Chemistry - Redox Reactions Question with Solution | TestHub

$\left(\text{I}\right)\text{ 2KMn}{\text{O}}_4+3{\text{H}}_2\text{S}{\text{O}}_4+5{\text{H}}_2{\text{O}}_2\to {\text{K}}_2\text{S}{\text{O}}_4+2\text{Mn}\text{S}{\text{O}}_4+8{\text{H}}_2\text{O}+5{\text{O}}_2$

$\left(\text{I}\text{I}\right)\text{ 2KMn}{\text{O}}_4+3{\text{Mn}\text{S}\text{O}}_4+2{\text{H}}_2\text{O}\to 5\text{Mn}{\text{O}}_2+\text{K}{}_2\text{S}{\text{O}}_4+{2\text{H}}_2\text{S}{\text{O}}_4$

$\left(\text{I}\text{I}\text{I}\right)\text{ Mn}{\text{O}}_{\mathrm{ 2}}+{\text{Na}}_2{\text{C}}_2{\text{O}}_4+2{\text{H}}_2{\text{S}\text{O}}_4\to \text{Mn}{\text{S}\text{O}}_4+\text{Na}{}_2\text{S}{\text{O}}_4+{2\text{CO}}_2+{\text{H}}_2\text{O}$

$\text{Reaction (III)}\Rightarrow {\text{mM of Na}}_2{\text{C}}_2{\text{O}}_4=\left(\text{10 mL}\times \text{0.2M}\right)=2\text{mM}=\text{mM }\text{of Mn}{\text{O}}_2$

$\text{Reaction (II)}\Rightarrow {\text{mM of} \text{ K}\text{MnO}}_4=\text{2}\times \frac{\text{2}}{\text{5}}=\frac{\text{4}}{\text{5}}\text{mM}$

$\text{Reaction (I)}\Rightarrow \text{mM of }{\text{ H}}_2{\text{O}}_2=\frac{4}{\text{5}}\times \frac{5}{2}=2\text{mM}$

$2{\text{mM of H}}_2{\text{O}}_2=\text{20 mL}\times {\text{M}}_{{\text{H}}_2{\text{O}}_2}$

${\text{M}}_{{\text{H}}_2{\text{O}}_2}=\text{0.1 M}$