## 14.1 (b)

Moderators: Chem_Mod, Chem_Admin

Ilan Shavolian 1K
Posts: 58
Joined: Fri Sep 29, 2017 7:03 am

### 14.1 (b)

pretty basic question but can someone help me understand how to find the oxidation numbers of the atoms in the C2H5OH and C2H4O?

Alyssa Parry Disc 1H
Posts: 53
Joined: Sat Jul 22, 2017 3:01 am

### Re: 14.1 (b)

You have to look at the other molecules and their typical oxidation states like in this one you would look at the oxidation states of H and O, Usually +1 and -2 respectively and then look at the total charge of both molecules, I think in this case its 0 for both and then you just do algebra. Like for the first one there are 6 H and 1 O so 6(+1) +1(-2) + 2X = 0, and then you would get X=-2 so thats the oxidation state for that carbon in that molecule C2H5OH. Then you do the same for the others.

Grace Han 2K
Posts: 33
Joined: Tue Nov 15, 2016 3:00 am

### Re: 14.1 (b)

C2H5OH
Oxygen is going to have an ON (Oxidation number) of -2 unless its in a peroxide.
Hydrogen is going to have an ON of +1 if its with nonmetals.
Now we have to solve for Carbon's ON
When adding up all of the ON for C2H5OH, it should equal zero.

____/2 (Carbon)+ 6x +1(Hydrogen) + -2(Oxygen)= O
Then ON of carbon is -2

Return to “Balancing Redox Reactions”

### Who is online

Users browsing this forum: No registered users and 1 guest