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^(2+) (aq) + 2e^(-) to Cu (s)`
2 Faradays or 193000 coulomb can deposite 63.5 grams of copper.
Quantity of electricity (Q) required to deposite 60 grams of copper `= 60/63.5 xx 193000 = 182362 coul. `
Time taken for the electrolysis (t) = `Q/i = (182362)/(3) = 60787 "sec" = 16.9 hrs`.