EK chemistry lecture 7 exam question, redox, lowering formal potential, endpoint

[MedGrl@2022]

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This question is taken from EK’s Chemistry Lecture 7 Exam:

Fe(2+) is titrated with dichromate Cr2O7(2-) according to reaction 1:

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

Chromium (III) is green in color, and dichromate solution is orange. The orange is not intense enough to be used as an indicator. Instead the endpoint of the titration can be indicated by the redox indicator diphenylamine sulfonic acid (DAS), which changes from colorless to violet when oxidized. The color change observed during the titration is from green to violet. After all the Fe(2+) ions have been oxidized, the dichromate ion oxidizes DAS. However, the formal potential of the Fe(2+) solution must be lowered in order to match the endpoint with the equivalence point of the titration. This is accomplished by addition of H2SO4 and H3PO4 immediately before titration.


143. If the formal potential of the solution is not lowered, which of the following will be the result of the titration when DAS is used as the indicator?

A. The solution will turn violet before the equivalence point is reached.
B. The solution will turn violet after the equivalence point is reached.
C. Dichromate ion will not be able to oxidize DAS.
D. Dichromate ion will reduce DAS to its colorless form.

I was wavering between A and B and I chose B, even though my gut said A. I guess I am having some issues understanding what they meant by lowering the formal potential of the solution and how this results in matching the endpoint with the equivalence point. Is the formal potential Eº=1.33V? Do the SO4(2-) and the PO4(3-) ions interact with the Fe(2+) ions and prevent them from being oxidized somehow?

This is EK’s explanation “A is correct. The equivalence point is defined as the point when the Fe(2+) is completely oxidized. Changing the formal potential won’t change that definition. The Endpoint is when the indicator changes color. This should be at the equivalence point, but it does not have to be. The passage says that the formal potential is lowered so that the equivalence point and endpoint will coincide. This indicates that the indicator will change color at a low potential. If the formal potential isn’t lowered, the indicator will change color early (still at the low potential) before the equivalence point is reached. The passage states that the indicator changes from colorless to violet; choice A. Dichromate ion will still oxidize DAS, but it will do so before oxidizing Fe(2+), so choice C is incorrect. Dichromate ion is an oxidizing agent and will not reduce anything, so choice D is incorrect.”

After reading their explanation, I still don’t understand how lowering the formal potential makes the endpoint coincide with the equivalence point. It seems to me that the passage states that “After all the Fe(2+) ions have been oxidized, the dichromate ion oxidizes DAS. “ Therefore, the equivalence point would be after all the Fe(2+) is completely oxidized then DAS would start turning violet because it is oxidized. Thus, endpoint and equivalence point should already coincide, right?

Also, what did EK mean in its answer response “Dichromate ion will still oxidize DAS, but it will do so before oxidizing Fe(2+), so choice C is incorrect.” Doesn’t this go against the passage since they state that “After all the Fe(2+) ions have been oxidized, the dichromate ion oxidizes DAS. “?

Thank you for all your help! 🙂

 

[polyatomic]

It sounds like there's some bad wording in the explanation. That or I'm misinterpreting it.

I'd approach this conceptually...

1) Why would DAS be oxidized after Fe? Probably because it takes more "energy" to oxidize DAS than Fe. Since Fe is easier to oxidize, it will be oxidized at lower potentials.

2) Let's pretend DAS requires 5.00 V to be oxidized, and Fe takes 2.00 V to be oxidized. If we have a potential of 6.00 V, then, even though DAS is harder to oxidize, we'll still end up oxidizing it because we have enough potential for it. But, if we could lower that potential to say 4.00 V, then we'll end up oxidizing Fe at first since it's the easier route.

3) They say in the question that you must add acid in order to lower the potential to AVOID oxidizing DAS till after Fe is completely oxidized. Therefore, this IMPLIES that if you don't add acid, you don't lower the potential, and DAS will be oxidized too early.

4) The equivalence point is the point when you have completely converted Fe(2+) to Fe(3+). Therefore...DAS will be oxidized (solution turns violet) prior to this IF the potential is too high. (Choice A)

5) If you see violet after the equivalence point, then that means DAS was oxidized after all of Fe(2+) was converted to Fe(3+), which is what would happen if the potential was lowered. So this cannot be correct, because the question is addressing the problem of not lowering the potential.

I'm not sure why EK is saying DAS will be oxidized before Fe(2+)...but perhaps they mean it will be oxidized before Fe(2+) is COMPLETELY oxidized, which would make more sense.

 

[slz1900]

I might be way off, but my thinking would be that acids make it harder for DAS to get oxidized since they can donate protons to the solution, so they effectively act as a buffer between your Fe2+ and the DAS. Putting acid into solution pushes the reaction to the right since there are protons on the reactant side. If you don't put it in, DAS gets oxidized before the reaction has gone to completion. I guessed B too since I had no clue what they meant by lowering the formal potential, but I think they said it to be tricky.

Edit: Massive brainfart

 

[polyatomic]

I might be way off, but my thinking would be that acids make it harder for DAS to get oxidized since they can donate protons to the solution, so they effectively act as a buffer between your Fe2+ and the DAS. Look at the reaction with Fe2+. There are protons on the left, so if you add acid to the solution you are driving the reaction toward the reactant side where no color change is happening. If you leave the acid out, your reaction goes to the product side quicker, so a color change happens faster. I guessed B too since I had no clue what they meant by lowering the formal potential, but I think they said it to be tricky.

If you're adding strong acid, then you're adding more protons. Because protons are the reactants, this would drive your reaction to the right to form products.

 

[slz1900]

If you're adding strong acid, then you're adding more protons. Because protons are the reactants, this would drive your reaction to the right to form products.

Yep, I changed it 😳