Worksheet - Electrochemistry Problems - AP level

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General note: I kept all the digits on my calculator; I rounded off to the final answer at the end of each problem.

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¯ ---> Cu

therefore, 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

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¯ ---> Ni

therefore, 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

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

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¯ ---> Cu

therefore, 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

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¯ ---> Ag <--- that's the half-reaction for the deposition of silver

therefore, 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 A

Problem #6: A metal cup of surface area 200. cm2 needs to be electroplated with silver to a thickness of 0.200 mm. The density of silver is 1.05 x 104 kg m¯3. The mass of a silver ion is 1.79 x 10¯25 kg and the charge is the same magnitude as that on an electron. How long does the cup need to be in the electrolytic tank if a current of 12.5 A is being used?

Solution:

1) Determine the volume of silver that gets electroplated:

First, convert values to m:

(200. cm2) (1 m2 / 1002 cm2) = 0.0200 m2
(0.200 mm) (1 m / 1000 mm) = 0.000200 m

(0.0200 m2) (0.000200 m) = 0.00000400 m3

Here's an alternate way to think about the area conversion:

Think of the area as 200. cm x 1 cm (which equals 200. cm2

Convert each cm value to m

[(200. cm) (1 m / 100 cm)] x [(1 cm) (1 m / 100 cm)]

2) Determine the mass, then the moles of silver:

(0.00000400 m3) (1.05 x 104 kg m¯3) = 0.0420 g

0.0420 g / 107.8682 g/mol = 0.000389364 mol

Note that a g/atom value is provided in the problem. That number would have been used to get the mass of one mole of silver. Like this:

(1.79 x 10¯25 kg) (1000 g / kg) (6.022 x 1023 mol¯1) = 107.7938 g/mol

3) Moles of electrons required:

Ag+(aq) + e¯ ---> Ag(s)

0.000389364 mol of Ag+ plated out requires 0.000389364 mol of electrons

4) Determine Coulombs of charge that was transfered:

(0.000389364 mol) (96485 C/mol) = 37.56778 C

5) Determine time required to transfer charge:

12.5 A = 12.5 C/s

37.56778 C / 12.5 C/s = 3.00 s


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