The correct option is
A 2, 5
Steps for Balancing redox reactions:
- Identify the oxidation and reduction halves.
- Find the oxidising and reducing agent.
- Find the n-factor of oxidising and reducing agents.
- Cross multiply the oxidising and reducing agent with the n-factor of each other.
- Balance the atoms other than oxygen and hydrogen.
- Balance oxygen atoms.
- Balance hydrogen atoms.
For basic medium:
If x oxygens are less on one side than the other side add x
H2O units to that side. As soon as we add x
H2O units for balancing oxygen, we add generally 2x (or whatever suitable)
H+ ions on the opposite side to balance hydrogen.Then after that we add equal (equal to number of
H+ added) number of
OH− ions to the both sides and combine
H+ and
OH− ions to form
H2O. Then we cancel
H2O which is common to both sides and which can be eliminated thus finally
OH− is left on one side and the equation is balanced. Then we can verify whether the equation is balanced or not by checking the balance of charge on both sides.
formula used for the n-factor calculation,
nf=(|O.S.Product−O.S.Reactant|×number of atoms
Taking the given equation and following the above mentioned steps:
H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g)
Oxidation state of O in
H2O2=−1
Oxidation state of Oin
O2=0
Oxidation state of Cl in
Cl2O7=+7
Oxidation state of Cl in
ClO−2=+3
Clearly,
H2O2 is undergoing oxidation and
Cl2O7 is undergoing reduction.
using the formula of n-factor given above,
nf of
H2O2=1
nf of
Cl2O7=4
Cross multiplying these with
nf of each other.
we get,
4H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g)
Balancing the main elements on both sides, generally oxygen is not balanced in this step but we will balance as it is undergoing oxidation in this reaction.
4H2O2(aq)+Cl2O7(aq)→2ClO−2(aq)+4O2(g)
Adding the
H2O to balance the oxygen,
LHS of the equation contains 15 oxygens while RHS contains only 12 oxygens, so adding 3
H2O to RHS.
4H2O2(aq)+Cl2O7(aq)→2ClO−2(aq)+4O2(g)+3H2O
adding
H+ to balance hydrogen,
As LHS of the equation contains 8 hydrogens while RHS contains 6 hydrogens, so adding 2
H+ to RHS.
4H2O2(aq)+Cl2O7(aq)→2ClO−2(aq)+4O2(g)+3H2O+2H+
Now adding
OH− to both sides to combine with
H+ and make it
H2O,
4H2O2(aq)+Cl2O7(aq)+2OH−→2ClO−2(aq)+4O2(g)+3H2O+2H++2OH−
4H2O2(aq)+Cl2O7(aq)+2OH−→2ClO−2(aq)+4O2(g)+5H2O
This is the final balanced equation.
We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.
4H2O2(aq)+Cl2O7(aq)+2OH−→2ClO−2(aq)+4O2(g)+5H2O
So, 2, 5 moles of
OH− and
H2O respectively.