Disproportionation products of one mole of Mn | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $Cl_2$ is passed through hot aqueous $NaOH$, the products formed have $Cl$ in different oxidation states. These are indicated as

In the following reaction,assign the oxidation state for the underlined atom in the products of complete hydrolysis at $R.T.$:
$H_2\underline{S}_2O_6 + H_2O \longrightarrow H_2SO_3 + H_2SO_4$

$2CuI \to Cu + CuI_2$,the reaction is

Which of the following is a disproportionation redox reaction?

When $H_2S$ is passed through $Hg_2^{2+}$ solution,the products obtained are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module