Question

How many minutes will it take to electroplate 0.443 moles of copper from a solution containing Cu²⁺(aq) with a current of 3.88 amps?

Answer 1

PRINCIPLE:

Step1 : Calculate of the Quantity of electricity required to deposit 0.443 moles of copper using Faraday's Second law of electrolysis

Step 2: Calculate the time corresponding to this quantity of electricity

METHOD

Step 1: Calculation of the Quantity of electricity required to deposit 0.443 moles of copper

Cu⁺⁺(aq) + 2e = Cu (s)

Equivalent mass of Cu = Atomic mass of Cu / Number of electrons involved in the reduction

                                = 63.54 / 2 = 31.77

According to Faraday's Second law of electrolysis

1 Faraday ie 96500 Coulombs of electricity will deposit 1 gram equivalent of copper ie 31.77 g of Cu

1 mol of Cu = 63.54 g of Cu

Number of moles of Cu = 31.77g / 63.54 =0.5 mol

0.5 mol of Cu is deposited by 96,500 Coulombs of electricity

Therefore, 0.443 moles of copper is deposited by 96500 Coulombs x 0.443 mol / 0.5 = 85,499 Coulomb

Step 2: Calculation of time corresponding to 85,499 Coulomb electricity

The quantity of electricity Q is defined as  

Q= I x t    

I = current in ampere

t = time in seconds

t = Q / I

t = time in seconds =?

Q = 85,499 Coulomb

I = 3.88 amps

Therefore t = 85,499 Coulomb / 3.88 amp = 22,035.8 s  

( Couomb = amp.s. Hence, Coulomb/ amp = amp. s./amp = s )

Time in seconds = 22,035.8 s

Time in minutes = 22,035.8 /60 = 367.26 min

It will take 367.26 minutes to electroplate 0.443 moles of copper from a solution containing Cu²⁺(aq) with a current of 3.88 amps

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

ALTERNATE METHOD to calculate of the Quantity of electricity:

Cu⁺⁺(aq) + 2e = Cu (s)

1 mol Cu is equivalent to 2e and hence 2 Faraday or 2x 96500 Coulomb

1 mol of Cu is deposited by 2 x 96,500 Coulombs of electricity

Therefore, 0.443 moles of copper is deposited by 2 x 96500 Coulombs x 0.443 mol = 85,499 Coulomb