The correct option is B 3ClO−+2CrO−2+2OH−→3Cl−+2CrO2−4+H2O
ClO−+CrO−2→Cl−+CrO2−4
Oxidation state of Cl in ClO−=+1
Oxidation state of Cl in Cl−=−1
Oxidation state of Cr in CrO−2=+3
Oxidation state of Cr in CrO2−4=+6
CrO−2 is undergoing oxidation and is the reducing agent.
ClO− is undergoing reduction and is the oxidising agent.
nf of ClO−=2
nf of CrO−2=3
Cross multiplying these with nf of each other.
we get,
3ClO−+2CrO−2→Cl−+CrO2−4
Balancing the elements other than oxygen and hydrogen on both sides,
3ClO−+2CrO−2→3Cl−+2CrO2−4
Adding the H2O to balance the oxygen,
3ClO−+2CrO−2+H2O→3Cl−+2CrO2−4
adding H+ to balance hydrogen,
3ClO−+2CrO−2+H2O→3Cl−+2CrO2−4+2H+
Now adding OH− to both sides to combine with H+ and make it H2O,
3ClO−+2CrO−2+H2O+2OH−→3Cl−+2CrO2−4+2H++2OH−
3ClO−+2CrO−2+H2O+2OH−→3Cl−+2CrO2−4+2H2O
Eliminating the unrequired common H2O which is present on the both sides,
3ClO−+2CrO−2+2OH−→3Cl−+2CrO2−4+H2O
Charge on both sides is -7
So, this is the final balanced equation.