Redox reactions in Basic Medium

Asked by vk1994 | 24th Jul, 2010, 10:38: PM

Expert Answer:

  The oxidation number of CHLORINE reduces from (+7) in Cl 2O7 to (+3) in ClO2 - so it is cathodic half reaction while oxidation number of oxygen increases its from (-1) in H2O2 to (0) in O2 so it is anodic half reaction

ANODIC Half Reaction

H2O2(aq) ---> O 2(g)

In order to balance the Oxidation Number put the Electrons exchanged on right hand side

H2O2(aq) ---> O 2(g) + 2e-

Since the reaction is carried out in basic medium, balance the charge and hydrogen atoms by adding OH   on left hand side and water on right hand side

H2O2(aq) +2 OH- (aq)--> O 2(g) + 2e -+ 2 H2O(aq)
It results balanced.

CATHODIC Half .Reaction

Cl2O7(aq) ---> ClO2-(aq)

first balance the CHLORINE ATOMS

Cl2O7(aq) --->2ClO2-(aq)

Then balance the Oxidation Number by adding electrons on left hand side

Cl2O7(aq) +8 e  ---> 2ClO2-(aq)

Now add water molecules in order to balance the Oxygen Atoms
Cl2O7(aq) +8 e  ---> 2ClO2-(aq) + 3 H2O(aq)

Now to balance the hydrogen atoms add OH - ions on right hand side and water on left hand side(since electrons are on left) and water on left

Cl2O7(aq) +8 e   + 3 H2O(aq)---> 2ClO2-(aq) + 6 OH-(aq)
It results balanced.

Now, add the Half-Reactions in order to remove the elctrons

H2O2(aq) +2 OH- (aq)--> O 2(g) + 2e -+ 2 H2O(aq)
Cl2O7(aq) +8 e   + 3 H2O(aq)---> 2ClO2-(aq) + 6 OH-(aq)
Multiply the first equation by 4

4 H2O2(aq) +8 OH- (aq)--> 4O 2(g) + 8e -+ 8 H2O(aq)

Adding this to other half reaction

Cl2O7(aq) +8 e   + 3 H2O(aq)---> 2ClO2-(aq) + 6 OH-(aq)

Net reaction is

4 H2O2(aq) + Cl2O7(aq) + 2 OH-(aq) --->4O2(g) + 2 ClO2 -(aq) + 5 H2O(aq)

 

Answered by  | 26th Jul, 2010, 01:38: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.