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Redox Reactions

Question
CBSEENCH11006777

Balance the equation by half reaction method (ion-electron method):
straight I subscript 2 space plus space HNO subscript 3 space space rightwards arrow space space space HIO subscript 3 space plus space NO subscript 2 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.
                                           
Oxidation half-reaction:          space space stack bold I subscript bold 2 with bold 0 on top bold space bold space bold rightwards arrow bold space bold space bold H bold I with bold plus bold 5 on top bold O subscript bold 3
Reduction half-reaction:     bold space bold space bold H bold N with bold plus bold 5 on top bold O subscript bold 3 bold space bold rightwards arrow bold space bold space stack bold space bold N with bold plus bold 4 on top bold O subscript bold 2
2. Balancing the oxidation half reaction.
(i) The balance I atoms are done by multiplying HIO3 by 2.
                              0             +5
                              straight I subscript 2 space space rightwards arrow space space 2 HIO subscript 3
(ii) Add 10 electrons towards R.H.S. in order to balance the changes on iodine atoms. 
                                         space space stack bold I subscript bold 2 with bold 10 on top bold space bold rightwards arrow bold space bold space bold 2 bold H bold I with bold plus bold 5 on top bold O subscript bold 3 bold space bold plus bold space bold 10 bold e to the power of bold minus
(iii) Balance the O atoms by adding six H2O molecules towards L.H.S.
                           space space stack bold I subscript bold 2 with bold 0 on top bold plus bold 6 bold H subscript bold 2 bold O bold space bold space bold rightwards arrow bold space bold space bold space bold 2 bold H bold I with bold plus bold 5 on top bold O subscript bold 3 bold space bold plus bold space bold 10 bold e to the power of bold minus
(iv) Balance H atoms by adding ten H+ towards 
R.H.S.
                  space space stack bold I subscript bold 2 bold space with bold 0 on top bold plus bold space bold 6 bold H subscript bold 2 bold O bold space bold space bold rightwards arrow bold space bold space bold space bold 2 bold H bold I with bold plus bold 5 on top bold O subscript bold 3 bold space bold plus bold space bold 10 bold e to the power of bold minus bold space bold plus bold space bold 10 bold H to the power of bold plus
bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold Balanced bold space bold oxidation bold space bold half bold space bold reaction bold right parenthesis
3. Balancing the reduction half reaction. 
 (i) Balancing of N is not required as the number of each N is one on both the sides. 
                  +5                   +4
                HNO3         rightwards arrow   NO2
(ii) Add one electron towards L.H.S. in order to balance the charges on the nitrogen atom. 
                 +5                     +4
                  HNO3 + e-  rightwards arrow   NO2
(iii) Balance O atoms by adding one H2O molecule towards R.H.S.
                 5                   4
               HNO3  + e-     rightwards arrow  NO2 + H2O
(iv)  Balance H atoms by adding one H+ towards
L.H.S.
         +5                           +4
        HNO3 + H+ + e-     rightwards arrow   NO2 + H2O
                                     (Balance reduction half reaction)
4. Multiply balanced reduction half-reaction by 10 to equate electrons and add both the half reactions. 
              space space space space space space straight I subscript 2 space plus space 6 straight H subscript 2 straight O space space rightwards arrow space space space 2 HIO subscript 3 space plus space 10 straight H to the power of plus space plus space 10 straight e to the power of minus
HNO subscript 3 space plus space straight e to the power of minus space plus space straight H to the power of plus space space rightwards arrow space space space NO subscript 2 space plus space straight H subscript 2 straight O right square bracket space cross times space 10
minus negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative
straight I subscript 2 space plus space 10 HNO subscript 3 space space rightwards arrow space space space 2 HIO subscript 3 space plus space 10 NO subscript 2 space plus space 4 straight H subscript 2 straight O
minus negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative
This is a balanced redox reaction.