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Question

The given redox reaction

Br2+OHBrO3+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
nBr2nBrO3=32
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B
nBr2nBrO3=3
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C
nOHnH2O=2
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D
nOHnH2O=3
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Solution

The correct option is C nOHnH2O=2
0Br2+OH+5BrO3+Br+H2O

Oxidation half : Br2BrO3
Bromine changes its oxidation state from 0 to +5

Reduction half : Br22Br.
Bromine changes its oxidation state from 0 to 1

Taking the oxidation half reaction :
Br2BrO3
Balancing Br
Br22BrO3

Balancing O by adding H2O
Br2+6H2O2BrO3

Balancing H by adding H+
Br2+6H2O2BrO3+12H+

Adding similar amount of OH to both sides and combining H+ and OH to form H2O
Br2+6H2O+12OH2BrO3+12H++12OH
Br2+6H2O+12OH2BrO3+12H2O

Cancelling the extra H2O
Br2+12OH2BrO3+6H2O

Balancing the charge on RHS by adding 10e
Br2+12OH2BrO3+6H2O+10e ......(i)


Taking the reduction half :
Br2Br

Balancing Br
Br22Br

Balancing charge by adding 2e to LHS
Br2+2e2Br......(ii)

Multiplying eq(ii) by 5 and then adding to eq (i)
5Br2+10e10Br
Br2+12OH2BrO3+6H2O+10e
On adding
__________________________________________
6Br2+12OH2BrO3+10Br+6H2O

nBr2nBrO3=62=3

nOHnH2O=126=2

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