Oxidation of C2H2O4

Moderators: Chem_Mod, Chem_Admin

Ranger Saldivar 2D
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

Oxidation of C2H2O4

Postby Ranger Saldivar 2D » Sat Jan 30, 2016 1:54 am

I was looking in the book and noticed that to oxidize C2H2O4 you need to identify the charges on the molecules. As far as I know C has a 4+ charge so why is it +3 on the left. Also once the oxidation takes place on the left it has a 4+ charge is there something I'm misssing?

Alexander Vong 3I
Posts: 56
Joined: Fri Sep 25, 2015 3:00 am

Re: Oxidation of C2H2O4

Postby Alexander Vong 3I » Sat Jan 30, 2016 2:41 am

The oxidation number of an atom is derived from the bonds it makes from other atoms. An atom's charge is influenced by differences in electronegativity. For example, every C-H bond will decrease the oxidation state by 1, and every carbon bond to a more electronegative element will increase the oxidation state by 1.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest