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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
Species Oxidation State
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.