[granite350]

its a solid so the ON would be zero

[Albert]

2 H3AsO4 + 4 e- + 4 H+ -> As2O3 + 5 H2O

[Borek]

its a solid so the ON would be zero

State of aggregation doesn't matter. Do what I asked you to do.

[granite350]

albert whe i worked out H3AsO4 i get 5 electrons because O is -2 times 4 which is 8 and H is -1 times 3 which is -3 so i get 5 e-

[granite350]

for the reduction par is this correct 2NO3- + 2H + 2 e- -->N2O3 + 2H2O

[Albert]

Hydrogen is +1!

[granite350]

8-3 = 5 just wondering because if it is 4 e it would make the simplification much easier

[Albert]

for the reduction par is this correct 2NO3- + 2H + 2 e- -->N2O3 + 2H2O

Why only 2 electrons? How many NO₃^-s can you see?

[granite350]

I mean for H3AsO4=5 electrons

[granite350]

NO3- i see +5 O=-6  central atom is 5

wonder what i'm doing wrong???

the final answer i has was As2O3 + 5H2O+4NOH3-  + 2 H  + 4e-  --> 2 H3AsO4 +4 H +4e- +2N2O3+4H20

[Albert]

NO3- i see +5 O=-6-1=7

 

NO3-  oxygen = (-2)*3 = -6
         nitrogen = (-6) - (-1) = +5

[granite350]

but to balance out the reduction part i need +2 electrons

wow i am confused

[Albert]

but to balance out the reduction part i need +2 electrons

Who said that?!

You need to balance the half reaction (reduction). The charges aren't balanced and remember you have 2 NO3- groups: two electrons are not enough...

[Borek]

wow i am confused

You are confusing yourself IMHO. Try to follow simple steps, don't try to do everything at the same time.

[granite350]

but N2O3 is +3

this is why i keep thinking i have to add 2 electrons to the other side.

this is where i get confused.