Three elctrolytic cells A,B,C containing solutions of $Z n S O_{4}, A g N O_{3}$ and $C u S O_{4}$ respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver deposited at the cathode3 of cell B. How long did the current flow ? What mass of copper and zinc were deposited ? [Atomic masses; Cu = 63,5, Zn = 65.3; Ag = 108]

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Solution

(a) $A g^{+} (a q) + e^{-} 1 m o l \to A g (s) 1 m o l (108 g)$
108 g silver are deposited by electricity = 96500C
1.45 g silver are deposited by electricity
$= \frac{(96500 C) \times (1.45 g)}{(108 g)} = 1295.6 C$
Current(I) = 1.5 amperes
$T i m e (t) = \frac{Q}{I} = \frac{(1295.6 C)}{(1.5 A)} = 863.7 s = 14 m i n 39 s$
(b) $C u^{2 +} (a q) + 2 e^{-} 2 F (2 \times 96500 C) \to C u (s) 63.5 g$
On passing electricity of $(2 \times 96500 C); C u$ deposited
$= \frac{(63.5) \times (1295.6 C)}{(2 \times 96500 C)} = 0.426 g Z n^{2 +} (a q) + 2 e^{-} 2 F (2 \times 96500 C) \to Z n (s) 65.3 g$
Deposited Cu = 0.426g
Deposited Zn = 0.438g

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