[lazyndnkid]

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method.

c) As2O3 + NO3-  yields  H3AsO4 +  NO

my attempt:
  (5H20 + As2O3   yields  2H3AsO4  +  4H+  +  2e-) 3
  (3e-  +  4H+ + NO3-   yields  NO +  2H2O) 2
+____________________________________________
11 H2O +  3As2O3  + 2NO3-   yields   6H3AsO4 +  4H+  +  2NO

[Borek]

Check if the charge is balanced in half-reactions.

[lazyndnkid]

the thing is, I do not know what to look at to determine the charges and the # of e- needed to added.
can you help?

[lazyndnkid]

ok nevermind. I googled and now understand the whole concept. Thank you Borek for pointing out the mistake.
So now do I just delete this topic or leave it here?

[Alpha-Omega]

The skeletal half-reactions are:

As2O3  ----> H3AsO4
NO3¯  ---->N2O3

Balance all non-H and non-O atoms:
As2O3  ----> 2 H3AsO4
2 NO3¯ ---->  N2O3

Balance the O atoms by placing water molecules:
As2O3  +  5 H2O  ----> 2 H3AsO4
2 NO3¯  ----> N2O3 + 3 H2O

Balance the H atoms by placing H+ ions:
As2O3  +  5 H2O  ---->2 H3AsO4  + 4 H+
2 NO3¯ + 6 H+ ----> N2O3 + 3 H2O

Balance the charge by adding e¯’s:
As2O3  +  5 H2O  ----> 2 H3AsO4  + 4 H+  +  4 e¯
2 NO3¯ + 6 H+  + 4 e¯ ----> N2O3 + 3 H2O

[Borek]

ok nevermind. I googled and now understand the whole concept. Thank you Borek for pointing out the mistake.
So now do I just delete this topic or leave it here?

Leave it as it is, someone may learn from your mistake

See http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc for more on balancing and stoichiometry calculations.