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

Oct 23, 2017

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

$H O N O \left(a q\right) + M n {O}_{2} \left(s\right) + {H}^{+} \rightarrow M {n}^{2 +} + {H}_{2} O \left(l\right) + N {O}_{3}^{-}$

Explanation:

How do we get to this?

Well, $\text{manganese(IV) oxide}$ is reduced to $M n \left(I I\right)$:

$M n {O}_{2} \left(s\right) + 4 {H}^{+} + 2 {e}^{-} \rightarrow M {n}^{2 +} + 2 {H}_{2} O \left(l\right)$ $\left(i\right)$

And, $\text{nitrous acid}$ is oxidized to $\text{nitrate ion}$:

$H O N O \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow N {O}_{3}^{-} + 3 {H}^{+} + 2 {e}^{-}$ $\left(i i\right)$

And we just add $\left(i\right) + \left(i i\right)$ together......

$H O N O \left(a q\right) + \cancel{{H}_{2} O \left(l\right)} + M n {O}_{2} \left(s\right) + \cancel{4} {H}^{+} + \cancel{2 {e}^{-}} \rightarrow M {n}^{2 +} + \cancel{2} {H}_{2} O \left(l\right) + N {O}_{3}^{-} + \cancel{3 {H}^{+} + 2 {e}^{-}}$

To give finally.....

$H O N O \left(a q\right) + M n {O}_{2} \left(s\right) + {H}^{+} \rightarrow M {n}^{2 +} + {H}_{2} O \left(l\right) + N {O}_{3}^{-}$