Answer: D
Total charge corresponds to n(Cu) + n(Zn) + n(Ni) mol of divalent metal. The cathode deposits that many moles of Cu, while the anode adds only n(Cu) to the solution. Net loss from solution = n(Zn) + n(Ni) = 0.6923 + 0.2754 = 0.9677 mol, i.e. 61.5 g of Cu.