Can I have help please?

Question

How many minutes are required to deposit 1.35 g Cr from a Cr³⁺(aq) solution using a current of 2.50 A? (F = 96,500 C/mol)

Answer 1

Cr³⁺(aq) + 3e- ==> Cr(s)

You need 3 moles of electrons per mole of Cr

Moles of Cr = 1.35 g Cr x 1 mol Cr / 52.0 g = 0.02596 moles Cr

Moles of electrons = 0.02596 mols Cr x 3 mol e- / mol Cr = 0.0788 moles of electrons

Coulombs (C) required = 0.0788 mols e- x 96,500 C / mol e- = 7604 C

Since 2.50 amp (A) = 2.50 C / sec, we can find the number of seconds needed...

7604 C x 1 s / 2.50 C = 3042 seconds

3042 s x 1 min / 60 s = 50.7 minutes required