Depression in Freezing point and Elevation in Boiling point

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Depression in Freezing Point: Depression in freezing point means freezing point of solution is always less than that of its pure solvent due to lowering of vapour pressure of solution. Depression in freezing point have the formula as given below, ∆T f = i × K f × m                                       eq.. (1) K f = cryoscopic constant This equation tells us that depression of freezing point depends only on   molality of solutions, so it is a colligative property. m= 1000W 2 /W 1 ×M 2                           putting the value of molality in we get          ∆T f   = i × K f × 1000W 2 /W 1 ×M 2        ...
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Balancing Redox Reactions Oxidation number method(part2)

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How to balance redox reactions by Oxidation number method?

Balancing Redox Reactions by Oxidation number method:

The key to the oxidation number method of balancing redox reactions is to realize that net change in the total of all oxidation numbers must be zero. Any increase in oxidation number for the oxidized atoms must be matched by corresponding decrease in oxidation state for the reduced atoms.

Following are the steps for balancing redox reactions by oxidation number method:

1.   1. Write down the skeletal equation of redox reaction. Let’s take the following reaction for example

Br2 + NaOH→ NaBr + NaBrO3 + H2O


2.   2. Assign oxidation number to all atoms including O and H atoms if present in both reactants and products.

Br2 + NaOH→ NaBr + NaBrO3 + H2O

↑      ↑    ↑   ↑      ↑    ↑      ↑ ↑   ↑      ↑   ↑

 0     +1 -2 +1    +1 -1   +1 +5 -6    +1 -2

3.   3. Now see which atoms have undergone change in oxidation number.

   Br2 + NaOH→ NaBr + NaBrO3 + H2O           (reduction)

0-1                               
                


    Br2 + NaOH→ NaBr + NaBrO3 + H2O          (oxidation)

0+5

 

4.  4.  We should write Br2 twice on left side because it has undergone both oxidation and reduction.


  Br2 +Br2 + NaOH→ NaBr + NaBrO3 + H2O

0-1

5. Now balance the bromine atoms which have undergone reduction

 
  Br2 +Br2 + NaOH→ 2NaBr + NaBrO3 + H2O            ( 2 electrons gained)

0-1(2)


    Br2 + Br2 + NaOH→ NaBr + 2NaBrO3 + H2O              (10 electrons lost)

             0  +5(2)  

 

5.   6. Now equate the number of electrons gained and lost by multiplying with suitable digit.

Multiply the electrons gained by 5, in short reduction equation by 5.

5Br2 +Br2 + NaOH→ 10NaBr + 2NaBrO3 + H2O  

 In short : 6Br2 + NaOH→ 10NaBr + 2NaBrO3 + H2O  

 Now balance sodium atoms as follows

  6Br2 + 12NaOH→ 10NaBr + 2NaBrO3 + H2O  

  Now balance O and H atoms

  6Br2 + 12NaOH→ 10NaBr + 2NaBrO3 + 6H2O  

 The above equation is the balanced form of equation showing reaction                 between Br2 and NaOH.

  However, the correct stoichiometric form of above equation is

  3Br2+6NaOH→5NaBr+NaBrO3+3H2O

 

Let’s take another example:

Unbalanced equation is  Cu +HNO3→Cu(NO3)2 + NO2+H2O


 

Cu +HNO3→Cu(NO3)2 + NO2+H2O

↑      ↑ ↑ ↑         ↑ ↑ ↑          ↑ ↑     ↑ ↑

0  +1 +5 -2    +2 +5 -2    +4 -2  +1  -2


 Cu +HNO3→Cu(NO3)2 + NO2+H2O            (oxidation)

0+2


Cu + HNO3→Cu(NO3)2 + NO2 +H2O            (reduction)

    +5+4            

We have to write nitrogen twice


Cu+HNO3+HNO3→Cu(NO3)2 + NO2+H2O       (1 electron gained)

              +5+4           

  





 Cu +HNO3→Cu(NO3)2 + NO2+H2O                      (2 electrons lost)            

0+2


Balance nitrogen atoms for HNO3 which have not undergone oxidation or reduction

Cu + 2HNO3 +HNO3→Cu(NO3)2 + NO2+H2O    

Now balance the electrons gained and lost by multiplying with suitable digit 


Cu + 2HNO3 +2HNO3→Cu(NO3)2 +2NO2+H2O       (2 electrons gained)

                      +5+4  

Cu + 4HNO3 →Cu(NO3)2 + 2NO2+H2O    

Now balance hydrogen and oxygen atoms

Cu + 4HNO3 →Cu(NO3)2 + 2NO2+2H2O   

This is the final balanced equation.

 

 

        

For further details visit Balancing Redox Reactions by both Ion-Electron and Oxidation number method (part3)

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