Two electrolytic cells are connected in series containing C u S O 4 solution and molten A l C l 3 . If in electrolysis 0.4 moles of ‘ C u ’ are deposited on cathode of first cell. The number of moles of ‘ A l ’ deposited on cathode of the second cell is

Step-by-step solution

According to Faraday’s second law of electrolysis: m c u m A l = E c u A A l 0.4 × 63.5 m A l = 63.5 2 27 3 m A l = 7.2 g ⇒ No. of moles of A l = n A l = 7.2 27 = 0.27 Hint: E c u = A t o m i c m a s s o f C u 2 and E A l = A t o m i c m a s s o f A l 3