Question

In a titration $H _2 O _2$ is oxidized to $O _2$ by $MnO _4^{-} .24\, mL$ of $0.1 \,M \,H _2 O _2$ of require $16 \,mL$ of $0.1 \,M \,MnO _4^{-}$ solution. Hence, $MnO _4^{-}$changes to

• A. $MnO _4^{2-}$
• B. $MnO _2$
• C. $MnO _4^{2-}$
• D. $Mn _2 O _7$

Answer 1

Answer: (B)

$H _2 O _2 \rightarrow 2 H ^{+}+ O _2+2 e ^{-} 0.1\, M =0.2\, N$
$MnO _{4}^{-} \Rightarrow Mn ^{x+} 0.1 \, M =0.1(7= x ) N$
$N_1 V_2=N_2 V_2$
$0.2 \times 24=0.1(7- x ) 16$
$(7- x )=3$
$x=4$
Thus, change in oxidation number of $MnO _4^{-}$is 3 . Thus, $MnO _4^{-}$changes to $MnO _2$