Problem 20.91 A Cr3+(aq) solution is electrolyzed, using a current of 8.00 A . P

Question

Problem 20.91

A Cr3+(aq) solution is electrolyzed, using a current of 8.00 A .

Part A

What mass of Cr(s) is plated out after 1.50 days?

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Part B

What amperage is required to plate out 0.210 mol Cr from a Cr3+ solution in a period of 8.60 h ?

Problem 20.91

A Cr3+(aq) solution is electrolyzed, using a current of 8.00 A .

Part A

What mass of Cr(s) is plated out after 1.50 days?

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Part B

What amperage is required to plate out 0.210 mol Cr from a Cr3+ solution in a period of 8.60 h ?

Explanation / Answer

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 =1.50 x 10^6 coulombs
one faraday = 96500 coulombs
1. 50 x 10^6 / 96500=15.54 faradays

15.54 faradays ( 1 mole Cr/ 3 faradays) = 5.16 moles Cr

mass Cr = 5.16mol x 51.9961 g/mol

a) Mass of Cr = 268.65 g


8.60 hours = 30960 s
faradays = 0.210 x 3 = 0.63
coulombs = 96500 x 0.63= 60795
A = 75270 coulombs/ 30960 s

b) lA = 1.96

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