Question

For the reaction $M^{x +} + MnO_{4}^{-} \rightarrow MO_{3}^{-} + Mn^{2 +},$ if 1 mol of $\text{MnO}_{4}^{-}$ oxidises 1.67 mol of $M^{x +}toMO_{3}^{-}$ , then the value of x in the reaction is

• A. 5
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

$\left(\text{M}\right)^{\text{x} +} \rightarrow \left(\left[\overset{+ 5}{\text{M}} \left(\text{O}\right)_{3}\right]\right)^{- 1} + \left(5 - \text{x}\right) \, \, \left(\text{e}\right)^{-}$ (OHR)
$\left[\overset{+ 7}{\text{Mn}} \text{O}_{4}\right]^{-1} + 5 \bar{\text{e}} \rightarrow \text{Mn}^{2 +}$ (RHR)
Equivalent of $\text{MnO}_{4}^{-} \equiv $ Equivalents of M^x+
1 mole x 5 = 1.67 mole (5 - x)
$5 = \frac{5}{3} \left(5 - \text{x}\right) \, \, \, ⇒ \, \, \, 3 = 5 - \text{x}$
$∴ \, \text{x} = 2$ .