[polyfractal]

I'd like to preface this post with a small disclaimer:  I'm an amatuer chemist, learning this for fun.  For any of my questions, if its easier to point me to an online source or term to search for, by all means do so.  No need to re-explain chemistry to me if its available elsewhere, I'm just not sure what to search for.  I thoroughly enjoy chemistry and am doing a lot of learning myself, but my formal education in chemistry consists of basic High School chem.  Also, if this should go to High School chem forum (I'm guesing its a fairly basic question) please move it.  I wasnt sure where to place it.

Anyhow, couple of question regarding oxidation and oxidizing agents.  These apply to all oxidants, but the ones I'm interested in is Potassium Permanganate (KMnO4) and Hydrogen Peroxide.

-How can you determine how much of an oxidant is needed to oxidize another substance?
-How can you determine the rate of oxidation?
-On a similar note, where can I find (or how to determine) how the pH affects the the oxidation rate?  I've vaguely heard that KMnO4 oxidizes slower under acidic conditions than basic, but I've yet to find a reliable source to confirm that.

On the pH question:  I'm thinking that an acidic solution would slow down oxidation.  An acidic solution would have an excess of H+ ions, forming H3O+.  These positive H3O+ ions would steal electrons from the solute and thus distribute where the oxidant attacked, which would slow down the rate at which the solute is oxidized.  Is this reasonable thinking?  Sorry if this is really off target.

Thanks for all the help, its much appreciated.  

[Borek]

How can you determine how much of an oxidant is needed to oxidize another substance?

Reaction equation plus stoichiometric calculations.

How can you determine the rate of oxidation?

Only experimentally.

On a similar note, where can I find (or how to determine) how the pH affects the the oxidation rate?  I've vaguely heard that KMnO4 oxidizes slower under acidic conditions than basic, but I've yet to find a reliable source to confirm that.

Never heard about. I know that in at least some reactions Mn²⁺ ions catalyze oxidation, thus reaction goes slowly at the beginning, then gains momentum.

Sorry if this is really off target.

IMHO completely off Low pH is paramount - take a look at the reaction equation and you will know why.

[polyfractal]

Pheww!  Spent all day researching, reading and learning.  I've learned so much today its crazy (oxidation states, half reaction equations, relearned my stoicheometry, etc).

Anyway, it looks like I was rather wrong.  Completely opposite infact.  I've been toying with various equations today.  For instance, MnO₄^- and ethanol:


Acid:  MnO₄^- + 5 C₂H₅OH + 3 H⁺  =  Mn²⁺ + 5 C₂H₅O + 4 H₂O

It looks like oxidizing in an acid will reduce MnO₄^- to Mn²⁺, oxidize your solute and generate H₂O.  MnO₄^- strips five H from the ethanol and uses an additional 3 H+ from the acidic solution to create 4 H₂O's.  


Base: MnO₄^- + 3 C₂H₅OH + 3 H⁺  =  MnO₂ + 3 C₂H₅O + H₂O + OH^-
If you oxidize in a base, MnO₄^- reduces to Mn₂ (since it reduces differently in bases), your solute oxidizes, water is generated.  There is the potential for 5 hydrogens to be ripped off the ethanol, but the lack of extra H⁺ ions means only three H⁺ can be pulled.  These three combine to create H₂O and OH^-.

Did I get all that correct?  Thanks, I've learned a ton today!

[cofi]

Acid:  MnO₄^- + 5 C₂H₅OH + 3 H⁺  =  Mn²⁺ + 5 C₂H₅O + 4 H₂O

wrong... ethanol should be oxidized to CH₃COOH...
try to write the two reactions separately:

reduction of MnO₄^- to Mn²⁺:
MnO₄^- + 8 H⁺ + 5 e^- ->  Mn²⁺ + 4 H₂O

oxidation of C₂H₅OH to CH₃COOH:
C₂H₅OH + H₂O -> CH₃COOH + 4H⁺ + 4 e^-

now, you have to multiply the first equation by 4, and the second one by 5, so that the number of accepted electrons is the same as the number of released electrons...
now you can sum the two equations...

4 MnO₄^- + 32 H⁺ + 20 e^- + 5 C₂H₅OH + 5 H₂O ->  4 Mn²⁺ + 16 H₂O + 5 CH₃COOH + 20 H⁺ + 20 e^-

or, nicer:

4 MnO₄^- + 12 H⁺ + 5 C₂H₅OH ->  4 Mn²⁺ + 11 H₂O + 5 CH₃COOH

when you are working in basic solution, everything is the same, just in the end you have to cancel all the H⁺ by adding equal number of OH^- to both sides of equation... it's OK because
you can write H⁺ like H₂O - OH^-... think like mathematician

generally, writing redox half-reactions is like this:

1. write the skeleton, the part which is important:
MnO₄^- ->  Mn²⁺

2. then you add H₂O to balance the number of oxygen atoms:
MnO₄^- ->  Mn²⁺ + 4 H₂O

3. after that, you have to balance the number of hydrogen atoms by adding H⁺:
MnO₄^- + 8 H⁺ ->  Mn²⁺ + 4 H₂O

4. finally, you add electrons (e^-) to one side of the equation to balance charge:
MnO₄^- + 8 H⁺ + 5 e^- ->  Mn²⁺ + 4 H₂O