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

1 Answer
May 19, 2017

Answer:

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.