Question

What are the oxidation and reduction half-reactions for `Cr+F_2 -> CrF_3`?

Answer 1

Oxidation: `Cr->Cr^(3+)+3e^(-)`
Reduction: `F_2+2e^(-)->2F^-`

Explanation:

The empirical formula `CrF_3` suggests that fluorine has more electron tendency than chromium. Thus it has a negative oxidation state number.
The fluoride ion has a `-1` charge as a halogen anion.
Each chromium cation binds to `3` fluoride anion, so it has a `+3` ionic charge.
Keep in mind that oxidation is losing electrons, and reduction is gaining electrons.
And then you get the answers.