Answer:
FeI₂(aq) + F₂(g) → I₂(l) + FeF₂(aq)
F + FeI2 → FeF2 + I 2
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
