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$\mathrm{KMnO}_4$ acts as an oxidising agent in acidic medium. The number of moles of $\mathrm{KMnO}_4$ that will be needed to react with one mole of sulphide ions in acidic solution is
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The correct answer is:
$\frac{2}{5}$
$\frac{2}{5}$
Reaction of $\mathrm{KMnO}_4$ with sulphide ions in acidic medium is as follows :
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$
$$
\frac{\left[\mathrm{H}_2 \mathrm{~S}+[\mathrm{O}] \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{S}\right] \times 5}{2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{~S}}
$$
5 moles of $S^{2-}$ ions react with 2 moles of $\mathrm{KMnO}_4$ thus, 1 mole of $\mathrm{S}^{2-}$ ion will react with $\frac{2}{5}$ moles of $\mathrm{KMnO}_4$.
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$
$$
\frac{\left[\mathrm{H}_2 \mathrm{~S}+[\mathrm{O}] \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{S}\right] \times 5}{2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{~S}}
$$
5 moles of $S^{2-}$ ions react with 2 moles of $\mathrm{KMnO}_4$ thus, 1 mole of $\mathrm{S}^{2-}$ ion will react with $\frac{2}{5}$ moles of $\mathrm{KMnO}_4$.
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