Question

$100\mathrm{mL}\mathrm{of}0.01\mathrm{M}$ ${\mathrm{KMnO}}_4$ oxidised 100 mL ${\mathrm{H}}_2{\mathrm{O}}_2$ in acidic medium. The volume of same ${\mathrm{KMnO}}_4$ required in a strong alkaline medium to oxidize $100\mathrm{mL}$of same ${\mathrm{H}}_2{\mathrm{O}}_2$ is:

• A. $500$
• B. $\frac{100}{3}$
• C. $\frac{300}{3}$
• D. $\frac{500}{3}$

Answer 1

The correct option is D

$\frac{500}{3}$

Step 1: Given data:

$100\mathrm{mL}$ ${\mathrm{H}}_2{\mathrm{O}}_2$ is oxidized in acidic medium by = $100\mathrm{mL}\mathrm{of}0.01\mathrm{M}$ of ${\mathrm{KMnO}}_4$.

Also,we need to oxidise a strong alkaline medium of $100\mathrm{mL}$of same ${\mathrm{H}}_2{\mathrm{O}}_2$ .

The volume of ${\mathrm{KMnO}}_4$ required for both the cases is the same.

Step 2: Writing reduction equations of both cases:

The reduction equations of both cases can be given as:

Case 1: In the acidic medium the following reaction takes place: ${{\mathrm{MnO}}_4}^{-}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{+2}$

Case 2: In the basic medium the following reaction takes place: ${{\mathrm{MnO}}_4}^{-}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2$

Step 3: calculation of volume:

As the same number of milliequivalents of ${\mathrm{KMnO}}_4$is required in both acidic and basic mediums for oxidation of 100ml of ${\mathrm{H}}_2{\mathrm{O}}_2$.

Hence,

$$\begin{gathered} \Rightarrow milliequivalentinacidicmedium=milliequivalentinbasicmedium \\ \Rightarrow 100\times 5=V\times 3 \\ \Rightarrow V=\frac{500}{3} \end{gathered}$$

Therefore,$\frac{500}{3}$ ml of ${\mathrm{KMnO}}_4$ is required in a strongly alkaline medium to oxidize 100 mL of the same ${\mathrm{H}}_2{\mathrm{O}}_2$.