How can #"MnO"_4^(2-)(aq)# disproportionate to form #"MnO"_4^(-)(aq)# and #"MnO"_2(s)#?
Dioxygen gas is reduced to hydroxide...........
But manganate ion is normally produced under BASIC conditions, and so we add
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.
We can use standard electrode potentials (
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
However, these values refer to standard conditions. Chemists are able to alter reaction conditions to drive the reaction in the direction they want.
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:
To get the electrons to balance we multiply
To do this take 10 ml of
Add a little manganese(IV) oxide (
Filter the solution. You should see green
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
The net effect is Mn(VI)
When a species is simultaneously oxidised and reduced like this it is termed "disproportionation".
This all illustrates why many redox reactions are pH sensitive.