Answer pls 16. Three electrolytic cells A, B and C containing electrolytes ZnSO4, AgNO3 and CuSO4 respectively were connected in series. A steady current of 1.50 ampere was passed through them until 1.45 g of Ag were deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? (At. wts. of Cu=63.5, Zn=65.3, Ag=108) Share with your friends Share 0 Vartika Jain answered this Dear Student, Ag+ + e- → AgNow, 1F of e- deposits 1 mol of Agi.e. 96500 C deposits 108 g of AgSo, 1.45 g of Ag is deposited by 96500108×1.45 = 1295.6 Cand, we know that Q = Itor, t=Qi=1295.61.50=863.7 sSince the connection is in series, then the amount of charge is same. So, Cu2+ + 2e- → CuHere, 2F e- deposits 1 mol of Cui.e. 2×96500 C deposits 63.5 g CuSo, 1295.6 C will deposit 63.52×96500×1295.6=0.4263 gSimilarly, Zn2+ + 2e- → ZnHere, 2F e- deposits 1 mol of Zni.e. 2×96500 C deposits 65.4 g CuSo, 1295.6 C will deposit 65.42×96500×1295.6=0.44 g 1 View Full Answer