Kaplan MCAT General Chemistry Review

Isotypes of the same element have the same number of protons. Alpha decay results in the loss of two protons and two neutrons, while beta decay (?-decay) causes the gain of one proton. Thus, two beta decays must occur for each alpha decay to ensure that the number of protons in the daughter nucleus is equal to the number of protons in the parent nucleus. The answer is (C), 1:2.

PASSAGE III

23. A

In order to balance the equation, we must first combine the half-reactions. This requires writing all of the products and all of the reactants for both reactions in one equation:

LiCoO₂ + xLi⁺ + xe^- + 6C Li_1-xCoO₂ + xLi⁺ + xe^- + Li_xC₆

If a certain compound/particle is on both the left and the right sides of the equation, then it is appearing as both a reactant and a product. Since there is no net change in that specific species, we can omit it from the net reaction; for this reason, we can eliminate the terms “xLi⁺” and “xe^-” from both sides of this equation. This leaves us with (A) as the correct choice.

24. B

You should know that the overall potential of a galvanic cell (E°_cell) is equal to the sum of the potentials at the cathode and the anode. The overall potential of a discharging cell is always positive, but not necessarily greater than 1; for this reason, (B) is the only viable option. You can also arrive at this conclusion by realizing that the reaction in a battery is always spontaneous, since the battery must supply energy. Spontaneous reactions always have a negative G, which you can plug into the free energy equation (G = -nFE°_cell). To get a negative G from the free energy equation, you must have a positive E°_cell.

25. D

The first thing to note here is that lithium acts as an electrolyte in this reaction, meaning that it is not oxidized or reduced; that rules out (A) and (B) immediately. By definition, reduction happens at the cathode and oxidation happens at the anode. However, because this is a reversible reaction (which you should know based on the fact that the reaction is at equilibrium), we cannot use these definitions to distinguish between the species that is oxidized and the species that is reduced. The easiest way to answer this question is by noting that the carbon in the anode is neutral on the left side of the forward reaction and is negatively charged on the right side. This means that carbon is reduced, allowing us to select (D) as the correct answer. You can verify this by checking the cathode side: Because the cathode is made of CoO₂^-, in which Co has a +3 charge (since oxygen almost always carries a -2 charge, two oxygen atoms add up to a -4 charge; to create an overall -1 charge, Co must be +3), Co³⁺ must be oxidized in the conversion of LiCoO₂ to Li_1-xCoO₂. The product of this reaction, because it contains less +1 charge from lithium, must contain a more positive charge from cobalt (further confirming that the cobalt is reduced).

26. C

Because of the complexity of this question, the first step should be to eliminate as many choices as possible. There is no indication in the passage that E°_cell is positive for the forward reaction (A). The assumption about reaction kinetics in (B) is correct, but it’s likely that the effect of a change in x will be different for each reaction. One reaction will shift to the left and the other will shift to the right, but the shifts will not be exactly identical to one another; therefore, K_eq will change and (B) is incorrect. A galvanic cell gradually discharges while transporting electrons from the anode, where oxidation occurs, to the cathode, where reduction occurs. In the forward reaction, oxidation occurs at the anode and reduction occurs at the cathode. This means that the forward reaction does not represent a discharging cell, so the reverse reaction must be favored when the battery is in use. Because the reverse reaction takes precedence over the forward reaction, K_eq is low when the cell is discharging.

27. B

LiCoO₂ is usually present in the battery since it is on the left side of the cathode reaction. The presence of the Li(CoO₂)₂ complex is more important for this question. It is clear that Li(CoO₂)₂ is a form of the Li(CoO₂)_1/(1-x) complex, so we can calculate x by writing the equation 1/(1-x) = 2. Simple algebra yields the fact that ½ of the battery’s initial energy is remaining. Because the initial energy was equal to 100 J, the current energy equals 50 J.

28. D

According to the passage, the system contains only lithium as a salt in a solvent. Because it is not part of the cathode or the anode, it always retains its +1 charge. All three items are correct.

29. A