To deposit 31.75 g Cu from CuSO at a current of 5.00 A requires how much time? (M(Cu) = 63.5)
- A
5360 s
- Bcheck_circle
19300 s ( with z=2)
- C
9650 s
- D
1930 s
Explanation
Moles Cu = 31.75/63.5 = 0.500. Moles e⁻ = 1.00. Charge q = (1.00)(96500) = 96500 C. t = q/I = 96500/5.00 = 19300 s ≈ 5.36 hours.