# How can "MnO"_4^(2-)(aq) disproportionate to form "MnO"_4^(-)(aq) and "MnO"_2(s)?

Apr 7, 2017

$\text{Potassium manganate}$, ${\text{K"_2"MnO}}_{4}$, is a green salt, produced from ${\text{MnO}}_{2}$ under basic conditions..........

#### Explanation:

${\text{MnO}}_{2}$ is oxidized to ${\text{MnO}}_{4}^{2 -}$:

${\stackrel{+ I V}{\text{Mn""O"_2 +2"H"_2"O" rarr stackrel(+VI)"Mn""O"_4^(2-) +4"H}}}^{+} + 2 {e}^{-}$ $\left(i\right)$

Dioxygen gas is reduced to hydroxide...........

$\frac{1}{2} {\text{O"_2 +"H"_2"O" + 2e^(-) rarr "2HO}}^{-}$ $\left(i i\right)$

And $\left(i\right) + \left(i i\right) :$

${\stackrel{+ I V}{\text{MnO"_2 +1/2"O"_2 +"H"_2"O" rarr stackrel(VI+)"MnO"_4^(2-) +2"H}}}^{+}$

But manganate ion is normally produced under BASIC conditions, and so we add $2 \times {\text{HO}}^{-}$ to EACH SIDE.

${\stackrel{+ I V}{\text{MnO"_2 +1/2"O"_2 +"H"_2"O" +2"HO"^(-) rarr stackrel(+VI)"MnO"_4^(2-) +2"H"^(+)+2"HO}}}^{-}$

to give.........

$\stackrel{+ I V}{\text{MnO"_2 +1/2"O"_2 +2"HO"^(-) rarr stackrel(+VI)"MnO"_4^(2-) +"H"_2"O}}$

Is this balanced with respect to mass and charge? All care taken, but no responsibility admitted.

Apr 9, 2017

You can make Mn(VI) from Mn(VII) and Mn(IV) in alkaline conditions.

#### Explanation:

We can use standard electrode potentials ( $\textsf{{E}^{\circ}}$) to set this up. In alkaline conditions we have:

$\textsf{\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "E^@" } \left(V\right)}$

$\textsf{\text{ "" "" "" "MnO_4^(-)+erightleftharpoonsMnO_4^(2-)" "" "" "" } \textcolor{w h i t e}{.} + 0.56}$

$\textsf{2 {H}_{2} O + M n {O}_{4}^{2 -} + 2 e r i g h t \le f t h a r p \infty n s M n {O}_{2} + 4 O {H}^{-} \text{ "" } + 0.59}$

When listed -ve to +ve like this a general rule to predict the outcome of a redox reaction is to say "bottom left will oxidise top right."

On this basis we can say that $\textsf{M n {O}_{4}^{-}}$ will not be able to oxidise $\textsf{M n {O}_{2}}$ to sf(MnO_4^(2-) because the $\textsf{{E}^{\circ}}$ value is not +ve enough.

However, these values refer to standard conditions. Chemists are able to alter reaction conditions to drive the reaction in the direction they want.

Although $\textsf{+ 0.59 > + 0.56}$ the values are very close. If we raise the concentration of $\textsf{O {H}^{-}}$ Le Chatelier's Principle tells us that the 2nd 1/2 reaction should be driven to the left.

This will push out more electrons which will reduce the E value which makes the reaction thermodynamically feasible.

The two 1/2 reactions now become:

$\textsf{M n {O}_{2} + 4 O {H}^{-} \rightarrow 2 {H}_{2} O + M n {O}_{4}^{2 -} + 2 e \text{ "" "" } \textcolor{red}{\left(1\right)}}$

$\textsf{\text{ "" "MnO_4^(-)+erarrMnO_4^(2-)" "" "" "" "" "" "" } \textcolor{w h i t e}{. .} \textcolor{red}{\left(2\right)}}$

To get the electrons to balance we multiply sf(color(red)((2)) by 2 then add to sf(color(red)((1))rArr

$\textsf{M n {O}_{2} + 4 O {H}^{-} + 2 M n {O}_{4}^{-} + \cancel{2 e} \rightarrow 2 {H}_{2} O + 3 M n {O}_{4}^{2 -} + \cancel{2 e}}$

To do this take 10 ml of $\textsf{0.01 \textcolor{w h i t e}{x} M}$ $\textsf{K M n {O}_{4}^{-}}$ solution and 5 ml of 1 M NaOH solution.

Add a little manganese(IV) oxide (sf(MnO_2)) and shake for a few minutes.

Filter the solution. You should see green $\textsf{M n {O}_{4}^{2 -}}$ that looks like this:

This ion is only stable at high pH conditions. If you try adding a solution of dilute sulfuric acid then the process is reversed.

You see the purple color of sf(MnO_4^(-) return.

The net effect is Mn(VI) $\rightarrow$ Mn(VII) + Mn(IV)

When a species is simultaneously oxidised and reduced like this it is termed "disproportionation".

This all illustrates why many redox reactions are pH sensitive.