Electrolysis

In 1832–1833, Michael Faraday’s studies of electrolysis led to this conclusion.

The amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

This is Faraday’s Law of Electrolysis. A quantitative unit of electricity is now called the faraday.

One faraday is the amount of electricity that corresponds to the gain or loss, and therefore the passage, of 6.022 x 1023 electrons, or one mole of electrons.

1 faraday = 6.022 x 1023 e = 96,485 C

Example

Calculate the mass of copper metal produced during the passage of 2.50 amperes of current through a solution of copper (II) sulfate for 50.0 minutes.

Plan
The half-reaction that describes the reduction of copper (II) ions tells us the number of moles of electrons required to produce one mole of copper metal. Each mole of electrons corresponds to 1 faraday, or 9.65 x 104 coulombs, of charge. The product of current and time gives the number of coulombs.

   

Given:
I = 2.50 A
t = 50.0 min = 300 s
1 F = 1 mole electrons
           9.65 x 104 C 		 Cu2+ (aq) + 2 e- → Cu (s)   

Multiple Steps

First Calculate the amount of charge (number of Coulombs)
Q = I * t
    = 2.50 A x 300 s
    = 750 C

Second calculate the moles of electrons (using Faraday's Constant)
mole Electrons = Q * 1/F
                        = 750 C     x        1 mole e-      
                                                 9.65 x 104 C
                       = 0.00789 mol e-

Third Calculate moles of Cu (s) (using mole ratio)
mole Cu = mole e- x mole ratio
              = 0.00789 mol e- x 1 mol Cu
                                           2 mol e-
              = 0.00395 mol Cu

Finally, calculate the mass of Cu
mass Cu = mole Cu x Molar Mass
              = 0.0097 mol Cu x 63.55 g Cu
                                              1 mol Cu
              = 0.247 g Cu

One Step

mass Cu = I x t x   1   x mol ratio x MCu
                             F

              = 2.50 A   x   300   s x        1 mol e-        x   1 mol Cu   x   63.55 g Cu   
                                                     9.65 x 104 C          2 mol e-          1 mol Cu

              = 0.247 g Cu