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pkmnmushy

Im gping to help you like this: In ZnF2 , you have a lattice of Zn2+ ions and F− ions.

So you can re-write your equation as:

Zn→Zn2++2e−
F2+2e−→2F− 
I think with this you may know how to proceed. May this help you a lot

Answer : \(F_2\) reactant is reduced and \(Zn\) reactant is oxidized.

Explanation :

Redox reaction : It is a type of reaction in which an oxidation and reduction reaction takes palace simultaneously.

Oxidation : The loss of electrons or increase in the oxidation number is an oxidation.

Reduction : The gain of electrons or decrease in the oxidation number is a reduction.

The given balanced redox reaction is,

\(F_2+Zn\rightarrow ZnF_2\)

The oxidation number of \(F_2\) = 0

The oxidation number of \(Zn\) = 0

The oxidation number of Zn in \(ZnF_2\)

x + 2(-1) = 0

x = +2

The oxidation number of \(F_2\) in \(ZnF_2\) = -1

That means \(F_2\) is reduced due to decrease in the oxidation number from 0 to (-1) and 'Zn' is oxidized  due to increase in the oxidation number from 0 to (+2).

Therefore, \(F_2\) reactant is reduced and \(Zn\) reactant is oxidized.

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