Question

Use the half-reaction method to balance the following redox reaction in acidic solution: As2O3(s) + NO3(aq)- ---> H3AsO4(aq) + N2O3(aq) How do you balance a redox reaction in acidic conditions?

Answer 1

Well you gots nitrate reduced to `N_2O_3`....

Explanation:

`2stackrel(+V)NO_3^(-) +6H^+ +4e^(-) rarr (O=)N-stackrel(+)N(=O)(-O)^(-)+3H_2O(l)`

...i.e. we gots `N(+III)` in `N_2O_3`....(I think that `N_2O_3` has a `N-N` bond not a bridging `N-O-N`...if anyone knows otherwise please correct me. The formal oxidation states are `N(+II)` and `N(+IV)`, i.e. `N(+III)` average.....

And meanwhile arsenic(III) oxide is oxidized to arsenate....

`As_2O_3(s) +5H_2O(l) rarr 2AsO_4^(3-)+10H^+ + 4e^(-)`

And so we adds the former to the letter to remove the electrons as virtual particles....

`As_2O_3(s) +cancel(5)2H_2O(l) +2NO_3^(-) +cancel(6H^+) +cancel(4e^(-))rarr 2AsO_4^(3-)+cancel(10)4H^+ + cancel(4e^(-))+(O=)N-stackrel(+)N(=O)(-O)^(-)+cancel(3H_2O(l))`

...to give...

`As_2O_3(s) +2H_2O(l) +2NO_3^(-) rarr 2AsO_4^(3-)+4H^+ +(O=)N-stackrel(+)N(=O)(-O)^(-)`

Please don't assume that I have done my sums right.....And note that when I do a redox equation I ALWAYS assume acidic conditions. If basic conditions are required I would simply add `4xxHO^-` to both sides.....