Question

Three electrolytic cells A, B, C containing solutions of $\mathrm{ZnSO}{}_4,\mathrm{AgNO}{}_3$ and $\mathrm{CuSO}{}_4$ 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 $Cu=6.35$
$\mathrm{Ag}{}^{+}+$ $\underset{\text{1 F}}{\mathrm{e}{}^{-}}\to$$\underset{\text{1 mol}}{\mathrm{Ag}}$
$\because 108g$ of $Ag$ is deposited by $=96500C$
$1.45g$ of $Ag$ will be deposited by
$=\frac{96500\times 1.45}{108}C=1295.6C$
Since, the reaction related to deposition of $Cu$ is
$\mathrm{Cu}{}^{2+}+$ $\underset{\text{2F}}{2\mathrm{e}{}^{-}}\to$$\underset{\text{63.5 g}}{\mathrm{Cu}}$
$\because 2\times 96500C$ electricity deposits $63.5g$ of $Cu$
$\therefore 1295.6C$ electricity will departs of
$Cu=\frac{63.5\times 1295.6}{2\times 96500}=0.42g$
Since, the reaction related to deposition of zinc is
$\mathrm{Nn}{}^{2+}+$ $\underset{\text{2F}}{2\mathrm{e}{}^{-}}\to$$\underset{\text{65.4 g}}{\mathrm{Zn}}$
$\because 2\times 96500C$ electricity deposits $65.4g$ of $Zn$.
$\therefore 1295.6C$ electricity will deposit of
$Zn=\frac{65.4\times 1295.6}{2\times 96500}=0.44g$