Question

A current of 5.18 A is pass through a Cr(NO3)2 solution for 1.20 hours. How much chromium is plated out of the solution? __g


I have no idea how to approach this. Here's what I tried to do logically but I'm not sure if it works out:

I know that the conversions are 1 A *s = 1 C and 1 mol e- = 96485 C which is given by my textbook.

one problem im having is I'm not sure whether to use Cr 3+ or Cr 2+ [to know number of electrons. I used 2... for no reason in particular it was just a wild guess]

1.20 hours = 72 minutes = 4320 seconds

96485 x 2 coulombs [or 2e-?] plates 51.9961g/mol Cr
5.18 A x 4320 seconds =
51.9961/(96485 x 2) x 5.18 A x 4320 g Cr = 24.118 g?

Answer 1

First, the 2 is right. You look at Cr(NO3)2 so Cr is +2 (you know NO3 is -1) and it must be going to zero to plate.
You also are right that 96,485 x 2 Coulombs will plate about 52g.
So you have 5.18 x 1.2 x 60 x 60 = about 22,378 C. Therefore, you will plate
52 x (22,378/96,486*2) = about 6 g.
If your equation were
51.996 x 5.18 x 1.2 x 60 x 60/(96485 x 2)= about 6 you would be ok.

Answer 2

Whoops your right I totally calculated "51.9961/(96485 x 2) x 5.18 A x 4320 g Cr = 24.118 g?" incorrectly. Thank you!