Chemistry - Redox Reactions Question with Solution | TestHub

Acidic medium:

$Mn{O}_4^{-}+5e^{-}+8H^{+}⟶M{n}^{2+}+4H_{2}O$

$H_2{O}_2⟶O_2+2H^{+}+2e^{-}$

Alkaline medium:

$Mn{O}_4^{-}+3e^{-}+2H_{2}O⟶Mn{O}_2+4O{H}^{-}$

$H_2{O}_2⟶O_2+2H^{+}+2e^{-}$

In acidic medium:

moles of $KMn{O}_4=1M\times 0.1L=0.1$ moles

moles of $e^{-}$ from $KMn{O}_4=0.1\times 5=0.5$

moles of $H_2{O}_2=x$

moles of $e^{-}$ from $H_2{O}_2=2x$

$2x=0.5$

$x=0.25$ moles

In alkaline medium:

moles of $H_2{O}_2=0.25$ moles

moles of $e^{-}$ from $H_2{O}_2=0.25\times 2=0.5$ moles

moles of $KMn{O}_4=y$

moles of $e^{-}$ from $KMn{O}_4=3y$

$3y=0.5$

$y=\frac{0.5}{3}=\frac{1}{6}$ moles

Volume of $KMn{O}_4=\frac{1}{6}moles/1M=\frac{1}{6}L=\frac{1000}{6}ml=\frac{500}{3}ml$

Final Answer: The final answer is $\boxed{\frac{500}{3}\mathrm{ml}}$