#### Stoichiometric Electrochemical CalculationsHow much current or time is required to deposit some metal?How much metal gets deposited by a specified amount of time and current?

Here are the various "bits" involved in stoichiometric electrochemical calculations (say that three times as fast as you can):
(a) the amount of current, usually given in Amperes (make sure to use Coulombs per second (C/s) in the calculation)
(b) The Faraday Constant (96,485 C/mole e¯, many teachers & sources use 96,500)
(c) the mass of a substance (measured in grams)
(d) the molar mass of said substance (measured in g/mol)
(e) the amount of time (seconds, minutes, hours are the most-commonly used units)
(f) the half reaction with the substance as the product

Some calculation steps you will see:

(a) Coulombs per second times seconds equals Coulombs.
(b) Grams divided by grams per mol equals mole. This is moles of the substance.
(c) Coulombs divided by Coulombs per mole e¯ equals mol e¯. This is NOT moles of the substance.

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

Example #1: Calculate the quantity of electricity (Coulombs) necessary to deposit 100.0 g of copper from a CuSO4 solution.

Solution:

1) Determine moles of copper plated out:

100.00 g / 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.

(2) (1.573663 mol) = 3.147326 mol e¯ required

3) Convert moles of electrons to Coulombs of charge:

3.147326 mol e¯ x 96,485.309 C/mol = 3.038 x 105 C

Example #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 / 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.

(2) (0.6815076 mol) = 1.363015 mol e¯ required

3) Convert moles of electrons to Coulombs of charge:

(1.363015 mol e¯) (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/s):

1.31511 x 105 C / 3.450 C/s = 3.8119 x 104 sec

5) Convert to minutes:

3.8119 x 104 s / 60 s/min = 635.3 min

Example #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/s
1 hour = 3600 s
0.2500 C/s x 3600 s = 900.0 C

2) Determine moles of electrons in 900.0 C:

900.0 C / 96,485.309 C/mol of electrons = 9.327845 x 10¯3 mol e¯

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 #4: How many hours will it take to plate out copper in 200.0 mL of a 0.1500 M Cu2+ solution using a current of 0.200 amp?

Solution:

1) Determine moles of copper to be plated out:

(0.2000 L) (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.

(2) (0.03000 mol) = 0.06000 mol e¯ required

3) Convert moles of electrons to Coulombs of charge:

(0.06000 mol e¯) (96,485.309 C/mol e¯) = 5789.12 C

4) Convert to seconds required to deliver the Coulombs determined in step 3 (remember, 1 A = 1 C/s):

5789.12 C / 0.200 C/s = 28945.6 sec

5) Convert seconds to hours:

28945.6 sec / 3600 s/hr = 8.04 hours

Example #5: A constant electric current deposits 0.3650 g of silver metal in 1296 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 / 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.

(1) (0.00338376 mol) = 0.00338376 mol e¯ required

3) Determine Coulombs of charge that 0.00338376 mol e¯ represents:

(0.00338376 mol e¯) (96,485.309 C/mole¯) = 326.483 C

4) Determine current (remember that 1 A = 1 C/s):

326.483 C / 1296 s = 0.252 A

Example #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

Example #7: A constant current of 0.912 A is passed through an electrolytic cell containing molten MgCl2 for 14.5 h. What mass of Mg is produced?

Solution:

0.912 A = 0.912 C/s

14.5 hr = 52200 s

(0.912 C/s) (52200 s) = 47606.4 C

47606.4 C / 96485 C/mol = 0.493407 mol of electrons

Mg2+ + 2e¯ ---> Mg

Two moles of electrons per one mole of Mg deposited

0.493407 mol / 2 = 0.2467035 mol of Mg deposited

(0.2467035 mol) (24.305 g/mol) = 5.996 g

Round to three sig figs for a final answer of 6.00 g

Example #8: Using a current of 4.75 A, how many minutes does it take to plate out 1.50 g of Cu from a CuSO4 solution?

Solution #1:

1.50 g / 63.546 g/mol = 0.023605 mol of Cu plated out

Cu2+ + 2e¯ ---> Cu

0.023605 mol x 2 = 0.04721 mol of electrons required

0.04721 mol times 96485 C/mol = 4555.057 C

4.75 A = 4.75 C/s

4555.057 C / 4.75 C/s = 959 s

959 s / 60 s/min = 15.98 min

to three sig figs, 16.0 min

Solution #2:

1.50 g / 63.546 g/mol = 0.023605 mol of Cu ion

(0.023605 mol) (6.022 x 1023 ion/mol) = 1.4215 x 1022 ions

The charge of one electron is 1.602 x 10¯19 coulomb
Each copper ion needs two electrons to become copper.

(1.4215 x 1022 ions) (2 electrons/ion) (1.602 x 10^¯19 C/electron) = 4554.486 C

4.75 A = 4.75 C/s

4554.486 C / 4.75 C/s = 958.84 s = 959 s

And on to 16.0 min

Example #9: A vanadium electrode is oxidized electrically. Its mass decreases by 114 mg during the passage of 650. Coulombs. What is the oxidation state of the vanadium product?

Solution #1:

1) Determine moles of electrons passed:

 650. C ––––––––––– = 0.0067368 mol of e¯ 96485 C/mol

2) Determine moles of vanadium reacted:

 0.114 g –––––––––––– = 0.002237861 mol 50.9415 g/mol

3) Determine electrons per atom:

 0.0067368 mol of e¯ ––––––––––––––––––– = 3 <--- which means V3+, an oxidation state of 3+, is the answer 0.002237861 mol of V

Solution #2:

1) Determine the number of electrons in one Coulomb:

 6.022 x 1023 e¯/mol ––––––––––––––––––– = 6.24138 x 1018 e¯ / C 96485 C/mol

2) Determine how many electrons passed:

(650. C) (6.24138 x 1018 e¯ / C) = 4.0569 x 1021

3) Determine atoms of V that got oxidized:

(0.002237861 mol) (6.022 x 1023 atoms mol-1) = 1.34764 x 1021 atoms of V

4) Determine electrons per atom:

 4.0569 x 1021 e¯ ––––––––––––––––––– = 3 1.34764 x 1021 atoms of V

Example #10: What current is needed to deposit 0.480 g of chromium metal from a solution of Cr3+ in a period of 1.25 hr?

Solution:

1) Step-by-step style:

0.480 g / 51.9961 g/mol = 0.00923146 mol of Cr gets deposited

Cr3+ + 3e¯ ---> Cr

Three moles of electrons are required for every one mole of Cr deposited.

(0.00923146 mol) (3) = 0.02769438 mol of e¯ passed

(96485 C/mol) (0.02769438 mol) = 2672.1 C of charge passed

(1.25 hr) (3600 s/hr) = 2880 s

Given that 1 A = 1 C/s, we have:

2672.1 C / 2880 s = 0.928 A (to three sig figs)

2) Dimensional analysis style:

 1 mol Cr 3 mol e¯ 96485 C 1 0.480 g x ––––––– x ––––––– x ––––––– x ––––––– = 0.928 C/s = 0.928 A 51.9961 g 1 mol Cr 1 mol e¯ 2880 s

Example #11: A constant current is passed through an electrolytic cell containing molten MgCl2 for 17.0 h. If 4.73 L of Cl2 (at STP) is produced at the anode, what is the current in amperes?

Solution:

1) The entire calculation done dimensional analysis style:

 1 mol Cl2 2 mol e¯ 96485.34 C 1 1 hr 4.73 L x ––––––– x ––––––– x ––––––––– x ––––– x ––––––– = 0.665 C/s = 0.665 A 22.414 L 1 mol Cl2 mol e¯ 17 hr 3600 s

2) Each step described:

Liters of Cl2 produced to moles of Cl2 produced

Moles of Cl2 produced to moles of electrons required

Moles of electrons required to amount of charge (in Coulombs) required

Amount of charge required to amount of charge required per hour

Amount of charge required per hour to amount of charge required per second

1 C/s = 1 A

Example #12: If a current plates out 13.5 g of aluminium, what mass of magnesium would be plated out in the same time by the same current?

Solution:

1) Consider the electroplating of aluminium:

Al3+ + 3e¯ ---> Al

Mole ratio of e¯ to Al = 3 : 1

2) Moles of Al plated:

13.5 g / 27.0 g/mol = 0.500 mol

3) Moles of e¯ passed:

0.500 mol x 3 = 1.50 mol

4) Consider the electroplating of magnesium:

Mg2+ + 2e¯ ---> Mg

Mole ratio of e¯ to Mg = 2 : 1

5) Moles of e¯ passed:

1.50 mol

6) Moles of Mg plated:

1.50 mol / 2 = 0.750 mol

7) Mass of Mg plated:

0.750 mol x 24.3 g/mol) = 18.2 g

Example #13: Which of the following expressions is correct for the maximum mass of copper, in grams, that can be plated out by electrolyzing aqueous CuCl2 for 16.0 hours at a constant current of 3.00 A.

 (16.0) (3600) (3) (63.55) (2) (16.0) (3600) (3) (63.55) (a) ––––––––––––––––––––––– (c) –––––––––––––––––––– 96,485 96,485

 (16.0) (3600) (3) (63.55) (16.0) (60) (3) (96,485) (b) –––––––––––––––––––– (d) ––––––––––––––––––– (96,485) (2) (63.55) (2)

Solution using Dimensional Analysis:

1) Here's the start:

 16.0 hr ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer 1

Notice that it immediately lets you determine that (c) is not the correct answer. There are six entries in the answers for (a), (b) and (d), but only 5 values in (c). But, then again, it could be that only five steps will be needed and (c) turns out to be the correct answer.

2) We need the number of seconds, not hours. This is because the unit on Amperes is C/s.

 16.0 hr 3600 s ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer 1 1 hr

This step immediately dismisses (d) as the correct answer. Note that there is a 60, by the conversion from hours to seconds involves two values of 60 (60 minutes per hour and 60 seconds per minute). Also, not that 3600 s/hr is a well-accepted conversion and can be used without any explanation of it.

3) We now have the time the electrolytic cell operated, so the next step is to determine how much current (in Coulombs) passed in the time determined.

 16.0 hr 3600 s 3.00 C ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer 1 1 hr 1 s

4) The next step is to use the amount of current passed to determine the number of moles of electrons passed. The Faraday Constant is used in this step.

 16.0 hr 3600 s 3.00 C 1 mol e¯ ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer 1 1 hr 1 s 96,485 C

Note that this is the total moles of electrons in the current, NOT the moles of substance that got plated out.

5) The next step is to determine the moles of substance plated out. To do this, we must se the half-reaction that shows the ion being reduced to the metallic state. This is the half-reaction for this problem:

Cu2+(aq) + 2e¯ ---> Cu(s)

The key piece of information is that, for every ONE mole of Cu2+ reduced, TWO moles of electrons are required.

 16.0 hr 3600 s 3.00 C 1 mol e¯ 1 mol Cu2+ ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer choice (b) 1 1 hr 1 s 96,485 C 2 mol e¯

The lack of a 2 in the denominator eliminates answer choices (a) and (c). This means that (b) is shown to be the correct answer. I will finish the dimensional analysis sequence with the last step.

6) The last step is to use the atomic weight of copper to convert moles to grams:

 16.0 hr 3600 s 3.00 C 1 mol e¯ 1 mol Cu2+ 63.55 g ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = answer choice (b) 1 1 hr 1 s 96,485 C 2 mol e¯ 1 mol

Example #14: A steady current of 10.0 amperes is passed through an aluminum-producing cell for 15.0 minutes. Which of the following is the correct expression for calculating the number of grams of aluminum produced?

 (10.0) (15.0) (96,485) (10.0) (15.0) (60) (27.0) (a) ––––––––––––––––– (c) ––––––––––––––––––– (27.0) (60) (96,485) (3)

 (10.0) (15.0) (27.0) (27.0) (96,485) (b) ––––––––––––––––– (d) ––––––––––––––––– (60) (96,485) (10.0) (15.0) (60) (3)

Solution:

1) Choices (a) and (b) both have five parts to the answer while choices (c) and (d) have six. Based on experience, we know that six different amounts are required, not five. We set (a) and (b) aside.

2) Rather than using dimensional analysis (the typical means to answer this type of question), let us analyze the units involved. We know that grams MUST be the final answer. I propose to start wil choice (c). Here it is:

 (10.0) (15.0) (60) (27.0) ––––––––––––––––––– (96,485) (3)

3) The units on 10.0 are C/s (Coulombs per second). Notice where the unit for seconds goes:

 (C) (15.0) (60) (27.0) ––––––––––––––––––– (s) (96,485) (3)

That's right, the unit goes in the denominator.

4) The unit on 15.0 is minutes:

 (C) (min) (60) (27.0) ––––––––––––––––––– (s) (96,485) (3)

4) The unit on 60 is s/min:

 (C) (min) (s) (27.0) ––––––––––––––––––– (s) (min) (96,485) (3)

By the way, the 60 does not affect significant figures since 60 seconds are in one minute by definition (not by a measurement using instruments).

5) I want to do the Faraday Constant next:

 (C) (min) (s) (mol e¯) (27.0) ––––––––––––––––––––––– (s) (min) (C) (3)

The mol e¯ unit was in the denominator (C/mol e¯) of the Faraday Constant and the entire unit was in the overall denominator. Since mol e¯ was in the denominator of the denominator, it moves to the numerator.

6) That 27.0 is the atomic mass of Al, with the unit of g/mol Al:

 (C) (min) (s) (mol e¯) (g) ––––––––––––––––––––––– (s) (min) (C) (mol Al) (3)

7) That 3 is 3 mol e¯ per mol Al. This is based on the following reduction half-reaction:

Al3+(aq) + 3e¯ ---> Al(s)

 (C) (min) (s) (mol e¯) (g) (mol Al) –––––––––––––––––––––––––––– (s) (min) (C) (mol Al) (mol e¯)

8) Everything cancels except grams. Choice (c) is the correct answer.