Experiment of KMnO4

[jlnWind]

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Hey guys,
my class did not do the experiment with KMnO4
if anyone could please enlighten me on the steps taken
and then mention the results with colour changes id be greatly appreciative!

thanks in advance i guess

[brenton1987]

This one?
"Choose equipment, perform a first-hand investigation to demonstrate and gather first-hand information about the oxidising strength of KMnO₄"

For this experiment I would gather a set of beakers and fill them with various metal salt solutions. Then I would add permanganate to them and observe.

The permanganate will oxidise the metal cations and in turn it will be reduced.

MnO₄^- + e^- -->MnO₄^2- (0.56V)
MnO₄^- + 3e^- + 4H⁺ --> MnO₂+ H₂O (1.69V)
MnO₄^- + 5e^- + 8H⁺ --> Mn²⁺ + 2H₂O (1.51V)

You can tell which one it is because MnO₄^- is dark purple, MnO₄^2- is green, MnO₂ is a black/brown precipitate and Mn²⁺ is very very pale pink.

Basically if you are getting a lot of precipitate or very pale pink solutions then permanganate is a good oxidant.

[jlnWind]

Wouldn't by any chance know what solutions id use?

[brenton1987]

Iron (II) solution. (Nitrate).

MnO₄^- + 5 Fe²⁺ + 8 H⁺ --> Mn²⁺ + 5 Fe³⁺ + 4 H₂O

MnO₄^- is purple
Fe²⁺ is colourless
Mn²⁺ is colourless
Fe³⁺ is pale yellow

A change in solution colour from purple to pale yellow indicates that the permanganate has been reduced while the iron has been oxidised.

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Iodide solution. (Sodium).

2 MnO₄^- + 10 I^- + 16 H⁺ --> 2 Mn²⁺ + 5 I₂ + 8 H₂O

MnO₄^- is purple
I^- is colourless
Mn²⁺ is colourless
I₂ is pale yellow

A change in solution colour from purple to pale yellow indicates that the permanganate has been reduced while the iodide has been oxidised.

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Chloride Solution. (Sodium).

2 MnO₄^- + 10 Cl^- + 16 H⁺ --> 2 Mn²⁺ + 5 Cl₂ + 8 H₂O

MnO₄^- is purple
Cl^- is colourless
Mn²⁺ is colourless
Cl₂ is a noxious gas

A change in solution colour from purple to colourless indicates that the permanganate has been reduced while the chloride has been oxidised. Gas will also be present.

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Methanoic Acid.

2 MnO₄^- + 6 H⁺ + 5 HCOOH --> 2 Mn²⁺ + 8 H₂O + 5 CO₂

MnO₄^- is purple
HCOOH^- is colourless
Mn²⁺ is colourless
CO₂ is a gas

A change in solution colour from purple to colourless indicates that the permanganate has been reduced while the methanoic acid has been oxidised. Gas will also be present.

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Theres plenty more solutions out there. Just look at the table of reduction potentials to see what else permanganate can oxidise.
(Oxidation of Cu, Cu⁺, H₂C₂O₄, Au⁺, Pb, Mn, Ni, Sn, Sn²⁺, Zn)

[jlnWind]

hey thanks a bunch i really appreciate it