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Free MCAT Practice Question, Round 2 – Chemistry

  • by Allison Chae
  • Jun 10, 2013
  • MCAT Blog, MCAT Chemistry, MCAT Long Form

MCAT Practice Chemistry Question

Our previous Chemistry question asked about a common fundamental topic on the MCAT – trends on the periodic table.

For this question, we’ll go to the other end of the difficulty spectrum and look at a redox question. Electrochemistry and redox reactions are one of MCAT students’ least favorite topics on the test. Most freshman chemistry classes don’t get around to electrochemistry until the end of the school year, and by then who’s really paying attention anymore? Plus, if your professor drops the lowest grade and you’re already doing well, why not just ignore the electrochem test?

Unfortunately, those really common attitudes can end up hurting you on the MCAT, since electrochemistry is fair game on Test Day. Try your hand at this sample problem:

Item 19

Given these reduction potentials:

Ag^{2+}(aq) + 2e^{-} rightarrow Ag(s)          E^{o}= 0.80 V

Fe^{3+}(aq)+3e^{-}rightarrow Fe(s)         E^{o}=-0.036V

If a strip of solid Ag is placed into a beaker of iron (III) nitrate, which of the following will occur?

A) Fe ions will precipitate.

B) Ag ions will precipitate.

C) Both Fe and Ag ions will precipitate.

D) No reaction will occur.



To understand this question, we’ve got to look at what reaction could occur given the setup of the problem, and then ask ourselves if that reaction would occur.

The problem states that we’ve got a strip of solid Ag. That means that we’re starting with Ag(s) and the only thing that could happen to those atoms would be for them to dissolve into the solution, forming ions. So we can eliminate choices (B) and (C) since we don’t even have any silver ions that could precipitate.

Next, the problem tells us that we’re starting with iron (III) nitrate, so we do have iron ions that could precipitate – the question is, would they?

For the iron ions to precipitate, we would need a redox reaction that reduces the iron into solid iron, and that oxidizes the silver into silver ions. Remember to balance the number of electrons transferred! The overall reaction would look like this:

2Fe^{3+}(aq)+3Ag(s)rightarrow 2Fe(s)+3Ag^{2+}(aq)

Notice that we’ve also balanced charge. The left and right side of the reaction both have a total ionic charge of 6+.

To figure out if this reaction would occur, we need to remember a few facts:

1. Voltage and ΔG always have opposite signs for a given reaction.

2. A negative ΔG indicates that a reaction is spontaneous (positive is non-spontaneous).

3. When calculating the voltage for a reaction you do NOT multiply by the stoichiometric coefficients in the reaction.

4. Finally, if you reverse a redox reaction, you reverse the sign on the given voltage.

So, taking those facts together, we can conclude that this reaction is non-spontaneous and will not happen.

The reduction of iron generates a voltage of -0.036V and the oxidation of silver generates a voltage of -0.80V. The total reaction would have a voltage of -0.836 (simply add the two numbers together, but do NOT multiply by the 2 in front of the iron or the 3 for silver).

This negative voltage indicates a positive ΔG, and thus a non-spontaneous reaction.

Thus, the correct answer is (D).