Chemistry - Redox Reactions Question with Solution | TestHub

$\underset{\underset{\text{25 ml}}{\text{Bleach}}}{{\mathrm{H}}_2{\mathrm{O}}_2}+\underset{\underset{30\text{ ml}}{0.\text{5M}}}{\text{2}{\text{I}}^{-}}+\underset{\underset{\text{4 N}\left(10\text{ ml}\right)}{\left(\text{from C}{\text{H}}_3\text{COOH}\right)}}{\text{2}{\text{H}}^{+}}\to {\mathrm{I}}_{2\mathrm{ }}+\text{2}{\text{H}}_2\mathrm{O}$...(i)

${\mathrm{I}}_2+\underset{0.\text{25 N}\left(\text{48 ml}\right)}{\text{2N}{\text{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3}\to \text{N}{\text{a}}_2{\mathrm{S}}_4{\mathrm{O}}_6+\text{2NaI}$...(ii)

m. mol of${\mathrm{I}}_2=\frac{1}{2}$(m. Moles of$\text{N}{\text{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3$)

$=\frac{1}{2}\times (0.\text{25}\times \text{48})=\text{6 m}.\text{mol}$

Using equation (i)

1 m. Mol of${\mathrm{I}}_2\equiv 1$m. Mol of${\mathrm{H}}_2{\mathrm{O}}_2$

m.mol of${\mathrm{H}}_2{\mathrm{O}}_2=6$m.mol

Molarity of${\mathrm{H}}_2{\mathrm{O}}_2=\frac{6}{25}=0.24\mathrm{M}$