Question

For the reaction $M^{x+}+Mn{O}_4^{-}\to M{O}_3^{-}+M{n}^{2+},$ if 1 mol of ${\text{MnO}}_4^{-}$ oxidises 1.67 mol of $M^{x+}toM{O}_3^{-}$ , then the value of x in the reaction is

• A. 5
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

${\left(\text{M}\right)}^{\text{x}+}\to {\left(\left[\stackrel{+5}{M}{\left(\text{O}\right)}_3\right]\right)}^{-1}+\left(5-\text{x}\right){\left(\text{e}\right)}^{-}$ (OHR)
${\left[\stackrel{+7}{\mathrm{Mn}}{\text{O}}_4\right]}^{-1}+5\bar{\text{e}}\to {\text{Mn}}^{2+}$ (RHR)
Equivalent of ${\text{MnO}}_4^{-}\equiv$ Equivalents of M^x+
1 mole x 5 = 1.67 mole (5 - x)
$5=\frac{5}{3}\left(5-\text{x}\right)\Rightarrow 3=5-\text{x}$
$\therefore \text{x}=2$ .