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

2 Answers

"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.

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:

www.sciencephoto.com

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.