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What is the current strength in …

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What is the current strength in ampere required to deposit 1.27 gram of copper from a solution of copper sulphate in 1.5 hours

Posted by Neeraj Thakur 5 years, 6 months ago

CBSE > Class 12 > Chemistry

2 answers

Aakash Kumar 5 years, 6 months ago

1.5 hr=5400 s

Aakash Kumar 5 years, 6 months ago

Applying Faraday law of electrolysis, Cu^²+ 2e-=Cu So 2 moles of electrons required to produce 1 mole of copper =63.5 g So, To produce 1.27 g we require 1.27/63.5×2 moles of electrons =0.05 moles of electrons =0.05 ×96500 C(as 1 mole of electrons has 96500 C) We know Q=It=>I=Q/t=0.05×96500 C=4825 C/5400 s=0.8935 A (0.9A approx) ☺️happy

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