# Permanganate ion is reduced to MnO_2, and ammonia is oxidized to nitrate ion. Can you write a balanced chemical equation to represent the redox process?

Apr 22, 2017

This is a good one...............

#### Explanation:

You really have to make a meal out of these redox equations.

Permanganate ion (STRONGLY COLOURED PURPLE) could be reduced to almost colourless $M {n}^{2 +}$....

${\stackrel{+ V I I}{M n {O}_{4}}}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i\right)$

But here it is reduced to $M n {O}_{2}$.

${\stackrel{+ V I I}{M n {O}_{4}}}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow {\stackrel{+ I V}{\text{MnO}}}_{2} + 2 {H}_{2} O$ $\left(i i\right)$

Ammonia is OXIDIZED to nitrate anion.......(there is no associated colour change; ammonia pen and inks very badly, whereas ammonium ion/salt is odourless, but I do not suggest you use this method to differentiate them!)

stackrel(-III)"NH"_3 +3H_2O rarr (stackrel(+V)"NO"_3)^(-) +9H^(+)+ 8e^(-) $\left(i i i\right)$

Now if I have done my sums right; $\left(i\right)$, $\left(i i\right)$, and $\left(i i i\right)$ ARE STOICHIOMETRICALLY BALANCED WITH RESPECT TO MASS AND CHARGE. If they are not, then they cannot be accepted as a representation of chemical reality. I think they are balanced, and so the final redox equation eliminates the electrons, and we take the sum...........$8 \times \left(i\right) + 5 \times \left(i i i\right)$:

$8 M n {O}_{4}^{-} + 5 N {H}_{3} + \cancel{15 {H}_{2} O} + \cancel{64} 19 {H}^{+} \rightarrow 8 M {n}^{2 +} + 5 N {O}_{3}^{-} + \cancel{45 {H}^{+}} + \cancel{32} 17 {H}_{2} O$

And we cancel out common reagents to give (finally!):

$8 M n {O}_{4}^{-} + 5 N {H}_{3} + 19 {H}^{+} \rightarrow 8 M {n}^{2 +} + 5 N {O}_{3}^{-} + 17 {H}_{2} O$

Even despite the whack coefficients this is stoichiometrically balanced AS IS ABSOLUTELY REQUIRED...........

But you specified the reduction of permanganate to $M n {O}_{2}$, so this is ...........$3 \times \left(i\right) + 5 \times \left(i i i\right)$:

$8 M n {O}_{4}^{-} + 3 N {H}_{3} + 5 {H}^{+} \rightarrow 8 M n {O}_{2} \left(s\right) + 3 N {O}_{3}^{-} + 7 {H}_{2} O$

But I note (well eventually I did!) that you specified BASIC conditions; all I have to do is add $5 \times H {O}^{-}$ to each side of the equation............and eliminate the waters..........

$8 M n {O}_{4}^{-} + 3 N {H}_{3} \rightarrow 8 M n {O}_{2} \left(s\right) + 3 N {O}_{3}^{-} + 5 H {O}^{-} + 2 {H}_{2} O$

Manganese metal has a particularly rich redox chemistry. For another example of this redox manifold, see here. See here for redox reactions in general.