Cell $$B$$: $$\\displaystyle Ag^+ + e^- \\rightleftharpoons Ag$$ at cathode.$$1$$ mole ($$108$$ g) of Ag is deposited by $$96500$$ $$C$$.$$1.45$$ g of Ag will be deposited by $$\\displaystyle \\dfrac {96500 \ imes 1.45}{108} = 1295.6 \\: C$$.Now,$$\\displaystyle Q = It $$$$\\displaystyle 1295.6 = 1.5 \ imes t $$$$\\displaystyle t = 864 \\: s$$Cell $$A$$: $$\\displaystyle Zn^{2+} + 2e^- \\rightarrow Zn$$$$2$$ moles of electrons ($$\\displaystyle 2 \ imes 96500$$ C of current) produces $$1$$ mole ($$63.5$$ g) of zinc.$$1295.6$$ $$C$$ of electricity will deposit $$\\displaystyle \\dfrac {65.3}{2 \ imes 96500} \ imes 1295.6 = 0.438$$ g of zincCell $$C$$: $$\\displaystyle Cu^{2+} + 2e^- \\rightarrow Cu$$$$2$$ moles of electrons ($$\\displaystyle 2 \ imes 96500$$ $$C$$) of current will produce $$1$$ mole ($$63.5$$ g) of $$Cu$$. $$1295.6$$ $$C$$ of current will deposit $$\\displaystyle \\dfrac {63.5 \ imes 1295.6}{2 \ imes 96500} = 0.426 \\: g$$ of copper