electrons (be reduced - choice B is correct) and form lead solid. Chloride is a relatively inert anion and is less likely to participate in side-reactions or form a solid (eliminate choices C and D).

With the following standard redox potentials in mind, reaction of which of the following pairs will proceed with the greatest spontaneity? Al3+ + 3 e− → Al(s) E° = -1.68 V Fe2+ + 2 e− → Fe(s) E° = -0.44 V Cu+ + e− → Cu(s) E° = 0.52 V Ag+ + e− → Ag(s) E° = 0.80 V Question 24 Answer Choices A. Al3+ and Ag(s) B. Fe2+ and Al(s) C. Ag+ and Cu(s) D. Cu+ and Ag(s)

B. The combination giving the most positive electrical potential represents the most spontaneous reaction (choice B is correct). To determine the overall potential of any given pair, the reduction and oxidation half-reactions are added. To find the oxidation half-reactions, the reduction half-reaction in question can be reversed and its potential reversed in sign. In this instance to find the oxidation half-reaction, the reduction of aluminum above can be reversed to Al(s) → Al3+ + 3 e−, E° = 1.68 V, which can then be added to the reduction half-reaction Fe2+ + 2 e− → Fe(s), E° = -0.44 V. Note that while the oxidation and reduction half-reactions are multiplied by 2 and 3, respectively to appropriately balance the reactions, this does not impact their electrical potentials. The overall electrical potential can be found by adding the two half-reaction potentials: 1.68 V + (-0.44 V) = 1.24 V.

Methyl red (pKa = 5.1) undergoes a color change from red at a pH of approximately 4.1 to yellow at a pH of approximately 6.1. Which of the following best explains this transition? Question 29 Answer Choices A. The increased alkaline nature of the solution results in decomposition of the indicator into a yellow intermediate. B. Different proportions of the acid and conjugate base predominate at differing pHs. C. Acidic methyl group protons are being lost, resulting in a change in the length of the conjugated system of the acid. D. The weak acid forms a weak bond with its proton which only becomes strong enough to change the color at higher pHs.

B. The protonated and deprotonated form of most indicators generate a different color based upon the length of their conjugated systems. When the pH of a solution changes, the proportion of the acid and conjugate base forms of the indicator change and result in a color change (choice B is correct, D is incorrect). When the pH of the solution is equal to the pKa of the indicator, we would expect equal amounts of the protonated form of the indicator and its conjugate base (and a mixing of the two colors). As you move one pH unit from the pKa, you have 90% of one form and only 10% of the other (explaining why you see little color change beyond the pH values listed above). If an indicator decomposed, it would not likely be a reversible process (eliminate choice A). Methyl protons are very poor acids and will not dissociate near the pH values described here (eliminate choice C).

Why is the mass of a bound atom smaller than the sum of the constituent particles? Question 10 Answer Choices A. Because the relatively high-mass electrons are not considered in the calculation. B. The difference in mass accounts for the binding energy in the atom. C. It is difficult to measure the mass of the electrons because they are always moving. D. The masses of charged particles are variable depending on their environment.

B. There is a mass defect that results from the binding of the nucleons in an atom. The bound atom has a lower energy than its unbound particles (because it is more stable than free protons and neutrons), and thus a lower mass compared to the sum of its constituent particles, as given by Einstein's equation E = mc2.

Which of the following best describes a concentration cell? Question 10 Answer Choices A. It is an electrolytic cell where current flows once the system has reached equilibrium. B. It is an electrolytic cell with identical reactants and products. C. It is a galvanic cell with two identical electrodes. D. It is a galvanic cell where current flows once the system has reached equilibrium.

C. A concentration cell is best described as a galvanic cell with two identical electrodes. The potential difference between half-cells is due to differing concentrations of the solutions in contact with those electrodes. The potential of each half-cell can be determined using the Nernst equation. As the reaction proceeds toward equilibrium the concentrations change and the voltage difference ultimately reaches zero. When an electrochemical cell has no voltage, there is also no more current.

Which of the following is emitted in the conversion of 11C 11B? Question 1 Answer Choices A. Gamma radiation B. β- particle C. β+ particle D. α particle

C. A β+ particle is emitted in the conversion of 11C --> 11B. In this reaction, the atomic weight of the atom remains the same. Therefore, the emitted particle(s) must be weightless, so "α particle" can be eliminated. In going from 11C --> 11B, the atomic number decreases from 6 --> 5. Therefore, a proton is converted to a neutron, which requires the emission of a β+ particle, eliminating "β- particle." Gamma radiation is massless, but also has no charge, so it can be eliminated. This makes β+ particle the best answer choice.

Each of the following are true of redox titrations EXCEPT: Question 2 Answer Choices A. the reaction occurring during the titration involves the transfer of electrons between the titrant and analyte. B. at the half-equivalence point, the concentrations of the oxidized and reduced form of the analyte are equal. C. at the equivalence point, the number of moles of titrant added equals the number of moles of analyte at the start of the titration. D. at the equivalence point, the reaction between the analyte and titrant is complete.

C. At the equivalence point, the number of moles of titrant added equals the number of moles of analyte at the start of the titration is not always a true statement about a redox titration. By definition, at the half-equivalence point one-half of the original number of moles of analyte has reacted. For a redox titration that means half the number of moles has either been oxidized or reduced, depending on the titrant used. This means the oxidized form and the reduced form will both have the equivalent of half the original number of moles than at the start. Since both forms are in the same vessel, concentrations will also be equal. A redox titration is a method which uses redox reactions to analyze for an unknown concentration. All redox reactions involve electron transfer. Once the equivalence point is reached, the reaction between analyte and titrant is complete. Addition of more titrant will simply increase the concentration of the titrant without subsequent change to the solution. The final statement is only true if for each half-reaction, the same number of electrons is lost in the oxidation half-reaction as is gained in the reduction half-reaction. When this is not the case, the number of moles of titrant and analyte will not be equal, and they will have different coefficients in the overall balanced redox equation.

Given the half-reaction reduction potentials in the table below, which of the following reactions represents the overall equation for a lead-acid battery during its recharging cycle? A. PbO2(s) + Pb(s) + 2 H2SO4(aq) → 2 PbSO4(aq) + 2 H2O B. 2 Pb(OH)2(s) → PbO2(s) + Pb(s) + 2 H2O C. 2 PbSO4(aq) + 2 H2O → PbO2(s) + Pb(s) + 2 H2SO4(aq) D. PbO2(s) + Pb(s) + 2 H2O → 2 Pb(OH)2(s)

C. Given the half-reaction reduction potentials in the table below, the reaction that represents the overall equation for a lead-acid battery during its recharging cycle is 2 PbSO4(aq) + 2 H2O → PbO2(s) + Pb(s) + 2 H2SO4(aq). The lead-acid battery operates under acidic conditions not basic ones, so choices that contain Pb(OH)2 can be eliminated. During the recharging cycle, the battery acts as an electrolytic cell, not a voltaic cell. This means that the overall cell potential should be negative, not positive. The two half-reactions which when added together will contain both PbO2 and PbSO4 are the second and third equations in the table. To make these add up to a negative cell potential, the third equation, which has the largest Eo value, must be reversed. This will make PbSO4 a reactant and PbO2 a product.

Rechargeable batteries such as NiCad or lead-acid batteries can operate as both galvanic cells during the discharge cycle, or as electrolytic cells during the recharging cycle. Which one of the following statements accurately compares and contrasts these two cycles? Question 4 Answer Choices A. Reduction takes place at the positive electrode during recharging, and oxidation takes place at the negative electrode during discharging. B. Reduction takes place at the anode during recharging, but reduction takes place at the cathode during discharging. C. Oxidation takes place at the positive electrode during recharging, but oxidation takes place at the negative electrode during discharging. D. Oxidation takes place at the anode during recharging, but oxidation takes place at the cathode during discharging.

C. Rechargeable batteries such as NiCad or lead-acid batteries can operate as both galvanic cells during the discharge cycle, or as electrolytic cells during the recharging cycle. Oxidation takes place at the positive electrode during recharging, but oxidation takes place at the negative electrode during discharging accurately compares and contrasts these two cycles. Oxidation always occurs at the anode and reduction always occurs at the cathode, regardless of electrochemical cell type, so choices that change the location of the process can be eliminated. The main difference between a galvanic cell and an electrolytic cell are the signs of the anode and cathode, so the best answer choice is the one that defines oxidation as taking place at oppositely charged electrodes for the different cycles. For a galvanic cell, the sign of the anode is negative due to the spontaneous release of electrons during oxidation. For an electrolytic cell, the electrons must be "pushed" to the cathode to be available for reduction, making the cathode negative. Oxidation takes place at the negative electrode in a galvanic cell during the discharge cycle, and at the positive electrode in an electrolytic cell during the recharging cycle.

Given the Nernst equation below, solve for Keq for the oxidation-reduction reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) where E˚ = 1.10 V. Question 15 Answer Choices A. 1 × 10‒37 B. 1 × 1019 C. 1 × 1037 D. Insufficient information provided

C. The Keq for the oxidation-reduction reaction Zn(s) + Cu2+(aq) ? Zn2+(aq) + Cu(s) where E° = 1.10 V is 1 × 1037. At equilibrium, the electrical potential for a cell is equal to zero and Q = Keq, therefore we can rearrange the Nernst equation as follows:

The following two half-reactions occur in a voltaic cell: (1) Cr2O72-(aq) + 14 H+(aq) + 6e- 2 Cr3+(aq) + 7 H2O(l) (2) 6 I-(aq) 3 I2(s) + 6e- Which one of the following statements is true? Question 16 Answer Choices A. Reaction 2 is an oxidation and occurs at the cathode. B. Reaction 1 is an oxidation and occurs at the anode. C. Reaction 2 is an oxidation and occurs at the anode. D. Reaction 1 is a reduction and occurs at the anode.

C. The following two half-reactions occur in a voltaic cell: (1) Cr2O72-(aq) + 14 H+(aq) + 6e- --> 2 Cr3+(aq) + 7 H2O(l) (2) 6 I-(aq) --> 3 I2(s) + 6e- The statement, "reaction 2 is an oxidation and occurs at the anode" is true. The anode is always the site of oxidation, and the cathode is always the site of reduction in any cell, so "Reaction 1 is a reduction and occurs at the anode" and "reaction 2 is an oxidation and occurs at the cathode" are wrong. In Reaction 2, it is clear that each I- anion must lose one electron in order for neutral I2 to be formed. The process of losing electrons is called oxidation, so the answer is "reaction 2 is an oxidation and occurs at the anode".

Using the following half-reaction potentials, Br2(l) + 2e- --> 2Br-(aq) E° = 1.07 V F2(g) + 2e- --> 2F-(aq) E° = 2.87 V determine the cell potential and whether the following reaction as written is spontaneous under standard conditions: 2 Br-(aq) + F2(g) -->Br2(l) + 2 F-(aq) Question 17 Answer Choices A. E° = -1.80 V, spontaneous B. E° = +1.80 V, nonspontaneous C. E° = +1.80 V, spontaneous D. E° = -1.80 V, nonspontaneous

C. Using the following half-reaction potentials, Br2(l) + 2e- --> 2 Br-(aq) E° = 1.07 V F2(g) + 2e- --> 2 F-(aq) E° = 2.87 V the cell potential and whether the following reaction as written is spontaneous under standard conditions is E° = +1.80 V, spontaneous. 2 Br-(aq) + F2(g) --> Br2(l) + 2 F-(aq) Reactions with positive potentials (E values) are spontaneous; those with negative E values are nonspontaneous. Therefore, neither "E° = -1.80 V, spontaneous" nor "E° = +1.80 V, nonspontaneous" can be correct. The overall reaction given in the question is the sum of the following two half-reactions (note that the first half-reaction here has been flipped to an oxidation with the corresponding sign change in E°): 2 Br-(aq) --> Br2(l) + 2e- E° = -1.07 V F2(g) + 2e- --> 2 F-(aq) E° = +2.87 V Thus, the total cell potential is +1.80 V, and, being positive, signifies that the overall reaction is spontaneous.

When an element is oxidized, it will: Question 15 Answer Choices A. lose electrons, and its oxidation state will decrease. B. gain electrons, and its oxidation state will decrease. C. lose electrons, and its oxidation state will increase. D. gain electrons, and its oxidation state will increase.

C. When an element is oxidized, it will lose electrons, and its oxidation state will increase. Oxidation is defined as an increase in oxidation number and corresponds to the loss (or apparent loss) of electrons.

A conjugate species will differ from its parent molecule by which of the following? Question 8 Answer Choices A. A neutron only B. An electron only C. A proton only D. An electron and a proton

C. Conjugate pairs differ only by a proton (e.g. H2S/HS-, choice C is correct). Although the Lewis definition can also describe acids and bases, those involve the transfer of a pair of electrons (choices B and D are wrong). Changes in the number of neutrons result in the formation of a different isotope of the original compound (choice A is wrong).

Which of the following best expresses the flow of electrons in a voltaic cell? Question 7 Answer Choices A. Electrons nonspontaneously flow from the anode up their concentration gradient. B. Electrons nonspontaneously flow from the cathode down their concentration gradient. C. Electrons spontaneously flow toward the cathode due to potential difference.D. Electrons spontaneously flow toward the anode due to potential difference.

C. Electrons flow from anode to cathode (eliminate choices B and D) due to a difference in electrical potential in a galvanic cell (also known as a voltaic cell). Concentration gradients are frequently responsible for solute diffusion. However, electron flow is not driven by diffusion of electrons down a concentration gradient (eliminate choice A). In the case of voltaic cells, electron flow occurs spontaneously (choice C is correct).

Which particle would likely be ejected from a radionuclide that has a neutron/proton ratio that is too high? Question 3 Answer Choices A. Alpha particle B. Positron C. Beta particle D. Gamma ray

C. Gamma rays do not change the number of protons or neutrons, and thus would not affect the neutron/proton ratio (eliminate choice D). Alpha particles have the effect of reducing a nuclide by 2 protons and 2 neutrons. Depending on the original mass and atomic numbers of the parent nucleus, alpha decay can raise, lower, or not change the neutron/proton ratio (eliminate choice A). Loss of a positron effectively converts a proton into a neutron, thereby increasing the ratio in question and making the situation worse (eliminate choice B). Loss of a beta particle (an electron) effectively converts a neutron into a proton, thereby decreasing the neutron/proton ratio and making choice C the best answer.

A car part manufacturer wants to plate each muffler in her warehouse with 6.5 g of chrome (chromium) using an immersion technique with chromium solid and Cr(NO3)3. Given a constant current of 10 A, how long should the manufacturer immerse her mufflers to ensure the appropriate amount of chrome plating? Note: F = 96,500 C/mol e- Question 12 Answer Choices A. 10 minutes B. 30 minutes C. 1 hour Correct Answer (Blank) D. 3 hours

C. Given I = C/t, we can rearrange to get t = C/I where I is current (in amps), C is a measure of charge (in coulombs), and t is time (in seconds). (answer C is correct)

Which of the following pairs of titrant and analyte would most likely generate this titration curve? A. NaOH/HCl B. HCl/NaCH3COO (pKb = 9.3) C. NaOH/HF (pKa = 3.2) D. HCl/NaF (pKb = 9.8)

C. Here we are presented with a titration curve where we know neither the titrant (solution of known composition and concentration) nor the analyte (solution being analyzed). Since the solution begins in the acidic range, we can conclude the analyte to be an acid (choices B and D are incorrect). Furthermore, at the equivalence point (the most vertical section of the graph) the pH of the solution is above seven, indicating that a strong base was mixed with a weak acid. The half-equivalence point is approximately 3, and should match the pKa of the weak acid. The only answer choice that involves a strong base and a weak acid is NaOH and HF (choice C is correct).

A chemist attempts to assemble a galvanic cell but only has solid silver and silver chloride as reagents. He assembles a cell consisting of two idential beakers each containing a silver electrode and an equimolar solution of silver chloride, but observes no reaction. Which of the following would result in current flow? Increasing the temperature of one half-cell Adding pure water to one half-cell Adding equal quantities of pure water to both half-cells Question 30 Answer Choices A. I only B. II only C. I and II only D. I, II, and III

C. In order for current to flow, the electrical potential of the two half-cells must differ. Both changing the concentration of the reagents and changing the temperature (see the Nernst equation below) will result in changes in electrical potential and current flow (Roman numerals I and II are true, so choices A and B can be eliminated). However, if both half-cells were diluted equally, the electrical potentials would remain identical and no current flow would occur (Roman numeral III is false, eliminating choice D).

The chemical reaction occurring in a standard AA battery is as follows: Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + Zn(NH3)2Cl2(aq) + H2O(l) Which of the following represents the (-) pole on the battery? Question 33 Answer Choices A. MnO2(s) B. Mn2O3(s) C. Zn(s) D. NH4+(aq)

C. The anode in a galvanic cell possesses a negative charge. Ions in solution cannot serve as electrodes, so eliminate choice D. In the above reaction, zinc is being oxidized while manganese is being reduced. Zn(s) is therefore the anode (an ox) and the negative pole on the battery (choice C is correct). Note that determining the new oxidation state of zinc may appear difficult. However, we do not need to know an exact number, only realize that zinc now has a positive oxidation state.

How many moles of electrons are transferred per two moles of Cr3+(aq) in the redox reaction shown? 2 Cr3+(aq) + 6 Cl-(aq) → 2 Cr(s) + 3 Cl2(g) Question 19 Answer Choices A. 2 moles B. 3 moles C. 6 moles D. 12 moles

C. To find the number of moles of electrons transferred in the reaction, simply break the overall reaction into each half-reaction and insert the appropriate number of electrons. After appropriate multiplication, the oxidation half-reaction consists of 6 Cl-(aq) → 3 Cl2(g) + 6 e- and the reduction half reaction consists of 2 Cr3+(aq) + 6 e- → 2 Cr(s) (choice C is correct).

The oxidation states of chromium and oxygen in K2Cr2O7 are (respectively): Question 2 Answer Choices A. +8 and -1. B. +7 and -2. C. +6 and -2. D. +6 and -1.

C. The oxidation states of the atoms in dichromate anion can be determined using the typical oxidation state rules. Oxygen maintains an oxidation state of -2 (overall contribution of -14), so choices A and D can be eliminated. The potassium has an oxidation state of +1 (overall contribution of +2). This means that +12 must be canceled by chromium, resulting in an oxidation state of chromium of +6 (answer C is correct).

Squaric acid, shown below, has two acidic protons with pKa values of 1.5 and 3.4. If 10 mL of a 1 M sample of squaric acid is titrated with 10 mL of a 2 M solution of NaOH, the pH of the resultant solution will be in which of the following ranges? A. 0 - 4 B. 4 - 7 C. 7 - 11 D. 11 - 14

C. The pKa values given (1.5 and 3.4) indicate that while the first proton is a reasonably strong acid, the second is weaker, but still acidic. After neutralizing all the H+s in squaric acid with an equivalent amount of OH- from NaOH (a strong base), the pH of the solution will be greater than 7 (eliminate choices A and B). The final solution will be 0.5 M sodium squarate, a weak base with pKb = 10.6. This will result in a mildly basic solution between 7 and 11. For those who need proof: 10-10.6 = 10-10.6 = 5-11.6 = x2 x = [OH-] = ~ 10-6 pOH = ~6, so pH = ~8

Which statement does NOT accurately depict the characteristics of the anode in both galvanic and electrolytic cells? Question 4 Answer Choices A. Oxidation takes place at the anode. B. Anions are attracted to the anode. C. The sign of the anode is negative. D. Electrons flow away from the anode.

C. The sign of the anode is negative.

Given the pKa values of phenol (C6H5OH) and benzoic acid (C6H5CO2H) are 9.8 and 4.2, respectively, which of the following is the strongest base in water? Question 13 Answer Choices A. C6H5OH B. C6H5CO2H C. C6H5O- D. C6H5CO2-

C. This question requires us to understand the relationship of acids/bases and their conjugates. The original compounds (i.e. choices A and B) are very poor bases compared to their negatively charged conjugates. Thus, answers A and B are not correct. In general, the stronger the acid, the weaker the conjugate base, and the weaker the acid, the stronger the conjugate base. Here we are given pKas, so the compound with the highest pKa will be the weakest acid, and generate the strongest conjugate base, making C6H5O- our answer (answer C is correct).

What would be the predicted pH of a potassium malate (KC4H5O5) solution? Question 9 Answer Choices A. < 7 B. 7 C. > 7 D. Inadequate information to determine the pH

C. When confronted with an unfamiliar salt, we must evaluate each ion individually in order to determine the overall pH of the solution. Potassium is a group I metal and is therefore neutral in water (it fails to react with water). The anion malate is not a conjugate base of one of the strong acids you must recognize for the MCAT. You should therefore conclude it is the conjugate base of a weak acid, and is therefore a basic anion (it reacts with water by accepting a proton). Overall, this salt generates a basic solution with a pH > 7, so choice C is correct.

A voltaic cell is run for one hour at 2 A and 1.4 grams of solid iron are formed. Assuming a one-to-one mole ratio of solid iron to its oxidized form, iron in which oxidation state served as the reactant? (Faraday's constant = 96,500 C/mol e-) Question 12 Answer Choices A. Fe0 B. Fe+ C. Fe2+ D. Fe3+

D. A voltaic cell is run for one hour at 2 A and 1.4 grams of solid iron are formed. Assuming a one-to-one mole ratio of solid iron to its oxidized form, iron in its Fe3+ oxidation state served as the reactant. To determine the oxidation state of the reactant iron, we must find the number of moles of electrons transferred per mole of iron. This indicates that the iron began in the +3 oxidation state before accepting three equivalents of electrons to become solid iron.

Which of the following decay modes would be best in stabilizing a nucleus with too many neutrons? Question 8 Answer Choices A. Electron capture B. Positron emission C. Gamma radiation D. Beta emission

D. Beta emission would be best in stabilizing a nucleus with too many neutrons. Gamma radiation can be immediately eliminated because this decay mode does not change the number of particles in the parent nucleus. Nuclei with too many neutrons (or a neutron/proton ratio that is too high) can be stabilized by decreasing the number of neutrons or increasing the number of protons. Both positron emission and electron capture result in a reduction in the atomic number (number of protons), so have the opposite of the desired effect. Beta emission is the correct answer because a neutron in the parent nucleus is converted into a proton in the daughter nucleus via this process.

The radioactive isotope 32P has a half-life of 14 days. If 200 mg of 32P remains in a vial, approximately how much existed 60 days prior? Question 9 Answer Choices A. 1.6 g B. 0.8 g C. 3.2 g D. 3.9 g

D. If 200 mg of 32P remains in a vial, approximately 3.9 g existed 60 days prior. Since the question is asking for a mass going back in time, we have to double the mass at each half-life, rather than reduce it. The period of 60 days is between four and five half-lives (i.e., 60/14 is a little over 4) and thus represents a minimum of four doublings. Doubling the mass four times gives 200 mg 400 mg 800 mg 1600 mg 3200 mg, or 3.2 g over four half-lives. Since the total decay time represents slightly more than four half-lives, 3.9 g is the best answer since five half-lives would be 6400 mg.

In the earth's crust, iron is predominantly found as iron ore (a mixture of iron oxides) while gold is found in its neutral, metallic form. Which of the following statements is most likely true? Question 4 Answer Choices A. Iron oxides are composed of neutral iron bound to oxygen. B. Gold is more easily oxidized than iron in the primarily aerobic environment of Earth. C. The reduction potential of metallic gold is larger than that of cationic iron. D. The reduction potential of cationic gold is larger than that of cationic iron.

D. In the earth's crust, iron is predominantly found as iron ore (a mixture of iron oxides) while gold is found in its neutral, metallic form. It is most likely true that the reduction potential of cationic gold is larger than that of cationic iron. The answer choice, "Gold is more easily oxidized than iron in the primarily aerobic environment of Earth" may be eliminated because it is the opposite of what is stated in the question (i.e., iron is typically oxidized while gold is not). Taking neutral gold to a -1 oxidation state would require significantly more energy compared to reducing cationic iron. Iron oxides are the result of iron metal being oxidized to a cationic form by O2 from the atmosphere. If iron is oxidized and gold is not, gold must have a more negative oxidation potential, and hence cationic gold must have a more positive reduction potential than cationic iron.

In the galvanic cell Ni|Ni2+||Fe3+, Fe2+|Pt, what best describes the direction of current flow and flow of Na+ from the salt bridge? Question 17 Answer Choices A. Current flows from Ni to Pt, and Na+ flows toward Pt. B. Current flows from Ni to Pt, and Na+ flows toward Ni. C. Current flows from Pt to Ni, and Na+ flows toward Ni. D. Current flows from Pt to Ni, and Na+ flows toward Pt.

D. In the galvanic cell Ni|Ni2+||Fe3+, Fe2+|Pt, the current flows from Pt to Ni and Na+ flows toward Pt. In electrochemical cells, electrons flow from anode to cathode (therefore, due to the sign convention, current flows from cathode to anode). In this cell, Ni2+ is being oxidized and nickel is the anode, while Fe3+ is being reduced at a platinum cathode. Thus current flows from the Pt cathode to the Ni anode, allowing us to eliminate two answer choices. Na+ meanwhile travels from the salt bridge to the cathode to offset the charge imbalance generated by the current (cations migrate to the cathode).

In the galvanic cell Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s), the addition of excess Zn2+ to the solution at the Zn electrode will have what impact on the voltage? Question 14 Answer Choices A. Insufficient information provided B. No change C. Increase D. Decrease

D. In the galvanic cell Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s), addition of Zn2+ will decrease voltage. The electrochemical cell here gives this overall reaction: Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g) According to the Nernst equation, as Q increases (as with addition of a product) the voltage decreases. Alternatively, we can relate this to Le Chatelier's principle. As we add more product, the reaction proceeds in reverse to a greater degree. Therefore the forward reaction becomes less spontaneous (voltage decreases) as a product is added.

Which of the following cells generates the greatest amount of energy per Coulomb? Zn2+ + 2e- → Zn E˚ = ‒0.76 Al3+ + 3e- → Al E˚ = ‒1.66 Ag+ + e- → Ag E˚ = +0.80 Cu+ + e- → Cu E˚ = +0.52 Question 16 Answer Choices A. Al|Al3+||Zn2+|Zn B. Cu|Cu+||Ag+|Ag C. Ag|Ag+||Cu+|Cu D. Zn|Zn2+||Cu+|Cu

D. The cell that generates the greatest amount of energy per Coulomb is Zn|Zn2+||Cu+|Cu. Energy per Coulomb, or J/C, is another form of the unit for the volt, thus this question is asking us for the galvanic cell with the greatest potential difference. To determine the potential difference of each listed cell, we must sum the two half reactions comprising the cell. The cell with the greatest potential is Zn|Zn2+||Cu+|Cu with E°= 0.76 V + 0.52 V = 1.28 V.

A chemist creates a concentration cell with equal volume half-cells (Cu|Cu+ (0.030 M)||Cu+ (1.0 M)|Cu), allows it to react for several minutes, and analyzes the solutions before the system has reached equilibrium. What best describes the concentrations seen in the cell at this time? Question 11 Answer Choices A. Anode = 0.030 M Cu+; cathode = 1.0 M Cu+ B. Anode = 1.0 M Cu+; cathode = 0.030 M Cu+ C. Anode = 0.020 M Cu+; cathode = 1.01 M Cu+ D. Anode = 0.100 M Cu+; cathode = 0.93 M Cu+

D. The final concentrations seen in the cell will be 0.100 M Cu+ at the anode and 0.93 M Cu+ at the cathode. Concentration cells are galvanic cells with a potential difference between the half-cells due to differing concentrations. The anode is the site of oxidation, thus we can expect the concentration of Cu+ to increase. Conversely, the concentration of Cu+ should decrease at the cathode, the site of reduction. The reaction proceeds until both half-reactions have the same concentration, resulting in a voltage difference of zero. Therefore the only answer choice that would be possible is 0.100 M Cu+ at the anode and 0.93 M Cu+ at the cathode. Note that 1.0 M Cu+ at the anode and 0.030 M Cu+ at the cathode would not be possible as the concentration at the anode would not continue to increase (or the concentration at the anode decrease) once the voltage has fallen to zero.

The reduction potentials of the reactions below are +1.2 V and -0.8 V, respectively. MnO2(s) + 4 H+ + 2e- → Mn2+(aq) + 2 H2O(l) Zn2+(aq) + 2e- → Zn(s) What is the voltage of a MnO2/Zn(s) cell? Question 9 Answer Choices A. +0.4 V B. +0.8 V C. +1.6 V D. +2.0 V

D. The reduction potentials of the reactions below are +1.2 V and -0.8 V, respectively. MnO2(s) + 4 H+ + 2e- → Mn2+(aq) + 2 H2O(l) Zn2+(aq) + 2e- → Zn(s) The voltage of a MnO2/Zn(s) cell is +2.0 V. From the reactions given, it is clear that if one begins with zinc metal, it will lose two electrons to become Zn2+ (the reverse of the second reaction); thus, zinc gets oxidized. The total voltage of a cell is the sum of the voltages of the two half-reactions. Changing the sign of the potential for the given reduction half-reaction for Zn2+ [since the reaction that takes place in the cell is the oxidation of Zn(s)] yields +0.8 V. Adding this to the potential of the given reduction reaction for MnO2(s) gives an overall cell voltage of +0.8 + 1.2 = 2.0 V.

How many neutrons are in the daughter nuclide after 231Th undergoes beta emission? Question 7 Answer Choices A. 231 B. 91 C. 371 D. 140

D. There are 140 neutrons in the daughter nuclide after 231Th undergoes beta emission. The reaction described by the question is The choices 91 and 231 can be eliminated because this represents the number of protons (i.e., atomic number) and protons plus neutrons (i.e., mass number) of the daughter nuclide. Choice 371 can be eliminated because this is the sum of the atomic number and mass number, which has no physical significance. Choice 140 is the correct answer because to obtain the number of neutrons, one subtracts the atomic number (i.e., 91 protons) from the mass number (i.e., 231 protons and neutrons).

When a nucleus has too many neutrons in the nucleus, the predicted decay will occur with emission of which particle? Question 6 Answer Choices A. Neutron B. Alpha particle C. Positron D. Beta particle

D. Beta particle

Fully charged lead-acid batteries utilize metallic Pb0 as the anode, and PbO2 as the cathode. Recharging spent batteries using very high voltages has been observed to cause the electrolysis of water into H2 and O2 at the electrodes. Which of the following is true? Question 27 Answer Choices A. O2 and H2 evolution are observed on the surface of Pb. B. O2 and H2 evolution are observed on the surface of PbO2. C. O2 is evolved on the surface of Pb, and H2 is evolved on the surface of PbO2. D. O2 is evolved on the surface of PbO2, and H2 is evolved on the surface of Pb.

D. During recharge, the formation of Pb0 will be taking place at the cathode, and PbO2 at the anode. Excessively high voltages suggest an excess of free electrons at the cathode, and a strong driving force for the removal of electrons at the anode. The production of H2 from water (2 H2O + 2e- → H2 + 2 OH-) is a reduction reaction, and will require these highly energetic electrons. Therefore it will form on the surface of Pb (eliminate choices B and C). The conversion of water to O2 (2 H2O → O2 + 4e- + 4 H+) is an oxidation, requiring a large oxidative force, such as present at the surface of PbO2 at excessively high recharge voltages (eliminate choice A).

Given the two half reactions below, which species is the best oxidizing agent? F2 + 2e- → 2 F- Eo = +2.87 V Li+ + e- → Li Eo = -3.05 V Question 3 Answer Choices A. Li B. Li+ C. F- D. F2

D. F2

Given that 0.0040% of a 0.5 mole sample of HCN dissociates in one liter of solution, what is the approximate pKa of HCN? Question 20 Answer Choices A. 4.2 B. 5.2 C. 7.2 D. 9.2

D. Given 0.0040% of the acid has dissociated, we know that [CN-]/([HCN]-[CN-]) = 4 x 10-5. We can assume the [CN-] to be negligible compared to the [HCN], which allows us to remove [CN-] from the denominator. The [HCN] is 0.5 M, so we can calculate [CN-] to be 2 x 10-5M. These equilibrium concentrations can be substituted into the Ka expression to get a value for Ka, and from there we can find our pKa. (2 x 10-5)2/0.5 = 8 x 10-10 = Ka, which gives us a pKa between 9 and 10 (choice D is correct).

A type of rechargeable battery using vanadium species as redox couples cycles between VO2+ and VO2+ at one electrode and between V2+ and V3+ at the other. Hypothetically, if V3+ is produced during discharge, which of the following is consumed during recharge? Question 18 Answer Choices A. VO2+only B. Both VO2+ and V3+ C. V2+ only D. Both V3+ and VO2+

D. If V3+ is produced during discharge, this means that the V2+/V3+ couple must be the anode in discharge (V2+ → e- + V3+). As such, during recharging this electrode must be the cathode, so the reverse of the process must occur, meaning V2+ will be produced, not consumed (eliminate choice C). This also means the VO2+/VO2+ couple is at the anode during the recharging process, where oxidation occurs. Since vanadium in VO2+ is in the +4 oxidation state and in VO2+ is +5, the anodic reaction would consume VO2+ (2 OH- + VO2+ → VO2+ + e- + H2O) (eliminate choices A and B).

During nuclear decay, the time elapsed for each successive half-life interval: Question 5 Answer Choices A. will change during the decay process depending on the isotope. B. is double the previous one. C. is different than the previous one. D. is the same as the previous one.

D. is the same as the previous one.

Which of the following half-reactions results in the greatest release of energy? Question 22 Answer Choices A. Ca2+ + 2 e− → Ca(s) E° = -2.87 V B. Fe2+ + 2 e− → Fe(s) E° = -0.44 V C. Cu+ + e− → Cu(s) E° = 0.52 V D. NO3− + 4 H+ + 3 e− → NO(g) + 2 H2O E° = 0.96 V

D. Reaction potentials provide much of the same information as ΔG and Keq. As a reaction potential becomes more positive, it becomes more spontaneous and releases more energy. Therefore, we are looking for the half-reaction with the largest and most positive potential (answer D is correct).

MnO4-(aq) + 8 H+(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) It takes 35 mL of a 0.093 M solution of potassium permanganate to react completely with 10 mL of an aqueous solution of iron(II) in a redox titration, according to the reaction above. Which of the following represents the molarity of the iron(II) solution? Question 3 Answer Choices A. (5)(35)(0.093)/(10) B. (10)/(5)(0.093)(35) C. (35)(0.093)/(10) D. (5)(10)/(35)(0.093)

MnO4-(aq) + 8 H+(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) It takes 35 mL of a 0.093 M solution of potassium permanganate to react completely with 10 mL of an aqueous solution of iron(II) in a redox titration, according to the reaction above. Therefore, (5)(35)(0.093)/(10) represents the molarity of the iron(II) solution. The mathematical relationship at the equivalence point is nFeMFeVFe = nMnMMnVMn, where n is the number of moles of the respective compound in the balanced reaction, M is the molarity of the solution, and V is the volume of solution used in the titration. Solving for the requested value, we get: MFe = nMnMMnVMn/nFeVFe and substituting the appropriate values in gives: MFe = (5)(35)(0.093)/(1)(10)

Based on the following half-reaction potentials, Sn4+(aq) + 2e- Sn2+(aq) E° = -0.14 V Ag+(aq) + e- Ag(s) E° = +0.80 V Cr3+(aq) + 3e- Cr(s) E° = -0.74 V Fe2+(aq) + 2e- Fe(s) E° = -0.44 V which of the following is the strongest reducing agent? Question 19 Answer Choices A. Fe2+(aq) B. Cr(s) C. Ag(s) D. Sn2+(aq)

the strongest reducing agent is Cr(s). A reducing agent is the reactant that gets oxidized. Therefore, the species that is the best reducing agent is the one that is most easily oxidized. Flipping each of the half-reactions accompanying this question reveals that the oxidation of Cr(s) has a potential of +0.74 V, which is greater than the potential (+0.14 V) for the oxidation of Sn2+(aq). Thus, Cr(s) must be the better reducing agent.