The amount of current in Faraday is required for the reduction of 1 mol of Cr2O7–2 ions to Cr3+ is,

Oxidation number of Cr in Cr2O7-2    =  2(6) =12Oxidation number of cr in 2 Cr+3      =  2(3) = 6Total decrease = 12 - 6 = 61 mole of Cr2O7-2 requires 6 F to get reduced to Cr+3