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Question

When the following redox reaction
Cl2+OHClO3+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 A 3,6
0Cl2+OH+5ClO3+Cl+H2O

Oxidation half : Cl2ClO3
Chlorine changes its oxidation state from 0 to +5

Reduction half : Cl22Cl.
Chlorine changes its oxidation state from 0 to 1

Taking the oxidation half reaction :
Cl2ClO3
Balancing Cl
Cl22ClO3

Balancing O by adding H2O
Cl2+3H2O2ClO3

Balancing H by adding H+
Cl2+3H2O2ClO3+6H+

Adding similar amount of OH to both sides and combining H+ and OH to form H2O
Cl2+3H2O+6OH2ClO3+6H++6OH
Cl2+3H2O+6OH2ClO3+6H2O

Cancelling the extra H2O
Cl2+6OH2ClO3+3H2O

Balancing the charge on RHS by adding 4e
Cl2+6OH2ClO3+3H2O+4e ......(i)


Taking the reduction half :
Cl2Cl

Balancing Cl
Cl22Cl

Balancing charge by adding 2e to LHS
Cl2+2e2Cl......(ii)

Multiplying eq(ii) by 2 and then adding to eq (i)
2Cl2+4e4Cl
Cl2+6OH2ClO3+3H2O+4e
On adding
__________________________________________
3Cl2+6OH2ClO3+4Cl+3H2O

So the correct answer is 3,6

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