Chemistry - Redox Reactions Question with Solution | TestHub

The reaction is:

$2{\mathrm{CrO}}_4^{2-}+3{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+10{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{Cr}(\mathrm{OH}{)}_4^{-}+6{\mathrm{SO}}_4^{2-}+2{\mathrm{H}}^{+}$

In ${\mathrm{CrO}}_4^{2-}$, Cr is in +6 oxidation state. In $\mathrm{Cr}(\mathrm{OH}{)}_4^{-}$, Cr is in +3 oxidation state.

Change in oxidation state for Cr = $6-3=3$.

So, n-factor for ${\mathrm{CrO}}_4^{2-}$ is 3.

In ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$, S is in +2 oxidation state. In ${\mathrm{SO}}_4^{2-}$, S is in +6 oxidation state.

Change in oxidation state for S per atom = $6-2=4$.

Since there are 2 S atoms in ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$, total change = $2\times 4=8$.

So, n-factor for ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ is 8.

Using the equivalence concept:

Milliequivalents of ${\mathrm{CrO}}_4^{2-}$ = Milliequivalents of ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$

$(M\times V\times n{)}_{{\mathrm{CrO}}_4^{2-}}=(M\times V\times n{)}_{{\mathrm{S}}_2{\mathrm{O}}_3^{2-}}$

$0.154\times V\times 3=0.25\times 40\times 8$

$0.462V=80$

$V=\frac{80}{0.462}=173.159\approx 173$ mL

The volume of $0.154\mathrm{M}{\mathrm{CrO}}_4^{2-}$ required is 173 mL.