Karl Fischer Coulometric Moisture Analyzer Principle
1. In 1935, Karl Fischer first proposed the method of measuring moisture by volumetric analysis, which is the visual method in GB6283 "Determination of Moisture Content in Chemical Products". The visual method can only determine the water content of colorless liquid substances. Later, it developed into the electricity method. With the development of science and technology, the coulomb meter and the volumetric method were combined to launch the coulomb method. This method is the test method in GB7600 "Determination of Moisture Content in Transformer Oil in Operation (Coulometric Method)". The classification visual method and the electricity method are collectively referred to as the capacity method. The Karl Fischer method is divided into two methods: the Karl Fischer volumetric method and the Karl Fischer Coulomb method. Both methods are designated as standard analytical methods by many countries to calibrate other analytical methods and measuring instruments.
2. The Karl Fischer Coulomb method is an electrochemical method for determining moisture. The principle is that when the Karl Fischer reagent in the electrolytic cell of the instrument reaches equilibrium, inject the sample containing water, the redox reaction of water ginseng, iodine and sulfur dioxide, in the presence of pyridine and methanol, generate pyridinium hydriodate and pyridinium methylsulfate , the consumed iodine is electrolyzed at the anode, so that the oxidation-reduction reaction continues until the water is completely exhausted. According to Faraday's law of electrolysis, the iodine produced by electrolysis is proportional to the electricity consumed during electrolysis. The reaction is as follows:
H2O+I2+SO2+3C5H5N→2C5H5N?HI+C5H5N?SO3
C5H5N?SO3+CH3OH→C5H5N?HSO4CH3
During electrolysis, the electrode reaction is as follows:
Anode: 2I--2e→I2
Cathode: I2+2e→2I-
2H++2e→H2↑
It can be seen from the above reaction that 1 mole of iodine oxidizes 1 mole of sulfur dioxide and requires 1 mole of water. Therefore, it is the equivalent reaction of 1 mole of iodine and 1 mole of water, that is, the electricity for electrolyzing iodine is equivalent to the electricity for electrolyzing water. The electrolysis of 1 mole of iodine requires 2×96493 coulombs of electricity, and the electrolysis of 1 millimole of water requires 96493 milliocoulombs of electricity.
