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 #1:** Calculate the quantity of electricity (Coulombs) necessary to deposit 100.00 g of copper from a CuSO_{4} 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:

Cu^{2+}+ 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 10^{5}C

**Example #2:** How many minutes will take to plate out 40.00 g of Ni form a solution of NiSO_{4} 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:

Ni^{2+}+ 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 10^{5}C

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

1.31511 x 10^{5}C divided by 3.450 C/sec = 3.8119 x 10^{4}sec

5) Convert to minutes:

3.8119 x 10^{4}sec divided by 60 sec/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/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 #4:** How many hours will it take to plate out copper in 200.0 mL of a 0.0 M Cu^{2+} 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:

Cu^{2+}+ 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 #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 silvertherefore, 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

**Example #6:** A metal cup of surface area 200. cm^{2} needs to be electroplated with silver to a thickness of 0.200 mm. The density of silver is 1.05 x 10^{4} 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. cm

^{2}) (1 m^{2}/ 100^{2}cm^{2}) = 0.0200 m^{2}

(0.200 mm) (1 m / 1000 mm) = 0.000200 m(0.0200 m

^{2}) (0.000200 m) = 0.00000400 m^{3}Here's an alternate way to think about the area conversion:

Think of the area as 200. cm x 1 cm (which equals 200. cm^{2}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 m^{3}) (1.05 x 10^{4}kg m¯^{3}) = 0.0420 g0.0420 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 10^{23}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/s37.56778 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 MgCl_{2} for 14.5 h. What mass of Mg is produced?

**Solution:**

0.912 A = 0.912 C/s14.5 hr = 52200 s

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

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

Mg

^{2+}+ 2e¯ ---> MgTwo 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 CuSO_{4} solution?

**Solution #1:**

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

^{2+}+ 2e¯ ---> Cu0.023605 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 10

^{23}ion/mol) = 1.4215 x 10^{22}ionsThe charge of one electron is 1.602 x 10¯

^{19}coulomb

Each copper ion needs two electrons to become copper.(1.4215 x 10

^{22}ions) (2 electrons/ion) (1.602 x 10^¯^{19}C/electron) = 4554.486 C4.75 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 V ^{3+}, an oxidation state of 3+, is the answer0.002237861 mol of V

**Solution #2:**

1) Determine the number of electrons in one Coulomb:

6.022 x 10 ^{23}e¯/mol––––––––––––––––––– = 6.24138 x 10 ^{18}e¯ / C96485 C/mol

2) Determine how many electrons passed:

(650. C) (6.24138 x 10^{18}e¯ / C) = 4.0569 x 10^{21}e¯

3) Determine atoms of V that got oxidized:

(0.002237861 mol) (6.022 x 10^{23}atoms mol^{-1}) = 1.34764 x 10^{21}atoms of V

4) Determine electrons per atom:

4.0569 x 10 ^{21}e¯––––––––––––––––––– = 3 1.34764 x 10 ^{21}atoms of V

**Example #10:** What current is needed to deposit 0.480 g of chromium metal from a solution of Cr^{3+} 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 depositedCr

^{3+}+ 3e¯ ---> CrThree 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 MgCl_{2} for 17.0 h. If 4.73 L of Cl_{2} (at STP) is produced at the anode, what is the current in amperes?

**Solution:**

1) The entire calculation done dimensional analysis style:

1 mol Cl _{2}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 Cl _{2}mol e¯ 17 hr 3600 s

2) Each step described:

Liters of Cl_{2}produced to moles of Cl_{2}producedMoles of Cl

_{2}produced to moles of electrons requiredMoles 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:

Al^{3+}+ 3e¯ ---> AlMole 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:

Mg^{2+}+ 2e¯ ---> MgMole 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