[Xovan]

Hi I'm having a bit of trouble figureing out a certain problem.  Its in a basic solution.
AsO₄^3-  + NO₂^- -->  AsO₂^- + NO₃^-
The thing is to balance out the equation using the half reaction method.  I know its probably As losing 2e^- but what is gaining electrons?  If nothing then why is there 2 spectator ions?

How would adding H₂O and H⁺ change things?
I figure the change in configuration of NO₂^- and NO₃^- has something to do with it but I don't see what exactly.

edited for objectional title, and new thoughts

[wereworm73]

Actually, the arsenic atom gained 2 electrons.  Its oxidation state went from 5+ to 3+.  The nitrogen, on the other hand, lost 2 electrons, going from 3+ to 5+.

As for H⁺ & H₂O, look at the number of oxygens & total charge on each side of your equation.  It's not quite balanced, though you can balance this if you include the H⁺ & H₂O.

[Xovan]

Thanks for the correction.

Therefor...
AsO₄ + 4H⁺ + 2e^- --> AsO₂ + 2H₂O is reduced and As is the Oxidizing Agent
NO₂ + H₂O --> NO₃ + 2H⁺ + 2e^- is oxidized and N is the Reducing Agent
AsO₄+ 4H⁺ + 2e^- + NO₂ + H₂O --> AsO₂ + 2H₂O + NO₃ + 2H⁺ + 2e^-

right?

[wereworm73]

   You got it.

Though you just need to cancel out some things and leave the charges on.  So you can write...

AsO₄^3- + 4H⁺ + 2e^- + NO₂^- + H₂O --> AsO₂^- + 2H₂O + NO₃^- + 2H⁺ + 2e^-

...as this:

AsO₄^3- + 2H⁺ +  NO₂^-  --> AsO₂^- + H₂O + NO₃^-

[Xovan]

Thanks for the help.
Though I didn't write the charges because I was tired of writing sup /sup and sub /sub.


Thanks again for the help.

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method