The correct option is B Cl2(g)+2OH−(aq)→OCl−(aq)+Cl−(aq)+H2O(l)
0Cl2(g)+OH−(aq)→O+1Cl−(aq)+−1Cl−(aq)
nf=(|O.S.Product−O.S.Reactant|)×number of atom
(nf)OCl−=|+1−0|×1=1
(nf)Cl−=|−1−0|×1=1
Cross multiplying with nf values.
Cl2(g)→1 OCl−(aq)+1 Cl−(aq)
Elements except O and H are already balanced.
Now, balancing oxygen by adding H2O
Cl2(g)+H2O(l)→OCl−(aq)+Cl−(aq)
Balancing hydrogen by adding H+
Cl2(g)+H2O(l)→OCl−(aq)+Cl−(aq)+2H+(aq)
For balancing in basic medium adding 2OH− to both sides
Cl2(g)+H2O(l)+2OH−(aq)→OCl−(aq)+Cl−(aq)+2H+(aq)+2OH−(aq)
Cl2(g)+H2O(l)+2OH−(aq)→OCl−(aq)+Cl−(aq)+2H2O(l)
Canceling H2O
Cl2(g)+2OH−(aq)→OCl−(aq)+Cl−(aq)+H2O(l)
Balancing charge:
charge in reactant side =−2=−2
charge in product side =−1−1=−2
So the balanced equation is
Cl2(g)+2OH−(aq)→OCl−(aq)+Cl−(aq)+H2O(l)