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# Potentiometric Titration of Iron(II)

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By using samples of the same cerium solution throughout each trail and precise volumetric pipettes, accuracy, along with precision, the potential change of the reaction and the determination of the molar concentration of iron (II) ions in a sample of unknown concentrations were determined by potentiometric titration of the reaction between ferrous ammonium sulfate and ammonium cerium (IV) nitrate by titrating a ferrous ammonium solution with a cerium titrant.

The molarity of the Cerium (IV) solution was determined to be 0.0316M. The Cerium solution was titrated in the ferrocyaninde solution three times and the molarity was calculated using the formula M1V1=M2V2. The mean molarity of the three trials was determined to be 0.0253M with a standard deviation of 0.00024. In conclusion, the potential change of reaction and the determination of molar concentration of iron (II) were determined using basic titrations and calculations in electrochemistry within minimal standard deviation acceptance.

Questions:
1. What is the potential of the iron(III)/iron(II) couple you determined in this experiment? How does your value compare to the standard potential for this redox couple given in your text?
2. What is the potential of the cerium(IV)/cerium(III) couple you determined in this experiment? How does your value compare to the formal potential for this redox couple given in your text?
3. If potassium permanganate had been used as the titrant instead of cerium(IV), what would the balanced overall reaction be? At what cell voltage would the equivalence point be observed?

https://brainmass.com/chemistry/titration/potentiometric-titration-of-iron-ii-530827

#### Solution Preview

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1.
For titration run 1, the half-equivalence point is 15.8 / 2 = 7.9 mL. By expanding the graph of run 1, the corresponding potential is found to be 470 mV. I believe you said the reference voltage = 0.222 V (Make sure this is the correct value for the reference electrode). Hence the calculated potential of the iron couple = 0.470 + 0.222 = 0.692 V. The ...

#### Solution Summary

A number of problems based on the potentiometric titration of iron are solved.

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