Question

What is the ionic form of the following unbalanced equation `MnO_2 + HNO_2 -> Mn(NO_3)_2 + H_2O`?

Answer 1

You gots a redox reaction of the form....

`HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)`

Explanation:

How do we get to this?

Well, `"manganese(IV) oxide"` is reduced to `Mn(II)`:

`MnO_2(s) +4H^(+) + 2e^(-) rarr Mn^(2+) +2H_2O(l)` `(i)`

And, `"nitrous acid"` is oxidized to `"nitrate ion"`:

`HONO(aq) +H_2O(l) rarr NO_3^(-)+ 3H^+ + 2e^(-)` `(ii)`

And we just add `(i) + (ii)` together......

`HONO(aq) +cancel(H_2O(l)) +MnO_2(s) +cancel4H^(+) + cancel(2e^(-))rarr Mn^(2+) +cancel2H_2O(l)+NO_3^(-)+ cancel(3H^+ + 2e^(-))`

To give finally.....

`HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)`