Question

Fluorine gas reacts with aqueous iron (II) iodide to produce aqueous iron (II) fluoride and iodine liquid. What is the balanced chemical equation for this reaction? Group of answer choices F+ Fe2 I → Fe2 F + FeI2 → FeF2 + I 2 F + FeI2 → FeF2 + 2 I F2 + 2 Fe2I → 2 Fe2 F + I2

Answer 1

The balanced chemical equation for the reaction between fluorine gas and aqueous iron (II) iodide is F₂ + 2 FeI₂ → 2 FeF₂ + I₂, indicating the direct combination of these substances to form new products.

Step-by-step explanation:

The reaction between fluorine gas and aqueous iron (II) iodide to produce aqueous iron (II) fluoride and iodine liquid can be represented as a balanced chemical equation. When writing a balanced chemical equation, it is vital to ensure that the number of atoms of each element on the reactants side is equal to the number of atoms of that element on the products side, adhering to the law of conservation of mass.

The correct balanced chemical equation for this reaction is:

F₂ + 2 FeI₂ → 2 FeF₂ + I₂

Here, one mole of fluorine gas reacts with two moles of iron (II) iodide to produce two moles of iron (II) fluoride and one mole of iodine.

Answer 2

FeI₂(aq) + F₂(g) → I₂(l) + FeF₂(aq)

F + FeI2 → FeF2 + I 2

Step-by-step explanation:

Fluorine gas reacts with aqueous iron (II) iodide to produce aqueous iron (II) fluoride and iodine liquid

Consider that the fluorine gas is a dyatomic molecule.

The other reactant is the FeI₂. As the iron acts with +2 in the oxidation state, we must have 2 atoms of iodide.

Therefore the products are: FeF₂ and I₂

It is a redox reaction where the iodide is oxidized to iodine and the fluorine is reduced to fluoride

2I⁻ → I₂ + 2e⁻ Half reaction oxidation

F₂ + 2e⁻ → 2F⁻ Half reaction reduction

We sum both: 2I⁻ + F₂ + 2e⁻ → I₂ + 2e⁻ + 2F⁻ and the electrons are removed.

2I⁻ + F₂ → I₂ + 2F⁻ so now, we add the iron, and the balanced reaction is:

FeI₂ + F₂ → I₂ + FeF₂