Redox and Oxidation

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deniise_garciia
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Joined: Wed Nov 16, 2016 3:04 am

Redox and Oxidation

Postby deniise_garciia » Fri Feb 22, 2019 7:03 am

I still have trouble with redox and oxidation, would someone be kind enough to explain step by step how Lavelle did his in lecture on Wednesday, thanks!

Ray Huang 1G
Posts: 30
Joined: Fri Sep 28, 2018 12:20 am

Re: Redox and Oxidation

Postby Ray Huang 1G » Fri Feb 22, 2019 8:57 am

The best way to do it is to look up the rules for oxidation numbers. He essentially looked at each element on each side of the table and found their oxidation number. Then he looked at if the number of the same element went up or down (comparing the left side to the right side). Using this you can find if an element is oxidized or reduced.

sallina_yehdego 2E
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Joined: Sun Apr 29, 2018 3:00 am

Re: Redox and Oxidation

Postby sallina_yehdego 2E » Fri Feb 22, 2019 9:23 am

First, you should always see if the equation is balanced out or not. From there, you go to each compound/ion and find the charge of each one. From there, you compare the reactants with the products to see which are being reduced and which are being oxidized. The reducing agent is the one that gets oxidized (looses e-) and the oxidizing agent is the one that gets reduced (gains e-).

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: Redox and Oxidation

Postby Michael Novelo 4G » Fri Feb 22, 2019 12:11 pm

First make sure the equation is balanced. and then compare reactants to products to form the half-reactions: oxidation and reduction. The oxidation reaction occurs when the Reactants gain a higher + charge. To balance out you add the electrons from the positive charge gained to balance out, for example he wrote the oxidation of 5Fe^2+ --> 5Fe^(3+) + 5 electrons. The Fe has a charge of 15+ in the products and to balance out he adds 5 electrons to make its original 10+ charge. The reduction adds the electrons in its Reactants to reduce the + charge in the example he gave MnO4^(-1) + 5 electrons reduced the charge of Mn (which is +7 in the Reactants) and got Mn^2+ (which has a +5 charge) on the products side. So Oxidation you add electrons on the products side and for reduction we add electrons on the Reactants side to reduce the charge of an atom. When you add these two Oxidation and reduction you get the original equation since these are half reactions.


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