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JEE Advanced levelSingle correct optionDifficulty 3 of 3mt191

A 2.50 kg impure copper anode contains 96.0% Cu, 1.80% Zn, 0.65% Ni and the rest Ag and Au. It is dissolved completely at the anode and the same charge is passed through the cell, with pure copper deposited at the cathode (only is discharged; Zn and Ni dissolve as and ). Taking Cu = 63.5, Zn = 65, Ni = 59, the net mass of copper lost from the electrolyte is

  1. A17.5 g
  2. B44.0 g
  3. C2400 g
  4. D61.5 g

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.
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More questions on Refining methods: zone refining, van Arkel, Mond process, electrolytic refining