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  1. The concentration of $\ce{Fe^2+(aq)}$ can be determined by a redox titration using

A. $\ce{KBr}$
B. $\ce{SnCl2}$
C. $\ce{KMnO4}$ (basic)
D. $\ce{KBrO3}$ (acidic)

Can anyone please help me on this question? I know the answer is between C and D, since those two are the only redox reactions. However, I don't know how to determine between C and D, as they will both act as an oxidizing agent and Fe will be reduced in both cases (the correct answer is D).

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  • $\begingroup$ This question requires some background chemistry knowledge. First of all, think whether the presence of base will interfere or not. Will iron(II) precipitate out as a hydroxide? We wouldn't like to do titration in a heterogeneous mixture. The next to keep in mind is electrode potentials of permanganate and bromate half cells. $\endgroup$ – M. Farooq Jun 9 at 3:42
  • $\begingroup$ May I ask for clarifications to electrode potentials? do you mean I should calculate for the overall E value of the reaction? $\endgroup$ – Joshua Jun 9 at 3:57
  • $\begingroup$ Can you please elaborate on that part? I know that the E value of Mno4- is +1.51, and the one for BrO3- is +1.48. What does that suggest?? that Mno4- will more likely to be reduced by Fe+ than Bro3-?? please explain $\endgroup$ – Joshua Jun 9 at 4:04
  • $\begingroup$ Did you understand what is the reductive electrode potentials are? By the way, $\ce{Fe}$ won't reduce in any case (as you suggested). $\endgroup$ – Mathew Mahindaratne Jun 9 at 4:10
  • $\begingroup$ Yes i do. I guess I dont understand the role that acid and base play in redox reactions. When doing a titration, how do I know when to use an acidic medium and when to use a basic one? I dont understand how looking at half cells potentials can allow me to determine this $\endgroup$ – Joshua Jun 9 at 4:22
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Can you please elaborate on that part? I know that the E value of Mno4- is +1.51, and the one for BrO3- is +1.48. What does that suggest?? that Mno4- will more likely to be reduced by Fe+ than Bro3-?? please explain

The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. Hydroxides of most metals are insoluble. Write the equations : Fe(II) + hydroxide ions = iron (II) hydroxide. This is why alkaline permanganate is not a good choice. However, this is not the final answer because you would like to further confirm whether C or D is the correct choice.

Next determine the E_cell= E(cathode)-E(anode) > 0 V or not.

Now you need to oxidize Fe(II) to Fe(III), so the iron half cell will be the anode. As a mnemonic, anode and oxidation both start with vowels; cathode and reduction don't. You would like to reduce the permanganate in an alkaline medium, this half cell is the cathode.

Repeat the same for bromate half cell in acidic medium and Fe(II). Both (C) and (D) would turn out to be positive E_cells. However, the (C) is not an answer because you don't want iron to precipitate out as iron(II) hydroxide.


In real life analysis, Fe(II) can be routinely titrated with potassium permanganate but in highly acidic medium. For practice, check the E_cell after writing a balanced reaction.

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    $\begingroup$ right, I see now. THANK YOU $\endgroup$ – Joshua Jun 9 at 4:38

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