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I saw this experiment many times but still can't explain why.

The experiment is first prepare two solutions. The first one is to dissolve potassium permanganate crystal into distilled water to form potassium permanganate (VII) while the second one is to put sodium hydroxide and glucose (sugar) into distilled water.

Secondly, I pour the second solution into the first one. My observation is the purple solution will slowly turn into blue, then slowly turn into green and lastly turn into yellowish orange.

So, what really happen in the solution?

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This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is probably due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.

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  • $\begingroup$ Alternatively the blue colour might be Manganate(V) ( en.wikipedia.org/wiki/… ), which disproportionates to Maganate(VI) and MnO2, and then as in the answer $\endgroup$ – Ian Bush Jun 28 '18 at 6:38

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