Chromium can be electroplated from an aqueous solution containing sulfuric acid and chromic acid, H2CrO4. What current is required to deposit chromium at a rate of 1.25 g/min?

Question

1mol H2CrO4 has 51.996 grams

current= coulomb/sec

How do I convert to get coulombs from the problem?

Answer 1

You know 96,485 Coulombs is required to plate out 51.996/6 g Cr (Cr goes from +6 in H2CrO4 to zero in Cr) = 8.666 g Cr.
You want to plate 1.25 g/min which is 1.25/60 = 0.02083 g/sec.

96,485 C x 0.02083/8.666 = ??Coulombs.
1 ampere current = 1C/sec
so you will need ?? C/sec or that much current.
Check my thinking. Check my numbers.

Answer 2

That makes sense, just so I understand, how did you get 8.666 g Cr? What does the +6 to 0 mean?

Answer 3

Re-read my first response. Cr goes from +6 in H2CrO4 to zero in Cr metal. That's a change of 6 electrons. The equivalent weight of a material is molar mass/delta electrons = 51.996 (I took your number and didn't check it out) or 51.996/6 = 8.666 g as the equivalent weight of Cr in this reaction.