43
Faraday’s second law
Textbook: pp. 123–124
GoalKnow Faraday’s second law, the formula k = (1/F)(A/Z) and the Faraday constant.
New words
Faraday constant F · Faradey doimiysi Fmolar mass · molyar massavalence · valentlikchemical equivalent · kimyoviy ekvivalent
Explanation
By Faraday’s second law the electrochemical equivalent of a substance is directly proportional to its chemical equivalent A/Z: k = (1/F)·(A/Z). Here A is the molar mass of the substance (g/mol), Z is its valence and F is the Faraday constant, F ≈ 96 500 C/mol; it is the charge needed to release 1 mol of a monovalent ion. So a substance of higher valence releases less for the same charge. For example silver: A = 108, Z = 1; copper: A = 63.5, Z = 2. Combining the two laws gives m = (1/F)(A/Z)·I·t. For calculations keep watching the units g and mg: 1 g = 1000 mg.
Worked examples
Silver: k = (1/96500) · (108/1) ≈ 0.00112 g/C = 1.12 mg/C (close to the table value 1.118 mg/C).
Copper: k = (1/96500) · (63.5/2) ≈ 0.000329 g/C = 0.329 mg/C. Copper releases about 3.4 times less than silver because Z = 2 and A is smaller.
Class activity
Make a table in your notebook: write A and Z for Ag, Cu, Al and calculate A/Z (108; 31.75; 9). See that the largest A/Z gives the largest k.
Practice
1
What does Faraday’s second law state?
k is directly proportional to the chemical equivalent A/Z: k = (1/F)(A/Z).
2
If Z = 2 and A = 64, what is A/Z?
32
3
How many grams of a monovalent substance with A = 23 g/mol are released when 96 500 C passes?
23
4
Why is the k of aluminium smaller than that of copper?
For Al A/Z = 27/3 = 9, for Cu 63.5/2 = 31.75; a smaller A/Z gives a smaller k.