Hey guys! Stuck on some Redox Reactions. I tried my best to put as much work as I could, but I'm super stuck on these ones. Thanks for any help:

1) Balance the equation as written, what is the coefficient for O2?
NH4+ + O2-->NO2− + H2O (in basic solution)
Maybe answer:
I tried treating it as an acid so I know I have to change the H+ ions at the end but I don't remember how to do that.
2(NH4+2H20->NO2-+8H+6e-)
3(4e-+O2+4H+->2H2O)
2NH4++4H2O+3O2+12H+->6H2O+2NO2-+16H+
(What do I do from here? Sorry!)

2) Balance the equation as written, what is the coefficient for Cu2+?
Cu + HNO3-->Cu2+ + NO
Maybe work:
All I know is
Cu->Cu2++2e-
I can't figure out HNO3--> Sorry!

3)Balance the equation as written, what is the coefficient for O2?
CH3CH2OH + O2-->CH3CO2H + H2O (in acidic solution)
Maybe work:
I'm confused by this one, I know it is in acid so I can use the symbolic equation 2H++"O"->H2O somewhere, but I'm super lost! Sorry!

4) A 1.5000 g sample of impure iron (II) ammonium sulfate was found to require 27.03 ml of 0.01876 M potassium permanganate (in excess acid) to reach the end point. What is the percentage of Fe in the sample?
Maybe answer:
Fe(NH4)2(SO4)2 and KMnO4
Fe2+->Fe3++e-
MnO4-+8H++5e- -->Mn2+4H2O
Balanced equation for this reaction:
5Fe2++MnO4-+8H+->5Fe3++Mn2++4H20
Moles MnO4-=.02703l*.01876M=5.07*10^-4
Moles Fe2+=5*5.07*10^-4=.002535414 mol
Mass of Fe2+=.0025mols*55.85g/mol=.1416g
Percent of Fe=.1416/1.5g*100=9.440% (my answer)

2 answers

http://www.webqc.org/balance.php?reaction=Cu+%2B+HNO3%28dilute%29+%3D+Cu%28NO3%292+%2B+H2O+%2B+NO
Thank you. I got 2 and 3, but I still don't understand how to balance the first one. Can you offer any help with that? Also, if you could, could you check my last problem? I'm sorry, and thank you!