Question

Three electrolytic cells A, B, C containing solutions of $\ce{ZnSO4, AgNO3}$ and $\ce{CuSO4}$ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. What mass of Cu and Zn were deposited.

• A. Zn = 0.44 gm ; Cu = 63.5 gm
• B. Zn = 65.4 gm ; Cu = 63.5 gm
• C. Zn = 0.44 gm ; Cu = 0.427 gm
• D. Zn = 1.45 gm ; Cu = 1.45 gm

Answer 1

Answer: (C)

Atomic mass of $Zn = 6.54$
Atomic mass of $Ag = 108$
Atomic mass of $Cu = 6.35$
$\ce{Ag^+ +}$ $\underset{\text{1 F}}{\ce{e-}}\ce{->}$$\underset{\text{1 mol}}{\ce{Ag}}$
$\because 108\,g$ of $Ag$ is deposited by $= 96500 \,C$
$1.45\,g$ of $Ag$ will be deposited by
$=\frac{96500 \times 1.45}{108}C=1295.6\,C$
Since, the reaction related to deposition of $Cu$ is
$\ce{Cu^{2+} +}$ $\underset{\text{2F}}{\ce{2e-}}\ce{->}$$\underset{\text{63.5 g}}{\ce{Cu}}$
$\because 2 \times 96500\,C$ electricity deposits $63.5\, g$ of $Cu$
$\therefore 1295.6 \,C$ electricity will departs of
$Cu=\frac{63.5 \times 1295.6}{2 \times96500}=0.42\,g$
Since, the reaction related to deposition of zinc is
$\ce{Nn^{2+} +}$ $\underset{\text{2F}}{\ce{2e-}}\ce{->}$$\underset{\text{65.4 g}}{\ce{Zn}}$
$\because 2 \times 96500\,C$ electricity deposits $65.4\, g$ of $Zn$.
$\therefore 1295.6\, C$ electricity will deposit of
$Zn=\frac{65.4\times1295.6}{2\times96500}=0.44\,g$