The change in oxidation number and balancing a redox reaction
Calculating the change in oxidation number:
Following are the steps to follow
when you calculate change in oxidation number in a redox reaction. Once you
practice two to three times than that becomes super easy to calculate the
change in O.N.
Take an example of the following
reaction (First and easy method).
H2S + Br2 -----> HBr + S
Step 1: Write
the equation in the ionic form and do not balance it.
H++S2-+ Br2 -----> H++Br-+S
Step 2: Ignore
common ions on both side
S2- + Br2 -----> Br- + S
Step 3: Write the
O.N of elements at the top now
S2- + Br20 -----> Br- + S0
Step 4: Find
the O.N change of any element by:
For example, O.N change of S = [0 – (--2)] X 1 = +2
O.N change of Br = [–1 – 0] X 2 = –2
Balancing the change in oxidation number:
We must be clear that increase in oxidation number is called the oxidation and decrease in oxidation number is the reduction. The molecule in which the oxidation number of an element increases is the reducing agent or reductant and whose oxidation number decreases is the oxidising agent or oxidant. Consider the example of:
NH3 + O2
------> NO + H2O
Now to balance the change in oxidation number,
1. Multiply the molecule of
oxidizing agent (O2 on the reactant side), with
the change in oxidation number in the oxidation process.
i.e., O2 becomes 5O2
2. Multiply the
molecule of reducing agent (NH3 on the reactant side), with the change in oxidation number in the
reduction process.
i.e., NH3 becomes 4NH3
3. The reaction is written
including point 1 and 2 together as:
4 NH3 + 5 O2 ------> NO +
H2O
4. Now keep the reactant
side unaltered and make the number of atoms of different elements (whose
O.N has changed) on the product side equal to the number of atoms of the
same elements in the reactant side by multiplying suitable integer.
5. Thus the reaction with
balanced O.N change is:
4 NH3 + 5 O2 ------> 4 NO
+ 6 H2O
Balancing a chemical equation by oxidation number change method
Before balancing a redox reaction, we should have an idea of the
type of redox reactions:
1. Reactions involving neural medium
2. Reactions involving acidic medium
3. Reactions involving alkaline medium
Balancing a chemical equation by ion electron method method:
[Fe(H2O)2(C2O4)2]2-
+ MnO4- ------> Mn2+ + Fe3+ + CO2
Solution:
1.
Write the two half reactions for oxidation and reduction separately,
Oxn
Half: [Fe(H2O)2(C2O4)2]2-
-------> Fe3+ + CO2
Redn
Half: MnO4- ------>
Mn2+
2.
Balance the elements undergoing oxn and reduction. If balanced
already, skip the step.
3.
Balance elements other than oxygen and hydrogen
[Fe(H2O)2(C2O4)2]2-
-------> Fe3+ + 4CO2
4.
Balance oxygen atoms by adding water molecules on the deficient side first and
then add required number of H+ ions to balance hydrogen atom if
needed.
[Fe(H2O)2(C2O4)2]2-
------> Fe3+ + 4CO2 + 2 H2O
The
above equation is automatically balanced for H
MnO4-
------> Mn2+ + 4 H2O
And
then,
MnO4-
+ 8H+------> Mn2+ + 4 H2O
5.
Add required number of electrons to balance the charge
[Fe(H2O)2(C2O4)2]2-
------> Fe3+ + 4CO2 + 2 H2O + 5e-
MnO4-
+ 8H+ + 5e- ------>
Mn2+ + 4 H2O
6.
Multiply suitable integer in both equation, if required, to cancel the
electrons.
In
this case both equations are automatically balanced electronically.
7.
Now add both equations to get the final balanced equation
[Fe(H2O)2(C2O4)2]2- +MnO4- +8H+ ------>Mn2++Fe3++4CO2 +6H2O
To
balance a redox reaction given in basic medium, follow
all the steps upto step 3 and then,
4. Balance oxygen atoms by adding
required number of H2O molecules on deficient side.
5. For balancing H atoms, add a water
molecule for each H-atom on the deficient side and add an equal number of OH-
ion on the opposite side.
6. Multiply suitable integer in both equation, if required, to cancel the electrons.
7. Now add both equations to get the final balanced equation.
Do it yourself now:
Cr(OH)3 + IO3-
-------> I - + CrO42- (In basic medium)
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