coulombs = amperes x seconds.
C = 10A x 180 min x (60 s/min) = ??
It takes 96,500 coulombs to deposit 1 mol Cr/3 or 52.01/3 grams.
What mass of chromium could be deposited by electrolysis of an aqueous solution of Cr2(SO4)3 for 180.0 minutes using a constant current of 10.0 amperes? (One faraday = 96,500 coulombs.)
4 answers
Why did you divide the molar mass by 3?
You would use stoichiometry. Multiply the amps by the time (convert minutes to seconds, then you end up with Coulombs. The stoichiometry should look a little like this...
(amps x time)(1 F/96,500 coulombs)(1 mol e-/1 F)(3 mol e-/1 mol Cr)(Molar Mass of Cr/1 mol Cr) ... I got the 3 mol e- from balancing the half-cell reactions. I got my answer to come out to 19.4 grams Cr
(amps x time)(1 F/96,500 coulombs)(1 mol e-/1 F)(3 mol e-/1 mol Cr)(Molar Mass of Cr/1 mol Cr) ... I got the 3 mol e- from balancing the half-cell reactions. I got my answer to come out to 19.4 grams Cr
(15.0 C/s) ((180 min) (60 s/min)) = 162000 C
(162000 C) / (96485 C/mol e-) = 1.67 mol e-
Cr2(SO4)3 ---> 2Cr^3+ + 3SO4^2-
This reaction shows that for chromium we are transferring 3 moles of electrons.
(1.67 mol e-) (mol Cr / 3 mol e- transferred) = 0.559 mol Cr
(0.559 mol Cr) (51.996 g/mol) = 29.9 g Cr
(162000 C) / (96485 C/mol e-) = 1.67 mol e-
Cr2(SO4)3 ---> 2Cr^3+ + 3SO4^2-
This reaction shows that for chromium we are transferring 3 moles of electrons.
(1.67 mol e-) (mol Cr / 3 mol e- transferred) = 0.559 mol Cr
(0.559 mol Cr) (51.996 g/mol) = 29.9 g Cr