Long Answer Type
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.Â
The Mn3+ ions is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction.
Short Answer TypeLong Answer TypePermanganate (VII) ion, 
 in basic solution oxidises iodide ion, I- to produce molecule iodine (I2) and manganese (IV) oxide (MnO2). Write a balanced ionic equation to represent redox reaction.Â
Balance the following equations in basic medium by ion-electron method and oxidation number method:
(A) Ion-electron method:Â
1. Write the oxidation and reduction half-reaction by observing the changes in oxidation numbers.
Oxidiation half reaction:
    -1            0
  
Reduction half-reaction:
    +7          +3
    
2. Balancing the oxidation half reaction
  (i) Add 2 electrons towards R.H.S. to balance the charges
         
  (ii) Add 2 OH- ions towards L.H.S to balance the charges
     
(iii) Balance O atoms by adding two H2O molecules towards H
      
3. Balancing the reduction half reaction
  
  
 4. Multiply balanced oxidation half-reaction by 4 and add it to the balanced reduction half-reaction.
(B) Oxidation number method      Â
 (i) The skeleton equation along with oxidation number of each atom is
+7 -2 Â Â Â Â Â +1 -1 Â Â Â Â Â Â Â Â +3 Â Â Â Â Â Â Â 0
(ii)Â The ON of O atom increases by 1 per atom while that of CI decreases by 4 per atom.
Thus Cl2O7(g) acts as an oxidising agent while H2O2(aq) as the reducing agent
Total increase in O.N. of H2O2Â = 2 X 1=2
Total decrease in O.N. of Cl2O7Â = 4 x 2 = 8
(iii) Equalise the increase/decrease in O.N. by multiplying H2O2Â and O2Â by 4.
(iv) Balance Cl atoms by multiplying 
   
(v) Balance O atoms by adding three 
 towards R.H.S.
      
(vi) Balance H atoms by adding two 
 towards R.H.S. and two OH- towards L.H.S.
This is the balanced redox equation.
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