ithdrawn is labeled as
positive. The one receiving the electrons is labeled as negative. Electrolysis
Electrolysis of Molten Salts
Voltaic cells are based on spontaneous oxidation-reduction reactions. Conversely, it is possible to use electrical energy to cause nonspontaneous oxidation-reduction reactions to take place. For example, electricity can be used to decompose molten sodium chloride into its component elements, a very nonspontaneous process.
2 NaCl(l) ==> 2 Na(l) + Cl2(g)
Processes such as this, which are driven by an outside source of electrical
energy, are called electrolysis reactions and take place in electrolytic
cells. An electrolytic cell consists of two electrodes in a molten
salt or solution of electrolyte. The cell is driven by a battery or some
other source of direct current. The battery acts as an electron
pump, pushing electrons into one electrode and pulling them from the
other. The electrode from which the electrons are withdrawn is labeled as
positive. The one receiving the electrons is labeled as negative.
Na+(l) + e-==> Na(l) reduction half- reaction
Cl-(l) ==> 1/2 Cl2(g) + e- oxidation half- reaction
If we pass electricity through molten NaCl in a cell such as the one below, we observe yellow-green chlorine gas form at the anode and silvery, liquid metallic sodium at the cathode. The decomposition of NaCl(l) into its elements is a very nonspontaneous reaction, with a large positive value of free energy. The function of the battery or external source of electricity is to act as an electron pump, forcing electrons onto the cathode where the reduction of sodium ions occurs. The battery withdraws electrons from the anode where the oxidation of chlorine ions occurs. above takes place. So if we impose a voltage of something in excess of 4 V we can force electrons to flow in the opposite direction from the spontaneous reaction.
Electrolysis in Aqueous Solutions
If we impose a voltage in excess to an electrolytic cell containing aqueous NaCl, we do not get the same products as in the electrolysis of molten NaCl. Electrolyses in aqueous solution are generally more complicated than in molten salts, because in aqueous solutions there can be competing reactions at the anode and cathode.
Two processes that can always occur in aqueous solution are the reduction and oxidation of water itself.
Reduction: 2 H2O + 2 e-==> H2(g) + 2 OH-
Oxidation: 2 H2O ==> 4 H+ (aq)+ O2(g)
When the half-reactions shown above are the only ones feasible, the electrolysis of an aqueous solution produces H2(g) and O2(g) as its products. There are cases, however, where only one or maybe neither of those half-reactions occurs. For example consider the electrolysis of NaCl(aq). To obtain sodium metal by the reduction of Na+(aq), we have to overcome a reduction potential of -2.713 V. This is more difficult to accomplish than the reduction, so the the reduction of Na+(aq) is not going to happen (mother nature, like most of us is going to take the easy way).
To obtain chlorine gas by the oxidation of Cl-(aq) we must overcome a potential of -1.358 V which is not that different from the -1.229 V for the oxidation of water. Both oxidations can occur. From a consideration of oxidation potentials alone, we would predict that the oxidation of water would predominate, since it has the lower oxidation potential. In fact in concentrated solutions of NaCl, the oxidation of chloride predominates, because the overvoltage for the formation of O2(g) is much higher than that of Cl2(g). The overvoltage is the excess voltage above that calculated to bring about a reaction.
Stoichiometry of Electrolysis
The quantitative basis of electrolysis was established by Michael Faraday in 1831. His results were summarized in two laws, now known as Faraday's laws of electrolysis.
Next a quick refresher in electricity. The quantity of electric charge is measured in coulombs (C). Electric current, which describes the rate of electric charge is measured in amperes (A) sometimes shortened to amps.
1 ampere (A) = 1 coulomb (C) / 1 second (s)
The total electric charge involved in an electrolysis reaction is a product of the current and the time that it flows.
Electric charge (C) = current (C/s) x time (s)
One other quantity we need to be familiar with is the electric charge on one mole of electrons. This is known as a Faraday and has the value 96,485 C/mol.
Example
How many grams of copper are deposited at a platinum electrode in the electrolysis of CuSO4(aq) by 1.75 A of electric current in 2.30 h?
The reduction half-reaction will be Cu2+(aq) + 2 e-==> Cu(s)
The total electric charge involved is
Charge = 1.75 C/s x 2.30 h x 60 min/h x 60 s/min = 1.45 x 104 C
The number of moles of electrons is
? mol e- = 1.45 x 104 C x 1 mol e- / 96,485 C = 0.150 mol e-
According to the reduction half-reaction, 2 mol e- are required for every mol Cu
? mol Cu = 0.150 mol e- x 1 mol Cu / 2 mol e- = 0.0750 mol Cu
The mass of Cu will be the molar mass times the number of moles
? g Cu = 0.0750 mol Cu x 63.55 g Cu / 1 mol Cu = 4.77 g Cu