The number of electrons involved in the reduction of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ to ${\mathrm{Cr}}^{3+}$ in acidic medium is 

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+6{\mathrm{e}}^{-}⟶2{\mathrm{Cr}}^{3+}$In this reduction reaction six electrons are required for the reduction of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ into 2 ${\mathrm{Cr}}^{3+}$.