In acidic medium, potassium permanganate oxidizes H2O2 to O2 as per the equation given below 2MnO4- + 6H+ + 5H2O2 -> 2Mn2+ + 8H2O + 5O2 100 mL of 0.02 M KMnO4 oxidises 10 mL of X vol H2O2 completely. The value of X approximately is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q132 – AP-EAMCET 2026

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Home Chemistry Redox ReactionsQuestion #549

In acidic medium, potassium permanganate oxidizes

H_{2} O_{2}

O_{2}

as per the equation given below

2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}

100 mL

0.02 M K M n O_{4}

oxidises

10 mL

X vol H_{2} O_{2}

completely. The value of

X

approximately is

Clue 1 of 3: Stoichiometry and Redox Reactions

This problem involves a redox titration. Understand the balanced chemical equation to determine the mole ratio between the reactants, potassium permanganate ( $K M n O_{4}$ ) and hydrogen peroxide ( $H_{2} O_{2}$ ).

🧠 Full Solution Path

Step 1: Determine the moles of potassium permanganate (

K M n O_{4}

) reacted.✦ Active

The number of moles of $K M n O_{4}$ can be calculated using its molarity and volume.

Moles of K M n O_{4} = Molarity \times Volume (in L)

Moles of K M n O_{4} = 0.02 M \times (100 mL \times \frac{1 L}{1000 mL})

Moles of K M n O_{4} = 0.02 M \times 0.1 L = 0.002 mol

💡 Teacher's Secret Hint

Ensure consistent units for volume (convert mL to L) when calculating moles using molarity.

Step 2: Calculate the moles of hydrogen peroxide (

H_{2} O_{2}

) reacted using stoichiometry.○ Expand

From the given balanced chemical equation:

2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}

The mole ratio of $M n O_{4}^{-}$ to $H_{2} O_{2}$ is $2 : 5$ .

Moles of H_{2} O_{2} = Moles of K M n O_{4} \times \frac{5 mol H_{2} O_{2}}{2 mol K M n O_{4}}

Moles of H_{2} O_{2} = 0.002 mol \times \frac{5}{2} = 0.005 mol

💡 Teacher's Secret Hint

Pay close attention to the stoichiometric coefficients in the balanced equation to derive the correct mole ratio.

Step 3: Determine the molarity of the

H_{2} O_{2}

solution.○ Expand

The molarity of $H_{2} O_{2}$ is calculated using the moles of $H_{2} O_{2}$ reacted and its given volume.

Molarity of H_{2} O_{2} = \frac{Moles of H_{2} O_{2}}{Volume of H_{2} O_{2} (in L)}

Molarity of H_{2} O_{2} = \frac{0.005 mol}{10 mL \times \frac{1 L}{1000 mL}}

Molarity of H_{2} O_{2} = \frac{0.005 mol}{0.01 L} = 0.5 M

💡 Teacher's Secret Hint

Remember to convert the volume of $H_{2} O_{2}$ from mL to L before calculating its molarity.

Step 4: Calculate the volume strength (

X

) of the

H_{2} O_{2}

solution.○ Expand

The volume strength of $H_{2} O_{2}$ solution is related to its molarity by the formula:

Volume Strength = Molarity \times 11.2

This factor of $11.2$ arises because $1 mole$ of $H_{2} O_{2}$ yields $0.5 moles$ of $O_{2}$ upon decomposition ( $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ ), and $0.5 moles$ of $O_{2}$ occupy $0.5 \times 22.4 L = 11.2 L$ at STP.

X = 0.5 M \times 11.2 = 5.6 vol

The value of $X$ is approximately $5.6$ .

💡 Teacher's Secret Hint

The constant $11.2$ is used for converting molarity to volume strength for $H_{2} O_{2}$ at STP, based on the decomposition reaction $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ and the molar volume of a gas ( $22.4 L$ at STP).

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Chemistry Question 132 – AP-EAMCET 2026

Clue 1 of 3: Stoichiometry and Redox Reactions

🧠 Full Solution Path