Three electrolytic 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 ampere was passed through them until 1.45 g of silver were deposited at the cathode of cell B. How long did the current flow? What mass of copper and what mass of zinc were deposited in the concerned cells? (Atomic masses of Ag = 108, Zn = 65.4, Cu = 63.5)

Explanation

`Zn^(2+)(aq)+underset(2" mol")(2e^(-)) to underset(1" mol")(Zn(s))`
`Ag^(+)(aq)+underset(1" mol")(e^(-)) to underset(1" mol")(Ag(s))`
`Cu^(2+)(aq)+underset(2" mol")(2e^(-)) to underset(1" mol")(Cu(s))`
The charge Q on n moles of electrons is given by, Q=nF
Step I. Calculation of time for the flow of current.
For reaction (ii), n=1 mol
Thus, `" " Q=1" mol"xx96500" C " mol"^(-1) =96500" C"`
Molar mass of silver `=108" g mol"^(-1)`
Therefore, mass of 1 mole of silver `=1" mol"^(-1)=108 g`
108 g of Ag is deposited by electric charge `=96500" C"`
`:.` 1.45 g of Ag is current be passed for time t, so that
quantity of electricity used `=1.5" A"xxt`
Thus, `" " 1.5Axxt=1295.6" C"`
`t=((1295.6C))/((1.5A))=((1295.6" A s"))/((1.5" A"))=863.7" s"~~14" min "24" sec" " "(`:'`C=A" s")`
Step II. Calculation of mass of zinc deposited.
For reaction (i), No. of electron moles (n)`=2" mol"`
Thus, `" "` Charge on 2 electron moles (Q)`=2" mol"xx96500" C "mol^(-1)`
`=193000" C" =1.93xx10^(5)C`
Molar mass of copper =63.5 g `mol^(-1)`
Therefore, mass of 1 mole of Zn =1 molxx65 g `mol^(-1)`=65 g
`:. " " 1.93xx10^(5)C` of charge deposit Zn =65 g
1295.6 C of charge deposit zinc `=((65.3 g))/((1.93xx10^(5) C))xx(1295.6 C)`=0.438 g
Step III. Calculation of mass of copper deposited.
For reaction (iii), No of electron mols (n) =2 mol
Thus, `" "` Charge on 2 electron moles (Q)`=2 molxx96500 C mol^(-1)`
=193000 C `=1.93xx10^(5)C`
Molar mass of copper =63.5 g `mol^(-1)`
Therefore, `" "` mass of 1 mole of copper `=1 molxx63.5 g mol^(-1)`=63.5 g
`:. 1.93xx10^(5) C` of charge deposit copper =63.5 g
1295.6 C of charge deposit copper `=((63.5g))/((1.93xx10^(5)C))xx(1295.6 C)=0.426 g`