What will 1 mol of MnO_4^{2-} produce upon d | vedclass

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Question

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in neutral or acidic medium is given by: $3MnO_4^{2-} + 2H_2O 	o MnO_2 + 2MnO_4^-

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in neutral or acidic medium is given by: $3MnO_4^{2-} + 2H_2O 	o MnO_2 + 2MnO_4^- + 4OH^-$. To find the products for $1 \ mol$ of $MnO_4^{2-}$,divide the entire equation by $3$: $MnO_4^{2-} + \frac{2}{3}H_2O 	o \frac{1}{3}MnO_2 + \frac{2}{3}MnO_4^- + \frac{4}{3}OH^-$. Thus,$1 \ mol$ of $MnO_4^{2-}$ produces $\frac{2}{3} \ mol$ of $MnO_4^-$ and $\frac{1}{3} \ mol$ of $MnO_2$.

What will $1 \ mol$ of $MnO_4^{2-}$ produce upon disproportionation in a neutral aqueous medium?

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