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Topic: Oxidation & Reduction (in cellular resp.)  (Read 4378 times)

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Offline Jules18

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Oxidation & Reduction (in cellular resp.)
« on: September 29, 2009, 02:54:20 PM »
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2

Which molecule is reduced and which is oxidized in the above reaction? 
Is it correct to ask the question that way, or should I be asking which atoms are reduced/oxidized?   :-\


thanks for any replies,
Jules

Offline azmanam

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #1 on: September 29, 2009, 03:36:23 PM »
it's more precise to say atoms, though people typically say molecules and most don't notice the difference.  You should answer whichever question your instructor is asking :)

What do you think the answer is?
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Offline Jules18

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #2 on: September 29, 2009, 03:44:54 PM »
I know what my prof tells us the answer is supposed to be, I just have no idea why.


Offline renge ishyo

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #3 on: September 29, 2009, 06:52:26 PM »
A useful way to think about oxidation and reduction in organic compounds is that removal of Hydrogen (H2) from a compound oxidizes that compound while addition of hydrogen reduces a compound.

Offline Jules18

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #4 on: September 29, 2009, 09:28:18 PM »
but hydrogen is a cation in its ionized form, right?
that's why I'm confused. 

Offline renge ishyo

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #5 on: September 30, 2009, 01:23:10 AM »
Electrons are removed along with the hydrogen ions in the above case. It is not the same thing as when you abstract a proton from an organic molecule in a simple acid and base reaction, and this might be where the confusion is arising. In the acid base case, when the proton is removed as H+, the electrons that were bound to the hydrogen are left behind on the organic molecule afterwards leaving the organic molecule with a -1 charge.

In the present case, after removing 2H+ protons the organic molecule is not left with a - 2 charge afterwards; it is neutral. The only way this can happen is if the molecule lost 2 electrons along with the protons (and loss of electrons from a molecule is an oxidation). A common way you may see this type of oxidation written is to see protons leave as 2H+ + 2e- in biochemistry diagrams.

Later on you will learn that these biochemical reactions don't occur as easily as simple proton exchange reactions in general acid base chemistry. The body has special molecules such as NAD+ that can carry a proton and two electrons away at the same time (as NADH).

Offline Jules18

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #6 on: September 30, 2009, 04:56:33 PM »
Okay I get what you mean.
So, if it loses two e- and two p+ . but it's still considered oxidized, what is the definition of oxidation?  I thought that to be oxidized the net charge on the atom had to become more positive? 

Offline azmanam

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Re: Oxidation & Reduction (in cellular resp.)
« Reply #7 on: September 30, 2009, 05:07:00 PM »
...and this is where talking about atoms and molecules interchangeably gets confusing.

Formally, the carbon atoms in the sugar get oxidized to the +4 oxidation state (to a carbon dioxide molecule, the highest oxidation state of carbon representing complete combustion of the sugar under aerobic conditions).  Formally, oxygen gets reduced to the -2 oxidation state (to a water molecule).  It is the carbon atoms in sugar that are being oxidized and the oxygen atoms in molecular oxygen that are being reduced.

Overall, we can say that the sugar molecule is the reductant.  It is the molecule reducing molecular oxygen and therefore it is the molecule being oxidized.  Molecular oxygen is the oxidant.  It is the molecule oxidizing the sugar molecule and is the molecule being reduced.
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