Faraday′s Laws of Electrolysis Tutorial
- First Law :
The mass of a substance produced by electrolysis is proportional to the quantity of electricty used.
Example: electrolysis of liquid sodium chloride produces liquid sodium metal and chlorine gas.
Quantity of Electricity Used (Q) mass of sodium produced (m) Q/m 28,950 coulombs 6.9 g 28,950/6.9
Increasing the quantity of electricity (Q) produces more grams of sodium (m).
Q/m is always 4196 in this case.
Q/m is a constant.
Q is proportional to m
that is Q ∝ m
96,500 coulombs 23.0 g 96,500/23.0
482,500 coulombs 115.0 g 482,500/115.0
This means that in order to produce more substance electrolytically we must use more electricity.
- Second Law:
1 The amount of electricity in coulombs required to produce 1 mole of a substance is a simple whole number multiple of2 96 500
That is the amount of electricity (Q) in coulombs divided by 96,500 is a simple whole number
To produce 1 mole of underlined substance quantity of electricity required (Q) Q/96,500 Na+ + e- → Na 96,500 coulombs 96,500/96,500
In each example, Q/96,500 is a simple whole number.
If n represents that simple whole number,
and Q is the quantity of electricity in coulombs, then
n = Q/96,500
Cu2+ + 2e- → Cu 193,000 coulombs 193,000/96,500
Fe3+ + 3e- → Fe 289,500 coulombs 289,500/96,500
The quantity 96,500 is given the name the Faraday (or Faraday Constant) and the symbol F
F is equal to the quantity of electricity carried by one mole of electrons:
F = Avogadro′s Number × charge on electron in coulombs
= 6.022 × 1023 mol-1 × 1.602192 × 10-19 C
= 96,484 C mol-1 (usually rounded up to 96,500 C mol-1 in high school chemistry)
- Faraday′s Laws of Electrolysis Calculations:
Q = n(e-) × F
Q = quantity of electricity measured in coulombs (C)
n(e-) = moles of electrons used
F = the Faraday (Faraday constant) = 96,500 C mol-1
We can calculate the mass of a substance produced during an electrolysis experiment by:
i) calculating the moles of electrons used: n(e-) = Q/F
ii) using the moles of electrons to calculate the moles of substance produced using the balanced reduction (or oxidation) half reaction equation
iii) using the moles of substance to calculate the mass of substance:
mass = moles × molar mass