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

Cell B contains `AgNO_(3)` and reaction may be represented as :
`AgNO_3 hArr Ag^(+) + NO_(3)^(-)`
`Ag^(+) + e^(-) hArr Ag` (At cathode)
According to the equation ,
1 mol of 108 g of silver is deposited by 96500 C
1.45 g silver is deposited by = `(96500 xx 1.45)/(108)`
= 1295. 6 C
Now , `Q = I xx t`
`1295.6 = 1.5 xx t`
or `" " t = (1295.6)/(1.5) = 863` s .
The weight of copper and zinc can be calculated by using Faraday.s second law of electrolysis .
In cell A , the electrode reaction is
`Zn^(2+) + 2e^(-) to Zn`
2 mol of electrons or `2 xx 96500` C of current produce 1 mol or 65.3 g of Zn so that
`2 xx 96500` C of electricity deposit Zn = 65.3 g
1295.6 C of electricity deposit `Zn = (65.3)/(2 xx 96500) xx 1295.6`
`= 0.438` g .
In cell C , the electrode reaction is , `Cu^(2+) + 2e^(-) to Cu`
2 mol of electrons or `2 xx 96500 C` of current produce 1 mol or 63.5 g of Cu so that
`2 xx 96500` C of current deposit Cu = 63.5 g
1295. 6 C of current deposit Cu = `(63.5 xx 1295.6)/(2 xx 96500)`
`= 0.426` g .