Question

For the reaction
$M^{x^{+}}+Mn{O}_4^{-}⟶M{O}_3^{-}+M{n}^{2+}+\frac{1}{2}{O}_2$. If one mole of $Mn{O}_4^{-}$oxidises $1^{\ast }67$ moles of $M^{x^{+}}$to $M{O}_3^{-}$, then the value of $x$ in the reaction is

• A. 5
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

$\stackrel{\text{ +7}}{\mathrm{MnO}}{O}_4^{-}+5e^{-}\to M{n}^{2+}$
Since 1 mole of $Mn{O}_4^{-}$ accepts $5$ moles of electrons, therefore, $5$ moles of electrons are lost by $1.67$ moles of $M^{x+}$
$\therefore$ I Mole of $M^{x+}$ will lose electrons
$=5/1-67=3$ moles (approx.)
Since $M^{x+}$ changes to $M{O}_3^{-}$
(where O.N. of $M=+5)$ by accepting $3$ electrons
$\therefore x=+5—3=+2$.