Is this just a bad MCAT question or am I missing something?

[MedGrl@2022]

Hi I just took EK Lecture 7 Chemistry Exam and I had a question about a question.

So the reaction that they are referring to is Reaction 1:

6Fe(2+) + Cr2O7(2-) + 14H+ <-------> 2Cr(3+) + 7H2O + 6Fe(3+) Eº=1.33V

Question 139 is:

Which of the following is true concerning Reaction 1?

A. Cr is oxidized and Cr2O7(2-) is the oxidizing agent.
B. Cr is reduced and Cr2O7(2-) is the reducing agent.
C. Cr is reduced and Cr2O7(2-) is the oxidizing agent.
D. Cr is oxidized and Cr2O7(2-) is the reducing agent.

Obviously, Cr2O7(2-) is the oxidizing agent as it is reduced in reaction 1. But I was confused as to which Cr they are referring to. Since they did not say the Cr in Cr2O7(2-), I assumed that it was the Cr from 2Cr(3+), which is oxidized in the reverse direction. However, according to EK "C is correct. Cr2O7(2-) is reduced to 2Cr(3+). Although this doesn't look like reduction from the charges, Cr in the dichromate has an original oxidation state of +6." So EK was obviously referring to the Cr in Cr2O7(2-), however, how was I supposed to know this? If I get a question like this on the MCAT, am I supposed to assume that they are talking about the atom in the molecule? Or is this just a bad MCAT question and I shouldn't worry about it because real MCAT questions will be clearer?

Thank you for all your help.

You are all awesome!

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[sdm33]

I think you need to assume they are talking about the reactants, so Cr got reduced from +6 to +3
But why is Cr2O7(2-) is the oxidizing agent, I know Cr got reduced but O get oxidized ???

 

[MedGrl@2022]

I think you need to assume they are talking about the reactants, so Cr got reduced from +6 to +3
But why is Cr2O7(2-) is the oxidizing agent, I know Cr got reduced but O get oxidized ???

Well, since the arrows go both ways in this reaction, 2Cr(3+) is a reactant in the reaction that is in reverse.

Also, O's oxidation state doesn't change in this reaction only the oxidation states of Fe and Cr.

Cr(+6) + 6e- <----> Cr(3+)

Fe(2+) <-----> Fe(3+) + e-

 

[sdm33]

why O does not change?
May be because E is positive non spontaneous rxn , so it shift to the right.

 

[MedGrl@2022]

why O does not change?
May be because E is positive non spontaneous rxn , so it shift to the right.

O has a oxidation state of -2 the entire time.

Cr2O7(2-) <-------> 7 H2O

O likes to have an oxidation state of -2. This makes the Cr in Cr2O7(2-) have an oxidation state of +6. In H2O, H has an oxidation state of +1 and there are 2 of them in each molecule of water leaving O with an oxidation state of -2.

The only thing that can change O's oxidation state is H and F. The first oxidation rule is that H has an oxidation state of +1 (except when bonded to a metal: +1). Also, flourine is more electronegative than O and has a greater affinity than oxygen so if flourine is attached to an oxygen then it is more likely to hog the electrons.

Does this help? Do you see it now?

 

[sdm33]

yes , I was thinking -14 for O not seeing the 7. good stuff. thanks

 

[slz1900]

If they were referring to the product, it'd say Cr^3+

 

[chem4ever]

Hi I just took EK Lecture 7 Chemistry Exam and I had a question about a question.

So the reaction that they are referring to is Reaction 1:

6Fe(2+) + Cr2O7(2-) + 14H+ <-------> 2Cr(3+) + 7H2O + 6Fe(3+) Eº=1.33V

Question 139 is:

Which of the following is true concerning Reaction 1?

A. Cr is oxidized and Cr2O7(2-) is the oxidizing agent.
B. Cr is reduced and Cr2O7(2-) is the reducing agent.
C. Cr is reduced and Cr2O7(2-) is the oxidizing agent.
D. Cr is oxidized and Cr2O7(2-) is the reducing agent.

Obviously, Cr2O7(2-) is the oxidizing agent as it is reduced in reaction 1. But I was confused as to which Cr they are referring to. Since they did not say the Cr in Cr2O7(2-), I assumed that it was the Cr from 2Cr(3+), which is oxidized in the reverse direction. However, according to EK "C is correct. Cr2O7(2-) is reduced to 2Cr(3+). Although this doesn't look like reduction from the charges, Cr in the dichromate has an original oxidation state of +6." So EK was obviously referring to the Cr in Cr2O7(2-), however, how was I supposed to know this? If I get a question like this on the MCAT, am I supposed to assume that they are talking about the atom in the molecule? Or is this just a bad MCAT question and I shouldn't worry about it because real MCAT questions will be clearer?

Thank you for all your help.

You are all awesome!

This is a terrible MCAT question. When talking about which molecule got reduced and which got oxidized (or which is the reducing agent and which is the oxidizing agent), we only refer to the MOLECULES on the reactant side, not the individual atoms. If the answer choices had said Cr(3+), then I would have said that it was referring to the reverse reaction like you thought. However, the answer choices refer to Cr not Cr(3+) so none of the answers really work.

I got points off on a test in gen chem for writing that the reducing agent was Ni instead of writing the whole compound (with the Ni in it) that was on the reactant side (I forget what the compound was).

 

[advair250]

Cr207(2-) is obviously the oxidizing agent, because the Cr(+6) gets reduced to Cr3+. And the Cr of this compound acting as a oxidizing agent is reduced.

It always refers to reactants. No different then Gen chem.

 

[jlescher]

Don't worry about it. If you know oxidation reductions you will be fine for the real MCAT. Don't rack your brain on things like this. This type of ambiguity doesn't appear on the MCAT. All of the questions I had on my test were well written. I hated questions like this when I was studying and I specifically remember this one.

 

[advair250]

Since all the options have Cr207(2-), you HAVE to be looking at the left side of the reaction. This is a very very basic gen chem question. Try not to over think it.

 

[JLeBling]

Oxidizing agent always gets reduced.
Reducing agent always gets oxidized
REDucation CAThode
ANode OXidation