Question

Cr2O72- (aq) + NO2- (aq) --> Cr3+ (aq) + NO3- (aq)
What is the reducing agent in this reaction?

Answer 1

NO2-?

Answer 2

Yes...Do you know why..?

Answer 3

Well, I worked it out that Cr2O72- was reduced from 2- to 3+. So, it must have been reduced by the NO2-. Is my reasoning right?

Answer 4

Cr(VI) ---> Cr(III)
N(III) ---> N(V)

Answer 5

I am wrong.

Answer 6

Why?

Answer 7

Because Cr2O72- is oxidized.

Answer 8

I am very confused now...

Answer 9

Cr2O7(2-) + 6e ---> 2Cr(3+)
NO2(-) ---->NO3(-) + 2e
Ok..?
What's the problem?

Answer 10

So what has been reduced then?

Answer 11

Reduction means receiving electron and Oxidation means giving electron...
Here Cr is reduced...

Answer 12

So, Cr2O72- has lost electrons hasn't it?

Answer 13

No Cr2O7(2-) has received electron...
Cr2O7(2-) + 14H(+) + 6e- >> 2Cr(3+) + 7H2O
Do you know how to calculate oxidation number...?

Answer 14

So Cr2O72- is reduced, and No2- is oxidized? And No2- is the reducing agent?

Answer 15

That's right man...:)

Answer 16

okay :)