1. _Electrolysis of sodium chloride._ Fig. 32 represents a vessel in which the electrolyte is a solution of sodium chloride (NaCl). According to the dissociation theory the molecules of sodium chloride dissociate into the ions Na^{+} and Cl^{-}. The Na^{+} ions are attracted to the cathode owing to its large negative charge. On coming into contact with the cathode, the Na^{+} ions give up their positive charge and are then ordinary sodium atoms. They immediately decompose the water according to the equation Na + H_{2}O = NaOH + H, and hydrogen is evolved about the cathode. The chlorine ions on being discharged at the anode in similar manner may either be given off as chlorine gas, or may attack the water, as represented in the equation 2Cl + H_{2}O = 2HCl + O. 2. _Electrolysis of water._ The reason for the addition of sulphuric acid to water in the preparation of oxygen and hydrogen by electrolysis will now be clear. Water itself is not an electrolyte to an appreciable extent; that is, it does not form enough ions to carry a current. Sulphuric acid dissolved in water is an electrolyte, and dissociates into the ions 2 H^{+} and SO_{4}^{--}. In the process of electrolysis of the solution, the hydrogen ions travel to the cathode, and on being discharged escape as hydrogen gas. The SO_{4} ions, when discharged at the anode, act upon water, setting free oxygen and once more forming sulphuric acid: SO_{4} + H_{2}O = H_{2}SO_{4} + O. The sulphuric acid can again dissociate and the process repeat itself as long as any water is left. Hence the hydrogen and oxygen set free in the electrolysis of water really come directly from the acid but indirectly from the water. 3. _Electrolysis of sodium sulphate._ In a similar way, sodium sulphate (Na_{2}SO_{4}), when in solution, gives the ions 2 Na^{+} and SO_{4}^{--}. On being discharged, the sodium atoms decompose water about the cathode, as in the case of sodium chloride, while the SO_{4} ions when discharged at the anode decompose the water, as represented in the equation SO_{4} + H_{2}O = H_{2}SO_{4} + O [Illustration: Fig. 33] That new substances are formed at the cathode and anode may be shown in the following way. A U-tube, such as is represented in Fig. 33, is partially filled with a solution of sodium sulphate, and the liquid in one arm is colored with red litmus, that in the other with blue litmus. An electrode placed in the red solution is made to serve as cathode, while one in the blue solution is made the anode. On allowing the current to pass, the blue solution turns red, while the red solution turns blue. These are exactly the changes which would take place if sodium hydroxide and sulphuric acid were to be set free at the electrodes, as required by the theory.