In: Chemistry
A Cr3+(aq) solution is electrolyzed, using a current of 8.00 A .
What amperage is required to plate out 0.260 mol Cr from a Cr3+ solution in a period of 8.40 h ?
EXACT QUESTION SOLVED BEFORE IN DETAILS
A Cr^3 +(aq) solution is electrolyzed, using a current of 8.00 A.
a)What mass of Cr(s) is plated out after 2.40 days?
b)What amperage is required to plate out 0.260 mol Cr from a Cr^3 + solution in a period of 8.40 h?
2.40 days ( 24 hours/day) ( 60 min / hr) ( 60 s / min)= 2.07 x 10^5 s
The cathode half-reaction is
Cr3+ + 3e- = Cr
It states that for every 3 faradays consumed onr mole of Cr is formed
8.00 A = 8.00 colulomb/s
8.00 coulomb/s x 2.07 x 10^5 s =1.66 x 10^6 coulombs
one faraday = 96500 coulombs
1.66 x 10^6 / 96500=17.2 faradays
17.2 faradays ( 1 mole Cr/ 3 faradays) = 5.73 moles Cr
mass Cr = 5.73 mol x 51.9961 g/mol=297.9 g
8.40 hours = 30240 s
faradays = 0.260 x 3 = 0.780
coulombs = 96500 x 0.780=75270
A = 75270 coulombs/ 30240 s =2.49 A
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