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1) Judging the redox ability of different oxidation states
In the free energy-oxidation number diagram, the slope of each line segment is different
.
Investigation segment the IO .
The corresponding electrode reaction is
From ②-③, we get
Obviously
Therefore, the line segment the IO .
3 - -IO - slope molecules couple the IO .
3 - / the IO - standard free energy change electrode reaction rGm in △ [Theta] @ a negative slope denominator couple the IO .
3 - / the IO - electrode reaction The number of electron transfers z
Substituting △rGm Θ =-zFE Θ into the above formula, we get
k=FE Θ
The above formula shows that the slope k of the line segment in the free energy oxidation number graph is proportional to the standard electrode potential E Θ of the electrical pair composed of the oxidation states at both ends of the line segment
.
Therefore, the greater the slope of the line segment in the figure, the stronger the oxidizing ability of the oxidizing type of the electric pair; the smaller the slope of the line segment, the stronger the reducing ability of the reducing type of the electric pair
In the above-described iodine in alkaline medium free energy - FIG oxidation number, H .
3 the IO .
2) Judging the possibility of disproportionation reaction
Iodine in free energy - FIG oxidation number, the I 2 different slope segments on both sides of a single substance, the I 2 -I - the slope is greater than the IO - -I 2 slope, indicating couple I2 / the I - electrode potential is greater than the couple the IO - / the I 2 electrode potential
.
That is, the electrode potential of I2 as an oxidizing pair is greater than its electrode potential as a reducing type.
The I 2 + 2OH - = the I - + the IO - + H 2 O
This discussion is intuitively reflected in the free energy-oxidation number diagram.
According to the above discussion, IO - and I 2 are unstable in alkaline medium, and disproportionation reaction occurs; IO 3 -is stable, and disproportionation reaction does not occur