How do we represent the disproportionation of #"chlorate ion"# to give #"chloride"# and #"perchlorate"# ions?

1 Answer
anor277
Nov 7, 2017

Well this is a disproportionation reaction.....

Explanation:

#"Chlorate ion"# #Cl(V+)# is OXIDIZED to #"perchlorate"#, #Cl(VII+)#.

#ClO_3^(-) +H_2O(l) rarr ClO_4^(-) + +2H^+ + 2e^(-)# #(i)#

#"Chlorate ion"# #Cl(V+)# is REDUCED to #"chloride"#, #Cl(-I)#.

#ClO_3^(-) +6H^+ +6e^(-)rarr Cl^(-) + 3H_2O(l)# #(ii)#

Charge and mass are balanced in each half equation, so this is kosher, and to represent the disproportionation reaction we take...#3xx(i)+(ii)# in order to remove the electrons.....

#3ClO_3^(-) +ClO_3^(-) +cancel(6H^+ +6e^(-)+3H_2O(l))rarr 3ClO_4^(-) + Cl^(-)+cancel(6H^+ + 6e^(-) + 3H_2O(l))#

....to give finally....

#4ClO_3^(-) rarr 3ClO_4^(-) + Cl^(-)#

Are charge and mass balanced?

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