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?

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? (A) 863.73 s (B) 500 s (C) 275 s (D) 925 s

Tardigrade Admin · Accepted Answer

I = 1.5 A, W = 1.45 g Ag, t = ?, E = 108, n = 1 Using Faradayâs 1st law of electrolysis W = ZIt or, W = (E/F) It or, 1.45 g = (108/96500) × 1.5 t or, t = (1.45 × 96500/1.5 × 108) = 863.73 seconds

Q. Three electrolytic 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 $1.5$ amperes was passed through them until $1.45 g$ of silver deposited at the cathode of cell B. How long did the current flow?

$863.73 s$

$500 s$

$275 s$

$925 s$

$I = 1.5 A, W = 1.45 g A g, t = ?, E = 108, n = 1$
Using Faraday’s $1^{s t}$ law of electrolysis $W = Z I t$
or, $W = \frac{E}{F} I t$
or, $1.45 g = \frac{108}{96500} \times 1.5 t$
or, $t = \frac{1.45 \times 96500}{1.5 \times 108}$
$= 863.73$ seconds

Q. 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 1.45g of silver deposited at the cathode of cell B. How long did the current flow?

863.73s

500s

275s

925s