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Question

ClO2+SbO2ClO2+Sb(OH)6
The balanced form of the above reaction in basic medium using oxidation number method is:

A
2ClO2+SbO2+2OH2ClO2+Sb(OH)6+2H2O
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B
ClO2+SbO2+2H2O+OHClO2+Sb(OH)6
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C
2ClO2+SbO2+2H2O+2OH2ClO2+Sb(OH)6
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D
2ClO2+4SbO2+2OHClO2+4Sb(OH)6+3H2O
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Solution

The correct option is C 2ClO2+SbO2+2H2O+2OH2ClO2+Sb(OH)6
+4ClO2++3SbO2+3ClO2++5Sb(OH)6

ClO2 is an oxidising agent.
SbO2 is a reducing agent.

nf=(|O.S.ProductO.S.Reactant|)×number of atom

+3SbO2+5Sb(OH)6 oxidation
nf=(|+53|)×1=2

+4ClO2+3ClO2 reduction
nf=(|+34|)×1=1

Ratio of nf for oxidation to reduction is 2:1.

Cross mutiplying the oxidising and reducing agents with ratio of n-factors.
2ClO2+SbO2ClO2+Sb(OH)6

Balancing elements except O and H
2ClO2+SbO22ClO2+Sb(OH)6

Balancing oxygen by adding H2O
2ClO2+SbO2+4H2O2ClO2+Sb(OH)6

Balancing hydrogen by adding H+
2ClO2+SbO2+4H2O2ClO2+Sb(OH)6+2H+

Adding 2OH on both sides to balance in basic medium
2ClO2+SbO2+4H2O+2OH2ClO2+Sb(OH)6+2H++2OH

2ClO2+SbO2+4H2O+2OH2ClO2+Sb(OH)6+2H2O

Canceling H2O
2ClO2+SbO2+2H2O+2OH2ClO2+Sb(OH)6

Balancing charge:

charge in reactant side =12=3
charge in product side =21=3

So the balanced equation is
2ClO2+SbO2+2H2O+2OH2ClO2+Sb(OH)6

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