By passing H₂S gas in acidified KMnO₄ solution, we get

The reaction of H₂S with KMnO4 is a redox reaction in which both reduction and oxidation processes are taken place.

Reduction of Mn (VII) to Mn (II) is:

$\stackrel{+7}{\mathrm{Mn}}{O}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O} ....\left(1\right)$

Oxidation of S from -2 to 0 is:

${\mathrm{H}}_2\stackrel{-2}{\mathrm{S}}\to 2{\mathrm{H}}^{+}+\stackrel{0}{\mathrm{S}}+2{\mathrm{e}}^{-} ....\left(2\right)$

The final reaction is obtained by multiplying reaction 1 by 2 and reaction 2 by 5 and then adding both the reactions.

$2{\mathrm{MnO}}_4^{-}+5{\mathrm{H}}_2\mathrm{S}+3{\mathrm{H}}^{+}⟶2{\mathrm{Mn}}^{2+}+5\mathrm{S}+8{\mathrm{H}}_2\mathrm{O}$