Faraday and Stoichiometric Relationships in Electrolysis

Much of the early work in electrochemistry was performed by British scientist Michael Faraday. He determined that the amount of chemical change that occurs during electrolysis is directly proportional to the amount of electricity that is passed through the cell. Consider the following reduction of iron:

Fe +3(aq) + 3 e- à Fe (s)

The reduction of 1 mol of iron (III) requires 3 moles of electrons. If we required the reduction of 3 moles of iron (III) we would need 9 moles of electrons.

Faraday was honoured with the unit for 1 mol of electrons , the Faraday F.

To use the Faraday in calculations we must be able to relate it to electrical measurements that can be made in the laboratory. This involves the use of Coulombs and Amperes. We know that 1 Coulomb = 1 Ampere x 1 second (since 1 A = 1 C/s). Experimentally it has been determined that 1 F = 96, 500 C. We can see how this is used practically with the following example:

Question: How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00A is run through a solution of CuSO4 for 20 minutes?

Solution: 2.00 A x (20 minutes x 60 s/min) = 2.00A x 1200 s = 2400C

2400 C /(96 500 C/F ) = 0.0249 F

The reduction of Cu would occur at the cathode :

Cu+2 (aq)+ 2 e- à Cu (s)

Thus there are 2 moles of e- for every 1 mol Cu

Moles of Cu = 0.0249 F x (1 mol Cu /2 F ) = 0.0125 mol Cu

Mass Cu = 0.0125 mol x 63.5 g / mol = 0.794 g

Work : 1) How many moles of OH- (aq) will be produced at the cathode during hydrolysis with a current of 4.00 A for 200s?

2) How long in minutes will it take a current of 10.0 A to deposit 3.00 g of gold from a solution of AuCl3?

3) What current must be supplied to deposit 3.00 g of gold from a solution of AuCl3 in 20.0 minutes?

1) 8.29 x 10 -3 mol OH- 2) 7.35 minutes 3) 3.67 A