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प्रश्न
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?
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उत्तर १
\[\ce{Ag^+ + e^- -> Ag}\]
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
⇒ t = \[\ce{\frac{Q}{I}}\]
= \[\ce{\frac{1295.6}{1.5}}\]
= 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
उत्तर २
Given: W = 1.45 g,
I = 1.5 amp,
t = ?
Atomic mass of Ag = 107.9
Valency of Ag = 1
Equivalent mass of Ag = `107.9/1` = 107.9
Z = `107.9/96500`
According to Faraday’s first law of electrolysis,
W = ZIt
∴ 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)`
⇒ WZn = `(W_(Ag) * E_(Zn))/E_(Ag)`
= `(1.45 xx 65.38/2)/107.9` ...(At mass: Zn = 65.38)
= 0.439
Similarly,
WCu = `(W_(Ag) * E_(Cu))/E_(Ag)`
= `(1.45 xx 63.55/2)/107.9` ...(At mass: Cu = 63.55)
= 0.427 g
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