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NCERT Exemplar · Q30

Q.Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R): Oxidation state of manganese changes from +2 to +7 during the reaction.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
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Permanganate ions (MnOX4X−\ce{MnO4^-}) do oxidise iodide, but manganese moves from +7 to +2 (reduction, not oxidation), so the Reason inverts the change. Answer: (iii).

Why oxidising agents are themselves reduced

An oxidising agent accepts electrons from another species, causing that species to lose electrons (be oxidised). In accepting electrons, the oxidising agent itself is reduced — its oxidation state falls. Permanganate is a textbook oxidising agent precisely because manganese in the +7 state is electron-hungry and drops to a lower, more stable oxidation state when it grabs electrons.

The question hinges on tracking the oxidation-state change of manganese correctly.


Step-by-step analysis

  1. Identify the oxidation state of manganese in permanganate. In MnOX4X−\ce{MnO4^-}, oxygen is −2-2 each. Let manganese be xx:

x+4(−2)=−1  ⟹  x=+7.x + 4(-2) = -1 \implies x = +7.

Permanganate contains MnXVII\ce{Mn^{VII}}.

  1. What happens when permanganate oxidises iodide? Iodide ions, IX−\ce{I^-}, are oxidised to iodine, IX2\ce{I_2}:

2 IX−→IX2+2 e−.2\,\ce{I^-} \to \ce{I_2} + 2\,e^-.

Those electrons must go somewhere — they are accepted by the permanganate.

  1. Determine the final oxidation state of manganese. In acidic solution, permanganate is reduced to MnX2+\ce{Mn^{2+}}:

MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O.\ce{MnO4^- + 8 H^+ + 5 e^- -> Mn^{2+} + 4 H_2O}.

Manganese moves from +7 to +2.

  1. Check the Assertion.

    Permanganate does act as the oxidising agent: it oxidises iodide to iodine while itself being reduced. Assertion (A) is true.

  2. Check the Reason. …

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