Balance the below in acidic solution (redox reaction)?

#Cr^(3+)(aq)+ClO_4^(-)(aq)\toCr_2O_7^(2-)(aq)+ClO_2^(- )(aq)#

1 Answer
anor277
Jul 9, 2018

#underbrace(4Cr^(3+))_"green"+3ClO_4^(-)+8H_2O rarr underbrace(2Cr_2O_7^(2-))_"orange-red" +3ClO_2^(-)+16H^+ #

Explanation:

Well, chromic ion is oxidized...

#2Cr^(3+)+7H_2Orarr Cr_2O_7^(2-) +14H^++ 6e^(-)#

And perchlorate is reduced to chlorite...

#ClO_4^(-) + 4H^+ +4e^(-)rarr ClO_2^(-)+2H_2O#

The difference in oxidation numbers between chromium, and chlorine, which I leave to you to determine, is accounted for by electron transfer...

We take TWO of the former, and add THREE of the latter to retire the electrons, virtual particles of convenience.

#4Cr^(3+)+14H_2O+3ClO_4^(-) + 12H^+ +12e^(-)rarr 2Cr_2O_7^(2-) +28H^+ + 12e^(-)+3ClO_2^(-)+6H_2O#

And we cancel away to get...

#underbrace(4Cr^(3+))_"green"+3ClO_4^(-)+8H_2O rarr underbrace(2Cr_2O_7^(2-))_"orange-red" +3ClO_2^(-)+16H^+ #

This I think is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical reality. Whether it works or not is a different matter.

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