The correct option is C 2ClO2+SbO−2+2H2O+2OH−→2ClO−2+Sb(OH)−6
+4ClO2++3SbO−2→+3ClO−2++5Sb(OH)−6
ClO2 is an oxidising agent.
SbO−2 is a reducing agent.
nf=(|O.S.Product−O.S.Reactant|)×number of atom
+3SbO−2→+5Sb(OH)−6 oxidation
nf=(|+5−3|)×1=2
+4ClO2→+3ClO−2 reduction
nf=(|+3−4|)×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+SbO−2→ClO−2+Sb(OH)−6
Balancing elements except O and H
2ClO2+SbO−2→2ClO−2+Sb(OH)−6
Balancing oxygen by adding H2O
2ClO2+SbO−2+4H2O→2ClO−2+Sb(OH)−6
Balancing hydrogen by adding H+
2ClO2+SbO−2+4H2O→2ClO−2+Sb(OH)−6+2H+
Adding 2OH− on both sides to balance in basic medium
2ClO2+SbO−2+4H2O+2OH−→2ClO−2+Sb(OH)−6+2H++2OH−
2ClO2+SbO−2+4H2O+2OH−→2ClO−2+Sb(OH)−6+2H2O
Canceling H2O
2ClO2+SbO−2+2H2O+2OH−→2ClO−2+Sb(OH)−6
Balancing charge:
charge in reactant side =−1−2=−3
charge in product side =−2−1=−3
So the balanced equation is
2ClO2+SbO−2+2H2O+2OH−→2ClO−2+Sb(OH)−6