is converting ADP to ATP considered a oxidation-reduction reaction? like in this reaction: 1,3 bis phosphate gycerate + ADP -----> 3 phosphate glycerate + ATP
Does the oxidation method work for all compounds? I used the same to find the oxidation state of S in H2S2O4 and H2S2O6 but ended up getting the wrong result.
+Andy M It should, as long as the initial Lewis structure is correctly drawn and the method applied correctly. Using the Lewis structure method I get +6 and +5 for the oxidation states of S in H2SO4 and H2S2O6 respectively. In the Lewis structure for H2SO4, the four oxygen atoms will be bonded to the central sulfur. Because oxygen is more electronegative that sulfur, all shared electrons are assigned to the attached oxygen leaving sulfur with zero assigned electrons. As the neutral S atom has 6 electrons, we can conclude that it has lost all six electrons and now has an oxidation state of +6. Does that help?
You missed the 2! I was talking abt H2S2O4 nd not H2SO4. You wud probably get +5 for S in H2S2O4 but the answer given in my book is +3 . I tried the same for methanal nd got +2 instead of zero which supposedly is the right ans. The method works for h2s2o6 though.
+Andy M Oops! Sorry about that. The Lewis dot structure for H2S2O4 would have each sulfur bonded to two oxygen atoms and the other sulfur atom and then a lone pair of electrons also on the sulfur atom. The lone pair electrons are not shown on the structure that you would call up with Wikipedia. (sad face) So around the sulfur atom you would assign one electron from the pair it is sharing with the neighboring sulfur plus the two electrons from the sulfur lone pair. That gives a total of 3 assigned electrons. Sulfur normally has six valence electrons and so it is deficient by 3 electrons giving it a +3 oxidation state. For methal, CH2O, oxygen gets assigned all the electrons that are shared with C but because C is more electronegative than H the two shared electrons in each C-H single bond get assigned to the carbon. That's a total of 4 electrons which is the same as the valance of the neutral atom. That gives the carbon atom an oxidation state of 0 as you had indicated it should have. I hope that helps clarify and thanks for asking questions.
This video changed my life! Thank you.
nice explaination to oxidation number! thx!
Very Useful sir
Thanks
This is amazing. Thank you!!!
I don't know how I came through it but it is so good 👍
this was great for me to understand ox rules as well. like why O is almost always -2 in a compound
you make my live easier
Thank you
is converting ADP to ATP considered a oxidation-reduction reaction?
like in this reaction: 1,3 bis phosphate gycerate + ADP -----> 3 phosphate glycerate + ATP
Just what I was looking for
Does the oxidation method work for all compounds? I used the same to find the oxidation state of S in H2S2O4 and H2S2O6 but ended up getting the wrong result.
+Andy M
It should, as long as the initial Lewis structure is correctly drawn and the method applied correctly. Using the Lewis structure method I get +6 and +5 for the oxidation states of S in H2SO4 and H2S2O6 respectively. In the Lewis structure for H2SO4, the four oxygen atoms will be bonded to the central sulfur. Because oxygen is more electronegative that sulfur, all shared electrons are assigned to the attached oxygen leaving sulfur with zero assigned electrons. As the neutral S atom has 6 electrons, we can conclude that it has lost all six electrons and now has an oxidation state of +6. Does that help?
@@TrueBlueChemProf
how about this problem:
2HNO3 +.3H2S ----> 3S+2NO+4H2O
edited: OK it worked
Great explanation!
I finally found an easy method to find oxidation numbers!
Thank you!
Oh...nd ur welcome!
excellent
Only good video I found much better than khan academy
thank you so much! :D
You missed the 2!
I was talking abt H2S2O4 nd not H2SO4.
You wud probably get +5 for S in H2S2O4 but the answer given in my book is +3 .
I tried the same for methanal nd got +2 instead of zero which supposedly is the right ans.
The method works for h2s2o6 though.
+Andy M
Oops! Sorry about that.
The Lewis dot structure for H2S2O4 would have each sulfur bonded to two oxygen atoms and the other sulfur atom and then a lone pair of electrons also on the sulfur atom. The lone pair electrons are not shown on the structure that you would call up with Wikipedia. (sad face) So around the sulfur atom you would assign one electron from the pair it is sharing with the neighboring sulfur plus the two electrons from the sulfur lone pair. That gives a total of 3 assigned electrons. Sulfur normally has six valence electrons and so it is deficient by 3 electrons giving it a +3 oxidation state.
For methal, CH2O, oxygen gets assigned all the electrons that are shared with C but because C is more electronegative than H the two shared electrons in each C-H single bond get assigned to the carbon. That's a total of 4 electrons which is the same as the valance of the neutral atom. That gives the carbon atom an oxidation state of 0 as you had indicated it should have.
I hope that helps clarify and thanks for asking questions.
2:45 oxidation #
Nice lesson , thx!