Determine the equivalent weight of K₂Cr₂O₇ in the reaction, 

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+7{\mathrm{H}}_2{\mathrm{SO}}_4+6{\mathrm{FeSO}}_4⟶$${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{K}}_2{\mathrm{SO}}_4+3{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+7{\mathrm{H}}_2\mathrm{O}$

 

The given reaction may be written as, 

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{Fe}}^{2+}⟶2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

Thus, ${\mathrm{Cr}}^{6+}\text{ in }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ undergoes reduction to ${\mathrm{Cr}}^{3+},\text{ and }{\mathrm{Fe}}^{2+}$ gets oxidized to ${\mathrm{Fe}}^{3+}$. The reaction stoichiometry gives:

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\equiv {\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\equiv 2{\mathrm{Cr}}^{6+}$

That is 1 mol of potassium dichromate is equivalent to 2Cr⁶⁺ ions. Each Cr⁶⁺ ion gains 3 electrons to give Cr³⁺. Therefore, one molecule of K₂Cr₂O₇ gains six electrons to get reduced to two Cr³⁺ ions. As per definition, for this reaction, Equivalent weight of K₂Cr₂O₇.

$$\begin{gathered} =\frac{\text{ Molecular mass of }{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}{6} \\ =\frac{(2\times 39\times +2\times 52+7\times 16)}{6} \\ =\frac{78+104+112}{6}=\frac{294}{6}=49 \end{gathered}$$