Three electrolytic cells A,B,C containing solutions of `ZnSO_(4),AgNO_(3) and CuSO_(4)`, passed through them until 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Explanation

To calculate time
Weight of Ag deposited = 1.45 g
Current, I passed = 1.5 amperes
Quantity of electricity passed, Q = ?
We know that `underset("1 mole")(Ag^(+)(aq)) +underset("1 mole")(e^(-)) to underset(108g)underset(1"mole")(Ag(s))`
= 1 Faraday = 96500 coulombs
`" "[ :. "Atomic mass of Ag" = 108]`
108 g Ag is deposited by charge = 96500 coulombs
`:.` 1.45 g Ag is deposited by charge
`=(19600)/(108) xx 1.45=1295.6` coulombs.
But `Q= I xx t`
`:. 1295.6=1.5 xx t`
`:.t=(1295.6)/(1.5)=(863.7` second = 14.4 min
(b) To calculate the weight of Cu deposited
Let wt. of Cu deposited = x g
Using Faraday.s second law of electrolysis, we have
`("Mass of Cu")/("Mass of Ag")=("Equivalent mass of Cu")/("Equivalent mass of Ag"), (x)/(1.45) =(31.75)/(108)`