The amount of current in Faraday is required for the reduction of 1 mol of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{-2} \mathrm{ions}\mathrm{to} {\mathrm{Cr}}^{3+}$ is 

$C{r}_2{O}_7^{2-} +\hspace{0.17em}6e^{-}\to 2C{r}^{3+}$

for the given transformation of 1 mol of dichromate to Cr³⁺ required 6 mol of electrons. 

1 mol of electrons = 1F 

6 mol of electrons = 6F