Answer
16 hours
Work Step by Step
In 1.0 L, there are initially 1.5 moles of Cu$^{2+}$. This means:
1.5 mole of Cu$^{2+}$ $\times$ $\frac{2 mol e^{-}}{1 molCu^{2+}}$ $\times$ $\frac{96,485 C}{1 mol e^{-}}$ $\times$ $\frac{1s}{5.0C}$ $\times$ $\frac{1 min}{60 s}$ $\times$ $\frac{1 hr}{60 min}$ = 16 hours
