Question

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

Answer 1

`"Potassium manganate"`, `"K"_2"MnO"_4`, is a green salt, produced from `"MnO"_2` under basic conditions..........

Explanation:

`"MnO"_2` is oxidized to `"MnO"_4^(2-)`:

`stackrel(+IV)"Mn""O"_2 +2"H"_2"O" rarr stackrel(+VI)"Mn""O"_4^(2-) +4"H"^(+) +2e^(-)` `(i)`

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

`1/2"O"_2 +"H"_2"O" + 2e^(-) rarr "2HO"^(-)` `(ii)`

And `(i) + (ii):`

`stackrel(+IV)"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 `2xx"HO"^-` to EACH SIDE.

`stackrel(+IV)"MnO"_2 +1/2"O"_2 +"H"_2"O" +2"HO"^(-) rarr stackrel(+VI)"MnO"_4^(2-) +2"H"^(+)+2"HO"^(-)`

to give.........

`stackrel(+IV)"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.

Answer 2

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

Explanation:

We can use standard electrode potentials ( `sf(E^@)`) to set this up. In alkaline conditions we have:

`sf(" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "E^@" "(V))`

`sf(" "" "" "" "MnO_4^(-)+erightleftharpoonsMnO_4^(2-)" "" "" "" "color(white)(.)+0.56)`

`sf(2H_2O+MnO_4^(2-)+2erightleftharpoonsMnO_2+4OH^-" "" "+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 `sf(MnO_4^-)` will not be able to oxidise `sf(MnO_2)` to `sf(MnO_4^(2-)` because the `sf(E^@)` 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 `sf(+0.59>+0.56)` the values are very close. If we raise the concentration of `sf(OH^-)` 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:

`sf(MnO_2+4OH^(-)rarr2H_2O+MnO_4^(2-)+2e" "" "" "color(red)((1)))`

`sf(" "" "MnO_4^(-)+erarrMnO_4^(2-)" "" "" "" "" "" "" "color(white)(..)color(red)((2)))`

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

`sf(MnO_2+4OH^(-)+2MnO_4^(-)+cancel(2e)rarr2H_2O+3MnO_4^(2-)+cancel(2e))`

To do this take 10 ml of `sf(0.01color(white)(x)M)` `sf(KMnO_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 `sf(MnO_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) `rarr` 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.