[Epistimi]

I was given this extra assignment by my teacher (as a challenge), but I won't see him for another week to ask him for help. Here is the unbalanced reaction (takes place in an acidic solution):

[Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ (s) + MnO₄^- (aq)  Cr₂O₇^2- (aq) + CO₂ (g) + NO₃^- (aq) + Mn²⁺ (aq)

The way we've usually balanced redox reactions is the following:

• Find the oxidation states of all elements in the reaction (except hydrogen and oxygen, for the most part)
• Identify the elements that change oxidation states during the reaction
• Multiply the positive and negative changes with each other, such that the total positive and negative changes cancel out
• Calculate the change in charge and add a number of H⁺ or OH^- ions to the appropriate side depending on the pH of the solution
• Balance the number of hydrogen atoms by adding water molecules

More or less. In this particular reaction I'm not sure if I found the correct oxidation states for the elements in the first compound [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃:

First Cr: +3
First N: -3
First C: +4

Second Cr: +2
Second N: -3
Second C: +2

For the "first" of the two, my teacher told me the structure of the N₂H₄CO-molecule, and I think I was successful in finding the oxidation states. However, I don't really know how the complex(?) bonds with the Cr look. I suppose each N₂H₄CO-molecule partially gives an electron pair to the Cr-ion (I haven't taken this into account). As far as I can see, this would affect one of the N-atoms in each of the six molecules, and thus half of the N-atoms would have an oxidation state of -2 and the Cr-ion one of -3.

For the "second", I assumed that the CN was a cyano group CN^-. I looked it up, and as far as I can tell, the N has an oxidation state of -3 and the C one of +2 (N being the more electronegative of the two, the N and C each contributing three valence electrons to the triple bond, but the C-atom having an extra electron (= a free electron pair) on the opposite site). By trial and error I found that the first Cr must then have an oxidation state of +3 and the second Cr one of +2 (based on the ratio between them and the possible oxidation states of chromium), though I'm not so sure as the Cr-atoms aren't free ions, so their charges doesn't have to equal their oxidation states.

The oxidation states of the "simple" compounds are no problem.

As for balancing the equation I've tried a few things, but nothing really seems to work (perhaps owing to mistakes in determining oxidations states). I've never tried balancing a reaction where multiple elements that change oxidation states are found in the same compound.

One thing, though. In class, we balanced an equation where the nitrogen in a couple of nitrates changed oxidation states by different amounts. We then took the average of the changes. In this case, however, the Cr-atoms (for instance) are not in a 1:1 ratio, so I tried taking a weighted average instead, but this seemed to result in coefficients in the hundred thousands.

By the way, English isn't my first language and I go to school in Denmark, so some of the chemical terminology might not be correct (though I always appreciate being corrected). Also, the method for balancing these reactions seems to be a bit different from the one I stumbled upon trying to Google a solution to this.

I would very much appreciate some help on this (or a nudge in the right direction). Thanks.

[Borek]

Tricky.

Don't bother with balancing using any of the redox methods. Don't balance reaction as written, rather go for

[Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + KMnO₄ + H₂SO₄ K₂Cr₂O₇ + CO₂ + HNO₃ + MnSO₄ + H₂O + K₂SO₄

and balance using the algebraic method. Then convert back to net ionic.

Don't ask why - just be sure it will work

Some of the coefficients will be in the thousands range.

Your oxidation states of Cr are OK.

[curiouscat]

I get this. Could be wrong. I tried to use an half baked tool.

10 [Cr(N2H4CO)6]4[Cr(CN)6]3 + 2093 MnO4{-1 }  35 Cr2O72{-} + 420 CO2 + 660 NO3{-} + 2093 Mn{+2 } + -2312 H{+} + 3272 OH{-}

Using Borek's version I get

Edit: removed, please leave some work for the OP.

[Borek]

+ -2312 H{+} + 3272 OH{-}

I don't get these.

[Epistimi]

Thank you very much Borek. Using the algebraic method I get the following:

4 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1134 KMnO₄ + 1134 H₂SO₄ 14 K₂Cr₂O₇ + 168 CO₂ + 264 HNO₃ + 1134 MnSO₄ + 1194 H₂O + 7 K₂SO₄

I must say, though, that I don't really see where this "expanded" reaction comes from. I mean, why sulphuric acid and why potassium permanganate? The products, I suppose, follow from there, and I can sort of see that. Also, I'm not quite sure how to convert this back to net ionic.

In any case, thanks for your help. I really appreciate it.

[Borek]

Your equation is definitely wrong - potassium is not balanced (1134 on the left, but only 42 on the right). I suspect some math error (nothing unusual when dealing with this type of a problem).

Net ionic: write everything that is dissociated as ions and see what cancels out. Assume the Cr compound to be not dissociated.

[Epistimi]

Ah, I think I see where I went wrong. This, however, should be right:

10 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1176 KMnO₄ + 1729 H₂SO₄ 35 K₂Cr₂O₇ + 420 CO₂ + 660 HNO₃ + 1176 MnSO₄ + 1879 H₂O + 553 K₂SO₄

I added up all the elements, and as far as I can tell it's balanced. As for the net ionic:

10 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1176 K⁺ + 1176 MnO₄^- + 3458 H⁺ + 1729 SO₄^2- 70 K⁺ + 35 Cr₂O₇^2- + 420 CO₂ + 660 H⁺ + 660 NO₃^- + 1176 Mn²⁺ + 1176 SO₄^2- + 1879 H₂O + 1106 K⁺ + 553 SO₄^2-

The potassium ions cancel out:

10 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1176 MnO₄^- + 3458 H⁺ + 1729 SO₄^2- 35 Cr₂O₇^2- + 420 CO₂ + 660 H⁺ + 660 NO₃^- + 1176 Mn²⁺ + 1176 SO₄^2- + 1879 H₂O + 553 SO₄^2-

Sulphate cancels out:

10 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1176 MnO₄^- + 3458 H⁺ 35 Cr₂O₇^2- + 420 CO₂ + 660 H⁺ + 660 NO₃^- + 1176 Mn²⁺ + 1879 H₂O

Some hydrons cancel out:

10 [Cr(N₂H₄CO)₆]₄[Cr(CN)₆]₃ + 1176 MnO₄^- + 2798 H⁺ 35 Cr₂O₇^2- + 420 CO₂ + 660 NO₃^- + 1176 Mn²⁺ + 1879 H₂O

This seems to be as far as it can be simplified.

[Borek]

That's what I got as well.