Balancing of Equations:
- Ion-electron method
- Oxidation number method
[Concept involved that in any chemical reaction e- cannot be produced so no. of e- s in O.H. & R.H. should be same]
- Free halogen on reduction gives halide ion (F2 → F-)
- Alkali metals on oxidation give metallic ion with +1 oxidation state.
- In alkaline medium KMnO4 is reduced to K2MnO4 but in strong alkaline or neutral medium it reduces of Mn+4
- In acid solution KMnO4 is reduced to Mn2+
- H2O2 on reduction gives water and on oxidation gives oxygen.
- Dichromate ion in acid solution is reduced to Cr3+.
Lowest oxidation state → Basic (MnO)
Intermediate oxidation state → Amphoteric (Mn3O4, MnO2)
Highest oxidation state → Acidic (Mn2O7)
Rules For Assigning Oxidation Number:Intermediate oxidation state → Amphoteric (Mn3O4, MnO2)
Highest oxidation state → Acidic (Mn2O7)
- Oxidation number of free elements or atoms is zero.
- Oxidation number of allotropes is zero.
- Oxidation number of atoms in homo-nuclear molecules is zero.
- Oxidation number of mono-atomic ions is equal to the algebraic charge on them.
- Oxidation number of F in compounds is – 1
- Oxidation number of H in its compounds is +1, except in metal hydrides where it is-1
- Oxidation number of O is-2in its compound, but in and in peroxides it is 1 and -0.5 in KO2
- Oxidation number of alkali metals in their compounds + 1.
- Oxidation number of alkaline earth metals in their compounds is + 2.
- The sum of oxidation number of all the atoms in a molecule should be zero and in an ion equal to its charge.
Fe3O4 is FeO. Fe2O3
O.N. of Fe in FeO is + 2 ; O.N. of Fe in Fe2O3 is + 3.
Therefore average O.N. of three Fe atoms (+2 + 2x (+3))/3 = +(8/3)O.N. of Fe in FeO is + 2 ; O.N. of Fe in Fe2O3 is + 3.
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