Apparently the hardest reaction to balance? (Reposted with more information)?

Question

HNO_3 + K_2CrO_4 + Fe(NO_3)_2 -> KNO_3 + Cr(NO_3)_3 + Fe(NO_3)_3 + H_2O

Alright... this one's tricky. If you can solve it, you're a genius. Seriously. I hadn't checked my email earlier when I posted this Q, but my prof has emailed me some help which I will post in the 'added detail.' Thanks everybody.

My professor says:

My own approach is to strip away - put to one side, so to speak - all of the counter-ions that do not themselves get involved in oxidation/reduction, and focus on the species that are oxidized and reduced. Thus, in your problem I would look at:

CrO4^2- + Fe^2+ -> Cr^3+ + Fe^3+

The K^+ and the NO3^- are just counter ions. In solution, they are floating off somewhere else, they are not involved in the redox. So go ahead and balance the reaction I've written above using the half-reaction method. Then, when you are done identify the H^+ that you had to put in with the HNO3 in the given reaction (since HNO3 is a strong acid, every HNO3 put in gives you an H+) Finally, "put back" K's and NO3's
onto the balanced redox reaction so that all of the charges are accounted for. You should find that they basically all fall into place.

The sad thing is... I still don't get it.

Answer 1

Lancenigo di Villorba (TV), Italy

I distinguish two ELECTRODIC HALF-REACTIONS, an anodic and a cathodic ones, as follows :

Anodic)
Fe(NO3)2(aq) + HNO3(aq) ---> Fe(NO3)3(aq) + H+(aq) + e

Cathodic)
K2CrO4(aq) + 8 HNO3(aq) + 3e --->
---> 2 KNO3(aq) + Cr(NO3)3(aq) + 4 H2O(aq) + 3 NO3-(aq)

Now, I multiply (e.g. X 3, triplication) the ANODIC half-reaction in a manner I can find the same ELECTRON's NUMBER EXCHANGED by CATHODIC one.
By means of the sum of CATHODIC half-reaction and the TRIPLICATED ANODIC one I retrieve

3 Fe(NO3)2(aq) + K2CrO4(aq) + 8 HNO3(aq) --->
---> 3 Fe(NO3)3(aq) + 2 KNO3(aq) + Cr(NO3)3(aq) + 4 H2O(aq)

I hope this helps you.