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Redox Reactions

Question
CBSEENCH11006737

Determine the oxidation number of:
(i) Mn in KMnO (ii)Cr in K2Cr2O7    (iii) Cl in KClO?

Solution

 (i) KMnO4.
Let the oxidation number of Mn=x
Writing the oxidation number of each atom at the top of its symbol,
          +1             x           -2
          K             Mn           O4
The algebraic sum of the oxidation number of various atoms = 0
therefore space space space space space space plus 1 space plus space straight x space plus space 4 left parenthesis negative 2 right parenthesis space equals space 0
or space space space space space space space space space space space space space space space space space straight x minus 7 space equals space 0
or space space space space space space space space space space space space space space space space space space space space space space space space space straight x space equals space plus 7

(ii) straight K subscript 2 Cr subscript 2 straight O subscript 7
Let the oxidation number of Cr = x
Writing the oxidation number of each atom at the top of tis symbol,
                         +1             x            -2
                         K2             Cr2           O7
The algebraic sum of the oxidation number of various atoms = 0
therefore space space 2 left parenthesis plus 1 right parenthesis space plus 2 left parenthesis straight x right parenthesis space plus space 7 left parenthesis negative 2 right parenthesis space equals space 0
or space space space space space space space space space space space space space space space space space space space space space 2 straight x minus 12 space equals space 0
or space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight x space equals space plus 6
left parenthesis iii right parenthesis space KClO subscript 4 colon
Let the oxidation number of Cl = x
Writing the oxidation number of each atom at the top of its symbol,
                     +1         x            -2
                      K          Cl            O4
The algebraic sum of the oxidation number ofvarious atoms = 0
therefore space space space plus 1 plus straight x plus 4 left parenthesis negative 2 right parenthesis space equals space 0
or space space space space space space space space space space space space space straight x space minus 7 space equals space 0
or space space space space space space space space space space space space space space space space straight x space equals space plus 7