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Form 4 Chemistry lessons on electrochemistry

How Oxidation by potassium Manganate (VII) (KMnO4) take place

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Answer Text:
Oxidation by potassium Manganate (VII) ( $KMnO_4$ )
Procedure:
- Purple Potassium manganate (VII) is added into a solution containing iron (II) ions in a test tube.
- A few drops of concentrated sulphuric (VI) acid are added.
Observations:
- The purple solution (containing Manganate (VII) ions) turns to colourless (manganate (II) ions) i.e. the purple solution is decolourised;
Explanation:
- The Manganate (VII) ions which give the solution a purple colour are reduced to Manganese (II) ions which appear colourless.
This is a redox reaction.

Note:
-oxidation number of Manganate ions in $KMnO_4$ decreases from 7 to 2 (in $Mn^(2+)$ ) hence reduction; while oxidation number of iron increases from 2 (in $Fe^(2+)$ ) to 3 (in $Fe^(3+)$ ); hence oxidation;
-The presence of $Fe^(3+)$ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of $Fe(OH)_3$ ;