Answer:
468 h
Step-by-step 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.
- 1 kg = 1,000 g
- The molar mass of Cr is 52.00 g/mol
- 1 mole of Cr is deposited when 3 moles of electrons circulate
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 A = 1 c/s
- 1 h = 3,600 s
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:
![11.5kgCr * (1,000gCr)/(1kgCr) * (1molCr)/(52.00gCr) * (3mole^(-) )/(1molCr) * (96,468c)/(1mole^(-)) * (1s)/(38.0c) * (1h)/(3,600s) = 468 h](https://img.qammunity.org/2021/formulas/chemistry/college/m1aqsof92me1frlu1y2ec6zz7bzkpovarv.png)