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Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified KMnO₄ for complete oxidation?

A. FeC₂O₄

B. Fe(NO₂)₂

C. FeSO₄

D. FeSO₃

User Foxesque
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1 Answer

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Final answer:

The compound that requires the least amount of acidified KMnO₄ for oxidation is FeSO₄, as only the iron(II) ion is oxidized and the sulfate ligand does not require additional oxidizing agent.

Step-by-step explanation:

Assuming complete ionization, the compound that will require the least amount of acidified KMnO₄ for complete oxidation can be determined by comparing the stoichiometry of the redox reactions each compound will undergo with KMnO₄. Acidified KMnO₄ acts as an oxidizing agent, converting Fe(II) to Fe(III). In a redox reaction, the number of moles of electron transferred is an important factor to consider.

The number of electrons each iron compound will transfer when it is oxidized by KMnO₄ is critical. For FeC₂O₄ (oxalate) and FeSO₃ (sulfite), two electrons per mole of Fe(II) are involved, as both C₂O₄²⁻ and SO₃²⁻ can be further oxidized. In contrast, Fe(NO₂)₂ (nitrite) and FeSO₄ (sulfate) involve only one electron transfer per mole of Fe(II), as NO₂⁻ and SO₄²⁻ are not oxidizable under these conditions.

Considering these reactions, FeSO₄ will require the least amount of KMnO₄ because only the Fe(II) ion is oxidized, without any associated ligand that would require additional KMnO₄ for oxidation. Thus, the correct answer is C. FeSO₄.

User Ze Grisi
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