Question

$100$ mL of $0.01M$ $KMn{O}_4$ oxidized $100$ mL of $H_2{O}_2$ in acidic medium. The volume of the same $KMn{O}_4$ required in alkaline medium to oxidize $100$ mL of the same $H_2{O}_2$ will be :

($Mn{O}_4^{⊝}$ changes to $M{n}^{2+}$ in acidic medium and to $Mn{O}_2$ in alkaline medium)

• A. $\frac{100}{3}$ mL
• B. $\frac{500}{3}$ mL
• C. $\frac{300}{5}$ mL
• D. none of the above

Answer 1

Answer: (B)

$\frac{500}{3}$ mL

$Mn{O}_4^{⊝}+5e^{-}\to M{n}^{2+}$(acidic)
$Mn{O}_4^{⊝}+3e^{-}\to Mn{O}_2$(basic)
$100mL$ $H_2{O}_2\equiv 100\times 5NMn{O}_4^{⊝}\equiv V\times 3NMn{O}_2$
$V=\frac{500}{3}mL$