Question

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 ?

Answer 1

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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