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A solution of copper sulphate is electrolysed using a current strength of 3 amp to deposite 60 grams of copper. What is the time taken for the electrolysis? |
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Answer» Solution :Deposition ofcopper is given as : `Cu_((aq))^(2+)+2e^(-)rarrCu_((s))` 2 Faradays or 193000 coulomb can DEPOSIT 63.5 grams of copper. QUANTITY of ELECTRICITY (Q) required to deposit 60 grams of copper `=(60)/(63.5)xx19300 = 182362` coulomb. Time taken for the electrolysis `(t)=(Q)/(i)=(182362)/(3)=60787` sec `=16.9` hrs. |
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