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In a copper plating experiment in which copper metal is deposited a copper(II) ion solution, the system is run for 2.6 hours at a current of 12.0A. What mass of cooper is deposited?(F = 96,485 C/mol). Cu^2 (aq) + 2e^- Rightarrow Cu(s) Q = I Times T = 12 Times (2.6 Times 3600) = 123 Times 10^5 (1.164)(63.546)/2 = 36.88

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Answer: The amount of Cu deposited is approximately 37.0g

Explanation:

According to Faraday Law of electrolysis, the mass of substance deposited at the electrode during an electrolysis is directly proportional to quantity of electricity(Q) passing through the electrolyte.

Given that

Time of passing current = 2.6hr

T = 2.6×60×60 = 9360s

current passed I = 12.0A

Quantity of electricity Q = IT

Q= 9360 × 12= 112320 C

For the half cell reaction

Cu2+ + 2e- → Cu

Cu is a dipositive ion.

2F of electricity will be required for dipostive ion( 2moles of electron).

1F= 96485

2F=2 × 96485 = 192970C

192970C will deposit 1mole of Cu( 63.546g)

112320 will deposit Xg

Where X is the mass of Cu deposited when Q was passed.

X = (112320 × 63.546)/192970

X = 37.0g

Therefore, the mass of Cu deposited is 37.0g

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