Ratio of moles of Fe (II) oxidised by equal volumes of equimolar KMnO₄ and K₂Cr₂O₇ solutions in acidic medium will be:

${\mathrm{Fe}}^{2+}+{\mathrm{MnO}}_4^{-}⟶{\mathrm{Fe}}^{3+}+{\mathrm{Mn}}^{2+}$
n_f =7 n_f =5
Moles of Fe²⁺ x 1 by KMnO₄
=moles of ${\mathrm{MnO}}_4^{-}\times 5$ ...(1)
${\mathrm{Fe}}^{2+}+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}⟶{\mathrm{Fe}}^{3+}+{\mathrm{Cr}}^{3+}$
n_f =7 n_f =6
Moles of Fe²⁺ x 1 by K₂Cr₂O₇
=moles of ${\mathrm{MnO}}_4^{-}\times 6$ ...(2)
$\begin{array}{l}\left(1\right)+\left(2\right)\\ \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{5}{6}\end{array}$