[CHM 111] Chapter 4-CHM 111

[Andrew J. Pounds]

On 10/10/2017 01:27 PM,Jones wrote:
> Dr. Pounds,
> Hello again. In your powerpoint of chapter 4 about oxidation reduction reactions. H2 + F2——>2HF. H2 has a charge of zero on reaction side. And 1 on product side. How? Same goes for every compound/ element ever with these problems.

H_2 and F_2 on the reactant side are  ELEMENTS in their  STANDARD STATE, 
and therefore have oxidation numbers of zero.  On the product side in 
the compound HF H has a +1 oxidation state to balance out the F with an 
oxidation state of -1.


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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