Question

Ratio of moles of Fe(II) oxidized by equal volumes of equimolar $KMn{O}_4$ and $K_{2}C{r}_2{O}_7$ solutions in acidic medium. $K_P$ will be

• A. 5 : 3
• B. 1 : 1
• C. 1 : 2
• D. 5 : 6

Answer 1

The correct option is D

5 : 6

$Fe\left(II\right)\to Fe\left(III\right)+e^{–}$
Eq. mass of $Fe\left(II\right)$ = molar mass
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
Eq. mass of $KMn{O}_4=\frac{molarmass}{5}$
$C{r}_2{O}_7^{2-}+14H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
Eq. mass of $K_{2}C{r}_2{O}_7=\frac{molarmass}{6}$
Eq. of $KMn{O}_4$ in VL of xM = 5xV = mol of Fe(II)
Eq. of $K_{2}C{r}_2{O}_7$ in VL of xM=6xV=mol of Fe(II)
Hence, ratio of moles of Fe(II) oxidized by $KMn{O}_4$
and $K_{2}C{r}_2{O}_7=5:6$