When neutral or faintly alkaline $KMnO_4$ is treated with potassium iodide, iodide ion is converted into 'X'.

Question

When neutral or faintly alkaline $KMnO_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is:

Accepted Answer

Potassium permanganate ($KMnO_4$) acts as a strong oxidising agent. Its reaction product with iodide depends on the pH of the medium:
1.  **In acidic medium:** Iodide ($I^-$) is oxidised to iodine ($I_2$).
2.  **In neutral or faintly alkaline solution:** Iodide ($I^-$) is oxidised to iodate ($IO_3^-$). The manganese is reduced from +7 ($MnO_4^-$) to +4 ($MnO_2$).

The specific reaction equation provided in the NCERT text is:
$$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$$

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