Answer:
468 h
Explanation:
Let's consider the reduction of chromium (III) to chromium that occurs in the electrolytic purification.
Cr³⁺ + 3 e⁻ → Cr
We can establish the following relations.
The hours that will take to plate 11.5 kg of chromium onto the cathode if the current passed through the cell is held constant at 38.0 A is:
[tex]11.5kgCr \times \frac{1,000gCr}{1kgCr} \times \frac{1molCr}{52.00gCr} \times \frac{3mole^{-} }{1molCr} \times \frac{96,468c}{1mole^{-}} \times \frac{1s}{38.0c} \times \frac{1h}{3,600s} = 468 h[/tex]