Return to Electrochemistry Menu
General note: I kept all the digits on my calculator; I rounded off to the final answer at the end of each problem.
Example Problem #1: Calculate the quantity of electricity (Coulombs) necessary to deposit 100.00 g of copper from a CuSO4 solution.
Solution:
1) Determine moles of copper plated out:
100.00 g divided by 63.546 g/mole = 1.573663 mol
2) Determine moles of electrons required:
Cu2+ + 2e¯ ---> Cutherefore, every mole of Cu plated out requires two moles of electrons.
1.573663 mol x 2 = 3.147326 mol e¯ required
3) Convert moles of electrons to Coulombs of charge:
3.147326 mol e¯ x 96,485.309 C/mol = 3.0367 x 105 C
Example Problem #2: How many minutes will take to plate out 40.00 g of Ni form a solution of NiSO4 using a current of 3.450 amp?
Solution:
1) Determine moles of nickel plated out:
40.00 g divided by 58.6934 g/mole = 0.6815076 mol
2) Determine moles of electrons required:
Ni2+ + 2e¯ ---> Nitherefore, every mole of Ni plated out requires two moles of electrons.
0.6815076 mol x 2 = 1.363015 mol e¯ required
3) Convert moles of electrons to Coulombs of charge:
1.363015 mol e¯ x 96,485.309 C/mol = 1.31511 x 105 C
4) Convert to seconds required to deliver the Coulombs determined in step 3 (remember, 1 A = 1 C/sec):
1.31511 x 105 C divided by 3.450 C/sec = 3.8119 x 104 sec
5) Convert to minutes:
3.8119 x 104 sec divided by 60 sec/min = 635.3 min
Example Problem #3: What is the equivalent weight of a metal if a current of 0.2500 amp causes 0.5240 g of metal to plate out a solution undergoing electrolysis in 1 hour? (Comment: One mole of electrons will plate out one equivalent weight of metal.)
Solution:
1) Determine total Coulombs of charge delivered:
0.2500 A = 0.2500 C/sec
1 hour = 3600 seconds
0.2500 C/sec x 3600 sec = 900.0 C
2) Determine moles of electrons in 900.0 C:
900.0 C divided by 96,485.309 C/mol of electrons = 9.327845 x 10¯3 mol of electrons
3) Determine mass of metal plated out by one mole of electrons:
0.5240 g / 9.327845 x 10¯3 mol = 56.18 g per equivalent weight
Example Problem #4: How many hours will it take to plate out copper in 200.0 mL of a 0.0 M Cu2+ solution using a current of 0.200 amp?
Solution:
1) Determine moles of copper to be plated out:
0.2000 L x 0.1500 mol/L = 0.03000 mol
2) Determine moles of electrons required:
Cu2+ + 2e¯ ---> Cutherefore, every mole of Cu plated out requires two moles of electrons.
0.03000 mol x 2 = 0.06000 mol e¯ required
3) Convert moles of electrons to Coulombs of charge:
0.06000 mol e¯ x 96,485.309 C/mol = 5789.12 C
4) Convert to seconds required to deliver the Coulombs determined in step 3 (remember, 1 A = 1 C/sec):
5789.12 C divided by 0.200 C/sec = 28945.6 sec
5) Convert seconds to hours:
28945.6 sec divided by 3600 sec/hr = 8.04 hours
Example Problem #5: A constant electric current deposits 0.3650 g of silver metal in 12960 seconds from a solution of silver nitrate. What is the current? What is the half reaction for the deposition of silver?
Solution:
1) Determine moles of silver deposited:
0.3650 g divided by 107.8682 g/mol = 0.00338376 mol
2) Determine moles of electrons required:
Ag+ + e¯ ---> Agtherefore, every mole of Ag plated out requires one mole of electrons.
0.00338376 mol mol x 1 = 0.00338376 mol e¯ required
3) Determine Coulombs of charge that 0.00338376 mol e¯ represents:
0.00338376 mol e¯ times 96,485.309 C/mol = 326.483 C
4) Determine current (remember that 1 A = 1 C/sec):
326.483 C / 12960 sec = 0.0252 amp
Example Problem #6: Someday!