Disproportionation products of one mole of Mn | vedclass

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Question

(B) The disproportionation reaction of $MnO_4^{2-}$ in acidic medium is given by the following half-reactions:$2e^{-} + 4H^{+} + MnO_4^{2-} \longright

Vedclass · Accepted Answer

$\frac{2}{3} \ mol$ of $MnO_4^{-}, \frac{1}{3} \ mol$ of $MnO_2$

Vedclass · Answer

(B) The disproportionation reaction of $MnO_4^{2-}$ in acidic medium is given by the following half-reactions:$2e^{-} + 4H^{+} + MnO_4^{2-} \longrightarrow MnO_2 + 2H_2O$$(MnO_4^{2-} \longrightarrow MnO_4^{-} + e^{-}) 	imes 2$Adding these,the overall balanced equation is:$3MnO_4^{2-} + 4H^{+} \longrightarrow MnO_2 + 2MnO_4^{-} + 2H_2O$From the stoichiometry,$3 \ mol$ of $MnO_4^{2-}$ produces $1 \ mol$ of $MnO_2$ and $2 \ mol$ of $MnO_4^{-}$.Therefore,for $1 \ mol$ of $MnO_4^{2-}$,the products are $\frac{1}{3} \ mol$ of $MnO_2$ and $\frac{2}{3} \ mol$ of $MnO_4^{-}$.Thus,option $(B)$ is correct.

Disproportionation products of one mole of $MnO_4^{2-}$ in aqueous acidic medium are

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