KMnO_4 acts as an oxidising agent in acidic m | vedclass

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(D) In acidic medium,$KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$. The half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2

Vedclass · Accepted Answer

$2$

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(D) In acidic medium,$KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$. The half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Sulphide ion $(S^{2-})$ is oxidised to elemental sulphur $(S)$: $S^{2-} \rightarrow S + 2e^-$. To balance the electrons,we multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$: $2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$ $5S^{2-} \rightarrow 5S + 10e^-$ Adding these,the balanced equation is: $2MnO_4^- + 5S^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O$. From the stoichiometry,$5$ moles of $S^{2-}$ react with $2$ moles of $KMnO_4$. Therefore,$1$ mole of $S^{2-}$ reacts with $2/5$ moles of $KMnO_4$.

$KMnO_4$ acts as an oxidising agent in acidic medium. The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphide ion $(S^{2-})$ in acidic solution is . . . . . . (in $/5$)?

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