Three electrolytic cells A, B and C containing solutions of ZnSO₄(zinc sulphate), AgNO₃ (silver nitrate) and CuSO₄  (copper sulphate),  respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver is deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? (Atomic masses: Ag = 108, Zn = 65.4, Cu = 63.5, all in amu). 

       But quantity of electricity passed
                    = current x time
                    = 1295.6 C  = 1.5 A x time (in sec.)
       or time for which current is passed
               $= \frac{1295.6}{1.5 } \mathrm{seconds} = 863.7 \mathrm{s} = 14.40 \min .$
          The cathode reaction in copper sulphate cell is
                         $$\begin{gathered} {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-} \to \mathrm{Cu} \\ 2 \mathrm{mol} 1 \mathrm{mol} (63.5 \mathrm{g}) \\ (2\times 96500 \mathrm{c}) \end{gathered}$$
              2 x 96500 coulombs gives a deposit of 63.5 g of Cu.
          Therefore, 1295.6 coulombs will deposit of 
                       $= \frac{63.5 \times 1295.6}{2\times 96, 500} = 0.4263 \mathrm{g}$
Similarly,
                        $$\begin{gathered} {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-} \to \mathrm{Zn} \\ 2 \mathrm{mol} 1 \mathrm{mol} (65.4 \mathrm{g}) \\ (2\times 96500) \end{gathered}$$
               (cathode reaction in zinc sulphate cell)
Mass of zinc deposited
                         $=\frac{65.4 \mathrm{x} 1295.6}{2\times 96, 500} \mathrm{g} = 0.44 \mathrm{g}.$