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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 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

**Problem #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

**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 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

**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 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

**Problem #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

**Problem #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/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

**Problem #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

**Problem #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