When the following redox reaction Cl2+OH−→ClO−3+Cl−+H2O
is balanced by ion-electron method , what will be the coefficient of Cl2 and OH− respectively?
A
3,6
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B
2,4
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C
1.2
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D
5,5
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Solution
The correct option is A3,6 0Cl2+OH−→+5ClO−3+Cl−+H2O
Oxidation half : Cl2→ClO−3
Chlorine changes its oxidation state from 0 to +5
Reduction half : Cl2→2Cl−.
Chlorine changes its oxidation state from 0 to −1
Taking the oxidation half reaction : Cl2→ClO−3
Balancing Cl Cl2→2ClO−3
Balancing O by adding H2O Cl2+3H2O→2ClO−3
Balancing H by adding H+ Cl2+3H2O→2ClO−3+6H+
Adding similar amount of OH− to both sides and combining H+ and OH− to form H2O Cl2+3H2O+6OH−→2ClO−3+6H++6OH− Cl2+3H2O+6OH−→2ClO−3+6H2O
Cancelling the extra H2O Cl2+6OH−→2ClO−3+3H2O
Balancing the charge on RHS by adding 4e− Cl2+6OH−→2ClO−3+3H2O+4e− ......(i)
Taking the reduction half : Cl2→Cl−
Balancing Cl Cl2→2Cl−
Balancing charge by adding 2e− to LHS Cl2+2e−→2Cl−......(ii)
Multiplying eq(ii) by 2 and then adding to eq (i) 2Cl2+4e−→4Cl− Cl2+6OH−→2ClO−3+3H2O+4e−
On adding
__________________________________________ 3Cl2+6OH−→2ClO−3+4Cl−+3H2O