How is #NO_2# oxidized by #Cr_2O_7^(2-)# to give nitrate ion?

1 Answer
May 3, 2017

Answer:

Dichromate oxidizes #NO_2# up to nitrate anion, and is reduced to #Cr^(3+)#

Explanation:

#"Oxidation half equation:"#

#NO_2(g) +H_2O(l) rarr NO"_3^(-) +2H^+ + e^(-) #

#stackrel(+IV)Nrarrstackrel(+V)N#

Charge and mass are balanced so tick.

#"Reduction half equation:"#

#Cr_2O_7^(2-)(aq) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_2O #

#stackrel(VI+)"Cr"rarrstackrel(+III)"Cr"#

Charge and mass are balanced so tick.

And we add 6 of one, and one of t'other to eliminate the electrons.

#Cr_2O_7^(2-)(aq) +6NO_2(g) +2H^(+) rarr 2Cr^(3+) +H_2O +6NO"_3^(-) #

Are charge and mass balanced. Don't trust my arithmetic!

What we would see in this reaction is the deep orange/red of the dichromate anion dissipate to give greenish #Cr^(3+)#.