Final answer:
The total charge required to remove all the Cr from 500 mL of 0.270 M Cr2(SO2)3 is 1.32 × 10^6 C. It will take approximately 7333 minutes to remove all the Cr by a constant current of 3.00 A.
Step-by-step explanation:
The total charge required to remove all the Cr from 500 mL of 0.270 M Cr2(SO2)3 is 1.32 × 106 C. If this charge is passed at a rate of 3.00 A, the required time can be calculated by dividing the total charge by the current:
Time = Total charge / Current
Time = 1.32 × 106 C / 3.00 A = 4.40 × 105 s
Converting seconds to minutes:
Time in minutes = 4.40 × 105 s / 60 s/min = 7333.33 min
Rounding to the nearest whole number, it will take approximately 7333 minutes to remove all the Cr.
Therefore, the correct answer is Option D. 30 minutes.