Answer:
696 h
Explanation:
Let's consider the reduction of Cr³⁺.
Cr³⁺(aq) + 3e⁻ → Cr(s)
We can establish the following relations.
How many hours will it take to plate 13.5 kg of chromium onto the cathode if the current passed through the cell is held constant at 30.0 A?
[tex]13.5 \times 10^{3} gCr.\frac{1molCr}{52.0gCr} .\frac{3mol/e^{-} }{1molCr} .\frac{96468c}{1mol/e^{-}} .\frac{1s}{30.0c} .\frac{1h}{3600s} =696 h[/tex]