Chemistry - Redox Reactions Question with Solution | TestHub

Balanced reactions:

$2{\mathrm{Cu}}^{2+}+4{\mathrm{I}}^{-}⟶{\mathrm{C}{\mathrm{u}}_2\mathrm{I}}_2+{\mathrm{I}}_2$

${\mathrm{I}}_2+2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3⟶{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}$

From the stoichiometry, 1 mole of ${\mathrm{I}}_2$ reacts with 2 moles of $\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3$, and 1 mole of ${\mathrm{I}}_2$ is produced from 2 moles of ${\mathrm{Cu}}^{2+}$.

Therefore, 2 moles of ${\mathrm{Cu}}^{2+}$ react with 2 moles of $\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3$.

This means moles of ${\mathrm{Cu}}^{2+}$ = moles of $\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3$.

Moles of $\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3=Volume\times Molarity=50mL\times 0.2M=10mmol$.

So, moles of ${\mathrm{Cu}}^{2+}=10mmol$.

Molarity of $\mathrm{CuS}{\mathrm{O}}_4=\frac{Moles}{Volume}=\frac{10mmol}{100mL}=0.1M$.

The final answer is $\boxed{0.1M}$.