Chemistry - Redox Reactions Question with Solution | TestHub

Here's a brief solution to the problem:

The balanced chemical equation for the reaction between KMnO₄ and H₂O₂ in acidic conditions is:

$$2{\mathrm{KMnO}}_4+5{\mathrm{H}}_2{\mathrm{O}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2$$

From the stoichiometry of the reaction, 2 moles of KMnO₄ react with 5 moles of H₂O₂.

Moles of KMnO₄ used = Molarity × Volume (in L) = $0.2\mathrm{M}\times \frac{25}{1000}\mathrm{L}=0.005$ moles

Moles of H₂O₂ reacted = $\frac{5}{2}\times$ moles of KMnO₄ = $\frac{5}{2}\times 0.005=0.0125$ moles

This amount of H₂O₂ is present in 10 ml of the diluted solution. Therefore, in 1000 ml (1 L) of the diluted solution, the moles of H₂O₂ would be:

Moles of H₂O₂ in 1 L diluted solution = $\frac{0.0125moles}{10ml}\times 1000ml=1.25$ moles

The volume strength of H₂O₂ is defined as the volume of oxygen gas liberated at STP (Standard Temperature and Pressure) by 1 volume of H₂O₂ solution.

The decomposition of H₂O₂ is given by:

$$2{\mathrm{H}}_2{\mathrm{O}}_2(aq)\to 2{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{O}}_2(g)$$

From the equation, 2 moles of H₂O₂ gives 1 mole of O₂. Therefore, 1.25 moles of H₂O₂ will give $\frac{1.25}{2}=0.625$ moles of O₂.

Volume of O₂ liberated at STP = moles × 22.4 L = $0.625\times 22.4=14$ L

Since this volume of O₂ is liberated from 1 L of diluted H₂O₂ solution, the volume strength of the diluted H₂O₂ solution is 14.