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प्रश्न
Three electrolytic cells A, B, C containing solutions of \[\ce{ZnSO4}\], \[\ce{AgNO3}\] and \[\ce{CuSO4}\], 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?
उत्तर
\[\ce{Ag^+ + e^- -> Ag}\]
108 g Ag is deposited = 1 F = 96500 C
∴ 1.45 g Ag will be deposited = `96500/108 xx 1.45`
= 1295.6 C
Q = I × t
Or t = `"Q"/"I"`
= `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 = `(63.5 xx 1295.6)/(2 xx 96500)` = 0.4263 g
Similarly, \[\ce{Zn^{2+} + 2e^- -> Zn}\]
Mass of zinc deposited = `(65.4 xx 1295.6)/(2 xx 96500)`
= 0.44 g
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