Three conductivity cells A, B and C containing solutions of zinc sulphate, silver nitrate and copper sulphate respectively are connected in series. A steady current of 1.5 amperes is 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 what mass of zinc got deposited in their respective cells ? (Atomic mass : Zn = 65.4 u, Ag = 108 u, Cu = 63.5 u)

Explanation

We are given that
Current (I) = 1.5 A
Amount of Ag deposited = 1.45 g
We know that
F = 96500 C , `Ag^(+)+e^(-)rarrAg`
Now,
108 g of Ag are deposited by `nF=1xx96500=96500C`
1.45 g of Ag will be deposited by `=(96500)/(108)xx1.45=1295.6C`
Using the relation :
`Q=Ixxt`
Or `t=(Q)/(I)=(1295.6)/(1.5)=863.7s`
Now, `Cu^(2+)+2e^(-)rarrCu`
i.e., `2xx96500C` deposit Cu = 63.5 g
1295.6 C deposit `Cu=(63.5)/(2xx96500)xx1295.6=0.426g`
Also,
`Zn^(2+)+2e^(-)rarrZn`
i.e., `2xx96500C` deposit Zn = 65.3 g
1295.6 C deposit `Zn=(65.3)/(2xx96500)xx1295.6=0.438g`