Read the following redox reaction and determine which reactant is reduced and which is oxidized.

f2 + zn → znf2

this equation shows that f2 is 1. (reduced or oxidized) and zn is 2. (reduced or oxidized) .

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.

D. is the answer hope that you