- #1
PFuser1232
- 479
- 20
According to my chemistry textbook, the I- ion is discharged at the anode. Yet, according to standard electrode potential values, the OH- ion should be discharged at the anode. Thoughts on this please?
Electrolysis of aqueous sodium iodide involves passing an electric current through a solution of sodium iodide in water. The electric current breaks down the sodium iodide into its individual components, sodium and iodine ions, which then migrate towards the positive and negative electrodes respectively.
Aqueous sodium iodide is used in electrolysis because it allows for the flow of electricity through the solution. The presence of water allows for the sodium iodide to dissociate into ions, making it an excellent conductor of electricity.
The products of electrolysis of aqueous sodium iodide are sodium and iodine ions. The sodium ions migrate towards the negative electrode and the iodine ions towards the positive electrode. At the electrodes, the ions gain or lose electrons to form sodium atoms and iodine atoms, which then combine to form sodium iodide molecules.
Several factors can affect the rate of electrolysis of aqueous sodium iodide. These include the concentration of the solution, the distance between the electrodes, the strength of the electric current, and the surface area of the electrodes. Higher concentrations, shorter distances, stronger currents, and larger surface areas generally lead to faster rates of electrolysis.
Electrolysis of aqueous sodium iodide has several practical applications, including the production of sodium hydroxide, chlorine gas, and hydrogen gas. It is also used in the purification of metals, the production of batteries, and in certain industrial processes such as electroplating and water treatment.