Three electrolytic cells A, B,C containing solutions of ZnSO _{4}, AgNO _{3} and CuSO _{4}, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Three electrolytic cells A, B,C containing solutions of ZnSO _{4}, AgNO _{3} and CuSO _{4}, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

According to the information given in the question the reaction will be,

    \[A g_{(a q)}^{+}+e^{-} \rightarrow A g_{(s)}\]


i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by =\frac{96487 \times 1.45}{107} C

    \[=1295.43 C\]


We have been given in the question,
Current =1.5 A
Time =1295.43 / 1.5 s =863.6 s=864 s =14.40 min
There is copper sulphate in the solution as well therefore,


    \[C u_{(a q)}^{2+}+2 e^{-} \rightarrow C u_{(s)}\]


i.e., 2 \times 96487 C of charge deposit =63.5 g of Cu
Therefore, 1295.43 C of charge will deposit \frac{63.5 \times 1295.43}{2 \times 96487}
=0.426 g of Cu

Finally, we have zinc sulphate,


    \[Z n_{(a q)}^{2+}+2 e^{-} \rightarrow Z n_{(s)}\]


i.e., 2 \times 96487 C of charge deposit =65.4 g of Zn

Therefore, 1295.43 C of charge will deposit \frac{65.4 \times 1295.43}{2 \times 96487}

    \[=0.439 g \text { of } Zn\]