Question

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

Answer 1

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+)`.