Kaplan MCAT General Chemistry Review

In 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. To balance charge, electrons are transported from the anode to the cathode, meaning that option III is also correct. Option II is incorrect for two reasons. First, it’s unlikely that the anions would be O^2- rather than OH^-. Second, these anions would flow to the anode, not the cathode.

2. A

Oxidation occurs at the anode, and reduction occurs at the cathode. Since Cu is at the anode, it must be oxidized. Its standard reduction potential is +0.52 V, so its standard oxidation potential is -0.52 V. The half-reaction potentials in a feasible galvanic cell must add up to a value greater than 0, so we can answer this question by adding the oxidation potential of copper to the reduction potentials of the other metals and finding a reasonable match.

Mercury has a reduction potential of 0.85 V, which is enough to outweigh the potential contributed by copper (-0.52 V). The cell’s emf would be +0.33 V. Zinc and aluminum both have negative reduction potentials, so the overall potential of the cell will be even lower than that of copper (-1.28 V and -2.18 V, respectively). Zinc and aluminum would only be viable options if we were dealing with an electrolytic cell.

3. A

The oxidizing agent is the species that is reduced in any given equation. In this problem, two H⁺ ions from H₃N are reduced to one neutral H₂ atom. H₃N is not the reducing agent because the H⁺ ions and the N^3- ions are independent of one another in solution.

4. B

First, let’s balance the equation, first making sure all the atoms are present in equal quantities on both sides.

Cr₂O₇^2- + 14 H⁺ + e^-2 Cr ²⁺ + 7 H₂O

Now, let’s adjust the number of electrons to balance the charge. Currently, the left side has a charge of +12 (-2 from dichromate and +14 from protons). The right side has a charge of +4 (+2 from each chromium cation). To decrease the charge on the left side from +12 to +4, we should add 8 electrons.

Cr₂O₇^2- + 14 H⁺ + 8 e^- 2 Cr²⁺ + 7 H₂O

5. A

Both half-reactions are written as reductions, so one of the two must be reversed to perform oxidation. If a reaction represents a spontaneous process, then the overall potential must be positive. The only way to create a spontaneous oxidation-reduction system is by reversing Reaction 1 to produce a positive net potential.

2 H₂O O₂ + 4 H⁺ + 4 e^- -1.23 V (oxidation)

The half-reactions are each already balanced for mass and charge. Before we add them together, we must make sure each of them transfers the same number of electrons. Multiply all of Reaction 2’s coefficients by 2 so that both reactions transfer 4 moles of electrons.

2 H₂O O₂ + 4 H⁺ + 4 e^- -1.23 V (oxidation)

2 PbO₂ + 8 H⁺ + 4 e^- 2 Pb²⁺ + 4 H₂O +1.46 V (reduction)

Now, add the two reactions together.

2 H₂O + 2 PbO₂ + 8 H⁺ + 4 e^- O₂ + 4 H⁺ + 4 e^- + 2 Pb²⁺ + 4 H₂O

Finally, let’s erase duplicate instances of H⁺, e^-, and H₂O.

2 PbO₂ + 4 H⁺ O₂ + 2 Pb²⁺ + 2 H₂O

This yields the reaction in (A).

6. B

This reaction is simply the reverse of the net balanced equation we found in question 5. Since that equation referred to a spontaneous system, this equation must refer to a nonspontaneous system. Nonspontaneous systems have negative E_cell values, so we can eliminate (A) and (D). Here, Reaction 1 represents reduction, so its given reduction potential should remain the same (+1.23 V). Reaction 2 represents oxidation, so its given reduction potential needs to be multiplied by -1 (to yield -1.46 V). The sum of those two values is -0.23 V.

7. C

In 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 (B) and (D). You should also know that 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 copper; this means that iron has a lower reduction potential.

8. A

To answer this question, you must know that a hydride ion is composed of a hydrogen nucleus with two electrons, thereby giving it a negative charge and a considerable tendency to donate its extra electron. This means that LiAlH₄ is a strong reducing agent.

9. D

This answer comes directly from the equation relating Gibbs free energy and E°_cell. G = -nFE°_cell, where n is the number of moles of electrons transferred and F is the Faraday constant, 96,487 J•V^-1mol^-1.

10. B