When potassium manganate disproportionates in a low ph it produces potassium permanganate and manganese dioxide. But would the produces separate to form a layer or something and if so is there a simple way to separate them? And how much of each produce would you obtain if say we had 50g of the manganate.
My guess is that the maganese dioxide would be floating on top of the potassium permanganate. Would this be right?
Colin
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When potassium manganate disproportionates in a low pH [ph is not the same thing] it produces potassium permanganate and manganese dioxide.
You get potassium permanganate solution [mauve] and brown-black insoluble manganese dioxide. This will form a precipitate if allowed to settle. So to separate them you need to filter .. but CARE .. you can't use ordinary filter paper as it you pour the mixture through, KMnO4 is such a strong oxidising agent it reacts with the filter paper turning it brown forming more MnO2 and you'll lose your KMnO4!
So how do you solve this dilemma? The simplest, to get samples of each, it to let the precipitate settle and either decant off the mauve supernatent solution or remove it using a dropper [teat pipette].
Alternatively you can filter it through a bed of glass wool.
Anonymous
I do not think oxidation number is a factor influencing oxidizing power. Oxidation power depends on the ion's ability to accept electrons (in the oxidizing agent). Anyway, Potassium Permanganate is a more powerful oxidizing agent than potassium dichromate as Manganate (VII) has a higher standard electrode potential value (+1.52V) than Dichromate (VI) (at +1.33V), both in an acidic medium.