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Redox Reactions

Question
CBSEENCH11006773

Balance the equation by half reaction method:

straight I subscript 2 space plus space OH to the power of minus space space space space space rightwards arrow space space space space IO subscript 3 superscript minus space plus space straight I to the power of minus space plus space straight H subscript 2 straight O

Solution
1. Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                                 0       +5
Oxidation half-reaction:                  space straight I subscript 2 space space rightwards arrow space space space straight I space space straight O subscript 3 superscript minus space space
                                              0
Reduction half-reaction:               space straight I subscript 2 space space rightwards arrow space space straight I to the power of minus
2.  Balancing the oxidation half reaction.
(i) Balance 1 atoms by multiplying space IO subscript 3 superscript minus  by 2.
                          0       +5
                          space straight I subscript 2 space space rightwards arrow space space 2 straight I space space straight O subscript 3 superscript minus
(ii) Add 10 electrons towards R.H.S. to balance the charges on iodine atoms,
                            0         +5
                           straight I subscript 2 space space space rightwards arrow space space space space 2 straight I space space space straight O subscript 3 superscript minus space space plus space 10 straight e to the power of minus
(iii) Since the medium is alkaline, 6H2O molecules are added towards R.H.S. and 12 OH-ions towards L.H.S. in order to balance oxygen and hydrogen atoms.
                       space space straight I subscript 2 space plus space 12 OH to the power of minus space space rightwards arrow space space space space 2 IO subscript 3 superscript minus space plus space 6 straight H subscript 2 straight O space plus space 10 straight e to the power of minus
space space space space space space space space space space space space space space space space space space space space space left parenthesis Balanced space oxidation space half space reaction right parenthesis
3. Balancing the reduction half reaction. 
 (i) Balancing 1 atoms by multiplying I- by 
2.                     0       +5
                     space space space straight I subscript 2 space space space rightwards arrow space space space space 2 straight I to the power of minus
(ii) Add two electrons towards L.H.S. in order to balance the charges on iodine atoms
                     space space space straight I subscript 2 space plus space 2 straight e to the power of minus space space space rightwards arrow space space space space 2 straight I to the power of minus
space space space space space space space space space space space space space space space space space space space space space space space left parenthesis Balanced space reduction space half space reaction right parenthesis
4. Multiplying reduction half-reaction by 5 to equate the electrons and add both the half reactions.
space straight I subscript 2 space plus space 12 OH to the power of minus space space space space rightwards arrow space space space space 2 IO subscript 3 superscript minus space plus space 6 straight H subscript 2 straight O space plus space 10 straight e to the power of minus
space straight I subscript 2 space plus space 2 straight e to the power of minus space space space space rightwards arrow space space space 2 straight I to the power of minus right square bracket space cross times space space 5
minus negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative
space 6 straight I subscript 2 space plus space 12 OH to the power of minus space space space rightwards arrow space space space 10 straight I to the power of minus space plus space 2 IO subscript 3 superscript minus space plus space 6 straight H subscript 2 straight O
space or space
3 straight I subscript 2 space plus space 6 OH to the power of minus space space rightwards arrow space 5 straight I to the power of minus space plus space IO subscript 3 superscript minus space plus space 3 straight H subscript 2 straight O
This is a balanced redox reaction.