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
