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


1. Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                                 0       +5
Oxidation half-reaction:                  
                                              0
Reduction half-reaction:               

2.  Balancing the oxidation half reaction.
(i) Balance 1 atoms by multiplying   by 2.
                          0       +5
                          
(ii) Add 10 electrons towards R.H.S. to balance the charges on iodine atoms,
                            0         +5
                           
(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.
                       
3. Balancing the reduction half reaction. 
 (i) Balancing 1 atoms by multiplying I- by 
2.                     0       +5
                     
(ii) Add two electrons towards L.H.S. in order to balance the charges on iodine atoms
                     

4. Multiplying reduction half-reaction by 5 to equate the electrons and add both the half reactions.

This is a balanced redox reaction.
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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

1. Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                           
Oxidation half-reaction:          

Reduction half-reaction:     

2. Balancing the oxidation half reaction.
(i) The balance I atoms are done by multiplying HIO3 by 2.
                              0             +5
                              
(ii) Add 10 electrons towards R.H.S. in order to balance the changes on iodine atoms. 
                                         
(iii) Balance the O atoms by adding six H2O molecules towards L.H.S.
                           
(iv) Balance H atoms by adding ten H+ towards 
R.H.S.
                  

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            NO2
(ii) Add one electron towards L.H.S. in order to balance the charges on the nitrogen atom. 
                 +5                     +4
                  HNO3 + e-     NO2
(iii) Balance O atoms by adding one H2O molecule towards R.H.S.
                 5                   4
               HNO3  + e-       NO2 + H2O

(iv)  Balance H atoms by adding one H+ towards
L.H.S.
         +5                           +4
        HNO3 + H+ + e-        NO2 + H2O
                                     (Balance reduction half reaction)
4. Multiply balanced reduction half-reaction by 10 to equate electrons and add both the half reactions. 
              
This is a balanced redox reaction.

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Balance the following equation by half reaction method in acidic medium:
MnO subscript 4 superscript minus space plus space Br to the power of minus space space space rightwards arrow space space space Mn to the power of 2 plus end exponent space plus space Br squared


(i) Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                      -1              0
oxidation half reaction:      Br-        Br2
Reduction half-reaction:
      
2. Balancing the oxidation half reaction.
(i) Balance Br atoms by multiplying Br- by 2.
              
(ii) Add 2 electrons towards R.H.S. in order to balance the charges on bromine atoms.
                   
3. Balancing the reduction half reaction. 
(i) Balancing of Mn is not required as the number of each Mn is one on both the sides. 
                      
(ii) Add 5 electrons towards L.H.S. in order to balance the charge on manganese atoms. 
              
(iii) Balance H atoms by adding 8H+ towards 
L.H.S.
           

4. Multiply balanced oxidation half-reaction by 5 and balanced reduction half-reaction by 2 to equate electrons and add both the half reactions.

This is a balanced redox equation.
                       
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Balance the following equation by half reaction method:
straight H subscript 2 straight S space plus space HNO subscript 3 space space rightwards arrow space space space NO space plus space straight S space plus space straight H subscript 2 straight O
 


(1) Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
                                      -2             0
Oxidation half-reaction:   H2S       S
                                     
                                      +5           +2
Reduction half-reaction:     HNO3    NO

(2) To balance the oxidation half reaction. (i) Balancing of S is not required as the number of each S is one on both the sides.
               2-             0
           H2S         S
(ii) Add two electrons towards R.H.S. to balance the charges on sulphur atom.
                     2-           0
                    
(iii) Balance H atoms by adding two H+ towards 
R.H.S.
           2-        0
              + 
   (Balanced oxidation half reduction)

(3) To balance the reduction half reaction
(i) Balancing of N is not required as the number of each N is one on both the sides.
                          + 5           +2
                         HNO3      NO
(ii) Add 3 electrons towards L.H.S. in order to balance the charges on the nitrogen atom. 
            +5                  +2
        HNO3   + 3e-    NO
(iii) Balance the O atoms by adding two H2O molecules on R.H.S.
                                       +2
           HNO3 + 3e-       NO + 2H2O
(iv) Balancing H atoms by adding 3H+ on L.H.S.
                             (Balancing reduction half reaction)

4. Multiplying oxidation half-reaction by 3 and reductive half-reaction by 2 to equate the electrons and adding both the half reactions. 
      

This is a balanced redox reaction.
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Write correctly balanced half reaction and the overall equations for the skeleton equation:
          MnO subscript 4 superscript minus space plus space straight H subscript 2 straight C subscript 2 straight O subscript 4 space space rightwards arrow space space space stack Mn to the power of 2 plus end exponent space plus space CO subscript 2 with left parenthesis in space acid space solution right parenthesis below

1. Write the oxidation and reduction half-reactions by observing the changes in oxidation number and write these separately.
                                        +3            +4 
Oxidation half reaction:        


Reduction half reaction:     +7                +2
                                      

2. Balancing the oxidation half reaction.
 (i) Balance C atoms by multiplying CO2 by 2.
                   +3                     +4
                    
(ii) Add 2 electrons towards R.H.S. in order to balance charges on carbon atoms.
                   
(iii) Balance H atoms by adding two H+ (reaction occurs in the acidic medium) towards R.H.S.


3. Balancing the reduction half reaction.
   (i) Balancing of Mn is not required as the number of each Mn is one on both the sides.
                   +7
                  
(ii) Add 5 electrons towards L.H.S. in order to balance the charges on manganese atoms. 
                       
(iii) Balance H atoms by adding eight H+ towards
L.H.S.
        
                                      (Balanced reduction half reaction)
4. Multiply balanced oxidation half reaction by 5 and balanced reduction half reaction by 2 to equate the electrons and add both the half reactions. 
                              
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