Question

How do we represent the reduction of nitrate ion to `NO`, and the corresponding oxidation of `"arsenic sulfide"` to `"arsenate ion"`, `AsO_4^(3-)`?

Answer 1

`4NO_3^(-) + 3As_2S_3 +16H_2O rarr 4NO + 6AsO_4^(3-) + 9S^(2-)+32H^+`

Explanation:

Well `"nitrate"` `N(V+)` is reduced to `NO` `N(+II)`....

`NO_3^(-) +4H^(+) + 3e^(-) rarr NO + 2H_2O` `(i)`

And `"arsenide (+III)"` is OXIDIZED to `"arsenate (+V)"`....

`As_2S_3 +8H_2O rarr 2AsO_4^(3-) + 3S^(2-)+ 16H^+ +4e^(-)` `(ii)`

Mass and charge for balanced for `(i)` and `(ii)` which means that the reactions are so far kosher....and we adds `(i)` and `(ii)` in such a way that the electrons are removed from the final equation....

We take `4xx(i)+3xx(ii)`....

`4NO_3^(-) +cancel(16H^(+)) + 3As_2S_3 +cancel(24)16H_2O+ cancel(12e^(-)) rarr 4NO + cancel(8H_2O)+6AsO_4^(3-) + 9S^(2-)+cancel (48)32H^+ +cancel(12e^(-))`

To give finally....

`4NO_3^(-) + 3As_2S_3 +16H_2O rarr 4NO + 6AsO_4^(3-) + 9S^(2-)+32H^+`

The which I think is balanced with respect to mass and charge.....but check my arifmetik and my spilling....

Nitrate ion is in fact quite a potent oxidant, but often in aqueous solution, its oxidizing potential is masked......