Generally, the flow of current is from the cell with the more negative cell potential (Oxidation Rxn) to the cell with the more positive cell potential (Reduction Rxn). In this case
#Cr^o(s) rightleftharpoons Cr^(+3) + 3e^-# => Oxidation Rxn
#MnO_4^(-)(aq) + 8H^+(aq) + 5e^(-) rightleftharpoons Mn^(+2)(aq) + 4H_2O(l)# => Reduction Reaction
To balance charge transfer, multiply the oxidation rxn by 5 and the reduction rxn by 3 and add reactions =>
#5Cr^o(s) rightleftharpoons 5Cr^(+3) + 15e^-#
#3MnO_4^(-)(aq) + 24H^+(aq) + 15e^(-)
rightleftharpoons 3Mn^(+2)(aq) + 12H_2O(l)#
Net Cell Rxn:
#5Cr^o(s) + 3MnO_4^(-)(aq) + 24H^+(aq) rightleftharpoons5Cr^(+3)(aq) + 3Mn^(+2)(aq) + 12H_2O(l)#
#E^o(cell)#= #E^(o)#(Reduction) - #E^o#(Oxidation) = 1.51v - (-0.74v) = 2.25v
