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?

Explanation

Deposition of copper 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.