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Question

The following redox reaction :
ClO+CrO2Cl+CrO24
when balanced in a basic medium by oxidation number method is :

A
2ClO+3CrO2+2OH3Cl+CrO24+2H2O
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B
3ClO+2CrO2+2OH3Cl+2CrO24+H2O
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C
ClO+2CrO2+3OH2Cl+3CrO24+2H2O
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D
3ClO+2CrO2+OH3Cl+3CrO24+2H2O
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Solution

The correct option is B 3ClO+2CrO2+2OH3Cl+2CrO24+H2O

ClO+CrO2Cl+CrO24
Oxidation state of Cl in ClO=+1
Oxidation state of Cl in Cl=1
Oxidation state of Cr in CrO2=+3
Oxidation state of Cr in CrO24=+6

CrO2 is undergoing oxidation and is the reducing agent.
ClO is undergoing reduction and is the oxidising agent.

nf of ClO=2
nf of CrO2=3

Cross multiplying these with nf of each other.
we get,
3ClO+2CrO2Cl+CrO24


Balancing the elements other than oxygen and hydrogen on both sides,
3ClO+2CrO23Cl+2CrO24

Adding the H2O to balance the oxygen,


3ClO+2CrO2+H2O3Cl+2CrO24
adding H+ to balance hydrogen,


3ClO+2CrO2+H2O3Cl+2CrO24+2H+

Now adding OH to both sides to combine with H+ and make it H2O,

3ClO+2CrO2+H2O+2OH3Cl+2CrO24+2H++2OH

3ClO+2CrO2+H2O+2OH3Cl+2CrO24+2H2O

Eliminating the unrequired common H2O which is present on the both sides,
3ClO+2CrO2+2OH3Cl+2CrO24+H2O

Charge on both sides is -7
So, this is the final balanced equation.


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