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

2 Answers

Answer:

#"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:

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:

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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.