Chemistry - Redox Reactions Question with Solution | TestHub

Acidified ${\mathrm{KMnO}}_4$ is decolourised by ${\mathrm{H}}_2{\mathrm{O}}_2$ in following way. $$\begin{array}{rl}& 2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+3{\mathrm{H}}_2\mathrm{O}+5\mathrm{O}\\ & 5{\mathrm{H}}_2{\mathrm{O}}_2+5\mathrm{O}\to 5{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2\\ & 2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4\\ & +8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2\\ & \end{array}$$ Thus, for 2 moles of ${\mathrm{KMnO}}_4$ required ${\mathrm{H}}_2{\mathrm{O}}_2=5$ For 1 mole of ${\mathrm{KMnO}}_4$ required ${\mathrm{H}}_2{\mathrm{O}}_2=5/2$ mole.