Q.Permanganate(VII) ion, MnO4–, in basic medium, oxidises iodide ion (I–) to produce molecular iodine (I2) and manganese dioxide (MnO2). Write a balanced ionic equation to represent this reaction.
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Start your 14-day free trial to unlock the full solution →In basic solution, permanganate () oxidises iodide () to iodine () and is itself reduced to manganese dioxide (). Using the half-reaction method, the balanced ionic equation is
The Concept: Permanganate in Basic Medium
Permanganate is a powerful oxidising agent, but its reduction product depends critically on the pH of the solution. In acidic medium, it goes all the way down to (colourless). In basic medium, the reduction stops at , a dark brown solid, because is unstable in base and would immediately precipitate and oxidise further.
A common mistake is to assume permanganate always reduces to . In basic solution, the product is , not — the colour change is from purple () to brown (), not to colourless.
Step-by-Step: The Half-Reaction (Ion-Electron) Method
1. Write the skeletal ionic equation.
2. Split into the two half-reactions.
Oxidation half:
Reduction half:
3. Balance atoms other than O and H.
Oxidation half needs 2 iodide ions to give 1 :
Reduction half already has 1 Mn on each side.
4. Balance O and H — first as if in acidic medium, then convert to basic.
For the reduction half, balance O by adding to the right:
Balance H by adding to the left:
Convert to basic medium: add to both sides, combine into on the left, then cancel common to both sides:
5. Balance charge by adding electrons.
Oxidation half: left charge , right charge — add to the right:
Reduction half: left charge , right charge — add to the left: …
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