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 21-1 Electrolysis
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.

Current x
time

Number of coulombs

Mole of electrons passed

Mass of Cu deposited

_ 

Solution
The equation for the reduction of copper(II) ions to copper metal is

Cu2+   + 2e → Cu (reduction, cathode)

mCu = I x t x 1/F x mol ratio x MCu

        = 2.5  C    x (5.0 min x     60 s   )    x     1 mol e   x     2 mol Cu    x  63.55 g Cu 
                   s                             min               96,478 C          1 mol e            mol Cu

        =  2.47 g Cu (about the mass of a copper penny)