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03/10/2020 4:26 pm
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Through a solution of CuSO4 a current of 3 amperes was passed for 2 hours. At cathode 3 g of Cu2+ ions were discharged. The current efficiency is [At. wt. of Cu = 63.5]
(a) 33.3%
(b) 42.2%
(c) 48.7%
(d) 54.4%
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03/10/2020 4:30 pm
(b) 42.2%
Explanation:
According to law of electrolysis,
Mass deposited (m) = Z i t
or i = m x 96500/t x Z
Here, m = 3g, t = 2 × 60 × 60 = 7200 sec
z = Eq. wt/96500 ; Eq. wt = At. wt/Oxidation number
∴ i = (3 x 96500 x 2)/(63.5 x 7200)
= 1.266 A
Efficiency of current
= Current used/ Total current passed x 100
= 1.266/3 x 100
= 42.2%