Q30 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Chemistry

Disproportionation products of one mole of $\mathrm{MnO}_4^{-2}$ in aqueous acidic medium are

  • A
    $\frac{1}{3} \mathrm{~mol}^{\mathrm{a}} \mathrm{MnO}_4^{-}, \frac{2}{3} \mathrm{~mol}$ of $\mathrm{MnO}_2$
  • Correct answer
    B
    $\frac{2}{3} \mathrm{~mol}^2 \mathrm{MnO}_4^{-}, \frac{1}{3} \mathrm{~mol}$ of $\mathrm{MnO}_2$
  • C
    $\frac{1}{3} \mathrm{~mol}_{\text {of }} \mathrm{Mn}_2 \mathrm{O}_7, \frac{1}{3} \mathrm{~mol}$ of $\mathrm{MnO}_2$
  • D
    $\frac{2}{3} \mathrm{~mol}$ of $\mathrm{Mn}_2 \mathrm{O}_7, \frac{1}{3} \mathrm{~mol}$ of $\mathrm{MnO}_2$

Solution

$2 e^{-}+4 \mathrm{H}^{+}+\mathrm{MnO}_4^{2-} \longrightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$ $\left(\mathrm{MnO}_4^{2-} \longrightarrow \mathrm{MnO}_4^{-}+e^{-}\right) \times 2$ Overall reaction, $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{MnO}_2+2 \mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}$ $\because 3 \mathrm{~mol} \mathrm{MnO}_4^{2-}$ gives $1 \mathrm{~mol} \mathrm{MnO}_2$ $\therefore 1 \mathrm{~mol} \mathrm{MnO}{ }_4^{2-} \longrightarrow \frac{1}{3} \mathrm{~mol}$ of $\mathrm{MnO}_2$ Moreover, $3 \mathrm{~mol}$ of $\mathrm{MnO}_4^{2-}$ gives $2 \mathrm{~mol} \mathrm{MnO}_4^{-}$ $\therefore 1 \mathrm{~mol} \mathrm{MnO}_4^{2-} \longrightarrow \frac{2}{3} \mathrm{~mol} \mathrm{MnO}_4^{-}$ Hence, option (b) is correct.

Hence, the correct answer is option: 'B'.

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