Br2+OH−→BrO−3+Br−+H2O
is balanced by ion electron method. Choose the correct statement(s) regarding stoichiometric ratios nanb :
Here a,b represents different species in the reaction.
A
nBr2nBrO−3=32
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B
nBr2nBrO−3=3
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C
nOH−nH2O=2
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D
nOH−nH2O=3
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Solution
The correct option is CnOH−nH2O=2 0Br2+OH−→+5BrO−3+Br−+H2O
Oxidation half : Br2→BrO−3
Bromine changes its oxidation state from 0 to +5
Reduction half : Br2→2Br−.
Bromine changes its oxidation state from 0 to −1
Taking the oxidation half reaction : Br2→BrO−3
Balancing Br Br2→2BrO−3
Balancing O by adding H2O Br2+6H2O→2BrO−3
Balancing H by adding H+ Br2+6H2O→2BrO−3+12H+
Adding similar amount of OH− to both sides and combining H+ and OH− to form H2O Br2+6H2O+12OH−→2BrO−3+12H++12OH− Br2+6H2O+12OH−→2BrO−3+12H2O
Cancelling the extra H2O Br2+12OH−→2BrO−3+6H2O
Balancing the charge on RHS by adding 10e− Br2+12OH−→2BrO−3+6H2O+10e− ......(i)
Taking the reduction half : Br2→Br−
Balancing Br Br2→2Br−
Balancing charge by adding 2e− to LHS Br2+2e−→2Br−......(ii)
Multiplying eq(ii) by 5 and then adding to eq (i) 5Br2+10e−→10Br− Br2+12OH−→2BrO−3+6H2O+10e−
On adding
__________________________________________ 6Br2+12OH−→2BrO−3+10Br−+6H2O