Question

Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, 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. How long did the current flow? What mass of copper and zinc were deposited?

Answer 1

\[\ce{Ag^+ + e^- -> Ag}\]

108 g Ag is deposited = 1 F = 96500 C

∴ 1.45 g Ag will be deposited = \[\ce{\frac{96500}{108} \times 1.45}\]

= 1295.6 C

Q = I × t

Or \[\ce{t = \frac{Q}{I}}\]

= \[\ce{\frac{1295.6}{1.5}}\]

= 863.7 s

= 14 min 24 s

\[\ce{Cu^{2+} + 2e^- -> Cu}\]

i.e., 2 × 96500 C deposits Cu = 63.5 g

So 1295.6 C will deposit Cu = \[\ce{\frac{63.5 \times 1295.6}{2 \times 96500}}\]

= 0.4263 g

Similarly, \[\ce{Zn^{2+} + 2e^- -> Zn}\]

Mass of zinc deposited = \[\ce{\frac{65.4 \times 1295.6}{2 \times 96500}}\]

= 0.44 g

Answer 2

Atomic mass of Ag = 107.9

Valency of Ag = 1

Equivalent mass of Ag = `107.9/1`

= 107.9

According to Faraday’s first law of electrolysis,

W = ZIt

Given, W = 1.45 g, I = 1.5 amp, t = ?, Z = `107.9/96500`

∴ `t = W/(Z xx I)`

= `1.45/(107.9/96500 xx 1.5)`

= `(1.45 xx 96500)/(107.9 xx 1.5)`

= 864.5 seconds

= 14.41 min

According to Faraday’s second law of electrolysis,

`(W_(Zn))/W_(Ag) = E_(Zn)/E_(Ag)`

or `W_(Zn) = (W_(Ag) * E_(Zn))/E_(Ag)`

= `(1.45 xx 65.38/2)/107.9`  ...(At mass: Zn = 65.38)

= 0.439

Similarly, 

or `W_(Cu) = (W_(Ag) * E_(Cu))/E_(Ag)`

= `(1.45 xx 63.55/2)/107.9` ...(At mass: Cu = 63.55)

= 0.427 g