Question

${Cr}_2{O}_4^{2-}$ can be oxidised to ${CO}_2$ by $Mn{O}_4^{-}$ in acidic as well as in basic medium. Moles ratio of $Mn{O}_4^{-}$ used per mol of ${Cr}_2{O}_4^{2-}$ is $\frac{a}{b}$ then, determine the value of $a$ and $b$.

• A. $a=3,b=5$
• B. $a=2,b=5$
• C. $a=1,b=3$
• D. None of the above

Answer 1

The correct option is A $a=3,b=5$

The reactions are as follows:
$5{Cr}_2{O}_4^{2-}+2Mn{O}_4^{-}\underset{}{\overset{H^{+}}{\to }}10{CO}_2+2{Mn}^{2+}$
${3Cr}_2{O}_4^{2-}+2Mn{O}_4^{-}\underset{}{\overset{{OH}^{-}}{\to }}6{CO}_2+2Mn{O}_2$
$1{Cr}_2{O}_4^{2-}=\frac{2}{5}Mn{O}_4^{-}\left(acidic\right)$
$1{Cr}_2{O}_4^{2-}=\frac{2}{3}Mn{O}_4^{-}\left(basic\right)$
Ratio of moles of $Mn{O}_4^{-}$ per mole of $C{r}_2{O}_4^{2-}=3:5$.
Therefore, the value of $a$ and $b$ are $3$ and $5$ respectively.