Question

Describe the method of preparation of potassium permanganate. How does acidic potassium permanganate react with the following? Give ionic equations: (i) H$_2$S (ii) Fe (iii) iodide ion.

Hint:

In acidic medium, \(\mathrm{MnO_4^-}\) is a very strong oxidising agent and is always reduced to \(\mathrm{Mn^{2+}}\). Balance such redox equations by combining the standard permanganate reduction half-reaction with the appropriate oxidation half-reaction of the substrate.

Answer 1

Preparation of Potassium Permanganate (KMnO$_4$):
Step 1 (Oxidative fusion): Natural MnO$_2$ (pyrolusite) is fused with KOH in air / with an oxidiser (e.g., KNO$_3$) to form potassium manganate.
\[ \boxed{2\,\mathrm{MnO_2}+4\,\mathrm{KOH}+ \mathrm{O_2}\;\longrightarrow\;2\,\mathrm{K_2MnO_4}+2\,\mathrm{H_2O}} \]
Step 2 (Conversion to permanganate): The green manganate (\(\mathrm{MnO_4^{2-}}\)) is oxidised to the purple permanganate (\(\mathrm{MnO_4^-}\)) by chlorine/ozone or by electrolysis. One convenient route is:
\[ \boxed{2\,\mathrm{K_2MnO_4}+ \mathrm{Cl_2}\;\longrightarrow\;2\,\mathrm{KMnO_4}+2\,\mathrm{KCl}} \] (Alternatively, \(\mathrm{MnO_4^{2-}}\) disproportionates in neutral/acidic medium: \(3\,\mathrm{MnO_4^{2-}}+2\,\mathrm{H_2O}\to2\,\mathrm{MnO_4^-}+\mathrm{MnO_2}+4\,\mathrm{OH^-}\).)
Reactions of acidic KMnO$_4$ (strong oxidant; Mn(VII) $\to$ Mn$^{2+$):}
Overall reduction half‐equation in acid: \(\mathrm{MnO_4^-+8\,H^++5\,e^- \to Mn^{2+}+4\,H_2O}\).
(i) With H$_2$S (to sulfur):
\[ \boxed{2\,\mathrm{MnO_4^-}+6\,\mathrm{H^+}+5\,\mathrm{H_2S}\;\longrightarrow\;2\,\mathrm{Mn^{2+}}+8\,\mathrm{H_2O}+5\,\mathrm{S}\downarrow} \] (ii) With Fe$^{2+$ (to Fe$^{3+}$):}
\[ \boxed{\mathrm{MnO_4^-}+5\,\mathrm{Fe^{2+}}+8\,\mathrm{H^+}\;\longrightarrow\;\mathrm{Mn^{2+}}+5\,\mathrm{Fe^{3+}}+4\,\mathrm{H_2O}} \] (iii) With iodide ion, I$^-$ (to iodine):
\[ \boxed{2\,\mathrm{MnO_4^-}+10\,\mathrm{I^-}+16\,\mathrm{H^+}\;\longrightarrow\;2\,\mathrm{Mn^{2+}}+5\,\mathrm{I_2}+8\,\mathrm{H_2O}} \]