This equation shows that F₂ is REDUCED and Zn is OXIDIZED.
Oxidation shows an increase in oxidation number from the reactant side to the product side while Reduction shows a decrease in oxidation number from the reactant side to the product side.
F₂ + Zn → ZnF₂
Reactant Side:
Species Oxidation State
F₂ 0
Zn 0
And that is because elements in their free uncombined state and diatomic state usually has an oxidation number of zero.
With the Product side however, one must apply a two rules of redox to solve for the oxidation state of Zn.
1) The sum of all oxidation numbers in a neutral compound is zero
2) The oxidation state of Fluorine in a compound is always -1
From this bit of information, then we can deduced that fluorine decreases (is reduced) from a 0 oxidation state in the reactant to a -1 oxidation state which automatically suggests that zinc was oxidized. However, this can be proven using the knowledge we have:
oxidation state of ZnF₂ = 0
let oxidation state of Zn = x and F = -1 oxidation state
∴ x + (-1)2 = 0
⇒ x - 2 = 0
⇒ x = 2
Therefore, it is proven. Zinc increases (oxidized) from an oxidation state of on the reactant side to an oxidation state of 2 on the product side.
