Question

200 ml of 0.5 m cuso4 solution to deposit 40% cu 0.2 ampere current should be passed for how many seconds​

Answer 1

volume of CuSO4 solution = 200ml

molarity of solution = 0.5M

so, no of moles of CuSO4 in solution = volume of solution in L × molarity

= (200/1000) × 0.5

= 0.1 mol

so, number of mole of deposited Cu = 40 % of 0.1 mol = 40/100 × 0.1 = 0.04 mol

we know, Cu²+ + 2e- ⇔Cu

2 moles of electrons are required to deposit 1 mole of Cu.

so, number of moles of electrons required to deposit 0.04 mol of Cu = 2 × 0.04 = 0.08 mol

from Faraday's law,

one mole of electrons is equivalent to 96500 C charge.

so, 0.08 mol of electrons = 0.08 × 96500C = 7720 C

given, Current , I = 0.2A

charge , Q = 7720C

so, time = charge /current = 7720/0.2 = 38600 sec

hence, time = (38600/3600)hr = 10.72hrs