Applications of Quantitative Aspects of Electrolysis

In the text Quantitative Aspects of Electrolysis, you saw some mathematical relationships that are established between the quantities involved in an electrolysis process, such as the electric current (i), the amount of electric charge (Q) needed for the process to occur and the time (t) that lead to. It was also discovered the amount of electrical charge that is carried when there are 1 mole of electrons or, according to Avogadro's constant, 6.02. 10²³ electrons.

Briefly, the relationships are:

Now here are three examples of how you can use this information to solve practical electrochemistry problems. It is important to note that here we use the value 96486 C. However, in most chemical literature, the rounded value 96500 C is used.

1st Example: Consider an electroplating in which a part has been coated with silver. At the end of this electrolytic process, the amount of charge used for the Ag ions⁺ if they reduced the Ag it was 0.05 faraday. Knowing that the molar mass of silver is equal to 108 g/mol, say what was the mass of silver deposited in this process?

Resolution:

Ag⁺ _(here) + and- → Ag_(s)
↓ ↓
1 mol e-1 mol
↓ ↓
1 faraday 108 g
0.05 faraday m
m = 5.4 g

2nd Example: Let's say we are carrying out the electrolysis of an aqueous solution of nickel sulphate (NiSO₄), applying an electrical current equal to 0.10 A for 386 seconds. What will be the mass of nickel that will be obtained at the cathode? (Given: molar mass of Ni = 58.7 g/mol)

Resolution:

Ni²⁺  + 2e- → Ni_(s)
↓ ↓
2 mol e-1 mol
↓ ↓
2 (96486 C) 58.7g

To make a rule-of-three relationship and find the mass that was formed in this case, we first need to find the amount of electric charge (Q):

Q = i. t
Q = 0.10. 386
Q = 38.6C

So we have:

2 (96486 C) 58.7g
38.6 cm
m = 2265.82C. g
192972 C
m = 0.01174 g or 11.74 mg

3rd Example: We have three electrolytic vats connected in series and subjected to a current of 5 A for a time of 32 minutes and 10 seconds. In the first vat, we have a CuSO solution₄; on the second, we have a FeCl solution_3; and in the third, we have an AgNO solution₃. Determine the masses of each of the metals deposited on the electrodes of the three wells. (Molar masses: Cu = 63.5 g/mol, Fe = 56 g/mol, Ag = 108 g/mol).

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Resolution:

First, let's pass the time value to seconds:

1 minute 60 seconds
32 minutes t
t = 1920 + 10 seconds = 1930 seconds

With this data, we can determine the amount of electrical charge Q:

Q = i. t
Q = 5. 1930
Q = 9650 C

Now, we use rules of three for each of the half-reactions that occur in the three vats to find out the respective masses of deposited metals:

1st Cuba: 2nd Cuba: 3rd Cuba:

Ass²⁺  + 2e- → Cu_(s) Faith³⁺ _(here) + 3 and- → Fe_(s) Ag⁺ _(here) + and- → Ag_(s)
↓ ↓ ↓ ↓ ↓ ↓
2 mol e-1 mol 3 mol e- 1 mol 1 mol e-1 mol
↓ ↓ ↓ ↓ ↓ ↓
2. (96486 C) 63.5 g 3. (96486 C) 56 g 96486 C 108 g
9650 C m 9650 C m 9650 C m
m ≈ 3.175 g of Cu_(s)m ≈ 1.867 g of Fe_(s)m = 10.8 g of Ag_(s)

By Jennifer Fogaça
Graduated in Chemistry

Would you like to reference this text in a school or academic work? Look:

FOGAÇA, Jennifer Rocha Vargas. "Applications of the Quantitative Aspects of Electrolysis"; Brazil School. Available in: https://brasilescola.uol.com.br/quimica/aplicacoes-dos-aspectos-quantitativos-eletrolise.htm. Accessed on June 28, 2021.