How Do You Calculate the Mass Percent of Iron in an Ore Using Redox Titration?

[Flammadeao]

Is anyone here able to help me or point me in the right direction?

Iron ores often involve a mixture of oxides and contain both Fe 2+ and Fe 3+ ions. Such an ore may be analyzed for its iron content by dissolving it in an acidic solution, reducing all of the iron to Fe 2+ ions and then titrating with a standard solution of potassium permanganate until there is just a tinge of pingk color in the solution containing the iron ions. (hint: the permanganate ion is intensely purple and Mn 2+ is colorless.) After the titration was complete, permanganate would be reduced and Fe 2+ would be oxidized.

A sample of iron ore weighing 0.3500g was dissolved in acidic solution, and all of the iron was reduced. The resulting solution was titrated with a 1.621x10^-2 Molar KMnO4 Solution. The titration endpoint was reached when 41.56 mL of the KMnO4 solution was used. Determine the mass percent of iron in the iron ore sample.

What I am thinking is that I need to create two half reactions with FeO and Fe2O3 --> Fe 2+. So I go through that and I end up with half reactions of :

2H+ + FeO -> Fe2+ + H2O
2e- + 6H+ + Fe2O3 -> 2Fe2+ + 3H2O

and then combine it to form

8H+ + 2e- + FeO + Fe2O3 --> 3Fe2+ + 4H2O

From what I've learned you are supposed to be able to cancel out the electrons when you combine the two half reactions, however in this case I only have electrons on one of the half reactions and thus I have no hope of canceling them out. Am I thinking about this correctly or is there something else I should be doing?

 

[mrjeffy321]

Your thinking is on the right track, but you are only considering one of the half-reactions there.

A sample of iron ore weighing 0.3500g was dissolved in acidic solution, and all of the iron was reduced.

I think the question’s wording here is a little confusing, but let's assume that it means all of the Iron ions in solution have been converted (reduced) to Fe+2 (aq), and not all the way back to Iron metal (which is another, however illogical, interpretation of that statement).
So you have a solution of Fe+2 (aq)

When finding the half-reactions, don’t worry about the reaction which occurs to dissolve the Iron Oxide, just assume that the Iron starts out already in solution in the form of Fe+2 ions.

The half-reactions we are talking about here make up a RedOx (Reduction-Oxidation) reaction.
One half reaction will be the reduction portion ("Gain electrons, Reduction"), the other half reaction will be the oxidization portion (Loose electrons, Oxidization).

What is oxidized in the reaction?
What is reduced?
Put these two reactions together to get the complete RedOx reaction, and the electrons should/will cancel out with the proper coefficients.

 

[Blueshift5]

I can offer some guidance and suggestions for solving this problem. First, it is important to understand that the goal of this titration is to determine the mass percent of iron in the sample. To do this, we need to calculate the amount of iron present in the sample and then divide it by the total mass of the sample (0.3500g).

To do this, we need to use the balanced chemical equation for the reaction between Fe 2+ and KMnO4. This will allow us to determine the moles of KMnO4 used in the titration, which can then be used to calculate the moles of Fe 2+ present in the sample. From there, we can use the molar mass of Fe 2+ to calculate the mass of iron present in the sample.

In terms of the half reactions, you are on the right track. However, in order to cancel out the electrons, you need to make sure that the number of electrons on each side of the equation is equal. In this case, you can multiply the first half reaction by 3 and the second half reaction by 2 to get a balanced equation. This will allow you to cancel out the electrons and combine the two half reactions into one overall reaction.

I hope this helps guide you in the right direction. If you have any further questions, please don't hesitate to ask. Good luck with your problem!