Question

Write balanced redox equations for the following reaction. Unless stated otherwise, assume that the reaction takes place in acidic or neutral conditions. Cr2O72− + HNO2 → Cr3+ + NO3− How do you do this?

Answer 1

Well, we use the method of half-equations....which utilizes oxidation numbers, and introduces electrons as virtual particles of convenience...

`Cr_2O_7^(2-)+3HNO_2(l)+5H^+rarr 2Cr^(3+)+3NO_3^(-) +4H_2O(l)`

Explanation:

Dichromate is reduced...

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

Nitrous acid is oxidized...

`Hstackrel(+III)NO_2(l)+H_2O(l)rarr stackrel(+V)NO_3^(-)+3H^+ +2e^-`

And so we take ONE of the former, and THREE of the latter to eliminate the electrons....

`Cr_2O_7^(2-)+14H^+ +6e^(-)+3HNO_2(l)+3H_2O(l)rarr 3NO_3^(-)+9H^+ +6e^(-) +2Cr^(3+)+7H_2O(l)`

We cancel common reagents....

`Cr_2O_7^(2-)+3HNO_2(l)+5H^+rarr 3NO_3^(-) +2Cr^(3+)+4H_2O(l)`

The which I think is balanced with respect to mass and charge (as indeed it MUST be), but you should check this...

And what would we see? The deep-red colour of dichromate would dissipate to give greenish `Cr^(3+)`...

There are other examples of the method of half-equations on these boards. You can get very good at them...