*Please note: you may not see animations, interactions or images that are potentially on this page because you have not allowed Flash to run on S-cool. To do this, click here.*
The process of decomposing a compound using electricity is called electrolysis.
2H2O(l) → 2H2(g) + O2(g)
Electrolysis is a most important industrial process with a wide application. However, the largest application is that of the manufacture of chlorine and sodium hydroxide from concentrated aqueous sodium chloride in the Chlor-alkali industry.
The cell consists of two conducting rods called electrodes, dipped into a compound in a molten state or in solution, that is able to conduct electricity called the electrolyte.
Electrolysis only works with direct current (the charge flows in only one direction). Direct current causes one electrode to take on a positive charge called the anode. The second electrode takes on a negative charge called the cathode.
Note: Anions (negative ions) are attracted to the anode. Cations (positive ions) are attracted to the cathode.
An example of electrolysis:
If we place molten sodium chloride into the electrolysis cell and allow current to flow, the following reactions occur at the electrodes:
At the cathode: Reduction
Na+(l) + e- → Na(l)
At the anode: Oxidation
2Cl-(l) → Cl2(g) + 2e-
These two half equations give the overall reaction:
2Na+(l) + 2Cl-(l) → Na(l) + Cl2(g)
A redox reaction has occured.
In the example above, of molten sodium chloride, there was one cation Na+ and one anion Cl-. Therefore, the products were easily calculated.
However, if a compound is in an aqueous solution, then we have two more ions to deal with, OH- and H+.
So how do we decide with ions will appear at the electrodes and which remain in solution?
In the case of the cathode, we must decide which of the two cations, H+ or Na+ is reduced most easily back to their atoms. Hydrogen ions are most easily reduced so the following cathode reaction occurs:
2H+(aq) + 2e- → H2(g)
In the case of the anode, we must decide which of the two anions, OH- or Cl- will be most easily oxidised back to atoms. Chloride ions are most easily oxidised, hence the following anode reaction:
2Cl-(aq) → Cl2 + 2e-
This leaves Na+ (aq) and OH-(aq) in solution i.e. sodium hydroxide.
We can refer to the electrochemical series, redox series or reactivity series for this information.
In 1832, Michael Faraday deduced that:
The quantity of electricity passed is proportional to the amount of substance discharged at the electrode.
Quantity of electricity (charge) = current x time
The units used are, coulombs for charge, amps for current and seconds for time.
Note 1: One mole of electrons has a charge of 96500C. This quantity of charge is called the Faraday constant, F. Thus, the Faraday constant is related to Avagadro's constant, L, and the charge on the electron, e.
F = L x e
Note 2: The number of moles of electrons required to discharge 1 mole of ions is equal to the charge on the ion.
For example, ions with a double charge, such as Cu2+ - it will take two moles of electrons to deposit one mole of copper.
This industry involves the production of chlorine and the alkali sodium hydroxide from the electrolysis of concentrated aqueous sodium chloride (brine).
The electrolysis cell used for this reaction is called the diaphram cell.
As previously stated, four ions are present Na+, Cl-, OH-, H+.
At the cathode:
2H+(aq) + 2e- → H2(g)
At the anode:
2Cl-(aq) → Cl2(g) + 2e-
The ions remaining in solution are OH- and Na+.
In the diaphragm cell, a porous asbestos partition, is placed between the electrodes to prevent the sodium hydroxide making contact with the chlorine (sodium hydroxide tends to concentrate near the cathode).
Purified brine solution is fed into the anode side and the level kept above that of the cathode. This allows the sodium chloride solution to seep into the cathode compartment and also prevents OH- ions migrating to the anode.
Log in here