3.2 - Faraday's Law
Module 6
Lesson 3.2 Faraday's Law
Key Concepts
Michael Faraday (1791-1867) discovered that the mass of an element produced or consumed at an electrode was directly proportional to the length of time that the cell operated, provided that the current was constant. In addition, he discovered that 9.65x104 C of charge is transferred for every mole of electrons that flows in the cell. This value of 9.65x104 C is known as Faraday's constant (F) and can be used to convert electric charge to number of moles of electrons using the following formula:
\({ n_\mathrm{e^-} = \dfrac{Q}{F} } \)
In the previous section you learned that Q=It.
Since Q = It
\({ n_\mathrm{e^-} = \dfrac{It}{F} } \)
For example, consider the following problem.
An electroplating cell operates for 35 minutes with a current of 1.9 A. Calculate the amount, in moles, of electrons transferred.
Remember that time must be in seconds!
\(t = \mathrm {35~ min \times \dfrac{60 ~s}{1~ min} = 2.1 \times 10^3~ s } \)
\({ n_\mathrm{e^-} = \dfrac{It}{F} } \)
\( \mathrm { = \dfrac{(1.9~ \frac{C}{s} )(2.1 \times 10^3~ s)}{9.65 \times 10^4 \frac{C}{mol}} } \)
\( \mathrm { = 0.041 mol } \)
The amount of electrons transferred is 0.041 mol.
Check Your Understanding
Go to your textbook and complete Practice Questions 6 and 7 on page 654 and Section 14.4 Question 1 on page 657. Click the link below to check your work.
Page 654 Practice Question 6
\( \mathrm { t = \dfrac{n_eF}{I} } \)
\( \mathrm { = \dfrac{(0.146~ mol)(9.65 \times 10^4 \frac{C}{mol})}{1.24 \frac{C}{s} } } \)
\( \mathrm { = 1.14 \times 10^4 s } \)
\( \mathrm { = (1.14 \times 10^4~ s) \times \dfrac{1~ h}{3600~ s} = 3.16~ h } \)
It took 3.16 h.
Page 655 Practice Question 7
\( \mathrm { t = (20 ~min) \times \dfrac{60 s}{1 ~min} = 1.2 \times 10^3 s } \)
\( \mathrm { I = \dfrac{n_eF}{t} } \)
\( \mathrm { = \dfrac{(0.015~ mol)(9.65 \times10^4 \frac{C}{mol})}{1.2 \times 10^3 s} } \)
The current required is 1.2 A.
Page 657 Section 14.4 Question 1
\( \mathrm { t = 15.0 ~min \times \dfrac{60 s}{1~ min} = 900 s } \)
\( \mathrm { n_e = \dfrac{It}{F} } \)
\( \mathrm { = \dfrac{(0.300 \frac{C}{s})(900 s)}{9.65 \times 10^4 \frac{C}{mol}} } \)
\( \mathrm { = 0.002 80 mol } \)
The amount of electrons transferred is 0.002 80 mol.