Question

# In the following reaction : H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g)  calculate the moles of OH− and H2O respectively present in a balanced equation in a basic medium :

A
3,6
B
4,6
C
2,5
D
4,4

Solution

## The correct option is C 2,5H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g) Oxidation state of O in H2O2=−1 Oxidation state of O in 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=2 nf of Cl2O7=8 simplest ratio of nf of H2O2:Cl2O7  is 1:4 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, 4H2O2(aq)+Cl2O7(aq)→2ClO−2(aq)+4O2(g)+3H2O Adding H+ to balance hydrogen, 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 are present in balanced equation.

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