MCAT General Chemistry Review
Chapter 12: Electrochemistry
Answers and Explanations
1. CIn the oxidation–reduction reaction of a metal with oxygen, the metal will be oxidized (donate electrons) and oxygen will be reduced (accept electrons). This fact allows us to immediately eliminate choices (B) and (D). A species with a higher reduction potential is more likely to be reduced, and a species with a lower reduction potential is more likely to be oxidized. Based on the information in the question, iron is oxidized more readily than those metals; this means that iron has a lower reduction potential.
To determine the standard electromotive force of a cell, simply subtract the standard reduction potentials of the two electrodes. In this case, the cathode is zinc because it is being reduced; the anode is silver because it is being oxidized. Thus,
E°cell = E°red,cathode − E°red,anode = − 0.763 − 0.337 = − 1.10 V
While we must multiply the silver half-reaction by two to balance electrons, the actual value for the reduction potential does not change. Remember that the standard reduction potential is determined by the identity of the electrode, not the amount of it present.
3. AOxidation occurs at the anode, and reduction occurs at the cathode. Because Cu is the anode, it must be oxidized. The reduction potential of the cathode cannot be less than that of the anode for a galvanic cell. Therefore, mercury, choice (A), must be the cathode. In a concentration cell, the same material is used as both the cathode and anode; however, this question assumes equal concentrations. If both electrolyte solutions have the same concentration, there will be no oxidation–reduction reaction and, therefore, no anode or cathode. This eliminates choice (B).
4. CIn an electrolytic cell, ionic compounds are broken up into their constituents; the cations (positively charged ions) migrate toward the cathode, and the anions (negatively charged ions) migrate toward the anode. In this case, the cations are H+ ions (protons), so option I is correct. Electrons flow from anode to cathode in all types of cells, meaning that option III is also correct. Option II is incorrect for two reasons. First, it is unlikely that the anions in any cell would be O2– rather than OH–. Second, and more significantly, these anions would flow to the anode, not the cathode.
5. DThis answer comes directly from the equation relating Gibbs free energy and E°cell. ΔG° = −nFE°cell, where n is the number of moles of electrons transferred and F is the Faraday constant, To determine n, one must look at the balanced half-reactions occurring in the oxidation–reduction reaction.
6. BSalt bridges contain inert electrolytes. Ionic compounds, such as choices (A), (C), and (D), are known to be strong electrolytes because they completely dissociate in solution. Choice (B) cannot be considered an electrolyte because its atoms are covalently bonded and will not dissociate in aqueous solution. Choices (B) and (C) may appear similar, but there is an important distinction to be made. Choice (C) implies that Mg2+ and SO32− are the final, dissociated ionic constituents, while choice (B) implies that neutral SO3 would have to be dissolved in solution.
7. BPotential, as measured by E°cell, is dependent only on the identity of the electrodes and not the amount present. Similarly, the equilibrium constant depends only on the identity of the electrolyte solutions and the temperature. However, as the electrode material is increased, the surface area participating in oxidation–reduction reactions is increased and more electrons are released, making statement II correct.
8. DE°cell is dependent upon the change in free energy of the system through the equation RT ln Keq = nFE°cell. The temperature, T, appears in this equation; thus, a change in temperature will impact the E°cell.
9. BIf this were a galvanic cell, the species with the more positive reduction potential (cadmium) would be reduced. The cathode is always reduced in an electrochemical cell, so sodium could not be the cathode in such a galvanic cell, eliminating choice (A). Sodium would be the cathode in an electrolytic cell, however, which would make cadmium the anode. Thus, the answer is choice (B). Note that we do not have to determine E°cell because we already know the answer. However, the E°cell would be −2.71 − (−0.40) = −2.31 V for an electrolytic cell, and +2.31 V for a galvanic cell, eliminating choices (C) and (D).
10.DThere are only two equations involving standard change in free energy in electrochemical cells: ΔG° = −nFE°cell and ΔG° = –RT ln Keq. Substituting E°cell = E°red,cathode − E°red,anode into the first equation and distributing the negative sign gives choice (D). Choice (A) would be the opposite of ΔG°. Setting the two equations equal to each other, we get RT ln Keq = nFE°cell. Solving for E°cell, we get which is the opposite of choice (B). Choice (C) incorrectly solves the algebra.
11.DA spontaneous electrochemical reaction has a negative ΔG. Using the equation ΔG° = –RT ln Keq, Keq > 1 would result in ln Keq > 0, which means ΔG° < 0. A negative electromotive force, choice (A), or equilibrium state, choice (B), would not correspond to a spontaneous reaction. Concentration cells can be spontaneous; however, if the concentration cell had reached equilibrium, it would cease to be a spontaneous reaction, eliminating choice (C). When an answer choice may be true, but does not have to be—it is the wrong answer on Test Day.
12.AA change in pH has a direct correlation to the hydrogen ion (H+) concentration. Decreasing the pH increases the H+ concentration, which means the concentration of products has increased in the oxidation of sulfur dioxide. This means it would be harder to liberate electrons, thus decreasing the emf. One could also view this decrease in oxidation potential as an increase in reduction potential. If E°red,anode increases, then E°cell must decrease according to E°cell = E°red,cathode − E°red,anode.
13.CAn electrolytic cell is nonspontaneous. Therefore, the ΔG° must be positive and E°cell must be negative, eliminating choices (B) and (D). The change in entropy may be positive or negative, depending on the species involved, eliminating choice (A). According to the equation ΔG° = –RT ln Keq, Keq < 1 would result in ln Keq < 0, which means ΔG° > 0.
14.DCompared to other cell types, lead–acid batteries have a characteristically low energy density, choice (D). While choice (A) is a true statement, the incomplete dissociation of sulfuric acid does not fully explain the low energy density of lead–acid batteries. Choice (C) is likely to be an opposite; the more easily the electrodes dissociate, the easier it is to carry out oxidation–reduction reactions with them.
15.CDuring the recharge cycle, Ni–Cd cells will accept current from an outside source until the Cd and NiO(OH) electrodes are pure; at this point, the reaction will stop because Cd(OH)2 runs out and no more electrons can be accepted. Choices (A) and (B) are both true statements, but they fail to explain why overcharging the battery (continuing to try to run current into the battery even when the electrodes are reverted to their original state) is not a problem with Ni–Cd batteries. Finally, surge current refers to the initial burst of current seen in some batteries; once charged, the surge current will not increase even if the power source continues to be run because no additional charge will be stored on the electrodes, eliminating choice (D).