Write the ionic equations for the oxidising action of potassium permanganate for its reaction with I- (iodide) in both acidic and alkaline solutions.
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Solution and Explanation
Answer:
In acidic solution (iodide is oxidised to iodine):
2 MnO4- + 16 H+ + 10 I- → 2 Mn2+ + 8 H2O + 5 I2
Here each Mn gains 5 electrons (purple MnO4- becomes colourless Mn2+).
In alkaline solution (iodide is oxidised to iodate):
2 MnO4- + H2O + I- → 2 MnO2 + 2 OH- + IO3-
Here MnO4- is reduced to brown MnO2 (Mn goes from +7 to +4), and iodide is oxidised to iodate IO3-.
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- Write the ionic equations for the oxidizing action of $\mathrm{MnO_4^-}$ in acidic medium for :
(I) $\mathrm{Fe^{2+}}$ ion
(II) $\mathrm{C_2O_4^{2-}}$ ion - A black-brown coloured solid (A), when fused with KOH in the presence of air, produces a dark green compound (B), which on electrolytic oxidation in alkaline medium gives a dark purple compound (C). Identify (A), (B) and (C) and write the reactions involved.
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