Chemistry - Redox Reactions Question with Solution | TestHub

$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2\mathrm{C}{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$
${\mathrm{a}}_1=6$
${\mathrm{a}}_2=1$
-factor of
-factor of
$\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$
$\underset{\underset{{\mathrm{a}}_1=6}{\left({\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7\right)}}{{\mathrm{a}}_1{\mathrm{M}}_1{\mathrm{V}}_1}=\underset{\underset{{\mathrm{a}}_2=1}{\left(\mathrm{F}\mathrm{e}\mathrm{S}{\mathrm{O}}_4\right)}}{{\mathrm{a}}_2{\mathrm{M}}_2{\mathrm{V}}_2}$
${\mathrm{M}}_1=0.1\text{M}$
${\mathrm{M}}_2=0.5$M
${\mathrm{V}}_1=?$
${\mathrm{V}}_2=35$mL
So${\mathrm{V}}_1=\frac{1\times 0.5\times 35\text{mL}}{6\times 0.1}$
$=29.2$mL.